re is set by the current in each BJT. The transconductance of the stage comes from 2*re (because the signal sees two in series) but it can be doubled by using a current mirror at the output.
2mA total means 1mA per BJT, so re=25. If you took LTP output just from one collector then gm=1/(2*re). A current mirror doubles this to gm=1/re.
2mA total means 1mA per BJT, so re=25. If you took LTP output just from one collector then gm=1/(2*re). A current mirror doubles this to gm=1/re.
Hello and thank you both sreten and DF96.
For the usual input differential stage in a power amp without current mirror gm=1/2*re. Addition of the current mirror doubles gm=1/re. Now, if I want to degenerate back to gm=1/2*re I must make my calculations with re=25. Correct?
Sorry for the naive approach but I am really lost. I thought that gm doubles because mirroring reduces re to re/2.
My best regards
For the usual input differential stage in a power amp without current mirror gm=1/2*re. Addition of the current mirror doubles gm=1/re. Now, if I want to degenerate back to gm=1/2*re I must make my calculations with re=25. Correct?
Sorry for the naive approach but I am really lost. I thought that gm doubles because mirroring reduces re to re/2.
My best regards
The gain doubling comes from using signal current from both collectors.
Hi, you've lost me on that one without a differential VAS, rgds, sreten.
I thought the "standard" LTP, SE VAS and EFP was being discussed.
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This is the schematic
An externally hosted image should be here but it was not working when we last tested it.
Hi DF96.
I have tried D. Self and B. Cordell. Neither expalins why gm doubles with current mirror. In fact B. Cordell says that gm=1/(total resistance from base to base) of the LTP transistors. I have thus erroneously assumed that this due to halving of the re.
The schematic http://www.neilmcbride.co.uk/output-amp2.pdf
I have tried D. Self and B. Cordell. Neither expalins why gm doubles with current mirror. In fact B. Cordell says that gm=1/(total resistance from base to base) of the LTP transistors. I have thus erroneously assumed that this due to halving of the re.
The schematic http://www.neilmcbride.co.uk/output-amp2.pdf
I have just checked 'Self on Audio' and you are right, he doesn't say why the current mirror doubles output. Maybe he thought it was obvious. He does talk about equalising currents, which may be what sreten was referring to. Self is slightly misleading here, as it is not the current mirror itself which does this, but current mirror plus global negative feedback plus VAS current gain.
Talking about gm of an LTP always carries the risk of confusion, because you need to specify whether you are talking about one-sided output, balanced output or current-mirror output (which converts balanced to unbalanced). The thing to remember is that BJTs have a fairly high collector resistance, so nothing you do in the collector circuit will make much difference to what happens at the emitter. Hence a current mirror (in the collector) cannot change re (at the emitter).
Talking about gm of an LTP always carries the risk of confusion, because you need to specify whether you are talking about one-sided output, balanced output or current-mirror output (which converts balanced to unbalanced). The thing to remember is that BJTs have a fairly high collector resistance, so nothing you do in the collector circuit will make much difference to what happens at the emitter. Hence a current mirror (in the collector) cannot change re (at the emitter).
Dinstam
Just to be clear, I don't see an active current mirror in the above schematic, what are you referring to as a current mirror?
FYI: many texts will use symmetry and split the LTP right down the middle and apply halve the differential voltage to each input (and because of symmetry the node at the current source doesn't change, i.e. can be treated as ground, just remember that you are applying 1/2 the input voltage to each side and ignore common mode effects).
You should see the same answer with Bob's equation when considering as a single stage "seeing" the emitter loading (Re) of the inverter input transistor.
Examples: For the symmetry approach:
Apply Vin/2 to the single non-inverting side with Re to a virtual ground.
For the single stage method:
Apply Vin to the single non-inverting side with Re + Re (the second Re is the load from the inverting side)
Hope this helps
-Antonio
Just to be clear, I don't see an active current mirror in the above schematic, what are you referring to as a current mirror?
FYI: many texts will use symmetry and split the LTP right down the middle and apply halve the differential voltage to each input (and because of symmetry the node at the current source doesn't change, i.e. can be treated as ground, just remember that you are applying 1/2 the input voltage to each side and ignore common mode effects).
You should see the same answer with Bob's equation when considering as a single stage "seeing" the emitter loading (Re) of the inverter input transistor.
Examples: For the symmetry approach:
Apply Vin/2 to the single non-inverting side with Re to a virtual ground.
For the single stage method:
Apply Vin to the single non-inverting side with Re + Re (the second Re is the load from the inverting side)
Hope this helps
-Antonio
Hi Antonio,
yes, the schematic does not have a current mirror and it is very unbalanced. So, I want to replace the 2 resitors (R4,R5) with a current mirror to balance the LTP and gain some slew rate. To keep the same gm I have degenarate the emitters of Q1 and Q2. For this, and according to Bob Cordell, I need to know the re of Q1 and Q2.
Regards
Dimitrios
yes, the schematic does not have a current mirror and it is very unbalanced. So, I want to replace the 2 resitors (R4,R5) with a current mirror to balance the LTP and gain some slew rate. To keep the same gm I have degenarate the emitters of Q1 and Q2. For this, and according to Bob Cordell, I need to know the re of Q1 and Q2.
Regards
Dimitrios
When used in the collector of an LTP, a current mirror converts balanced to unbalanced, as I said. Two signal currents in antiphase (balanced) are converted to one signal current (unbalanced) of twice the amplitude of one of the balanced ones.
The best textbook for understanding solid state electronics is Horowitz and Hill.
The best textbook for understanding solid state electronics is Horowitz and Hill.
OK. Actually the current mirror helps to balance the quiescent currents, provided that the VAS does not draw any input current (or very little - usually true) and that there is global negative feedback down to DC (also usually true). On its own the current mirror would merely force the output voltage from the LTP to bottom at either an LTP collector or current-mirror collector (depending on the inherent imbalance in the LTP). That is why I say that Self is misleading when he says that the current mirror causes the balance - it is merely part of the mechanism.
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