Author Topic: Behaviour of CE amplifier with active load  (Read 733 times)

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Offline aneevuser

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Behaviour of CE amplifier with active load
« on: December 10, 2018, 01:20:26 pm »
I've been playing around with a common emitter amplifier with an active load. I made up the following circuit and have measured some of its characteristics, as shown in the attached image.

I'm not familiar with the use of active loads, and I'm somewhat unclear about what is going on here:

1) What determines the quiescent operating point of Vout?

Note that my Iref and Ic don't match too well, so I'm assuming that Vout is being developed as (Iref - Ic) * r_out, where r_out is the output resistance of either Q1 or Q2. But if so, what determines r_out? (bearing in mind that this is a DC analysis, and I can't assume a small signal model for Q1/Q2 will be appropriate).

2) I have read in a few places that this kind of arrangement needs some kind of negative feedback - I have none but even so the circuit works just fine as an amplifier, with an unloaded voltage gain of about 34.

In addition, it seem pretty stable, thermally at least - if I disturb Iref by heating up Q2 or Q3, I can move the operating point so that the circuit clips either at the top or bottom - but it always returns to the 3.7 V quiescent point at V_out.

OTOH, I believe that ideally quiescent V_out should be 0 V when I_ref=Ic - I've been unable to balance these nicely, and I guess this could be done via feedback from Vout to VB - but how is this implemented specifically?

So: is it vital to operate this arrangement with negative feedback? If yes, then where does the stability that I'm seeing come from?


Anyone familiar with this stuff? I'm having difficulty finding any kind of detailed analysis of this arrangement that answers my questions.
 

Offline StillTrying

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Re: Behaviour of CE amplifier with active load
« Reply #1 on: December 12, 2018, 12:23:46 pm »
I've been unable to balance these nicely, and I guess this could be done via feedback from Vout to VB - but how is this implemented specifically?

The collector of a transistor is a constant current source, so you've got 2 constant current sources connected together, and a very high output impedance.
You could connect the 50k between B-C to provide the negative feedback, it would give some DC stability, reduce the gain to about 12 and reduce the input and output impedances.
I don't think there's much point in that arrangement!
CML+  That took much longer than I thought it would.
 

Offline IanMacdonald

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Re: Behaviour of CE amplifier with active load
« Reply #2 on: December 12, 2018, 12:37:53 pm »
A: It is difficult to give such an arrangement a stable operating point unless it is in a feedback loop. You're basically 'facing off' one current source against another, and the one with slightly higher current delivery will win, pulling the output to its supply rail.

B: You need a small resistor in the current source collector otherwise there are likely to be parasitic oscillations. Reason (I'm told) is that the c-b capacitance of the current source forms a high Q reactive circuit.
 

Offline spec

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Re: Behaviour of CE amplifier with active load
« Reply #3 on: December 12, 2018, 06:58:30 pm »
Hi  aneevuser

Output voltage stabilized at approx 5V in attached schematic (change the value of the 22k resistor to change the stabilization point).
« Last Edit: December 12, 2018, 07:03:28 pm by spec »
 
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Offline aneevuser

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Re: Behaviour of CE amplifier with active load
« Reply #4 on: December 13, 2018, 02:40:34 pm »
Hi  aneevuser

Output voltage stabilized at approx 5V in attached schematic (change the value of the 22k resistor to change the stabilization point).

That's nice. I must say that's a biasing scheme that I never consider (too tricky to calculate the quiescent voltage), but I can see why it's useful here. However, as someone mentioned above, the feedback reduces the gain substantially, so I think that it's of more theoretical interest than practical. Having said that, it certainly controls Vc nicely.

I wonder though if there's any feedback approach that allows the stiff biasing that I started with, without throwing away the extra gain due to the current source? It's something I'll have to have a think about.

Anyway, I've started playing around with active loads in combination with differential amplifiers, where it seems rather easier to match up collector currents than with the CE amp. The improvements in gain and CMRR there are also pretty impressive.
 

Offline aneevuser

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Re: Behaviour of CE amplifier with active load
« Reply #5 on: December 13, 2018, 02:44:30 pm »
A: It is difficult to give such an arrangement a stable operating point unless it is in a feedback loop. You're basically 'facing off' one current source against another, and the one with slightly higher current delivery will win, pulling the output to its supply rail.

Right. That is indeed the fundamental problem. A good solution seems to be fairly tricky though.

Quote
B: You need a small resistor in the current source collector otherwise there are likely to be parasitic oscillations. Reason (I'm told) is that the c-b capacitance of the current source forms a high Q reactive circuit.

At what frequency is this supposed to occur? Can't say I've had any problems so far, but I'm only driving the circuit at 50KHz or thereabouts.
 

Offline aneevuser

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Re: Behaviour of CE amplifier with active load
« Reply #6 on: December 13, 2018, 02:47:36 pm »

You could connect the 50k between B-C to provide the negative feedback, it would give some DC stability, reduce the gain to about 12 and reduce the input and output impedances.
I don't think there's much point in that arrangement!

Yes, someone suggested similar below, and it works fine (except for the pesky gain...). How did you calculate the gain of 12 BTW? I've set up this circuit, and voltage gain is indeed almost exactly 12.
 

Offline StillTrying

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Re: Behaviour of CE amplifier with active load
« Reply #7 on: December 14, 2018, 01:12:38 pm »
"How did you calculate the gain of 12"

I simulated it. :)  This is about the best I tuned it.
CML+  That took much longer than I thought it would.
 


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