hi everybody,
I was looking for a bass preamp because, after building different kind of pedals, I wanted to try to build a bass head.
I founded this one
http://elektrotanya.com/PREVIEWS/63463243/23432455/vox/vox_bass100_preamp_sch.pdf_1.png
It comes from an old Combo Bass Amp 100W.
I think to use this pre-amp with a ClassD power-amp.
I started to study it.
There is a double reaction for the first three transistor. One is negative reaction between the emitter of the third transistor and the first one. It's good because Rin encrease and Rout decrease, etc. . Olso the second is a negative reaction between VOL and the emitter of the first transistor.
The middle controll act in this way:
imgur: the simple image sharer
imgur: the simple image sharer
There are two more transistor in the middle that have no reaction and they make a open loop gain for the signal
I don't understand how do the last two transistors work...
My questions are:
have anyone of you ever built or used this pre-amp?
how does it work the treble and bass controll? I'm not able to find the transfer function...
in wich way does the output signal work? there is a reaction between the last 2 transistor but the emitter of the last one is also connected to the bass and treble filter by a 4.7uF capacitor... so how does the signal works?
excuse me for my bad english
ciao
I was looking for a bass preamp because, after building different kind of pedals, I wanted to try to build a bass head.
I founded this one
http://elektrotanya.com/PREVIEWS/63463243/23432455/vox/vox_bass100_preamp_sch.pdf_1.png
It comes from an old Combo Bass Amp 100W.
I think to use this pre-amp with a ClassD power-amp.
I started to study it.
There is a double reaction for the first three transistor. One is negative reaction between the emitter of the third transistor and the first one. It's good because Rin encrease and Rout decrease, etc. . Olso the second is a negative reaction between VOL and the emitter of the first transistor.
The middle controll act in this way:
imgur: the simple image sharer
imgur: the simple image sharer
There are two more transistor in the middle that have no reaction and they make a open loop gain for the signal
I don't understand how do the last two transistors work...
My questions are:
have anyone of you ever built or used this pre-amp?
how does it work the treble and bass controll? I'm not able to find the transfer function...
in wich way does the output signal work? there is a reaction between the last 2 transistor but the emitter of the last one is also connected to the bass and treble filter by a 4.7uF capacitor... so how does the signal works?
excuse me for my bad english
ciao
I think by "reaction" you mean negative feedback.
I have never used this preamp. The Bass and Treble controls are what is called the Baxandall circuit, invented by Peter Baxandall. The original article was published in Wireless World in 1952. It is still used in many products, usually with an opamp providing the gain.
The Baxandall circuit needs to be driven by a low impedance, the two transistors between the Volume control and the Bass and Treble controls is a simple buffer to isolate the Volume control and provide a constant low impedance to the tone control network.
The first three transistor circuit is configured as a variable gain amplifier. One half of the Volume control changes the gain of input amplifier by changing the feedback network and the other half changes the attenuation of the output. This gives a log like taper to the Volume control's rotation with a linear taper potentiometer.
yes, I mean negative feedback 🙂
thankyou Loudthud, now I will study Baxandall circuit.
So don't you think that the fourth transistor is used for more open loop gain?
I have another important question:
between the Voltage Reference (50V) and the circuit there is a 3.3kHom resistor so what is the effective Voltage Reference of this circuit? Is it 35V (that is the value of the 1000uf capacitor)?
thankyou Loudthud, now I will study Baxandall circuit.
So don't you think that the fourth transistor is used for more open loop gain?
I have another important question:
between the Voltage Reference (50V) and the circuit there is a 3.3kHom resistor so what is the effective Voltage Reference of this circuit? Is it 35V (that is the value of the 1000uf capacitor)?
Yes, the fourth transistor provides a small amount of gain. Probably just enough to compensate for the loss in the Mid control circuit.
A rough calculation suggests the Voltage on the 1000uF capacitor will be about 25V.
thankyou again Loudthud.
but how did you calculate the Voltage?
I tried to do it but I just made a lot of noise whit the calculation...
ciao
but how did you calculate the Voltage?
I tried to do it but I just made a lot of noise whit the calculation...
ciao
Ciao quasi paesano 🙂
Tutti i Portegni siamo mezzo Italiani 😉
That Bass preamp works; it's not horrible but not very good either, there's much better and similar or even simpler if you wish.
Build the preamp from this one, it's functionally similar, but simpler and more modern ... and you will simulate it better 😉
Crate B10 Service Manual free download,schematics,datasheets,eeprom bins,pcb,repair info for test equipment and electronics
The preamp is fed from the +19V main rail, you can get them from the + rail used by your new power amp, with a dropping resistor, a 19 to 24V Zener and a smoothing capacitor, think 1000uFx25 to 35V .
Tutti i Portegni siamo mezzo Italiani 😉
That Bass preamp works; it's not horrible but not very good either, there's much better and similar or even simpler if you wish.
Build the preamp from this one, it's functionally similar, but simpler and more modern ... and you will simulate it better 😉
Crate B10 Service Manual free download,schematics,datasheets,eeprom bins,pcb,repair info for test equipment and electronics
The preamp is fed from the +19V main rail, you can get them from the + rail used by your new power amp, with a dropping resistor, a 19 to 24V Zener and a smoothing capacitor, think 1000uFx25 to 35V .
ciao caro!
thankyou for your suggestions but I'm sure to continue to study this preamp for the moment because have a lot of things that I could learn.
(How works BJT, double loop feedback, Baxandall filter).
ciao e grazie ancora 🙂
thankyou for your suggestions but I'm sure to continue to study this preamp for the moment because have a lot of things that I could learn.
(How works BJT, double loop feedback, Baxandall filter).
ciao e grazie ancora 🙂
I made the assumption that each stage (each transistor) except the first has it's output at about half the supply voltage. Next, replace each transistor with a resistor equivalent to the load resistor because that resistor would drop half the supply voltage at the same current. Now, add up the resistance in each branch of the circuit and calculate the equivalent resistance of all those in parallel. Lastly, use the voltage divider theorem to calculate the voltage on the 1000uF cap using the equavalent resistor of the preamp, the 3.3K resistor and the 50V supply voltage.
I ignored the bias current of the first and second stages and the current in the first transistor. The calculation is not totally accurate, but it should be within a volt or two. If you need more accuracy, use a modeling program.
Hello again!
I studied the circuit but I'm not able to understand how the double reaction works...
I drawn this block scheme
imgur: the simple image sharer
I think that the B1 block is formed by 4.7k capacitor, the one between the emitter of the first and the emitter of the third transistor, and the capacitor of 0.33uF + resistor of 1k, there are on the first transistor emitter.
I think B2 block is formed by Vol pot. + capacitor of 1uF + resistor of 330 Ohm and the capacitor of 0.33uF + resistor of 1k, there are on the first transistor emitter.
What do you think about that?
Is it right?
Can you suggest me somethings to read about this kind of reaction?
ciao
I studied the circuit but I'm not able to understand how the double reaction works...
I drawn this block scheme
imgur: the simple image sharer
I think that the B1 block is formed by 4.7k capacitor, the one between the emitter of the first and the emitter of the third transistor, and the capacitor of 0.33uF + resistor of 1k, there are on the first transistor emitter.
I think B2 block is formed by Vol pot. + capacitor of 1uF + resistor of 330 Ohm and the capacitor of 0.33uF + resistor of 1k, there are on the first transistor emitter.
What do you think about that?
Is it right?
Can you suggest me somethings to read about this kind of reaction?
ciao
another question still remain...
how do the inputs works?
how are they connected with female Jack?
ciao 🙂
how do the inputs works?
how are they connected with female Jack?
ciao 🙂
OK, if you want to learn BJT, fine with me.
1) when you post images (gif - jpg - png) here, do not just write the link, which shows as a few words in blue and forces you to open another window, but first click the "image" icon on top of this window and put the address there, so the image opens inside the post.
What it does is put IMG and /IMG code around the address.
Here without:
http://elektrotanya.com/PREVIEWS/63463243/23432455/vox/vox_bass100_preamp_sch.pdf_1.png
here with:
I ask you to open this schematic in some image editor you have in your computer, I suggest Irfanview, and add (in using the text tool) labels to every part, calling them left to right R1 , R2 , etc. , plus C1 C2 plus Q1 Q2 (transistors) and P1 .... etc for pots.
It's much faster to work that way.
Choose some letter type which is not too large or too small (slightly larger than the original drawing values) and repost it here, such as Vox Preamp-v01 or something.
1) when you post images (gif - jpg - png) here, do not just write the link, which shows as a few words in blue and forces you to open another window, but first click the "image" icon on top of this window and put the address there, so the image opens inside the post.
What it does is put IMG and /IMG code around the address.
Here without:
http://elektrotanya.com/PREVIEWS/63463243/23432455/vox/vox_bass100_preamp_sch.pdf_1.png
here with:
I ask you to open this schematic in some image editor you have in your computer, I suggest Irfanview, and add (in using the text tool) labels to every part, calling them left to right R1 , R2 , etc. , plus C1 C2 plus Q1 Q2 (transistors) and P1 .... etc for pots.
It's much faster to work that way.
Choose some letter type which is not too large or too small (slightly larger than the original drawing values) and repost it here, such as Vox Preamp-v01 or something.
I did it!
I wrote the circuit with Spice and gave a name to every component.
I included also Rsp because Spice said to me that there were some "floating points" and I resolved the problem using a 100G resistor.
I also included RinClassD for the same reason ,but I still don't know how much is the Rin of a ClassD amplifier...
I studied the polarisation and I had this
The problem is Vout... I simulate a sine wave, as you can see, but I have this kind of graphic
Is it normal? I don't think so...
Can you help me to understand why I have this kind of graphic?
ciao 🙂
I wrote the circuit with Spice and gave a name to every component.
I included also Rsp because Spice said to me that there were some "floating points" and I resolved the problem using a 100G resistor.
I also included RinClassD for the same reason ,but I still don't know how much is the Rin of a ClassD amplifier...
I studied the polarisation and I had this
The problem is Vout... I simulate a sine wave, as you can see, but I have this kind of graphic
Is it normal? I don't think so...
Can you help me to understand why I have this kind of graphic?
ciao 🙂
I use BC414BP and BC416AP of zetex.lib and I don't use normal BC414 and BC416...
Could be this the reason of the Vout graphic?
I can't find BC414 Library...
Could be this the reason of the Vout graphic?
I can't find BC414 Library...
1) congratulations on going beyond what was asked 🙂
2) something is wrong, and like on standard troubleshooting.we'll have to break it down in sections and check them one by one.
I never simulate, I'm old (school, he he) enough to just calculate values.
When I started (1969) not even calculators existed and I used a slide rule 😱
Can you split your circuit in sections?
If possible, let's start with the first gain block.
Disconnect/lift 1 leg of: C2 , C8 and R10 so as to isolate it from the rest of the circuit.
Re simulate.
You *should* have around 24V V+ , +12V on the net Nt:R3-R4-Q1b , +12V on the net Nt:Q2c-R7 , 0.6V less than that on Nt:Q3e-R8
All values estimated, of course 😉
2) something is wrong, and like on standard troubleshooting.we'll have to break it down in sections and check them one by one.
I never simulate, I'm old (school, he he) enough to just calculate values.
When I started (1969) not even calculators existed and I used a slide rule 😱
Can you split your circuit in sections?
If possible, let's start with the first gain block.
Disconnect/lift 1 leg of: C2 , C8 and R10 so as to isolate it from the rest of the circuit.
Re simulate.
You *should* have around 24V V+ , +12V on the net Nt:R3-R4-Q1b , +12V on the net Nt:Q2c-R7 , 0.6V less than that on Nt:Q3e-R8
All values estimated, of course 😉
ciao
I studied so hard the circuit and I understood why it didn't work...
I just wrote 1MOhm as Mega but for the simulator M=milli!
So I just changed the wrong value with the correct one, 1MEGOhm
I have new Vout Graphic
The orange line is Vin.
It seems to work but is not 500mV peak to much low for the Vout?
JMFahey and Loudthud, you told me that V+ is about 25V and I just have this value on the simulator but I still do not understand how do you find this value...
I tried to do it by myself but I allways had not the right value...
Can you explain me better how do you find it? Which equation did you find?
Thankyou again 🙂
ciao
p.s. fantastic rule!!! How did you use it?
I studied so hard the circuit and I understood why it didn't work...
I just wrote 1MOhm as Mega but for the simulator M=milli!
So I just changed the wrong value with the correct one, 1MEGOhm
I have new Vout Graphic
The orange line is Vin.
It seems to work but is not 500mV peak to much low for the Vout?
JMFahey and Loudthud, you told me that V+ is about 25V and I just have this value on the simulator but I still do not understand how do you find this value...
I tried to do it by myself but I allways had not the right value...
Can you explain me better how do you find it? Which equation did you find?
Thankyou again 🙂
ciao
p.s. fantastic rule!!! How did you use it?
Depends on the gain of the power amplifier that follows and its maximum output power. It is not unusual to find 30-33 dB or even higher gains in musicians' gear.
Besides - do you know input level or have you merely guessed it? From the looks of it you're still a fair bit away from clipping.
BC414/416 are rated at Vceo = 30 V. At turn-on, the transistors often have to sustain the full supply voltage (even if they don't in operation) hence it is advisable not have it go higher than that. Resistor Rv is chosen accordingly after you've determined the circuit's current draw at nominal voltage.
For the input signal I used a signal of 120mV of peak and a frequency of 100Hz
I just use this signal because MI bass from a normal bass-guitar is about 41Hz, the foundamental, and because usually there is a 150mV peak as output signal from the instrument when you pick the string... obviously after little time the peak decreases so I think that was ok to use 120mV peak signal
I understood what you said but my trouble is that I don't understand how can I find this value using math calculation... I have 50V reference and Rv=3.3KOhm... how do you find the V+ value with calculations?
thankyou sgrossklass 🙂
ciao
I just use this signal because MI bass from a normal bass-guitar is about 41Hz, the foundamental, and because usually there is a 150mV peak as output signal from the instrument when you pick the string... obviously after little time the peak decreases so I think that was ok to use 120mV peak signal
I understood what you said but my trouble is that I don't understand how can I find this value using math calculation... I have 50V reference and Rv=3.3KOhm... how do you find the V+ value with calculations?
thankyou sgrossklass 🙂
ciao
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