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Stage Line 500W amp repair project

Sounds like Q6 is shot.

The base and emitter terminals of a transistor have an internal diode connected between them. It is called the "base-emitter junction". A diode conducts current in one direction: when the anode (the blunt side) is positive relative to the cathode (the point and bar side). The diode clamps the voltage to about 0.7V.

Here is a diode with its terminals labelled.

View attachment 17491
Here's how a transistor is constructed in terms of diodes. If you use the diode test range on your multimeter and measure between every combination of leads on a transistor, both ways round, you will get readings that correspond to these diodes.

View attachment 17490

Here is that section of the circuit, with Q6's base-emitter diode shown in red.

View attachment 17492

When a diode's anode is more positive than its cathode, it conducts, and clamps the voltage across it to no more than about 0.7V. If you measure more than a few volts across a diode, when its anode is positive relative to its cathode, it is damaged.

So if you measure the base-emitter voltage of Q6, with the positive probe on the emitter (the anode of the diode), and the negative probe on the base, and you get a positive reading that's more than about 1V, Q6 is definitely damaged.

So check the polarity of the voltage you measured across the base-emitter junction of Q6, and if it's out of range, Q6 is damaged.

You can check the base-emitter voltage of any other transistors as well, but remember that Q6 is a PNP (the emitter arrow points towards the base); the polarity is reversed for NPNs (emitter arrow pointing away from base).

I hope this makes sense. Please ask if anything is unclear.

Edit: I should have pointed out: the arrow on the emitter points in the same direction as the arrow on the equivalent diode.

am i tsting this with the board powered up? or do i need the power off? and am i voltage testing or diode testing, ?

i will read this over a few times to be sure i understand and do some more testing tomorrow and get back to you,

also would any of the transistors i have already removed be any good as a replacement for Q6 as long as it was good?
 
OK thanks for that.

I'm only interested in the voltages at the marked points. So if I've indicated a wire that goes to the base of a transistor, just measure the base.

Those voltages in the bottom half of the circuit should be negative!

Can you also measure the voltage across R20, and the voltage between the base and emitter of Q6 please.


i am going to do this test agin, i am sure i didnt have a negative sign on the bottom half of the circuit, i did take note you said add this if it was there!
going to go back to post #115 and re read and retest to make sure i have done everything right
 
Sounds like Q6 is shot.

The base and emitter terminals of a transistor have an internal diode connected between them. It is called the "base-emitter junction". A diode conducts current in one direction: when the anode (the blunt side) is positive relative to the cathode (the point and bar side). The diode clamps the voltage to about 0.7V.

Here is a diode with its terminals labelled.

View attachment 17491
Here's how a transistor is constructed in terms of diodes. If you use the diode test range on your multimeter and measure between every combination of leads on a transistor, both ways round, you will get readings that correspond to these diodes.

View attachment 17490

Here is that section of the circuit, with Q6's base-emitter diode shown in red.

View attachment 17492

When a diode's anode is more positive than its cathode, it conducts, and clamps the voltage across it to no more than about 0.7V. If you measure more than a few volts across a diode, when its anode is positive relative to its cathode, it is damaged.

So if you measure the base-emitter voltage of Q6, with the positive probe on the emitter (the anode of the diode), and the negative probe on the base, and you get a positive reading that's more than about 1V, Q6 is definitely damaged.

So check the polarity of the voltage you measured across the base-emitter junction of Q6, and if it's out of range, Q6 is damaged.

You can check the base-emitter voltage of any other transistors as well, but remember that Q6 is a PNP (the emitter arrow points towards the base); the polarity is reversed for NPNs (emitter arrow pointing away from base).

I hope this makes sense. Please ask if anything is unclear.

Edit: I should have pointed out: the arrow on the emitter points in the same direction as the arrow on the equivalent diode.

i have a problem with the word relative!! can you describe it another way?? do you mean more than , less than,
sorry but please dont for get i am a beginner, well sort of now lol
 

KrisBlueNZ

Sadly passed away in 2015
am i tsting this with the board powered up? or do i need the power off? and am i voltage testing or diode testing, ?
Test with the board powered up, and measure voltage.
also would any of the transistors i have already removed be any good as a replacement for Q6 as long as it was good?
Yes, Q11 and Q22 were MPSA92s. If the old ones measure OK out of circuit you can use them. Test them on the diode range of your meter. You should measure two diodes inside them, as in the PNP part of the second image in post #120. The correct measurements for a PNP, using diode range, are:

Black probe to base, red probe to emitter: should measure a diode forward voltage (about 0.5~0.8V).
Black probe to base, red probe to collector: should measure a diode forward voltage.
Red probe to base, black probe to emitter: should measure open circuit (indication of "1 ." or "0L" or similar).
Red probe to base, black probe to collector: should measure open circuit.
Collector to emitter: should measure open circuit both ways round.
i am going to do this test agin, i am sure i didnt have a negative sign on the bottom half of the circuit, i did take note you said add this if it was there! going to go back to post #115 and re read and retest to make sure i have done everything right
OK, good.
i have a problem with the word relative!! can you describe it another way?? do you mean more than , less than,
sorry but please dont for get i am a beginner, well sort of now lol
OK. Voltages are always measured between two points. Like distance, which you would measure with a ruler. The distance between my nose and my chin is about 7 cm, but in electronics, the polarity (direction) is also important, so I would have to say that my nose is 7 cm higher than my chin. (This explanation always assumes that my head is upright.)

I can then say various things about the relationship between these two points:
  • My nose is higher (more positive) than my chin;
  • My nose is 7 cm (7 volts, for example) higher (more positive) than my chin;
  • My nose, measured relative to my chin, is +7 cm
  • There is 7 cm between my nose and chin, with my nose being higher (more postive).
In this case we're talking about differences in voltage, where both ends of the measurement are at specified points. This corresponds to the red-and-black boxes that I drew on the schematic in post #112. There are also some red boxes, where I want to know the voltage at that point, measured relative to the 0V ("zero volt") rail of the circuit.

The 0V rail is a standard reference point for voltages to be measured relative to (also called "with respect to" or "with reference to"). It's kind of like sea level, or ground level, for the whole circuit. It is the default reference point for voltages that are said to be "at" or "on" a certain point in the circuit. That's why the black lead of the multimeter is often connected to the 0V rail of the circuit when you're making voltage measurments.

For example, if the ground or floor is the reference point, my chin is at about +150 cm. Therefore my nose is at about +157 cm. So have another look at that part of the circuit:

Q6 B-E junction unmarked.png

If that circuit was working properly, Q6's emitter (top terminal) would be at about +54V (relative to the 0V rail). Since an active transistor has about 0.7V between its base and emitter, and in the case of a PNP transistor (which Q6 is), the base is 0.7V more negative than the emitter, the base voltage would be (54V - 0.7V) = 53.3V.

I could have asked you to measure the individual voltages on those two points, relative to the 0V rail. But it's more interesting, and more accurate, to measure the actual voltage across the points. Just like if you want to know the distance between my chin and my nose, you wouldn't measure from my feet to my nose, then measure from my feet to my chin, and subtract them; it's better to measure the small distance directly.

I'm not sure whether this is helping, but I think it should. The important things to remember are that voltage is always measured between two points; the polarity of the voltage is important; if you're asked to measure the voltage "at" or "on" a specific point in the circuit, you need to measure it relative to the 0V reference rail.
 

KrisBlueNZ

Sadly passed away in 2015
BTW, I don't know if you're worried that you won't be able to understand this stuff, but if you are, don't be. You're a mechanically minded person, and quantities in electronics work in very similar ways; they're just harder to see! (That's why we have oscilloscopes - the second most important piece of test equipment, after the multimeter.)

Your confusion and your question were totally reasonable, and if you don't understand something, you need to tell me so, so I can explain it better. I have absolutely no doubt that you will be able to understand voltage, current, resistance etc if I can explain them clearly enough.

And don't worry about taking my time to explain things. I want to improve my explanations and everything I write, and your responses to them, help me to clarify my ideas and improve my descriptions.
 
R20 reads 00.0
base and emittor of Q6 reads22.5--23.0

just did this test again.
when you said mesure across R20, did you mean black probe on one side and red on the other?
thats what i did to get 00.0

or did you mean black probe to T4 and mesure either side of R20 with the red probe?
that reads 80.1 VDC either side that way.
 
So if you measure the base-emitter voltage of Q6, with the positive probe on the emitter (the anode of the diode), and the negative probe on the base, and you get a positive reading that's more than about 1V, Q6 is definitely damaged.

So check the polarity of the voltage you measured across the base-emitter junction of Q6, and if it's out of range, Q6 is damaged.
.

just did this test again replaced my resistor fuses to be sure there good,

amp turned on and meter set to 200m,
red probe on the emitter and black probe on the bass of Q6, it reads0.07,

You can check the base-emitter voltage of any other transistors as well, but remember that Q6 is a PNP (the emitter arrow points towards the base); the polarity is reversed for NPNs (emitter arrow pointing away from base).
.
so for PNP and NPN is the pinout still the same or is it reversed E-B-C ====== C-B-E?

the first diagram in post #120 suggests there the same
 

KrisBlueNZ

Sadly passed away in 2015
when you said mesure across R20, did you mean black probe on one side and red on the other? thats what i did to get 00.0
That's right. "Measure across" a component means put one probe on each end of it.
just did this test again replaced my resistor fuses to be sure there good, amp turned on and meter set to 200m, red probe on the emitter and black probe on the bass of Q6, it reads0.07
So the meter was on the 200 mV range and it read 0.07? That's wrong too. On the 200 mV range, the display should have three digits, then a decimal point, then one more digit. So it shouldn't be able to display 0.07.
so for PNP and NPN is the pinout still the same or is it reversed E-B-C ====== C-B-E? the first diagram in post #120 suggests there the same
Yes, the pinouts are the same. And the emitter is always the one with the arrow.

Can you make some more measurements please?

epoint 270979 for voltage annotations 2.png
 
Here is my meter display set on 200m
As you can see it has two 0s a decimal point then another zero, maybe it's the way my meter works making all the readings seem wrong, or I am using it incorrectly??
E25F8131-61B1-45A5-A1A7-46F194C44EB8_zpsv1rltudb.jpg
 
Last edited:

KrisBlueNZ

Sadly passed away in 2015
You're using it right. But you said:
amp turned on and meter set to 200m, red probe on the emitter and black probe on the bass of Q6, it reads0.07,
On the 200 mV range there is only one digit to the right of the decimal point, so it couldn't read 0.07. That's what is confusing me.
 
just did this again and must have shorted something, them resistor fuses go fast!!!!
changed it for a new one,noticed the reading drasticaly changes when my arm tuches the chassis of the unit?
anyway with me not touching it with a part of my body i get -87.3 with fluctuation up and down,
meter is set to 200m and red probe was on emitter and black was on bass
hope this helps ya kris
will do the other mesurements you asked for now and post my results
 

KrisBlueNZ

Sadly passed away in 2015
OK, those numbers seem to make sense, and they're pointing to a problem earlier in the output stage. I need you to make some more measurements, I'm afraid.

epoint 270979 for voltage annotations 3.png
 

KrisBlueNZ

Sadly passed away in 2015
D'oh! I know what's going on. The amplifier is being disabled by the protection circuit. That's why there's no current flowing in any of the stages!

You'll need to temporarily bridge Q2's collector and emitter together. That will bypass the protection circuit, but the amplifier will still be protected by the resistor fuses.

When you've bridged Q2's collector and emitter together, power up and re-measure the voltages I asked for in post #112.

Sorry for the confusion!
 

KrisBlueNZ

Sadly passed away in 2015
Aha! That's better.

Those voltages across R30 and R32 show that current is flowing in the output stage. That's good! And the -1.8 mV measurement on the output is good - the DC voltage at the output should be close to zero.

epoint 270979 where are Q7 and PR1.jpg

Is the transistor indicated in the picture Q7? If not, can you find Q7 and show me where it is? I was expecting it to be attached to the heatsink so it can monitor the temperature of the output transistors, but maybe it isn't.

Also, where is PR1? This should be a screwdriver-adjustable preset potentiometer or "trimpot" that is used to set the bias current in the output stage. It doesn't seem to be present on the board! And nor is PR2, for the right channel! Is there any sign of modifications to the board in those areas? The trimpots aren't underneath, are they? Is there fresh solder on the connection points? Is it possible that the previous owner took them out and didn't replace them?
 
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