Help repairing Phase Linear Series II Model 200

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Okay. I bought a new amp, and now I'm a lot less frustrated and coming back to the broken one.

Powered up a single channel, and these are the voltage readings I get:

Q1 E=0.0 B=0.0 C=-71.2
Q2 E=+69.6 B=+69.6 C=+69.6
Q3 E=+69.6 B=-71.2 C=-71.2
Q4 E=+69.1 B=+69.1 C=+71.6
Q5 E=+68.8 B=+69.1 C=+71.2
Q6 E=+68.8 B=+69.1 C=+71.2
Q7 E=+69.2 B=+66.5 C=+69.7
Q8 E=+68.7 B=+68.5 C=+68.7
Q9 E=+68.5 B=+69.4 C=-70.1
Q10 E=+68.8 B=+68.8 C=-71.2
Q11 E=+68.8 B=+68.8 C=-71.2

Does this give any clue as to what's wrong?

Charles.
 
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Either Q1 is shorted base to emitter or the op amp is dead (or is missing its minus rail). Applying positive rail voltage to pin 2 should pull pin 6 more negative than zero.

This is why I suggested deliberately pulling the base of Q1 negative to begin with. The results of that will tell you which one.
 
Fine. If you don't want to cut a PC trace, then lift the base leg of the transistor. Hook a couple k ohm resistor between the base leg and the -15 supply (op amp pin4). If the output switches polarity the rest of the amp works and op amp is the culprit. Looks like it is anyway, because its input/output voltages are inconsistent. Those Bi-FETs won't do anything dumb like reverse phase its output when the positive common mode range is exceeded. It can do it on the negative half, but the electrolytic feedback cap will be reverse biased (and leak) if that happens, minimizing the voltage applied to the op amp.


Jumpering Q1 C-E with 100 ohms will probably smoke R7. If you do that, use about a 10k resistor, not 100. Something that can take the full 65-ish volts is needed.
 
Replaced Z1, powered it up on a dim bulb tester, and this is what I'm getting now:

Q1 E=3.8 B=13.5 C=64.0
Q2 E=62.5 B=62.5 C=62.5
Q3 E=64.1 B=64.1 C=62.6
Q4 E=62.3 B=62.6 C=64.0
Q5 E=62.1 B=62.4 C=64.4
Q6 E=62.1 B=62.4 C=64.4
Q7 E=62.0 B=59.6 C=63.1
Q8 E=62.1 B=62.9 C=62.0
Q9 E=62.0 B=62.8 C=-64.4
Q10 E=62.4 B=62.4 C=-64.7
Q11 E=62.4 B=62.4 C=-64.7

Charles.
 
Hi,
Can you double check the Q1 collector voltage? Your voltages table showed Q1 collector voltage = 64volts. Also Q3 base = 64 volts. If this is true then Q3 it is shorted between collector to base or cap C7 it is shorted. It is the only way Q1 collector can be positive. Q3 base it is tied to the negative rails -64 volts thru resistor R9 1K. Maybe I am wrong but Q3 base can not be 64 volts positive .
 
Check my corrected values. Q3 is negative, that was a mistake.

you're right about Q1. It's
Q1 E=3.8 B=13.5 C=-64.0

So here's the updated, corrected readout:

Q1 E=3.8 B=13.5 C=-64.0
Q2 E=62.5 B=62.5 C=62.5
Q3 E=-64.1 B=-64.1 C=62.6
Q4 E=62.3 B=62.6 C=64.0
Q5 E=62.1 B=62.4 C=64.4
Q6 E=62.1 B=62.4 C=64.4
Q7 E=62.0 B=59.6 C=63.1
Q8 E=62.1 B=62.9 C=62.0
Q9 E=62.0 B=62.8 C=-64.4
Q10 E=62.4 B=62.4 C=-64.7
Q11 E=62.4 B=62.4 C=-64.7
 
Hi,
I still do not understand why everything is reading +64 volts. We need to balance the transistors drivers bases Q4 and Q9 close to zero and then be able to adjust it with the R24. To do that we need Q3 to conduct and to that the Q1 base need be negative. That will make the collector of Q3 positive consequently Q3 will conduct. Lets do a test. Can you remove Z1 and jumper 4 and 6 with a 10K resistor? that will make the base of Q1 negative and repeat the readings. This was discussed in thread #35 by wg_ski.
 
Looks like Z1 is defect or it's pin 4 not a -12V.
And there is no protection against pin 2 going more negative than pin 4.
There should be a at least two diodes from pin 2 to pins 7 and 4 in a blocking direction.
Now you get the 64V from the output via R4 on poor pin 2 :-(
Mona
 
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