# Opamp Circuit query

Discussion in 'General Electronics Chat' started by webpower, Nov 1, 2013.

1. ### webpower Thread Starter New Member

Jul 13, 2012
16
2
Hi

I have built a simple basic opamp on a breadboard with just 6 bjt transistors using example shown on allaboutcitcuits volume 6, chapter 5 section 'Chapter 5: DISCRETE SEMICONDUCTOR CIRCUITS' called Simple Opamp

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.

There is something that I don't really understand. (Please view diagram on above specified page 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 slightly increase? Clearly this current is bleeding 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 to 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).

I do understand that Current mirrors are not perfect. I want to try and understand what is happening though because of this imperfection.

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)?

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

Last edited: Nov 1, 2013
2. ### #12 Expert

Nov 30, 2010
16,685
7,322
You are looking at the change in Vce causing a change in the current gain of Q6.
Any transistor with only a few tenths of a volt from collector to emitter has less gain than a transistor with several volts from Collector to emitter.

Look at this, figure 5
When the transistor has a whole volt from collector to emitter, the gain is 71
When the transistor has only .2 volts from collector to emitter, its gain is 33.
Is that enough to steer you?

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3. ### webpower Thread Starter New Member

Jul 13, 2012
16
2
Thanks that did. I was getting too bogged down. I now realize that with the voltage drops in both legs identical for a voltage follower, with a single diode drop on each leg means the other remaining transisters on each side has matching Collector Emitter voltages (As both legs must add up to Vcc). I just got this now. Knowing that makes this circuit now clear. The current difference thing threw me out. Its not (Now I realize) what I needed to know. Thankyou that helped a lot, as it made me look at this differently. Solved!

4. ### #12 Expert

Nov 30, 2010
16,685
7,322
Yeah. Forest for the trees.

Last edited: Nov 1, 2013