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Trying To Understand an OPAmp circuit

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webpower

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Hi Not sure if this is correct place to post this in. Apologies if not. I am still trying to understand the correct protocols of forums.

I have built a working test opamp on a breadboard with 6 bjt transistors using example shown on allaboutcitcuits volume 6, chapter 5 section ''

I have attached voltmeters to the bases of Q3 and Q4 and an ammeter between q6 emitter to ground.

I have the circuit running correctly as an opamp and currently have it set up as a voltage follower with the collector of Q4 connected to its own base.

Now I thought I understood how this circuit works. But there is something that I don't really understand. (Please view diagram on above website whilst reading this. The voltage follower circuit.)

From what I understand (I am of course assuming that I have this overview understood correctly!) Q5 and Q6 produce a fixed unchanging current in the collector of Q6 (eg current mirror) to establish a fixed quiescent current level.

Now on my bench if I slowly increase the voltage on Q3s base then as I approach 0.7V I then start to see a voltage on Vout (collector of Q4). On passing through 0.7v both Q4 and Q3 base voltages then are both matched very closley to about 1/100 of a volt. So it is then acting as a voltage follower as it should do.

But the (pn junction) transition on Q3 base from 0.43v to just over 0.7v causes the Q6 emitter current (showing on my ammeter) to very slightly increase? Clearly this current needs to bleed through somehow. But what makes itself get forced through Q6 collector?

Is it something to do with capacitor action at the collector of Q6? Maybe it is because the quiescent condition of the current mirror must change to allow for this current to flow from Q3 to build up at emittter of Q4? (I can see on my meter that current does change very slightly at emitter of Q6 going fractionally up, but then (as expected) with further Q3 base voltage increases it then just remains at this new slightly higher quiescent level right up until we get 0.7v below vcc(12volts) saturation.)

So my question is what is happening to make the Q6 emitter current slightly increase to this new slightly higher quiescent level, on what I thought was an unchangeable current mirror defined by Rpg (The resistor that sets the initial quiescent current level)

Sorry if this is confusing post. I hope someone can explain this better than my understanding.

Many thanks in advance.
 
A current mirror is not perfect. Its current changes a little when its voltage changes and even when the temperature changes.
On an IC opamp all parts are very close together on the same tiny chip so their temperatures are all the same. Also there are many more parts on a real opamp to prevent the current from changing when the voltage changes.
 
Thankyou audioguru that makes perfect sense. But it would be nice if I could understand what makes the current go up in the Q6 transistor (Just for interest on the OPamps inner workings). My thoughts were that there has to be an current increase at top of Q6 collector when Q3 gets switched on. So, does some of this extra current pass though Q6? Or is it the sudden extra voltage difference across Q6 that makes it conduct more (eg minority carriers)?
 
The "Early Effect" (look at it in Google) causes the collector current of a transistor to increase a little when its collector to emitter voltage changes even when the base current does not change.

Also, the current mirrors are very simple and therefore are not perfect.

The range of your input voltage (input common mode voltage range) actually turns off some of the transistors that must always be turned on. The input voltage change should only cause the transistors to conduct a little more or conduct a little less. Maybe the input must be from +2V to +10V.
 
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