  ### Author Topic: where is the signal going in this output stage? [SUCCESS]  (Read 3233 times)

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#### rodd

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« on: February 21, 2017, 12:25:37 am »
Hi folks,
I am trying to repair an ac calibrator.
Up to now I manage to have a signal arriving at the input of the output stage, but no signal at the output.
I made a simplified schematic of the circuit to help me but it made things worse, since I can't figure out how the circuit works.
I can measure a 5V pp signal at the output of the OP17 op amp (point A in schematics) but the output is a very noisy signal about 50mV pp.
Would anybody give me a tip on how this circuit works and why the output signal is so low?
« Last Edit: February 22, 2017, 07:37:09 am by rodd »

#### Andy Watson

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« Reply #1 on: February 21, 2017, 12:55:25 am »
The output works by pulling or pushing the high-voltage DC supply. The return path, i.e. the centre/middle point of this supply is the actual output - via the 100R.
Can you measure the DC potentials on the gates and sources of the output transistors?

#### rodd

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« Reply #2 on: February 21, 2017, 04:35:48 am »
Hi,
I will measure the voltages tonight when I get home and post it.
But I still do not understand how this circuit can work with a high impedance load or no load.

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#### PA0PBZ ##### Re: where is the signal going in this output stage?
« Reply #3 on: February 21, 2017, 05:41:37 am »
But I still do not understand how this circuit can work with a high impedance load or no load.

There's probably a low R inside the power supply, did you measure the impedance at the output?
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#### StillTrying

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« Reply #4 on: February 21, 2017, 09:37:51 am »
But I still do not understand how this circuit can work with a high impedance load or no load.

Think of the floating -220V 0V +220V DC supply as two 220V batteries connected together in series.

Take care measuring 220/440 VDC.
CML+  That took much longer than I thought it would.

#### rodd

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« Reply #5 on: February 21, 2017, 10:09:26 am »
Hi,
Thank you for trying to help.
The "ground" of the +/- 220 V power supply comes from the center tap of the 190VAC secondary winding (no resistor in series).
The voltages are:
VN0350N1: Vg=-3.6V and Vs=0V
VP0350N1: Vg= 4.1V  and Vs=0V

Do they make any sense?

Thanks,
Roger

#### StillTrying

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« Reply #6 on: February 21, 2017, 10:26:29 am »
The Vs's being near 0V will be correct as they're connected to 0V through 10 ohms.

Have you carefully checked there's +220V on one Vd and -220V on the other.
CML+  That took much longer than I thought it would.

#### Andy Watson

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« Reply #7 on: February 21, 2017, 10:40:32 am »
VN0350N1: Vg=-3.6V and Vs=0V
VP0350N1: Vg= 4.1V  and Vs=0V
Assuming that you have mis-typed the VN - VP designators, then yes, the voltages are in right ball-park. Since you have replaced the output transistors I would expect that the biasing needs adjustment (the 5k potentiometer), however, this would not prevent the output from working (but it might be distorted or the transistors might run too warm).

Can we assume that the fuses are intact and that the +- 220V is reaching the drains of the mosfets?
How are you measuring the signal waveforms - do you have access to an oscilloscope?
What is the signal level at both sides of the 100R in series with the output?

#### rodd

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« Reply #8 on: February 22, 2017, 02:36:49 am »
Hi again,

and no I have not misspelled the MOSFETs
The voltages around the FETs are:
N-channel:   Vd=298V ; Vg=-3,6V and Vs=0V
P-channel:    Vd=-298V; Vg=4.1V and Vs=0V

Now I see that the gate voltages seem to be reversed, but maybe that is why the non inverted output of the OP17 has two 8V zenners.

The zenner are ok but both inputs of the op amp are at 0V.

#### Kleinstein

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« Reply #9 on: February 22, 2017, 03:29:21 am »
The gate voltages are somewhat reversed, it should be positive with the N channel.

The zener diodes are for protection of the OP only, so under normal operation they just make an auxiliary +-8 V supply used to clamp the signal.

#### Armadillo ##### Re: where is the signal going in this output stage?
« Reply #10 on: February 22, 2017, 03:39:22 am »
The gate voltage is correct. Because the opamp is on +15 and -15v rails and will swing the voltage around to turn on the mosfet. Imagine between the 5K variable resistor and the 24 K the voltage is either +15 or -15v [well somewhat close to it].

If the opamp don't swing, the circuit don't work.

#### Kleinstein

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« Reply #11 on: February 22, 2017, 04:28:17 am »
The N channel Gate is between the OPs output and +15 V, the P-channel between the OPs output and -15 V, so in any case to N channel mosfet gate should always be more positive than the p-channel. I won't make sense to have them swapped the other way. To get them both off in between and true class B operation one could just drive both gates together from the OP.

#### Armadillo ##### Re: where is the signal going in this output stage?
« Reply #12 on: February 22, 2017, 05:13:14 am »
Let's discuss based on this diagram. #### rodd

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« Reply #13 on: February 22, 2017, 06:15:53 am »
It occurred to me that the gate voltages might be influenced by the AC signal that is also present, so I removed the OP17 from its socket and measured the voltages again.
The new results were:
N-FET: Vg= +0.6V
P-FET: Vg= -0.7V
The other voltages stayed the same.
The gate voltages are strange, because if I assume that there is no current in/out to the gate bias circuit ( two 7.5K resistors and two 2.4K resistors in series) it follows:
Rtotal = 20k
DV=15-(-15)=30V
I= 30/20K = 1.5mA
so, voltage drop across 7.5K resistor should be 7.5K * 1.5mA = 11.25V
and VGp =15 - 11.25 = 3.75V which seems ok.
What is going on here?

#### Andy Watson

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« Reply #14 on: February 22, 2017, 06:23:51 am »
What is going on here?
Something very strange!
Can you double check that the correct transistors are in the correct place - i.e. the N-channel and NPN transistors should be in the positive section and the P-channel and PNP in the negative section.

#### rodd

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« Reply #15 on: February 22, 2017, 06:35:26 am »
More new from the front!
If I understood the circuit, both transistors should be off, under normal operation.
The Vbe of both transistors is zero, but there is a voltage drop of 0.35V around the 330R resistors.
This means there is 1mA passing through them
(OP17 is out )

#### Andy Watson

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« Reply #16 on: February 22, 2017, 06:46:04 am »
Under normal conditions the bi-polar transistors are there to limit the current in the output fets - therefore they should not normally be turned on. With no signal present the gates are biased to approximately $15 V \times \left( \frac{2.4k}{2.4k + 7.5k} \right)$, I make that to be about 3.6V - either side of point "A". Since the gates and the bi-polar transistor should not be taking current, there should be no voltage dropped across the 330Rs.

Either something is very wrong with circuit or it might be possible that the amplifier is oscillating - can you re-check the voltages with your scope?

#### Kleinstein

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« Reply #17 on: February 22, 2017, 07:09:51 am »
Let's discuss based on this diagram. The designation of FET type in this diagram is wrong. The upper one should be N-channel, thus VN...., the lower one should be p-channel, thus VP..... So the original circuit is correct, not this version.

The voltage measurement without the OP looks much more reasonable - likely the 5 K trimmer is set to a rather low value. Having about +-0.7 V might suggest the FETs are not FETs, but NPN and PNP transistors ?

#### rodd

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« Reply #18 on: February 22, 2017, 07:20:53 am »
Hi folks,
Following the suggestions I double checked the active devices and used the opportunity to add sockets to the transistors.
I have found out that the guy that has tried to fix the instrument before has switched the position of the BJT's.
I corrected the mistake the and measured the gate voltages again (OP17 out).
The values are:
VN--> Vg= +1.8V
Vp--> Vg=-2.4V

It seems were are getting closer!!
I will return OP17 to the socket and see what is happening.
This may take a while.

#### rodd

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« Reply #19 on: February 22, 2017, 07:35:28 am »

IT WORKED!!!

I have an output signal at last!

Thank you for your patience and time.

There are things to fix yet but I am closer than I ever was.

I will post pictures when the job is done.

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