I know how to solder, I am aware of the possible need to use a heatsink. I'm using a solderless board as this is just an experiment at this point.
Right now the blackberry is outputting about 200mV (give or take). I came up with that by driving the speaker and hoking up the scope. The top of the waveform seemed to be around 200mV+-. However, I'm not sure what current that was at. Though I do know the speaker is rated as 3ohm. My understanding is with AC I can't use Ohms Law.... if I could I'd say (.2V/3O=.0666) I'm at 67mA.
This cheap speaker doesn't appear to give a maximum. However I have run it from a 30 watt system... I'm pretty sure (if I recall) for me to *hear* a difference I have to like *double* the power? So If the blackberry is at 200mV (peak) then if I could even double that or so.. Or heck drive it up to 1v (p-p).. that would be a good experiment in my book!
There is no hidden agenda here. I'm just playing with this trying to learn and have some fun! I've been thinking about it and I see why the idea of driving a 3 ohm speaker from a 10k ohm amp is dumb. That makes sense. I'd either need speakers with WAY more impedance or I need something to go between my high impedance amp and the low impedance speaker. I just don't know what...
Oh, I do have a PSU, but up until now I was just using a 9v batt. It doesn't matter to me either way which I use. I'm open to suggestions!
Well if you want to learn then I would suggest a really simple opamp and 2 transistors as a basic Class B amp and get that working. Then refine it and maybe move on to an all discrete design.
It can be built up step by step and will help you learn along the way.
An important question is what power supply you have available and what voltage. I see you had 9 volts in the earlier circuit. Are you just using a PP3 or do you have something better ?
It can be built up step by step and will help you learn along the way.
An important question is what power supply you have available and what voltage. I see you had 9 volts in the earlier circuit. Are you just using a PP3 or do you have something better ?
With apologies to Craig for just wandering slightly...
Has LTSpice got this right ? and if so I'm impressed. The currents and voltage across the transistor are so low in this circuit that I wondered if it would work as now redrawn. Under the "right" conditions you can use a transistor in a totally reversed biased state and it will still function with gain.
This circuit just had that look about it.
Has LTSpice got this right ? and if so I'm impressed. The currents and voltage across the transistor are so low in this circuit that I wondered if it would work as now redrawn. Under the "right" conditions you can use a transistor in a totally reversed biased state and it will still function with gain.
This circuit just had that look about it.
Attachments
I can do an opamp. Whatever gets me rolling! I would like it *basic* enough that it isnt all just "chip magic" though. Do you know of a schematic i can use that would fit the bill? This is where this thread pretty much started. Im just looking to get something working.
Im not sure what a pp3 is. So im guessing i dont have that. The design i was using earlier used a 9v battery. I have a power supply i built from an lm317 that goes to aroung 14v. I do have a few transformers around that may get employed down the road.
For a discrete design, it doesent get much simpler than the old elektor darlington amp: http://i.imgur.com/AjAWI.png
But great for learning.
And here is an opamp based test circuit.
(PP3 refers to the small 9 volt batteries commonly used in remotes etc which tbh are useless for this kind of work as they have so little capacity.)
This circuit would suit a common 741 type opamp. It runs in Class B so there are no bias worries (all that comes later
) but it's biggest problem is the limited 9 volt supply. It's a starting point though. If you run the simulation and zoom in on the centre part of the output signal (where it crosses the notional centre part) you should see a problem. Now reduce R5 to say 82 ohms and try again.
Within reason this amp will work on any supply from 9 volts or so up to about 30 but higher voltages would require higher rated transistors.
Can you see how the gain is determined and how you would alter it to suit ?
Can you work out the input impedance and how you would alter it ?
Do you notice anything about the relationship between Vin and Vout ?
Attachments
Where did you find the "TIP" transistors in LTSPICE? If I use generic NPN and PNPs this is what I'm getting.
notice the red arrows are where the probes are connected.
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Let's see. There is a lot to look at in this circuit for me.. but:
The first thing I see working my way from left to right is a RC filter. Looks like it is sending everything below 200hz to dirt?
Next iss a 10K filter (Rin) in series with the signal. Not sure why you would do that, seems like it would reduce the voltage on your signal line giving you more amplification you need to do?
I see a voltage divider (R2 and R3) giving 4.5v to the non-inverting input of the opamp.
The inverting side gets the signal and also feedback (via Rf). I would think Rf is the answer to your Q1 above (I numbered them) but I tried messing with the value and my sinewave turned ugly really fast, so I don't think I'm correct on that. I wonder if it has to do with the Rin resistor I couldn't figure out? perhaps some ratio going on there, like a voltage divider?
Next opamp output goes to the base of both of the transistors (q1, q2) and also has a resitor to the output (r5) not sure what r5 does. Though I did lower it to 82 and saw the small "hitch" in the sinewave go away...
The transistors output (q1, q2) goes to a cap(c2) which I would think is just to pull out a DC bias? then it goes to the 8ohm load (r4).
Not sure what C3 is doing, I'm guessing another filter of some sort?
Q2: how to see input Z? Hmm, I'm going to guess 10K as that is the resistor in series with the input signal?
q3: They look backwards, or inverted?
Thanks for the continued assistance! A couple of questions other than above:
1. in this circuit the LT1022 is a preamp?
2. in this circuit Q1 and Q2 are considered the power amp?
Thanks again!
Here and here, Just add these models to your SWcadIII\lib\cmp folder to the standard.bjt file,
open it with notepad, add the text to the standard.bjt file from the two files and save as standard.bjt
And then you will be able to see them in the drop down menu for select component
ttp://www.onsemi.com/pub_link/Collateral/TIP31.SP3
http://www.onsemi.com/pub_link/Collateral/TIP32.LIB
open it with notepad, add the text to the standard.bjt file from the two files and save as standard.bjt
And then you will be able to see them in the drop down menu for select component
ttp://www.onsemi.com/pub_link/Collateral/TIP31.SP3
http://www.onsemi.com/pub_link/Collateral/TIP32.LIB
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I think this is right. I changed Rin to 5k and Rf to 25K and things look the same as Rin10k, Rf50k!!! Yay, I feel like I figured something out. So if my input impedance guess is right, then this is how I would adjust the input impedance?
From the left...
R1 is just included for good practice. It defines a "ground" point for C1 rather than just leaving it floating (for example with nothing connected to the input).
The opamp... you're pretty much there on that one
It is configured in an inverting configuration. Imagine for the moment the transistors and 8 ohm load aren't there and that Rf connects directly to the opamp output. In order that the output can swing between the rails (between 0 and 9 volt), we bias the output to half supply voltage. That's where the 4.5 volts from R1 and R2 comes in. They can be any equal value resistors although making them high values means they do not draw much current (a good thing). The non inverting input pin (pin 2) does draw a tiny current from this resistor chain but it's so small we can discount it.
There is a golden rule for opamps with feedback, and that is that the output will do whatever it takes to make the difference between the inputs ZERO. That's the difference, not the absolute values.
So we have Rf connected between output and inverting input. And we have Rin connected to the inverting input. We know the voltage on pin 3 is biased at 4.5 volts so what must the ouput be to keep the difference between the opamp inputs zero ? Answer 4.5 volts.
If we now apply an input voltage via Rin then that tries to alter the voltage on pin 2 of the opamp. The opamp though tries to keep the difference between the inputs at zero at all times. Now as pin 3 is fixed at 4.5 volts the output pin has to alter it's voltage away from 4.5 volts to keep this balance. How much it needs to alter is determined by Rf. As Vin goes more positive Vout has to go more negative to keep this zero volts difference. And vice versa. So that is why the amp is called inverting and why the output is 180 degrees out of phase with the input.
So for an inverting opamp circuit we say the gain is Rf/Rin. So the gain here is 47k/10k which is 4.7 That is the voltage gain of the circuit.
You experimented with the values By keeping the ratio of Rf and Rin the same then the voltage gain stays the same. Rin determines the actual input impedance.
Q1 and Q2 are connected as "emitter followers" to provide current gain to enable a low value load to be driven. By including them in the feedback loop (by moving Rf from the opamp output to the new output point) the circuit as a whole still follows the above rules.
Because the transisotors need around 0.7 volts across the base and emitter junction to turn them on there is a kind of dead zone around the centre part of the waveform. That's the glitch in the output. The opamp can't swing its output fast enough across this dead zone and so there is some distortion. The resistor R5 allows the opamp itself to supply the load during this region with the transistors taking over again as the output level increases.
So the opamp provides all the voltage gain and the transistors provide current gain.
C1 and C2 are needed to block DC and allow AC to pass. They do form a filter. The input impedance of the main circuit is always the value of Rin. If C1 were to small in value then low frequencies would be attenuated. The same applies to C2 and the 8 ohm load.
C3 is for decoupling the power supply. It provides a low impedance path to AC signal (shorting them out) and also provides a reserve of energy which is important in battery operated equipment.
Edit...
If you have Java try this putting in different values. You will see why C2 needs to be so large with an 8 ohm load and why C1 can be much smaller with a 10K "load" or input impedance.
http://www.st-andrews.ac.uk/~www_pa/Scots_Guide/experiment/highpass/hpf.html
R1 is just included for good practice. It defines a "ground" point for C1 rather than just leaving it floating (for example with nothing connected to the input).
The opamp... you're pretty much there on that one
It is configured in an inverting configuration. Imagine for the moment the transistors and 8 ohm load aren't there and that Rf connects directly to the opamp output. In order that the output can swing between the rails (between 0 and 9 volt), we bias the output to half supply voltage. That's where the 4.5 volts from R1 and R2 comes in. They can be any equal value resistors although making them high values means they do not draw much current (a good thing). The non inverting input pin (pin 2) does draw a tiny current from this resistor chain but it's so small we can discount it.
There is a golden rule for opamps with feedback, and that is that the output will do whatever it takes to make the difference between the inputs ZERO. That's the difference, not the absolute values.
So we have Rf connected between output and inverting input. And we have Rin connected to the inverting input. We know the voltage on pin 3 is biased at 4.5 volts so what must the ouput be to keep the difference between the opamp inputs zero ? Answer 4.5 volts.
If we now apply an input voltage via Rin then that tries to alter the voltage on pin 2 of the opamp. The opamp though tries to keep the difference between the inputs at zero at all times. Now as pin 3 is fixed at 4.5 volts the output pin has to alter it's voltage away from 4.5 volts to keep this balance. How much it needs to alter is determined by Rf. As Vin goes more positive Vout has to go more negative to keep this zero volts difference. And vice versa. So that is why the amp is called inverting and why the output is 180 degrees out of phase with the input.
So for an inverting opamp circuit we say the gain is Rf/Rin. So the gain here is 47k/10k which is 4.7 That is the voltage gain of the circuit.
You experimented with the values
By keeping the ratio of Rf and Rin the same then the voltage gain stays the same. Rin determines the actual input impedance. Q1 and Q2 are connected as "emitter followers" to provide current gain to enable a low value load to be driven. By including them in the feedback loop (by moving Rf from the opamp output to the new output point) the circuit as a whole still follows the above rules.
Because the transisotors need around 0.7 volts across the base and emitter junction to turn them on there is a kind of dead zone around the centre part of the waveform. That's the glitch in the output. The opamp can't swing its output fast enough across this dead zone and so there is some distortion. The resistor R5 allows the opamp itself to supply the load during this region with the transistors taking over again as the output level increases.
So the opamp provides all the voltage gain and the transistors provide current gain.
C1 and C2 are needed to block DC and allow AC to pass. They do form a filter. The input impedance of the main circuit is always the value of Rin. If C1 were to small in value then low frequencies would be attenuated. The same applies to C2 and the 8 ohm load.
C3 is for decoupling the power supply. It provides a low impedance path to AC signal (shorting them out) and also provides a reserve of energy which is important in battery operated equipment.
Edit...
If you have Java try this putting in different values. You will see why C2 needs to be so large with an 8 ohm load and why C1 can be much smaller with a 10K "load" or input impedance.
http://www.st-andrews.ac.uk/~www_pa/Scots_Guide/experiment/highpass/hpf.html
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Mooly, you really need to remove the "simulation free zone" from your signature.
I know
I'll think of something eventually.
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