I have designed, made, improved and tested a very simple, transistor based distortion effect for guitars ( can be used for anything, I guess, but I have so far never tried to use distortion on anything else than a guitar ).
I have written a simple document which is written at a very basic level and explains why 2 and 2 is 4 and not five. I will be happy to have people who want to read the whole document, but, most of the people would only want to read the addendum where the schematics and the pictures are.
Here is the document. Please, inform whether the link can be open by anyone : Distortion Amplification - Google Drive
I have written a simple document which is written at a very basic level and explains why 2 and 2 is 4 and not five. I will be happy to have people who want to read the whole document, but, most of the people would only want to read the addendum where the schematics and the pictures are.
Here is the document. Please, inform whether the link can be open by anyone : Distortion Amplification - Google Drive
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Comprehensive.
I do think a simplified schematic should be pasted at section "2". Text like "..voltage divider Rd1 and Rd2 provides the necessary DC offset for the AC signal as well as a DC offset..." makes more sense with a picture, and not 25 pages later.
Guitar pickup impedance is 5K at bass and way over 100K at 3KHz. The guitar's Volume control moderates this to nearer 50K over most of the band (I should model this). The proposed 90K input impedance is "low" by conventional guitar-chain custom. Clean effects usually strive for hundreds of K. Several beloved effects have few-K impedance, cheap, but serves to cut highs before the distortion adds highs (does not work except direct from guitar).
You propose a bass-cut to reduce wall-power hum. Given a full-wave rectifier the "hum" starts at 100/120Hz and extends up the spectrum.
"20mA" LEDs do not have to run 20mA. Modern ones may be plenty bright at 2mA. The LED can eat un-filtered power. This eases load on filtering for audio stages.
I do think a simplified schematic should be pasted at section "2". Text like "..voltage divider Rd1 and Rd2 provides the necessary DC offset for the AC signal as well as a DC offset..." makes more sense with a picture, and not 25 pages later.
Guitar pickup impedance is 5K at bass and way over 100K at 3KHz. The guitar's Volume control moderates this to nearer 50K over most of the band (I should model this). The proposed 90K input impedance is "low" by conventional guitar-chain custom. Clean effects usually strive for hundreds of K. Several beloved effects have few-K impedance, cheap, but serves to cut highs before the distortion adds highs (does not work except direct from guitar).
You propose a bass-cut to reduce wall-power hum. Given a full-wave rectifier the "hum" starts at 100/120Hz and extends up the spectrum.
"20mA" LEDs do not have to run 20mA. Modern ones may be plenty bright at 2mA. The LED can eat un-filtered power. This eases load on filtering for audio stages.
I had the text open in Word and the schematic in my browser and just clicked back and forth. The attachments, schematic and pictures are, I am guessing, in the back as this is a school paper? I thought of the impedance rated at dc rising with frequency, easy to measure with an ohm meter and doesn't confuse guitar players too much, but I wasn't suppose to be going over the document. It almost made me forget to put a rosette in the top I was going to brace. I came up to look at a guitar as an example and went through more of the text. I am easily distracted. Like now, why did I come up here?
Maybe I missed it but a block diagram of how it works would be nice and diagrams of the input signal compared to output. 30 pages is a lot to read. From what I gathered it takes the guitar signal and turns it into a square wave with a comparator. What's it comparing? Is just zero crossing? It will probably sound more like a synth than a guitar, wich would be cool. Wondering if it will effect pitch. Would like to hear it.
Steven, thanks for sharing your work!
Did you try it with a guitar? Did you compare it with any popular guitar overdrives or distortions out there? And, most important of all, did you like the sound of your creation?
I experimented with the same basic idea shortly after starting to play electric guitar in the 1980's. I also experimented with adding a wee bit of positive feedback, so you had a Schmitt trigger rather than a straight comparator.
Personally, I didn't like the results of any of my experiments. Take a crude early Fuzz Face and make it even cruder, and that's what you get from a comparator: no touch sensitivity at all. When working well, it goes from dead silence to full-loudness wasp-in-a-tin-can buzz, with no shades of grey in between. Fizz or nothing.
That's when working well. When working badly, the comparator triggers intermittently on noise, so you get brief bursts of full rail-to-rail noise when the guitar is supposed to be quiet. This is why I experimented with adding positive feedback, which means the circuit is dead-quiet until the input signal exceeds a threshold voltage.
Dunno if Steven had the same issues with his circuit(s) that I had with mine, but that was my experience.
I should add that I don't necessarily other people will react to Schmitt-trigger / comparator fuzz-boxes the same way I did. Musical and artistic tastes vary very widely, and one person's horrible noise might very well be someone else's glorious music.
Thinking back, I never heard the words "overdrive pedal" in those days. I think the popular belief at that time was that hard, fuzzy-edged clipping was what you wanted for guitar.
This may actually have had something to do with the fact that electronic keyboards and synths were extremely popular at the time - there were probably guitarists who were desperate to make their guitars sound just like the inexpressive and cheesy monophonic synths of the era, which were all over every big radio hit song.
-Gnobuddy
Did you try it with a guitar? Did you compare it with any popular guitar overdrives or distortions out there? And, most important of all, did you like the sound of your creation?
That's what I gathered as well.
I experimented with the same basic idea shortly after starting to play electric guitar in the 1980's. I also experimented with adding a wee bit of positive feedback, so you had a Schmitt trigger rather than a straight comparator.
Personally, I didn't like the results of any of my experiments. Take a crude early Fuzz Face and make it even cruder, and that's what you get from a comparator: no touch sensitivity at all. When working well, it goes from dead silence to full-loudness wasp-in-a-tin-can buzz, with no shades of grey in between. Fizz or nothing.
That's when working well. When working badly, the comparator triggers intermittently on noise, so you get brief bursts of full rail-to-rail noise when the guitar is supposed to be quiet. This is why I experimented with adding positive feedback, which means the circuit is dead-quiet until the input signal exceeds a threshold voltage.
Dunno if Steven had the same issues with his circuit(s) that I had with mine, but that was my experience.
I should add that I don't necessarily other people will react to Schmitt-trigger / comparator fuzz-boxes the same way I did. Musical and artistic tastes vary very widely, and one person's horrible noise might very well be someone else's glorious music.
Thinking back, I never heard the words "overdrive pedal" in those days. I think the popular belief at that time was that hard, fuzzy-edged clipping was what you wanted for guitar.
This may actually have had something to do with the fact that electronic keyboards and synths were extremely popular at the time - there were probably guitarists who were desperate to make their guitars sound just like the inexpressive and cheesy monophonic synths of the era, which were all over every big radio hit song.
-Gnobuddy
Thanks a lot for sharing your project .
Incredibly well explained, down to the smallest detail, with solid Math behind every statement.
All we need now is some MP3 or YT Video showing how it sounds
Congratulations.
Thank you very much. I will try to do MP3's. However, I need to first readjust the pickup coils because I use plain steel 3d ( G ) and 4th ( D ) strings. The pickup coils have the third and the fourth pole up because they are made for wound strings. Thus, 3d and 4th sounds very strong which would be cheating because I want to provide you all with example which is as close to normal as possible. ( I use 5th and 5th string flat wounds but this I do not want to change just for the experiment because to change the strings takes some difficulty. )
I saw a video on YouTube how someone from Texas uses soldering iron to heat the 3d and the 4th pole of a pickup, thus, melting the wax which holds the pole and then pushing with the plastic handle of a screw driver and the poles go down. The person suggested to make the 3d and the 4th pin equal to the others, which are pretty much equal to each other.
I did NOT have 1N60P initially and I used 1N914 for the output diodes which gave around 0.65V to 0.7V at 1mA. I have just received and changed all diodes ( the saturation protection diode too ) to 1N60P ( 0.24V at 1mA ) and the sound is marvelous. I will soon build and post a power supply which is supposed to be not as noisy as the off the shelf AC DC adapter I currently use. The noise level is not very high event at high gains and, because of the 0.24V level, lower gains are also OK, although I prefer high gains with lots of distortion.
I seem to like what happened with this device, although, I have been saying out loud, this device does not pretend to be destined for super professional use.
One thing for sure : I put the old distortion in the original wrap and in the wardrobe. I will use this from now on and I have the idea to make another one with a transistor made differential amplifier.
The more simplified schematics, the better. This, however, requires some labour.
Thought the output impedance of a pickup is <= 10K. Read somewhere online. They say the new pickups have a significantly lower impedance than the old ones. Will use a Darlington at the differential amplifier for another project to ensure everyone is happy with the input impedance.
I have, however, examined the guitar I have ( Squier Bullet Stratocaster with neck and middle pickup in parallel ( as original ) to full power ( 0 resitance ) WITHOUT a distortion and with the original amplifier ( Fender, 10W, Made in China ). The internet claim a level of 10mV of the highs as opposed to 1V full strum. Anyway, the guitar with the distortion performs OK for the highs too. I cannot detect a loss of signal caused by a high output impedance and low input impedance. I can easily make the input resistors in the megaohm range. I can even have a triple Darlington for the input. I am afraid I would get more noise with higher resistors. I must be sure they are necessary.
I call all noise generated by the 60Hz and 50Hz mains " a 60 ( 50 ) Hz noise " because the reason for this noise is the 60Hz ( 50Hz ) of the mains. Yes, when rectified, the main frequency is 120Hz ( 100Hz ) and this is the ripple frequency. For 60Hz, the ripples' period is 8.33ms as opposed to the period of 16.67ms of the mains.
However, 60Hz proliferates not only through ripples but as noise, say through the oppositely wound pickups, which reduce but do not eliminate the noise, thus, to filter this 60Hz is good whenever possible. Obviously, to filter 120Hz is even better but not possible when the low E is around 84Hz. Some people use peak filters, I do not because there is a good signal there.
Yes, the rectified ripples are 120Hz ( 100Hz ). I was to build a transformerless power supply with a half wave rectifier ( a diode ), LM317 to scale 60V to 20V to 35V and either another LM317 or an L78xx. Then the ripples are at 60Hz. I decided to first build a transformer based power supply for this and other projects and then will decide what to do with the transformerless.
Interestingly, I have purchased 2 pieces of a light diode which requires only 1mA to shine nicely. Never tried them but wanted to use them in the power supply to save current. Along with the normal, 20mA, light diode which I would attach to the power supply directly through a diode ( half wave rectifier ), resistor and some capacitance. Decided not to use the light diode for rectifier too, although should be OK for those which can take >= 60V ( 110V, 120V ). I got these from the UK via eBay and I spent 4 pounds for one light diode. This is around $7US!
I have historically used 16mA for the so called 20mA diodes and they shine OK. Also, 16mA is the maximum current for the 5mA diodes, so, either one can be put on.
One must be careful with LM317 and LM377. These require minimum regulator current of 10mA. Thus, when a very low current is used by the circuit, 10mA through light diode would ensure the minimum current which would be taken from all of the ripple fighter capacitors.
Interesting. And very nasty. But thank you for this information.
I have found this :
Guitar Impedance Matching - Ironstone Electric Guitar Pickups
Impedance question gtr pickups >>> gtr amps - Gearslutz Pro Audio Community
:
Guitar amps have high impedance inputs. (100K to 1Meg)
Passive electric guitars have high impedance outputs (6K to 10K)
Active guitar pickups have relatively low impedance outputs (100)
Effect Pedals have relatively low impedance outputs (100)
Active solid state devices have relatively low impedance outputs (100)
Yes. You are right. The document could have been written better.
Do NOT read the document, just go to the schematics.
Here are the main points :
The guitar signal is AC. The device is supplied by a single supply ( dual and ground is better ). Thus, the signal goes through a capacitor to a voltage divider. The voltage divider gives 6V. Thus, the AC signal gets to circle around these 6V. The input impedance of the voltage divider is roughly the parallel of the two resistors which, in this case, is around 100K. Many replies state this is insufficient and suggest 1M to 2.7M as a good input impedance.
After the capacitor and the voltage divider, there is a Darlington which buffers the voltage divider. In this case this Darlington is a standard Darlington made of 2 transistors. A triple Darlington ( three transistors ) may be needed for the suggested megaohm input voltage divider impedance.
After the Darlington, there is a standard common emitter amplifier without AC feedback. This means the AC signal from the Darlington buffer goes through the base resistor Rb ( a potentiometer ), then the base emitter junction of the transistor, then the huge capacitor Ce and ground. The lack of any AC feedback allows for a higher gain yet lower quality of the signal ( which is not very much lower for a very good transistor with excellent linearity ). The signal gets amplified by the common emitter.
The signal exits the common emitter amplifier through a large capacitor which filters the DC and allows the AC to go through.
Then, the amplified signal can be used as is. Just a simple preamplifier so far. Nothing else. The signal can either be slightly preamplified or preamplified a lot. When the signal is heavily preamplified, the next stage ( the amplifier ) cuts the sound at a given level thus making distortion by preamplification which I call overdrive.
However, there is another output to the circuit. After the common emitter amplifier and before reaching this output, the signal goes through a fairly large resistor of 5.1K and then sees two diodes connected to the resistor and to ground in different directions. These diodes ensure the signal is cut at Vf level. Thus, the positive wave will reach Vf and not more and the negative wave will reach - Vf and not more. The amplitude of the new signal will be Vf. Germanium Schottky diodes 1N60P have Vf = 0.24V at If = 1mA, which is the case with this device. Thus, the signal will be a standard audio signal which should have an amplitude of, say, 5V but is cut at 0.24V. This way, almost square signal is provided to the amplifier with an amplitude of 0.24V.
Because Vf is so low ( 0.24V as opposed to 0.7V in standard Silicon diodes ), even lower volume signals are well cut, thus, well distorted.
So, the potentiometer adjusts Rb and thus the gain of the common emitter which is :
G = - ( Rc / Rb ) B = - ( 750 / Rb ) 60.
The amplifier amplifies and the signal can either be used as is or cut at +- 0.24V and thus made almost square.
As far as I can understand, distortion is when the standard audio signal is made square and overdrive is when the signal is NOT SPECIFICALLY made square but preamplified so much, so the amplifier makes this huge signal square because the huge signal cannot go over the power supply rails of the amplifier.
HOWEVER, I may be wrong. This is the way I understand the definitions of overdrive and distortion but I may be wrong. The readers will be able to explain these better.
This device does NOT use a comparator, per se, in order to " make the signal square ". However, I have inserted another schematic in the document where a standard, 100ns, LM comparator is used to make the signal square. The comparator has a positive feedback which provides hysteresis. Thus, the comparator should compare the audio signal to 0V ( ground ) BUT does not. Instead, the comparator compares the audio signal to the hysteresis level which is adjustable by a potentiometer and changes polarity.
Assume the input voltage to the negative input of the comparator is very negative and the potentiometer is turned to make the hysteresis levels +- 0.1V. Very negative voltage of the negative input of the comparator makes a very positive output of the comparator. When the output of the comparator is positive, the positive input will have the positive output divided by the potentiometer based voltage divider and this is 0.1V. Thus, because of this positive feedback, the output voltage of the comparator will stay positive until the audio signal applied to the negative input of the comparator is <= 0.1V. When > 0.1V, the output of the comparator becomes very negative and the voltage divider jumps to - 0.1V. The comparator output will stay very negative until the audio signal is >= - 0.1V. When the audio signal becomes more negative than - 0.1V, the comparator output switches to very positive and the voltage on the positive input of the comparator becomes 0.1V and so on and so forth.
This way, the comparator will provide a true square wave which means 100% distortion. However, the suggested, comparator based circuit does not allow for the level of distortion to be adjusted. Will always be 100%. Some guitar players do not like so much distortion. I am not a professional guitar player but I do like super heavy 100% distortion only or no distortion at all. I also like overdrive, which, with the Fender guitar amplifiers, is super heavy and nice. But as you all know, every guitar player likes different things.
We are all crazy but we are different crazy and to a different level!
Steven, thanks for sharing your work!
Did you try it with a guitar? Did you compare it with any popular guitar overdrives or distortions out there? And, most important of all, did you like the sound of your creation?
That's what I gathered as well.
I experimented with the same basic idea shortly after starting to play electric guitar in the 1980's. I also experimented with adding a wee bit of positive feedback, so you had a Schmitt trigger rather than a straight comparator.
Personally, I didn't like the results of any of my experiments. Take a crude early Fuzz Face and make it even cruder, and that's what you get from a comparator: no touch sensitivity at all. When working well, it goes from dead silence to full-loudness wasp-in-a-tin-can buzz, with no shades of grey in between. Fizz or nothing.
That's when working well. When working badly, the comparator triggers intermittently on noise, so you get brief bursts of full rail-to-rail noise when the guitar is supposed to be quiet. This is why I experimented with adding positive feedback, which means the circuit is dead-quiet until the input signal exceeds a threshold voltage.
Dunno if Steven had the same issues with his circuit(s) that I had with mine, but that was my experience.
I should add that I don't necessarily other people will react to Schmitt-trigger / comparator fuzz-boxes the same way I did. Musical and artistic tastes vary very widely, and one person's horrible noise might very well be someone else's glorious music.
Thinking back, I never heard the words "overdrive pedal" in those days. I think the popular belief at that time was that hard, fuzzy-edged clipping was what you wanted for guitar.
This may actually have had something to do with the fact that electronic keyboards and synths were extremely popular at the time - there were probably guitarists who were desperate to make their guitars sound just like the inexpressive and cheesy monophonic synths of the era, which were all over every big radio hit song.
-Gnobuddy
In regards to square, please, be kind to read the previous answer.
Yes, I did want to make the signal as square as possible with some control provided by the gain potentiometer.
As I mentioned in the previous reply, many guitar players say like you : they do not like extremely heavy 100% square waves. I am the opposite : I do love only super heavy 100% distortion or overdrive or no distortion at all, just the guitar connected to a simple amplifier. No effects. Also, I do not like other guitar effects except distortion and overdrive. I must admit the other effects are excellent, though. Thus, I like your idea for a Schmitt trigger. Generally, what I explained in the previous answer was a comparator with a variable hysteresis which is a Schmitt trigger. I would love to have one which is in the nanosecond range. There are some 10ns comparators, there are even 7ns comparators. Some of them are limited to 0V to 5V inputs.
I have not heard the term overdrive pedal but I have seen the term " Overdrive " button on the Squier SP10 10W amplifier which I possess.
I love this effect. I may switch to overdrive only instead of distortion OR overdrive. For now, I would like to play the one I have made just for the sake of. Do I like the effect I have made? Yes. I do not play the standard distortion any more. Only the one I have made. Is this a professional one? No. I have clearly stated so. Interesting yes, professional, no. For professional purposes, there are very many effects sold elsewhere.
However, in regards to professional distortions, I would like to share these thoughts with you all :
1. Get a distortion in a fully6 metal enclosure. Some are in plastic. Some in Aluminium or other metals. Get a metal one to decrease the electromagnetic interference.
2. Most commercial distortions require an external power supply, usually, 9V. Get a high quality one with big capacitors. You may be happier with a linear one with a transformer as opposed to a switching one. However, the switching one may give a better performance since the noise is in the very high frequencies only. I have ordered some switching adapters, I will see what they are. In case you can get the one you like in a grounded metal enclosure, this is much better. However, the manufacturers are afraid to make metal enclosure devices because they have to ground the enclosure and have to obey some standards for grounding, etcetera. I am not sure why, but I have never seen an inexpensive, metal enclosure, AC DC adapter.
Hope I have answered all questions. And yes, I would make the MP3 file with the sound of the distortion. Do NOT have high expectations.
I am not sure whether I would make the comparator based one but I have designed another one with a transistor based differential amplifier and a transistor based differential buffer and invertor to convert the differential output of the differential amplifier back to a single signal to ground for which one of the outputs of the differential amplifier is buffered, the other is inverted and thus subtraction is made to provide the single signal. The subtraction also subtracts the noise which is in phase for all inputs, I hope. In theory.
I have always been dreaming of making analogue things with transistors. However, because I graduated in 1990, I have used amplifiers only. I think most people who graduated in the 70's or after have used amplifiers only except some in the ultra high frequency business. I have never seen anyone to use transistors for audio. Only for power stages, although, there are amplifiers there too.
I think the amplifiers ( which are gorgeous and much better than anything ) destroyed the real electronics engineering, which, I would define, as transistor based.
Also because I have graduated in 1990, I have never seen nor studied tubes. I just know some analogy with transistors.
I also know some interesting facts with tubes ( a. k. a. valves ) :
1. The first electronic effect was some reverse current detected and written down by Edison while experimenting with light bulbs.
2. The first tube ( valve ) was a diode made in the UK in 1899, during Queen Victoria's reign.
3. The first triode was made in the USA ( not sure ) at the beginning of the 20th century.
4. The first differential amplifier was made with tubes ( valves ) most likely in Germany, most likely in the 30's. Not sure. when and where.
5. Some companies made huge design tube ( valve ) which was around 2 meters high, so a person can open a door and go inside to adjust and tests electrodes, etcetera. The, the person would go out and close the door. Then, a machine would suck the air to make vacuum. Then, the huge scale tube ( valve ) would be tested and, eventually, manufactured at a lower size.
6. In the 60's through 80's, computers have been made with acorn sized tubes to be resistive to an electromagnetic pulse ( mainly in an airplanes where lead case is a no no )
7. I THINK BUT THIS HAS NOT BEEN PROVEN : The first IC was made by tubes ( valves ) with common walls. Thus, 5 or 6 tubes ( valves ) can be positioned next to each other to save room. Not sure whether true or false. Never heard of. Just imagination. I can imagine tiny tube based reel to reel magnetophones used in WW2 for field interrogations and spy missions, brief case bugging, etcetera. Seen such only in the movies. Not sure how made.
Hope you have been entertained!
This is to inform all interested :
The schematics in the document show all diodes to be 1N60P which is correct to the design.
However, I did NOT have any 1N60P. I used a different Schottky diode for saturation protection and 1N914 standard Silicon diodes for the output. I have just received 1N60P. I have just installed them and played the updated device for around half of an hour.
This is why I have not previously replied nor posted.
I do so now and express the opinion the device works much better and does not require the next stage amplifier to be almost impossibly quieted down because of high, 0.65V amplitude levels. The amplitude now is 0.24V, which gives the amplifier the opportunity for more amplification.
I am very happy with the sound and will make MP3 files soon.
Please, be informed this is not a professional device and is not to replace the professional and the commercial ones.
********
Please, note : One of the goal I always have is to provide extremely clear explanations.
I think how to make explanations is science covered in linguistics and studied carefully by teachers in their pedagogy lessons. I am not sure whether this is the case but should be.
In this sense, here are the most important points in the EXPLANATIONS and not in the design of this device.
1. Input Capacitor
People say current can flow from high to low voltage potential only and not the other way around. OK. The guitar audio signal is AC with a maximal amplitude, say, 1V. The guitar signal is a voltage between the core of the output jack / socket and the surrounding tube of the said jack / socket, which can be called ground. Thus, the guitar signal is referenced to what can be called ground. The distortion takes this signal also to ground but a different ground : the distortion ground. When the two grounds are connected, the guitar signal is always referenced to ground and can be either negative or positive.
The distortion has a capacitor which connects the input ( I. e. the guitar output ) to a voltage divider which provides 6V towards ground. Thus, the right pin of the capacitor is connected to 6V and the left to an AC signal with amplitude of maximum 1V ( can be - 1V or 1V or anything in between ).
Now, here comes the point : the SAME people who say current can fly ONLY from high potential to low potential, ALSO SAY a signal with an amplitude of 1V can fly current in the two directions. How come? The signal is 1V maximum. The other side of the capacitor is 6V. Current can never fly from 1V to 6V! Thus, no AC signal can go through the capacitor because of the basic low of the potentials!
How can you all answer this?
I think the best answer is : Capacitor is such an animal which internally works the potentials out and allows AC to fly through and does not allow DC to fly through. In other words, capacitor is something which moves energy internally and makes potentials inside in order to allow AC to fly. In this example, when the capacitor sees 1V on the left side, the capacitor is intelligent enough to think : OK, someone ( the guitar ) wants to get 1V through. Yes, but I have 6V on the right. Then, I will get this one volt and elevate this one volt to 7V internally in order to have an internal elevation of 1V higher than what I have on the right pin, so I allow current to fly through.
And this is the so called principle of AC DC coupling. The capacitor imposes ( couples ) the AC over DC.
2. Emitter Capacitor Ce
People say Kirchhoff's law cannot be broken. People say there is some input voltage at Rb and this voltage must be repeated over this chain : Rb, BE junction ( base emitter junction of the transistor ) and Re || Ce. Then people say AC goes through Ce freely and does not make any potential between the emitter and ground. They say the **** AC **** potential of the emitter is 0.
So, therefore, the emitter is at 0V, thus, I can have AC between 0 and Vcc with an amplitude Vcc / 2! Thus, when I have a 15V supply and 4.6V DC with AC imposed over, I can get a range of 15V at the output! WRONG! Why? How come?
Kirchhoff's law cannot be broken. In front of Rb, there is 4.6V DC and AC imposed over. The BE of a transistor makes a DC only ( mainly, but consider only ) of 0.7V. Thus, the transistor should have 4.6V - 0.7V = 3.9VDC on the emitter with the AC signal super imposed over these 3.9V DC. However, there is a capacitor Ce. The AC goes through the capacitor and makes no voltage as the capacitor provides resistance of 0 Ohms for the AC. HOWEVER, the upper pin of the capacitor will remain at 3.9V and this voltage cannot be changed ( theoretically, under normal circumstances ). Thus, the transistor can pulsate from closed to open because of the AC current through the BE junction as much as wanted and I do not care of how much and why. The only thing I care is the emitter ( the upper pin of the capacitor ) will always see 3.9V under normal circumstances.
Thus, when the transistor is, FOR A MOMENT ONLY, fully open, the collector will have Uc = Ue + Ucesat. Ue has been proven to be 3.9V. Ucesat is, say, 0.1V. Therefore, Uc = 4V when the transistor is, FOR A MOMENT ONLY, fully open.
When the transistor is fully closed, the transistor will have Uce = Vcc ( 15V ) at the collector.
THUS, UNDER THE FORCE OF THE AC CURRENT THROUGH BE, THE TRANSISTOR WILL PULSATE IN SUCH A WAY, SO THE COLLECTOR VOLATGE Uc CHANGES FROM 4V TO 15V AND NOT FROM 0 TO 15V.
Thus, the middle point ( the DC ) of Uc is :
Ucdc = 4 + ( 15 - 4 ) / 2 = 9.5V
and the amplitude is :
Uca = 15 - 9.5 = 9.5 - 4 = 5.5V
AND NOT 7.5V.
So, electronics does not change neither Kirchhoff's, nor, Ohm's, nor any other law. Electronics only adjusts potentials and currents obeying these laws.
3. Something General
Yes, but, there is another question : the upper pin of Ce is at 3.9V. The lower pin is at 0V. The AC amplitude is 1V maximum. Therefore, the capacitor Ce will have a voltage which changes from 2.9V to 4.9V on the upper pin and 0V on the lower pin. When the explanation of point 1 is used, does this not mean there will be current which flies in the two directions of the capacitor freely and, because the capacitor, as mentioned, does not exert any resistance to AC, these currents will be huge, infinitely high. OK, the transistor cannot conduct from emitter to collector nor from emitter to base ( NPN ), therefore, the capacitor allows huge, infinite current from emitter to ground. Where does this current come from?
This is true. Theoretically there will be an infinitely high current from power supply through the Darlington buffer through Rb=0 through BE and through Ce. This infinitely high current will make another infinitely high current to fly from Vcc through Rc = 0 through CE ( Rce = 0 when fully open ) through Ce. This is also true.
Therefore, Rb and Rc cannot be 0. They are the only current limiters.
Practically, there is Rbe which is inside the transistor and is around 23 Ohms. So, when Rc = 750 Ohms and Rb = 0 Ohms, the gain is :
Gmax = - ( Rc / Rb ) * B = - ( 750 /23 ) * 60 = 1956.5 ~ 2000
Anyway, this point was not so important as point 1 and point 2 because, most people explain this point 3 but do not explain point 1 and point 2.
The schematics in the document show all diodes to be 1N60P which is correct to the design.
However, I did NOT have any 1N60P. I used a different Schottky diode for saturation protection and 1N914 standard Silicon diodes for the output. I have just received 1N60P. I have just installed them and played the updated device for around half of an hour.
This is why I have not previously replied nor posted.
I do so now and express the opinion the device works much better and does not require the next stage amplifier to be almost impossibly quieted down because of high, 0.65V amplitude levels. The amplitude now is 0.24V, which gives the amplifier the opportunity for more amplification.
I am very happy with the sound and will make MP3 files soon.
Please, be informed this is not a professional device and is not to replace the professional and the commercial ones.
********
Please, note : One of the goal I always have is to provide extremely clear explanations.
I think how to make explanations is science covered in linguistics and studied carefully by teachers in their pedagogy lessons. I am not sure whether this is the case but should be.
In this sense, here are the most important points in the EXPLANATIONS and not in the design of this device.
1. Input Capacitor
People say current can flow from high to low voltage potential only and not the other way around. OK. The guitar audio signal is AC with a maximal amplitude, say, 1V. The guitar signal is a voltage between the core of the output jack / socket and the surrounding tube of the said jack / socket, which can be called ground. Thus, the guitar signal is referenced to what can be called ground. The distortion takes this signal also to ground but a different ground : the distortion ground. When the two grounds are connected, the guitar signal is always referenced to ground and can be either negative or positive.
The distortion has a capacitor which connects the input ( I. e. the guitar output ) to a voltage divider which provides 6V towards ground. Thus, the right pin of the capacitor is connected to 6V and the left to an AC signal with amplitude of maximum 1V ( can be - 1V or 1V or anything in between ).
Now, here comes the point : the SAME people who say current can fly ONLY from high potential to low potential, ALSO SAY a signal with an amplitude of 1V can fly current in the two directions. How come? The signal is 1V maximum. The other side of the capacitor is 6V. Current can never fly from 1V to 6V! Thus, no AC signal can go through the capacitor because of the basic low of the potentials!
How can you all answer this?
I think the best answer is : Capacitor is such an animal which internally works the potentials out and allows AC to fly through and does not allow DC to fly through. In other words, capacitor is something which moves energy internally and makes potentials inside in order to allow AC to fly. In this example, when the capacitor sees 1V on the left side, the capacitor is intelligent enough to think : OK, someone ( the guitar ) wants to get 1V through. Yes, but I have 6V on the right. Then, I will get this one volt and elevate this one volt to 7V internally in order to have an internal elevation of 1V higher than what I have on the right pin, so I allow current to fly through.
And this is the so called principle of AC DC coupling. The capacitor imposes ( couples ) the AC over DC.
2. Emitter Capacitor Ce
People say Kirchhoff's law cannot be broken. People say there is some input voltage at Rb and this voltage must be repeated over this chain : Rb, BE junction ( base emitter junction of the transistor ) and Re || Ce. Then people say AC goes through Ce freely and does not make any potential between the emitter and ground. They say the **** AC **** potential of the emitter is 0.
So, therefore, the emitter is at 0V, thus, I can have AC between 0 and Vcc with an amplitude Vcc / 2! Thus, when I have a 15V supply and 4.6V DC with AC imposed over, I can get a range of 15V at the output! WRONG! Why? How come?
Kirchhoff's law cannot be broken. In front of Rb, there is 4.6V DC and AC imposed over. The BE of a transistor makes a DC only ( mainly, but consider only ) of 0.7V. Thus, the transistor should have 4.6V - 0.7V = 3.9VDC on the emitter with the AC signal super imposed over these 3.9V DC. However, there is a capacitor Ce. The AC goes through the capacitor and makes no voltage as the capacitor provides resistance of 0 Ohms for the AC. HOWEVER, the upper pin of the capacitor will remain at 3.9V and this voltage cannot be changed ( theoretically, under normal circumstances ). Thus, the transistor can pulsate from closed to open because of the AC current through the BE junction as much as wanted and I do not care of how much and why. The only thing I care is the emitter ( the upper pin of the capacitor ) will always see 3.9V under normal circumstances.
Thus, when the transistor is, FOR A MOMENT ONLY, fully open, the collector will have Uc = Ue + Ucesat. Ue has been proven to be 3.9V. Ucesat is, say, 0.1V. Therefore, Uc = 4V when the transistor is, FOR A MOMENT ONLY, fully open.
When the transistor is fully closed, the transistor will have Uce = Vcc ( 15V ) at the collector.
THUS, UNDER THE FORCE OF THE AC CURRENT THROUGH BE, THE TRANSISTOR WILL PULSATE IN SUCH A WAY, SO THE COLLECTOR VOLATGE Uc CHANGES FROM 4V TO 15V AND NOT FROM 0 TO 15V.
Thus, the middle point ( the DC ) of Uc is :
Ucdc = 4 + ( 15 - 4 ) / 2 = 9.5V
and the amplitude is :
Uca = 15 - 9.5 = 9.5 - 4 = 5.5V
AND NOT 7.5V.
So, electronics does not change neither Kirchhoff's, nor, Ohm's, nor any other law. Electronics only adjusts potentials and currents obeying these laws.
3. Something General
Yes, but, there is another question : the upper pin of Ce is at 3.9V. The lower pin is at 0V. The AC amplitude is 1V maximum. Therefore, the capacitor Ce will have a voltage which changes from 2.9V to 4.9V on the upper pin and 0V on the lower pin. When the explanation of point 1 is used, does this not mean there will be current which flies in the two directions of the capacitor freely and, because the capacitor, as mentioned, does not exert any resistance to AC, these currents will be huge, infinitely high. OK, the transistor cannot conduct from emitter to collector nor from emitter to base ( NPN ), therefore, the capacitor allows huge, infinite current from emitter to ground. Where does this current come from?
This is true. Theoretically there will be an infinitely high current from power supply through the Darlington buffer through Rb=0 through BE and through Ce. This infinitely high current will make another infinitely high current to fly from Vcc through Rc = 0 through CE ( Rce = 0 when fully open ) through Ce. This is also true.
Therefore, Rb and Rc cannot be 0. They are the only current limiters.
Practically, there is Rbe which is inside the transistor and is around 23 Ohms. So, when Rc = 750 Ohms and Rb = 0 Ohms, the gain is :
Gmax = - ( Rc / Rb ) * B = - ( 750 /23 ) * 60 = 1956.5 ~ 2000
Anyway, this point was not so important as point 1 and point 2 because, most people explain this point 3 but do not explain point 1 and point 2.
There is another question which is, usually, explained well but I would like to quickly go through an explanation which I think is good :
Why does a common collector buffer? How come the common collector takes only a slight current from the voltage source applied at the base of the common collector. Assume we have an ideal voltage source of 10V, a Vcc of 15V, a common collector transistor and a load resistor of 100 Ohms. Therefore, the voltage source will see the base emitter junction ( a diode ) which consumes 0.7V and then a 100 Ohms load which has 10V - 0.7 = 9.3V on the upper pin and 0V ( ground ) on the lower. Therefore, the circuit will consume 9.3V / 100 Ohms = 93mA from the voltage source. This is almost the same as without the transistor : the load will consume 10V / 100 Ohms = 100mA. How come people say the common collector buffers?
This is because the transistor is not a simple base emitter diode but is a base emitter diode AND something else : Bipolar transistor is such an animal which measures the current between the base and emitter and opens in such a way as to allow a new current to flow from collector to emitter which current is the measured base emitter current multiplied by a coefficient which belongs to the transistor internally ( the transistor is made this way ) and this coefficient is called a current gain of the transistor or, simply, Beta because of the Greek symbol with which this coefficient has been denoted.
Unlike a simple hydro or pneumatic valve, the transistor allows two current to fly through : the input current which is between the base and emitter and the output current which is through collector and emitter and these two currents ADD EACH OTHER UP at the emitter in accordance with the current law which says : when two currents meet at a fork, the two add each other up and the combined, added current continues through the fork.
So transistors and electronics do NOT break the current law. They, as previously mentioned, do NOT break the voltage law ( the Kirchhoff's law ).
So, how does the buffer buffer? Simple : VERY INITIALLY, some current gets to fly through the base emitter junction. Yes, the base emitter junction is a diode. Takes 0.7V. Then, yes, there is a 100 Ohm resistor. So, VERY INITIALLY, a current of 93mA will fly from the voltage source through the base emitter and through the load resistor of 100 Ohms. Kirchhoff's law will not be broken : the voltage source' voltage of 10V will be equal to the base emitter voltage ( 0.7V ) plus the voltage over the load resistor ( 9.3V ). However, this is VERY INITIALLY. Then, the transistor opens to allow a current of 93mA * B to fly through collector emitter and through the same load resistor of 100 Ohms. Assume a Beta of 200 ( which is common for some transistors ). Therefore, the VERY INITIAL current through the transistor will be 93mA * 200 = 18.6A! This current will meet the input current of 93mA at the emitter fork to make a combined current of 93mA + 18.6A = 18.693A ~ 18.7A! This current will meet the load resistance of 100 Ohms to make ( as per Herr Ohm ) 18.7A * 100 Ohms = 1870V! This looks like a power station! Now look what happens : there is a voltage source of 10V. There is a supply voltage Vcc of 15V. And then, there is an output voltage of 1870V. Where does this voltage come from?
Nowhere. Why? Because this voltage is not true. Why? Because, assume this voltage is true, therefore, the base has 10V and the emitter has 1870V. The emitter voltage is higher than the base voltage. THEREFORE, CURRENT DOES NOT FLY FROM BASE TO EMITTER BECAUSE CURRENT ONLY FLIES FROM HIGH TO LOW POTENTIAL. So, the input current is 0A. Therefore, the output current is 0A * B = 0A * 200 = 0A. So, the output current is also 0A. The input current and the output current meet at the emitter fork and add up to make 0A + 0A = 0A and this current meets the load resistance to make a voltage of 0A * 100 Ohms = 0V. The load resistor, as mentioned, has the upper pin connected to the emitter and the lower pin connected to 0V ( ground ) and a voltage of 0V over. Therefore, the emitter will have 0V.
What happened. There was 1870A just a few minutes back and now there is 0V. Which one is true? What kind of buffer is this to buffer 10V into 0V.
Yes, but look what happened now : the base has 10V and the emitter has 0V. AND current ALWAYS flies from high potential to low when there is nothing to be stopped by. The BE junction is like a diode which allows current to flow from B to E ( NPN ), therefor, there is nothing to stop the current to fly from base ( 10V ) to emitter ( 0V ). Therefore current will fly from base to emitter through the resistor and, again as per Herr Kirchhoff, the voltage over the load resistor will be 10V - 0.7V = 9.3V which will drive 93mA through the load resistor which come from the voltage source at the base. The transistor will open to allow 93mA * B = 93mA * 200 = 18.6A BUT such a huge current cannot fly because this current would make 1870V voltage over the load resistor and this voltage cannot come form anywhere because the transistor is NOT a power station and neither generates current nor voltage BUT ONLY CHANGES THE RESISTANCE BETWEEN C AND E just like a hydro valve which opens to an amount defined by the user to exert only so much resistance to the flow of water. Thus, the transistor will fully open to allow Vcc - Ucesat = 15 - 0.1 = 14.9V over the resistor ( the maximum possible voltage in this situation.
Yes, but this voltage of 14.9V is higher than the 10V at the base, thus, no current will fly. Yes, but then, the voltage over the resistor will be 0V. Yes, but then the voltage ( potential ) of the emitter will be come 0V and 10V on the base and 0V on the emitter means huge current would fly which means huge current will fly through CE junction, yes, but this means the voltage over the resistor is 14.9V and, therefore, no current will fly anywhere, yes, but then the voltage over the resistor is 0V and current will fly and then will not and then will and then : this way all the way for ever!
The explanation is this : this cycle happens with the speed of the electric current which is near the speed of light which can be accepted as infinite speed. So, terms as VERY IMMEDIATE and THEN happen immediately for 0s. Thus, on one hand, the voltage source and Kirchhoff's law would want to make a large current over the load resistor, BUT, on the other hand, this large current will make the transistor fully open and make the voltage over the resistor high which would stop the input current. The two hands meet at the emitter and start to negotiate. And the negotiation outcome is to allow ONLY SOME current to fly through the base emitter junction and the collector emitter junction which will add up and make ONLY SOME voltage over the load resistor.
How do I know this? Because Kirchhoff's law cannot be broken. REGARDLESS OF WHAT THE TRANSISTOR DOES AND DOES NOT, THE CURRENT THROUGH THE LOAD RESISTOR WILL BE SUCH, SO THE VOLTAGE OVER THIS RESISTOR IS 9.3V WHICH VOLTAGE WILL ADD WITH THE BASE EMITTER VOLTAGE OF 0.7V TO MAKE 10V WHICH IS THE VOLTAGE OF THE SOURCE AND THIS SOURCE VOLTAGE CANNOT BE CHANGED BY ANYTHING BECAUSE THE SOURCE IS IDEAL AND HAS 0 OHMS OUTPUT RESISTANCE WHICH MEANS THE SOURCE WILL MAINTAIN THESE 10V REGRADLESS OF THE CURRENT.
The important point is : because of the Kirchhoff's law, the voltage over the load resistor will always be 9.3V.
Ohm's law can never be broken too. Electronics does not and cannot break the Ohm's law. Therefore, when a resistor of 100 Ohms has a voltage of 9.3V over, the current through this resistor will always be 9.3V / 100 Ohms = 93mA.
Therefore, regardless of what the transistor does or does not, the transistor cannot break neither Kirchhoff's law nor the Ohm's law, nor any other law. Therefore, the voltage ( potential ) of the emitter will always be 9.3V and the current from the emitter ( fork ) through the resistor and to ground will always be 93mA!
Yes, but, as previously mentioned, the fork output current of the emitter is the addition of two currents, one is the base emitter current, a. k. a. the input to the transistor current and the other is the collector emitter current, a. k. a. the out current of the transistor.
Yes, but, as previously mentioned, transistor is such an animal which, when possible, opens just so much as to allow an output ( CE ) current to flow which is exactly equal to the input ( BE ) current multiplied by an internal coefficient ( which belongs to the transistor, I. e., this is how the transistor has been made ) called Beta.
So, the current through the resistor is known and is 93mA. Also known is this current is the addition of an input and output current. also known is the output current = the input current * Beta.
Therefore, known is :
|
| Ibe + Ice = 93mA
| Ice = B * Ibe
|
Oops! We know two deterministic and independent things of two things we do not know : which, in logic, means we know everything.
In mathematics : we have a system of two equation with two unknowns :
Ibe = x
Ice = y
=>
|
| x + y = 0.093
| y = B * x
|
=>
x + B * x = 0.093
=>
x * ( B + 1 ) = 0.093 => x * ( 200 + 1 ) = 0.093
=>
x = 0.093 / 201 = 0.0004626865671641791
=>
y = ( 0.093 / 201 ) * 200 = 0.093 * 200 / 201 = 0.0925373134328358
=>
Ibe = 463uA ~ 500uA
Ice = 0.0925 ~ 92.5mA
Therefore, the circuit will consume only current which is equal to the input voltage minus 0.7, al these divided by the load resistance and all these divided by the transistor's Beta! Because the transistor's Beta is rather high ( for most transistors except for the old and still in use high power ones at high currents, where Beta is around 10 ), in most case the consumed current is in the microampere range.
Thus, the circuit consumes only a half of a milliampere ( 500 microamperes )from the source and maintains the source' voltage ( minus 0.7V ) over a load which consumes nearly 100mA ( 93mA ). And this is a buffer.
To make the buffer buffer even more, another transistor can be added to the schematic in Darlington configuration to the first one. This way the input current of the old transistor (500uA ) will be the output current of the newly added transistor and the input current to the newly added transistor would be 500uA / Beta = 500uA / 200 = 2.5uA. The voltage consumed by the base emitter junction of the new transistor ( 0.7V ) adds to the voltage consumed by the base emitter junction of the old transistor ( 0.7V ) to make a combined voltage of 1.4V.
Thus, the circuit consumes only 2.5 microamperes from the source and maintains the source' voltage ( minus 1.4V ) over a load which consumes nearly 100mA ( 93mA ). Regardless of how the load changes, the circuit would consume negligibly low current from the source and maintain the same voltage minus a constant offset of 1.4V. And this is a buffer.
Then to make a stronger buffer, another transistor can be added to make a triple Darlington, then another to make a quadriple Darlington etcetera. However, every transistor added reduces the voltage range by 0.7V because these 0.7V must be subtracted from the source. In many cases this is not the problem. However, the higher the transistor count, the lower the input current, the higher the input resistance of the circuit but also the higher the sensitivity. And the higher the sensitivity, the higher the noise amplification or transmission. The higher the output impedance of the source, the higher the noise too.
And this is why I said I could easily increase the input impedance of the distortion but, because the source ( the guitar ) has high output impedance, I will also increase the noise.
Generally, the engineers like the 10% signal decrease : the input impedance must be 10 fold higher than the output impedance to lose only around 10% of the signal. Thus, in case the maximum output impedance from two coils in sequence ( the worst case scenario ) is 500K, the input impedance must be 5MOhm but, in case the MAXIMUM output impedance of the guitar is <= 10K, then 100K is OK. The rule is not to go much higher than 10 * the maximum output impedance of the worst ( in output impedance ) guitar, currently in mass production.
An the problem is I do not know how much this is as different sources online say different things. Even work, now, some say, the output impedance changes drastically over the audio frequency range despite the manufacturers efforts the change not to be so drastic as far as I can imagine.
Why does a common collector buffer? How come the common collector takes only a slight current from the voltage source applied at the base of the common collector. Assume we have an ideal voltage source of 10V, a Vcc of 15V, a common collector transistor and a load resistor of 100 Ohms. Therefore, the voltage source will see the base emitter junction ( a diode ) which consumes 0.7V and then a 100 Ohms load which has 10V - 0.7 = 9.3V on the upper pin and 0V ( ground ) on the lower. Therefore, the circuit will consume 9.3V / 100 Ohms = 93mA from the voltage source. This is almost the same as without the transistor : the load will consume 10V / 100 Ohms = 100mA. How come people say the common collector buffers?
This is because the transistor is not a simple base emitter diode but is a base emitter diode AND something else : Bipolar transistor is such an animal which measures the current between the base and emitter and opens in such a way as to allow a new current to flow from collector to emitter which current is the measured base emitter current multiplied by a coefficient which belongs to the transistor internally ( the transistor is made this way ) and this coefficient is called a current gain of the transistor or, simply, Beta because of the Greek symbol with which this coefficient has been denoted.
Unlike a simple hydro or pneumatic valve, the transistor allows two current to fly through : the input current which is between the base and emitter and the output current which is through collector and emitter and these two currents ADD EACH OTHER UP at the emitter in accordance with the current law which says : when two currents meet at a fork, the two add each other up and the combined, added current continues through the fork.
So transistors and electronics do NOT break the current law. They, as previously mentioned, do NOT break the voltage law ( the Kirchhoff's law ).
So, how does the buffer buffer? Simple : VERY INITIALLY, some current gets to fly through the base emitter junction. Yes, the base emitter junction is a diode. Takes 0.7V. Then, yes, there is a 100 Ohm resistor. So, VERY INITIALLY, a current of 93mA will fly from the voltage source through the base emitter and through the load resistor of 100 Ohms. Kirchhoff's law will not be broken : the voltage source' voltage of 10V will be equal to the base emitter voltage ( 0.7V ) plus the voltage over the load resistor ( 9.3V ). However, this is VERY INITIALLY. Then, the transistor opens to allow a current of 93mA * B to fly through collector emitter and through the same load resistor of 100 Ohms. Assume a Beta of 200 ( which is common for some transistors ). Therefore, the VERY INITIAL current through the transistor will be 93mA * 200 = 18.6A! This current will meet the input current of 93mA at the emitter fork to make a combined current of 93mA + 18.6A = 18.693A ~ 18.7A! This current will meet the load resistance of 100 Ohms to make ( as per Herr Ohm ) 18.7A * 100 Ohms = 1870V! This looks like a power station! Now look what happens : there is a voltage source of 10V. There is a supply voltage Vcc of 15V. And then, there is an output voltage of 1870V. Where does this voltage come from?
Nowhere. Why? Because this voltage is not true. Why? Because, assume this voltage is true, therefore, the base has 10V and the emitter has 1870V. The emitter voltage is higher than the base voltage. THEREFORE, CURRENT DOES NOT FLY FROM BASE TO EMITTER BECAUSE CURRENT ONLY FLIES FROM HIGH TO LOW POTENTIAL. So, the input current is 0A. Therefore, the output current is 0A * B = 0A * 200 = 0A. So, the output current is also 0A. The input current and the output current meet at the emitter fork and add up to make 0A + 0A = 0A and this current meets the load resistance to make a voltage of 0A * 100 Ohms = 0V. The load resistor, as mentioned, has the upper pin connected to the emitter and the lower pin connected to 0V ( ground ) and a voltage of 0V over. Therefore, the emitter will have 0V.
What happened. There was 1870A just a few minutes back and now there is 0V. Which one is true? What kind of buffer is this to buffer 10V into 0V.
Yes, but look what happened now : the base has 10V and the emitter has 0V. AND current ALWAYS flies from high potential to low when there is nothing to be stopped by. The BE junction is like a diode which allows current to flow from B to E ( NPN ), therefor, there is nothing to stop the current to fly from base ( 10V ) to emitter ( 0V ). Therefore current will fly from base to emitter through the resistor and, again as per Herr Kirchhoff, the voltage over the load resistor will be 10V - 0.7V = 9.3V which will drive 93mA through the load resistor which come from the voltage source at the base. The transistor will open to allow 93mA * B = 93mA * 200 = 18.6A BUT such a huge current cannot fly because this current would make 1870V voltage over the load resistor and this voltage cannot come form anywhere because the transistor is NOT a power station and neither generates current nor voltage BUT ONLY CHANGES THE RESISTANCE BETWEEN C AND E just like a hydro valve which opens to an amount defined by the user to exert only so much resistance to the flow of water. Thus, the transistor will fully open to allow Vcc - Ucesat = 15 - 0.1 = 14.9V over the resistor ( the maximum possible voltage in this situation.
Yes, but this voltage of 14.9V is higher than the 10V at the base, thus, no current will fly. Yes, but then, the voltage over the resistor will be 0V. Yes, but then the voltage ( potential ) of the emitter will be come 0V and 10V on the base and 0V on the emitter means huge current would fly which means huge current will fly through CE junction, yes, but this means the voltage over the resistor is 14.9V and, therefore, no current will fly anywhere, yes, but then the voltage over the resistor is 0V and current will fly and then will not and then will and then : this way all the way for ever!
The explanation is this : this cycle happens with the speed of the electric current which is near the speed of light which can be accepted as infinite speed. So, terms as VERY IMMEDIATE and THEN happen immediately for 0s. Thus, on one hand, the voltage source and Kirchhoff's law would want to make a large current over the load resistor, BUT, on the other hand, this large current will make the transistor fully open and make the voltage over the resistor high which would stop the input current. The two hands meet at the emitter and start to negotiate. And the negotiation outcome is to allow ONLY SOME current to fly through the base emitter junction and the collector emitter junction which will add up and make ONLY SOME voltage over the load resistor.
How do I know this? Because Kirchhoff's law cannot be broken. REGARDLESS OF WHAT THE TRANSISTOR DOES AND DOES NOT, THE CURRENT THROUGH THE LOAD RESISTOR WILL BE SUCH, SO THE VOLTAGE OVER THIS RESISTOR IS 9.3V WHICH VOLTAGE WILL ADD WITH THE BASE EMITTER VOLTAGE OF 0.7V TO MAKE 10V WHICH IS THE VOLTAGE OF THE SOURCE AND THIS SOURCE VOLTAGE CANNOT BE CHANGED BY ANYTHING BECAUSE THE SOURCE IS IDEAL AND HAS 0 OHMS OUTPUT RESISTANCE WHICH MEANS THE SOURCE WILL MAINTAIN THESE 10V REGRADLESS OF THE CURRENT.
The important point is : because of the Kirchhoff's law, the voltage over the load resistor will always be 9.3V.
Ohm's law can never be broken too. Electronics does not and cannot break the Ohm's law. Therefore, when a resistor of 100 Ohms has a voltage of 9.3V over, the current through this resistor will always be 9.3V / 100 Ohms = 93mA.
Therefore, regardless of what the transistor does or does not, the transistor cannot break neither Kirchhoff's law nor the Ohm's law, nor any other law. Therefore, the voltage ( potential ) of the emitter will always be 9.3V and the current from the emitter ( fork ) through the resistor and to ground will always be 93mA!
Yes, but, as previously mentioned, the fork output current of the emitter is the addition of two currents, one is the base emitter current, a. k. a. the input to the transistor current and the other is the collector emitter current, a. k. a. the out current of the transistor.
Yes, but, as previously mentioned, transistor is such an animal which, when possible, opens just so much as to allow an output ( CE ) current to flow which is exactly equal to the input ( BE ) current multiplied by an internal coefficient ( which belongs to the transistor, I. e., this is how the transistor has been made ) called Beta.
So, the current through the resistor is known and is 93mA. Also known is this current is the addition of an input and output current. also known is the output current = the input current * Beta.
Therefore, known is :
|
| Ibe + Ice = 93mA
| Ice = B * Ibe
|
Oops! We know two deterministic and independent things of two things we do not know : which, in logic, means we know everything.
In mathematics : we have a system of two equation with two unknowns :
Ibe = x
Ice = y
=>
|
| x + y = 0.093
| y = B * x
|
=>
x + B * x = 0.093
=>
x * ( B + 1 ) = 0.093 => x * ( 200 + 1 ) = 0.093
=>
x = 0.093 / 201 = 0.0004626865671641791
=>
y = ( 0.093 / 201 ) * 200 = 0.093 * 200 / 201 = 0.0925373134328358
=>
Ibe = 463uA ~ 500uA
Ice = 0.0925 ~ 92.5mA
Therefore, the circuit will consume only current which is equal to the input voltage minus 0.7, al these divided by the load resistance and all these divided by the transistor's Beta! Because the transistor's Beta is rather high ( for most transistors except for the old and still in use high power ones at high currents, where Beta is around 10 ), in most case the consumed current is in the microampere range.
Thus, the circuit consumes only a half of a milliampere ( 500 microamperes )from the source and maintains the source' voltage ( minus 0.7V ) over a load which consumes nearly 100mA ( 93mA ). And this is a buffer.
To make the buffer buffer even more, another transistor can be added to the schematic in Darlington configuration to the first one. This way the input current of the old transistor (500uA ) will be the output current of the newly added transistor and the input current to the newly added transistor would be 500uA / Beta = 500uA / 200 = 2.5uA. The voltage consumed by the base emitter junction of the new transistor ( 0.7V ) adds to the voltage consumed by the base emitter junction of the old transistor ( 0.7V ) to make a combined voltage of 1.4V.
Thus, the circuit consumes only 2.5 microamperes from the source and maintains the source' voltage ( minus 1.4V ) over a load which consumes nearly 100mA ( 93mA ). Regardless of how the load changes, the circuit would consume negligibly low current from the source and maintain the same voltage minus a constant offset of 1.4V. And this is a buffer.
Then to make a stronger buffer, another transistor can be added to make a triple Darlington, then another to make a quadriple Darlington etcetera. However, every transistor added reduces the voltage range by 0.7V because these 0.7V must be subtracted from the source. In many cases this is not the problem. However, the higher the transistor count, the lower the input current, the higher the input resistance of the circuit but also the higher the sensitivity. And the higher the sensitivity, the higher the noise amplification or transmission. The higher the output impedance of the source, the higher the noise too.
And this is why I said I could easily increase the input impedance of the distortion but, because the source ( the guitar ) has high output impedance, I will also increase the noise.
Generally, the engineers like the 10% signal decrease : the input impedance must be 10 fold higher than the output impedance to lose only around 10% of the signal. Thus, in case the maximum output impedance from two coils in sequence ( the worst case scenario ) is 500K, the input impedance must be 5MOhm but, in case the MAXIMUM output impedance of the guitar is <= 10K, then 100K is OK. The rule is not to go much higher than 10 * the maximum output impedance of the worst ( in output impedance ) guitar, currently in mass production.
An the problem is I do not know how much this is as different sources online say different things. Even work, now, some say, the output impedance changes drastically over the audio frequency range despite the manufacturers efforts the change not to be so drastic as far as I can imagine.
Sure it can be. To the intended circuit you want to analyze, add a "leak", but do not tell Mr Kirchhoff about it. This happens a LOT in real life. Solder blobs, flux stains, bad parts, older cardboard-base circuit boards stored in a damp shed since before you were born.
And, as you say, Kirchhoff can't really be fooled. When his math differs from observation, we konw to observe closer and find the sneak-path. (Basic to much forum debugging.)
If you don't know distortion/fuzz/overdrive effects, you should get out more. I admire the effort to re-invent fuzz from a clean sheet but a survey of the landscape is good too.
Transistors destroyed the real electronics engineering, which is tube based. Tubes are easily explained. Transistors work by magic. (Even though I now have a deeper understanding of crystal fundamentals, I still blame it on magic.)
1. The first electronic effect .... 2. The first tube ( valve ) ...3. The first triode ... 4. The first differential ... 5. Some companies made huge design tube ( valve ) which was around 2 meters high, so a person can open a door and go inside to adjust and tests electrodes, etcetera. The, the person would go out and close the door. ... 6. In the 60's through 80's, computers have been made with acorn sized tubes to be resistive to an electromagnetic pulse .... 7. I THINK BUT THIS HAS NOT BEEN PROVEN : The first IC was made by tubes ( valves ) with common walls. Thus, 5 or 6 tubes ( valves ) can be positioned next to each other to save room. ...
Interesting interpretation of history. Many of your factoids are easily checked, such as the Edison - Fleming - De Forest story.
I have never seen a hint of a walk-in tube chamber. I was just looking at a quart-size vaccum for tube research. But glass is so easy to blow that much-much work was just done in lamp-bottles.
There were "Acorn Tubes". As useful frequency rose from 100Kcps to 1Mcps to above 10Mcps, the lead length (see "Octal") became a limit. The first crack was to make the tube guts small and press them between two glass thimbles, Acorn Tube. This led almost directly to the "Miniature" tube with slightly longer leads and much easier socketing. Beyond that we find "coaxial", "lighthouse", and other odd names with key electrodes directly brazed to rings for broad (low inductance) contact. Hearing-aid tubes are another path.
Tube makers charged by the tube and had little incentive to make multi-unit types. Also tube reliability was an issue until late 1930s-- do you want to throw-out 4 good units because one unit went bad? However in Germany there was a tax per "tube". One company did put the whole radio in one bottle to cheat the tax. They offered a rebuilding service to overcome the reliability objection. The idea did not catch on. Dual/twin tubes did become common mid-1940s, limited by standard bases. In 1960s triple tubes simplified TV production. Most tube systems are not space-limited by the bottles but by other parts.
Multi-unit tubes are made individually then assembled 2 to a bottle. The killer advantage promised by Integrated Circuits is that all 4 or 7 or 47 transistors *and* related parts are made in ONE (multi-step) process. Difference between printing a newspaper and assembling newspaper clippings into a ransom note. Drastic reduction in solder joints, which then as now is the #1 cause of trouble.
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I am NOT interested in terms, thus, I do NOT even want to know what fizz, distortion and overdrive are. I am only interested in the principles. To give an example : there is not such a thing called distortion, there is a sound to rectangular wave converter. There is no such a thing as overdrive, there is sound wave to out of range converter. Terms are made by companies to sell products. Terms bring misunderstanding, miscommunication and misleading. Better use one sentence to say what you want to say than a stupid word, such as laser ( Light Amplification and Stimulated Emission of Radiation ). There is not such a thing. There is a light amplifier but there is not a laser. Some idiots in physics decided to use garbage and made up the word.
Solder points, leakage, etcetera, do NOT break Kirchhoff's law. The fact people do not consider them does not mean the law is broken.
Bipolar Junction Transistor design requires more engineering because of the current and voltage control. MOS, JFET's, FET's, tubes ( valves ), etcetera are only voltage controlled. Another problem from the point of view of LOGIC is : tubes ( valves ) require heat to release electrons which takes away the focus of the engineer to things which are NOT connected with the logic of the circuit. The same applies to everything electrical, such as capacitors, coils and combination thereof with resistors. These require attention and they are NOT part of electronics, yet, electronics does use them. The use of the logic of the circuit only is called electronic schemotechnology which concentrates ONLY on electronic analogue LOGIC to make the schematics work. Electrical technology deals with all electrical and non electronic things.
People use these terms incorrectly : electrical is electrical only and does not include electronics. Electronic is electronic only and does not include electrical. Too bad electronics engineers need to know some electrical. Electrical engineers do NOT need to know anything electronic but they do anyway.
I watched a documentary which showed the chamber where engineers go to make and fine tune the tube ( valve ) in order to test a given design concept. The was done so the engineers did not need to manufacture a tube just to see whether an idea is appropriate. The engineers would use the chamber to play with their designs. After the chamber gave them a satisfactory result, they would ask the manufacturers to make them a real size tube ( valve ) to test the real deal.
This was the opposite than aeroplanes : with the aeroplanes, they would make a tiny design, like a model plane, to test the aerodynamics in a tunnel. Then they would make the real plane. With tubes ( valves ) they would make a giant tube and then the real one which is very tiny as compared to the test chamber.
I was amazed by this documentary.
The information on Germany was very interesting. I can only add some things :
Germany made the most inexpensive radio in the world in order to encourage people to possess a radio in every home.
There were tube ( valve ) factories in Germany as well as Danmark, also used by Germany.
German tank crews used radio inside the tank for inside communications as well as for outside communications. The other countries used this way too.
The US had a military cell phone made of tubes ( valves ). The device was more than 50cm long and 25cm wide. There were buttons with numbers like on the normal phones. The phone was used in the fields.
A British engineer called Tommy Flowers made one of the first tube ( valve ) computers. The program and the data was on perforated scrolls. The computer was named Colosus and had 1850 tubes with a suggestion for 2000 for other models. Historians claim Colosus was used to break the Enigma code.
Hope this was interesting. Whatever is posted just to be interesting does not need to be checked for truthfulness and, in most cases, cannot because the Internet is not a good source of information, because anyone can say anything there. The same applies to documentaries, yet, documentaries are somewhat ( not a lot ) screened for truthfulness. Thus, the factoids cannot be checked online. Even when they seem to be true because other people say the same, they may steel not be because many people may say wrong things. Thus, whatever anyone says may or may not be true and CANNOT BE PROVEN. Many people make the mistake to check something online and think this is true because OTHER PEOPLE say so. But, in most cases, this is not true. The internet is like urban legends : may or may not be true but people repeat them. Thus, what other say is not important. What a given person says is and may or may not be true and cannot be proven. The important thing is a given person says so not whether true. And there are many reasons for one to say so, for example, for INTERESTING and for TO BE SAID. In case I have not said these, as you call them, factoids, no one would have ever gone to find them out. And here is the other problem of internet : too much information is not information. Even Mick Jagger sings of useless information. Thus :
Internet is NOT a good source of information because whatever said there may NOT be true and, in most cases, is not. Also, too much information is not information because the relevant information cannot be found. Search engines cannot find information, they can react only to basic things and whatever they find is also part of the internet, which means : may not be true ( and in most cases is not ) and there is too much of information found by the search engines which means the real information cannot be found.
Solder points, leakage, etcetera, do NOT break Kirchhoff's law. The fact people do not consider them does not mean the law is broken.
Bipolar Junction Transistor design requires more engineering because of the current and voltage control. MOS, JFET's, FET's, tubes ( valves ), etcetera are only voltage controlled. Another problem from the point of view of LOGIC is : tubes ( valves ) require heat to release electrons which takes away the focus of the engineer to things which are NOT connected with the logic of the circuit. The same applies to everything electrical, such as capacitors, coils and combination thereof with resistors. These require attention and they are NOT part of electronics, yet, electronics does use them. The use of the logic of the circuit only is called electronic schemotechnology which concentrates ONLY on electronic analogue LOGIC to make the schematics work. Electrical technology deals with all electrical and non electronic things.
People use these terms incorrectly : electrical is electrical only and does not include electronics. Electronic is electronic only and does not include electrical. Too bad electronics engineers need to know some electrical. Electrical engineers do NOT need to know anything electronic but they do anyway.
I watched a documentary which showed the chamber where engineers go to make and fine tune the tube ( valve ) in order to test a given design concept. The was done so the engineers did not need to manufacture a tube just to see whether an idea is appropriate. The engineers would use the chamber to play with their designs. After the chamber gave them a satisfactory result, they would ask the manufacturers to make them a real size tube ( valve ) to test the real deal.
This was the opposite than aeroplanes : with the aeroplanes, they would make a tiny design, like a model plane, to test the aerodynamics in a tunnel. Then they would make the real plane. With tubes ( valves ) they would make a giant tube and then the real one which is very tiny as compared to the test chamber.
I was amazed by this documentary.
The information on Germany was very interesting. I can only add some things :
Germany made the most inexpensive radio in the world in order to encourage people to possess a radio in every home.
There were tube ( valve ) factories in Germany as well as Danmark, also used by Germany.
German tank crews used radio inside the tank for inside communications as well as for outside communications. The other countries used this way too.
The US had a military cell phone made of tubes ( valves ). The device was more than 50cm long and 25cm wide. There were buttons with numbers like on the normal phones. The phone was used in the fields.
A British engineer called Tommy Flowers made one of the first tube ( valve ) computers. The program and the data was on perforated scrolls. The computer was named Colosus and had 1850 tubes with a suggestion for 2000 for other models. Historians claim Colosus was used to break the Enigma code.
Hope this was interesting. Whatever is posted just to be interesting does not need to be checked for truthfulness and, in most cases, cannot because the Internet is not a good source of information, because anyone can say anything there. The same applies to documentaries, yet, documentaries are somewhat ( not a lot ) screened for truthfulness. Thus, the factoids cannot be checked online. Even when they seem to be true because other people say the same, they may steel not be because many people may say wrong things. Thus, whatever anyone says may or may not be true and CANNOT BE PROVEN. Many people make the mistake to check something online and think this is true because OTHER PEOPLE say so. But, in most cases, this is not true. The internet is like urban legends : may or may not be true but people repeat them. Thus, what other say is not important. What a given person says is and may or may not be true and cannot be proven. The important thing is a given person says so not whether true. And there are many reasons for one to say so, for example, for INTERESTING and for TO BE SAID. In case I have not said these, as you call them, factoids, no one would have ever gone to find them out. And here is the other problem of internet : too much information is not information. Even Mick Jagger sings of useless information. Thus :
Internet is NOT a good source of information because whatever said there may NOT be true and, in most cases, is not. Also, too much information is not information because the relevant information cannot be found. Search engines cannot find information, they can react only to basic things and whatever they find is also part of the internet, which means : may not be true ( and in most cases is not ) and there is too much of information found by the search engines which means the real information cannot be found.
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