# Calculate part of amplifier circuit.

Discussion in 'Analog & Mixed-Signal Design' started by pwnstars, Oct 1, 2016.

1. ### pwnstars Thread Starter Member

Feb 29, 2016
44
1
Hello there.

Im trying to calculate what voltage i should expect to measure at the green arrows.. The voltages should be equal for the entire circuit to work..

I've put in some values in blue and red but im not sure how to get any further from here.

please point me in the right direction

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2. ### #12 Expert

Nov 30, 2010
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Bad person! You cut off the right side of the circuit so I can't see the output stage.
The string you're looking at, T14,15,16 creates the spacer between the bases of the output transistors.
R21,22, and the pot create the idle current in the output stage by holding the bases of the output transistors apart.
T14 and T16 will use up most of the power voltage and T15 is a temperature compensator.
So what's the voltage at the green arrows? Not much. Maybe 1 or 2 or 3 volts away from center ground.

3. ### pwnstars Thread Starter Member

Feb 29, 2016
44
1
So the voltages at the green arrow is just enough to hold the output transistors open? There's 4 output transistors so that will be +1.2V and -1.2V. Hmm

4. ### #12 Expert

Nov 30, 2010
16,293
6,804
I'm not worried that you're taking days to figure this out. Closed loop amplifiers are not simple, but they are one of my specialties. When you do see the light, it comes on like a car crash...screeching tires as you realize what you were doing wrong and the "bang" when you see the answer. It's a very satisfying feeling when you finally grasp the whole closed loop thing.

ANY part in the system can throw the whole system over a cliff. Most of the failures look exactly like a dozen other possibilities. It is only when you realize that everything comes back to the input pair that you can see which part is throwing a wrench in the works. It's just a big differential amplifier like an op-amp, but done in one part at a time instead of inside an IC. After you get it, you will look like magic because this isn't easy and most people never accomplish the leap into understanding closed loop amplifiers.

Keep up the good work. When you understand this, it's like money in the bank.

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5. ### #12 Expert

Nov 30, 2010
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From the collector of T15, go right to a base-emitter and 47 ohms.
When R24 (47 ohms) gets about 0.6 volts across it, that voltage raises the bases of T20,21 and drives them in parallel.
Both of them have 0.5 ohms to the output.
When R31 (0.5 ohms) gets 0.6 volts on it, T23 (way over on the right) shuts down the drive back at that unlabeled transistor in the first line of this post.

So, at idle you have one Vbe and a little current through 47 ohms as the voltage at the collector of T15, and double that is the spread across T15.

That's the link to the page with the best schematic.

• ###### IMG_20160923_165010.jpg
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6. ### pwnstars Thread Starter Member

Feb 29, 2016
44
1
Thanks for the encouragement and clarification.
Watching it as an op-amp makes totally sense :O

l'll look at the circuit again tomorrow and see if i can find the issue. Thinking about taking T17 and T18 base out of circuit to better compare my measurements without the "auto adjustment" occurring.

7. ### #12 Expert

Nov 30, 2010
16,293
6,804
You can break this down into sections like: input pair, biasing circuit, power output stage, current limiting circuits.
But you can't just disconnect wires. You have to provide the input of each stage with what it would expect if everything was working.
If you disconnect the base of T14, you need to give it a DC level between zero and +60V.
Let's see...that's about 1.65 volts below the top rail.
That will force T14 to act like a constant current source for the bias group.

If you take the bases of T17,18 loose, they want to be at about zero volts +/- 0.7V

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8. ### #12 Expert

Nov 30, 2010
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R34 (47k) is the feedback resistor. It's what you take out to defeat the auto-adjustment. Without that, the input pair is free to wander around. The usual result is that the output voltage will slam either high or low, depending on which way the input pair drifts. You want to center the output voltage by providing that Vcc - 1.65 volts at Base of T13. Then adjust base T14 around Vcc - 1.65 volts to center the output voltage.

9. ### pwnstars Thread Starter Member

Feb 29, 2016
44
1
Hmm dont they need +/- 1.4V? They need to feed the output transistors too.

My reason for unconnecting T17 and T18 is to make sure that i measure the same voltages at the green arrows. (+/-) That way i can figure out if the first stage is working correctly. (in full circuit i measure -1.2V on positive rail and -3.5V on negative rail.)

Next step ill take ur advice and uncircuit R34 and bring down VCC to +/-20V so i avoid magic smoke across R38 and R41, while comparing measurements.

Whats ur opinion?

10. ### #12 Expert

Nov 30, 2010
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I thought I was very clear about that in post #5. You only need one Vbe and a little bit at the base of T17 to get some current through the 47 ohm resistor (R24). The voltage across that 47 ohm resistor THEN drives the next two transistors if more than 13 ma output current is needed. Without a low ohm speaker on the output, you don't need more current and T19,20 never come on. With no AC signal, the output should be stuck at zero volts, so the last two drivers T19,20 never come on.

If you want to disable T17,18, you can do that by just disconnecting their bases left of the over-current protection lines. Connect the 47K resistor to the base of T15 instead of the output line. You should be able to null the inputs by adjusting the pot next to T15. At the same time, you can adjust to get the 1.4V spread across T15.

This is getting pretty deep. There is still a chance I could make a mistake, like both nullng the input transistors and getting the 1.4V spread on T15 at the same time. (Obviously, I believe I'm correct.) Return to this conversation, PRN.