Hello again,
as i've stated in my transformer post, i would like to continue my electronic adventures in hi-fi not by simple replicating but also learning. However, i am still electronic dumbass.
In this thread i would like to start analyzing function of the popular Rod Elliots P3A amp. Idea is to stop by every stage and try to understand, with your help and comments appreciated, how it works. I believe that there are many like me that might found the topic helpful and i am aware that for most of you it would appear like "bringing wood to the forest". But i also hope, that there are many geeks among you enjoying guiding self-educating hobbyist like me to the success of my goals.
One can admit, that there is a lot of books on the subject, but i must admit that books dont give answers to custom questions. I have asked many times. Besides, at some point an instructor is a must...
So here is the obligatory P3A and i will start with explainning what i know about it. Lets start with input stage.
I will pass the input RC filter and Rods custom SIM signals and jump to the interesting thing, the Long tailed pair (aka differencial pair). Purpose of this circuit is to do the first gain of input signal and to isolate the amp input from next stages (impedance?) On the non-inverting input is amp input and on inverting input is global feedback applied. Output (q4) is connected to q1 collector. On the bottom is the current sink circuit from q3 with r7 as current limit resistor and r8/d1 network for constant voltage drop among q3 base (the ability of green LED, offen two diodes are used).
Here are the questions:
1) Am i right? Havent i missed something important? Some inaccuracy?
2) I wonder where the bias for the trannies comes from. For q1, q3 i guess the current flows from ground to r3/r2/r8/d1 network biasing the q1 and q3 respectivelly. But q2 drives me crazy, is it r4? Hard to believe as c3 simply doesnt conduct DC current.
3) The r6 is the working load? Is it the place where the voltage drop occurs and is taken to the voltage amp stage?
4) Why is the constant current flowing through the circuit so important? How the overall signal procesing works then? (e.g. how is the feedback applied to the incoming signal)
5) A bit hard, how can one calculate proper values of the components? This one is not fair as it is quite longish and i already know how to bias one transistor. But some further info might come handy.
Beat me hard, i am not afraid to think about the circuit. I really want to understand the basics to be able to modify/design my own amp - thats my goal and dream. I hope, with your help, i can reach it!
Thanks for ANY help people...
as i've stated in my transformer post, i would like to continue my electronic adventures in hi-fi not by simple replicating but also learning. However, i am still electronic dumbass.
In this thread i would like to start analyzing function of the popular Rod Elliots P3A amp. Idea is to stop by every stage and try to understand, with your help and comments appreciated, how it works. I believe that there are many like me that might found the topic helpful and i am aware that for most of you it would appear like "bringing wood to the forest". But i also hope, that there are many geeks among you enjoying guiding self-educating hobbyist like me to the success of my goals.
One can admit, that there is a lot of books on the subject, but i must admit that books dont give answers to custom questions. I have asked many times. Besides, at some point an instructor is a must...
So here is the obligatory P3A and i will start with explainning what i know about it. Lets start with input stage.
An externally hosted image should be here but it was not working when we last tested it.
I will pass the input RC filter and Rods custom SIM signals and jump to the interesting thing, the Long tailed pair (aka differencial pair). Purpose of this circuit is to do the first gain of input signal and to isolate the amp input from next stages (impedance?) On the non-inverting input is amp input and on inverting input is global feedback applied. Output (q4) is connected to q1 collector. On the bottom is the current sink circuit from q3 with r7 as current limit resistor and r8/d1 network for constant voltage drop among q3 base (the ability of green LED, offen two diodes are used).
Here are the questions:
1) Am i right? Havent i missed something important? Some inaccuracy?
2) I wonder where the bias for the trannies comes from. For q1, q3 i guess the current flows from ground to r3/r2/r8/d1 network biasing the q1 and q3 respectivelly. But q2 drives me crazy, is it r4? Hard to believe as c3 simply doesnt conduct DC current.
3) The r6 is the working load? Is it the place where the voltage drop occurs and is taken to the voltage amp stage?
4) Why is the constant current flowing through the circuit so important? How the overall signal procesing works then? (e.g. how is the feedback applied to the incoming signal)
5) A bit hard, how can one calculate proper values of the components? This one is not fair as it is quite longish and i already know how to bias one transistor. But some further info might come handy.
Beat me hard, i am not afraid to think about the circuit. I really want to understand the basics to be able to modify/design my own amp - thats my goal and dream. I hope, with your help, i can reach it!
Thanks for ANY help people...
basic amp principle
I think the ltp is easier to understand as a kind of balance. You set it up trying to get equal currents through each half (half of the q3 current). Assuming q1 & q2 are matched, then there is balance if the two bases are at equal voltage. q1 base is at gnd (through R3 & R2, neglecting for the moment the base current). Q3 base then should also be at gnd through R5. Note that R5 and R3+R2 are (almost) equal to preserve the balance even if there is base current (which there is).
The trick now is to construct the rest of the amp so that the feedback via R5 automagically balances the whole amp (we'll come back to that later).
Now, as you say, R6 is the working load, providing the output voltage. But it also sets the base voltage for Q4. Since Q4 has no local feedback via an emitter resistor, this becomes quite critical. You want just enough voltage (.6 or something like that) across R6 that Q4 conducts enough to satisfy the voltage across R9 + R10 which will be close to V-.
So, you need just a bit more than 1mA through R6 (and Q1), for balance also through Q2, so 2mA through Q3. That means (with the R7 as shown) about 1.8V across D1. Seems to be OK.
Is this helping you?
Jan Didden
I think the ltp is easier to understand as a kind of balance. You set it up trying to get equal currents through each half (half of the q3 current). Assuming q1 & q2 are matched, then there is balance if the two bases are at equal voltage. q1 base is at gnd (through R3 & R2, neglecting for the moment the base current). Q3 base then should also be at gnd through R5. Note that R5 and R3+R2 are (almost) equal to preserve the balance even if there is base current (which there is).
The trick now is to construct the rest of the amp so that the feedback via R5 automagically balances the whole amp (we'll come back to that later).
Now, as you say, R6 is the working load, providing the output voltage. But it also sets the base voltage for Q4. Since Q4 has no local feedback via an emitter resistor, this becomes quite critical. You want just enough voltage (.6 or something like that) across R6 that Q4 conducts enough to satisfy the voltage across R9 + R10 which will be close to V-.
So, you need just a bit more than 1mA through R6 (and Q1), for balance also through Q2, so 2mA through Q3. That means (with the R7 as shown) about 1.8V across D1. Seems to be OK.
Is this helping you?
Jan Didden
still digging
OK, i see the input as balanced circuit and the purpose of feedback applied this way. q2 simply shorts signal according to input, so the amp will never have infinite gain. I also understand the 2mA current through q3, but i cant evaluate the need for 1mA through q1. I know its purpose, to obtain magical point of 0.65V where the q4 starts opening -> generating voltage swing.
What confuses me also is the role of r9+r10 youve mentioned. They are generating base voltage for q6 from -Ve rail, but i cant realize how and how is q4 afecting them - my explanation is that r6 biased q4 starts conducting through them so that voltage drop for driver stage q6 bias occurs.
Please continue...
OK, i see the input as balanced circuit and the purpose of feedback applied this way. q2 simply shorts signal according to input, so the amp will never have infinite gain. I also understand the 2mA current through q3, but i cant evaluate the need for 1mA through q1. I know its purpose, to obtain magical point of 0.65V where the q4 starts opening -> generating voltage swing.
What confuses me also is the role of r9+r10 youve mentioned. They are generating base voltage for q6 from -Ve rail, but i cant realize how and how is q4 afecting them - my explanation is that r6 biased q4 starts conducting through them so that voltage drop for driver stage q6 bias occurs.
Please continue...
basic amp stuff
Re: R9, R10: first focus on the DC balance conditions, forget C5 the bootstrap cap for the moment.
You want the stuff around Q9 to be around gnd (without signal) so the output can be really at 0 volts. Because of C3, all the DC at the output is fed back through R5 to the inverting input, for balance. That means that with the values for R9 & R10 and -35V supply, you need about 35/5.4k = 6mA through q4. This basically completes the amp. Your last statement says this OK.
Now you put in a small ac at IN. Suppose it goes positive initially, so q1 conducts more, Q2 conducts less (because sum of Iq1 & Iq2 = constant), R6 gets more current, q4 opens up more, R9, R10 get more current, the Q9 stuff rises, the output rises, and through R5 the increase comes back to q2 to increase its current to restore the balance.
Now, because of C3, for AC, the feedback is attenuated by the ratio R5 to R4, so the increase of ac at the output needs to be 22x as much as in the input to restore balance > closed loop gain is 22 x.
The ratio of C3 to R4 sets the lf cutoff point.
OK so far?
Jan Didden
Re: R9, R10: first focus on the DC balance conditions, forget C5 the bootstrap cap for the moment.
You want the stuff around Q9 to be around gnd (without signal) so the output can be really at 0 volts. Because of C3, all the DC at the output is fed back through R5 to the inverting input, for balance. That means that with the values for R9 & R10 and -35V supply, you need about 35/5.4k = 6mA through q4. This basically completes the amp. Your last statement says this OK.
Now you put in a small ac at IN. Suppose it goes positive initially, so q1 conducts more, Q2 conducts less (because sum of Iq1 & Iq2 = constant), R6 gets more current, q4 opens up more, R9, R10 get more current, the Q9 stuff rises, the output rises, and through R5 the increase comes back to q2 to increase its current to restore the balance.
Now, because of C3, for AC, the feedback is attenuated by the ratio R5 to R4, so the increase of ac at the output needs to be 22x as much as in the input to restore balance > closed loop gain is 22 x.
The ratio of C3 to R4 sets the lf cutoff point.
OK so far?
Jan Didden
bit more on balanced operation
I think that i am starting to realizing it. But one confusing detail comes to my mind right now. I cant figure out the q1/q2 biasing with balanced mode operation.
Quick example: following the overal operation with AC on input, the signal starts to be amplified after 0.65RMS on input. To that point - silence. To this i see that to have 0V dc offset (ground) there must be gnd on input. And i cant find, where is the bias hided.
Also, please tell how did you calculated the 1mA flowing through q1. Simple applying Ohms law doesnt work for me...
Thx.
I think that i am starting to realizing it. But one confusing detail comes to my mind right now. I cant figure out the q1/q2 biasing with balanced mode operation.
Quick example: following the overal operation with AC on input, the signal starts to be amplified after 0.65RMS on input. To that point - silence. To this i see that to have 0V dc offset (ground) there must be gnd on input. And i cant find, where is the bias hided.
Also, please tell how did you calculated the 1mA flowing through q1. Simple applying Ohms law doesnt work for me...
Thx.
The ltp emitters sit at about -0.6V, that's were input bias comes from.
The current: there is about 0.6V over R6, so that's were the 1mA comes from
The current: there is about 0.6V over R6, so that's were the 1mA comes from
basic amp
Right, the q3 sink "pulls" the emitters of q1 & q2 down until they each carry about 1mA.
Jan Didden
Right, the q3 sink "pulls" the emitters of q1 & q2 down until they each carry about 1mA.
Jan Didden
Good discussion so far; the various books on amplifier
design don't go into this amount of detail, usually
assuming that the reader already knows the basics of
circuit design. I don't, and my progress in that part
of the learning curve has been difficult.
I'm looking forward to the output stage, since the
complementary feedback configuration isn't that
common and in theory is a good example of a local
feedback loop linearizing the output transistors.
In practice, though, it's tricky and really an example
of design choices and tradeoffs.
design don't go into this amount of detail, usually
assuming that the reader already knows the basics of
circuit design. I don't, and my progress in that part
of the learning curve has been difficult.
I'm looking forward to the output stage, since the
complementary feedback configuration isn't that
common and in theory is a good example of a local
feedback loop linearizing the output transistors.
In practice, though, it's tricky and really an example
of design choices and tradeoffs.
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