Vave should be close to zero. You probably have a DC offset close to 3.24V.

 

Hi kiss,In post #338 he is using an X10 probe so the offset is more like 30 volts. I suspect some kind of reading error which is why I asked him to see what the scope displayed for the + and - 45 V rails.

Les.

 

Need to confirm with a voltmeter.

A 30V offset which is why you do most of the testing with a signal. But Vmin/max is 33, -39V as measured on the scope's screen. That really suggests x1.

Offsets are generally harder to track down. One needs to look at the input signal. Thisthing suppots bridging, so more troubles.

 

Comment:

For work, I purchased a 0.5 ohm 1000W adjustable resistor to make a dummy load for an arc lamp power supply.

For audio amplifier testing, I never used the loads for an extensive period of time. Probably 10-15 minutes. I had a 8 ohm 100W resistor.

I also bought a 1000W load for a 13.56 MHz RF transmitter from these https://mfjenterprises.com/collections/dummy-load guys. RF loads are 50 ohm. I used the oil filled can.

 

Turns out this scope hammers PP9 batterys,(need a more permanent solution to this) it was flat.
here is the scope reading again right as it started to clip, this is 500htz in and connected to the 6Ω speaker, scope lead switched to x10,

Aaahhh bugger, just noticed i had the positave connected to T34 not T10, does this matter to much ?

 

Much better. So you got 38.9*38.9/6 or about 250W and the DC offset should measure 80mV or sol

I strapped an AA battery pack to the back of my DSO150. You might want to invest in a wall wart. I forget what the supply restriction is.

Specifications:



So it looks like you could be within specifications.

You do need to look at quiescent current, no load, no signal over time. Look at the voltage across say R64 or R73 over time for about 20-30 minutes. It should stay relatively stable over time, 0V isn't good.

I do think you should get a thermal adhesive for Q29. Not thermal paste. There might be some adhesives that will break ther bond when twisted. Not sure.

Remember, that you use only a really thin film of grease for insulators that require grease. Sill-pads do not require grease.
They are usually a soft grey, compressable and not smoothe. Kapton/Polyimid is orange and does require grease. Mica is an old material.

If you have a way to measure the temperature of the outputs, use it. Don't use the finger test unless you have say a thero-pad between you and your finger. Your looking at 50V.

You could even use the diode mode on your meter and the B-E junction of the transistor. Your interested in relative differences and not calibrated differences. It should be -10mV per deg C. So, you can say, thansistor A is hotter than transistor B by 10 degrees C.

Then do the test again if you can't what you need.

I think the amp has a clip indicator, but don't test it using speakers.

You have to get the repair right before you can push it hard.

Hopefully, you can see the importance of initial troubleshooting without a load and a signal applied.
No load, no signal is a bad place to be.

 

That looks good. Take kiss's advice from post #346 and don't be tempted to rush into staring on the left hand channel.

Les.

 

I dont have any voltage across R64 or R73 ?
Or is this a case of leaving the amp on to warm up and see if the bias regulator is working and we are not going to end up with thermal runaway ?,

Again i am just guessing here, hopefully some day my guessing will be right

 

Much better. So you got 38.9*38.9/6 or about 250W and the DC offset should measure 80mV or sol

I strapped an AA battery pack to the back of my DSO150. You might want to invest in a wall wart. I forget what the supply restriction is.

Specifications:

View attachment 247962

So it looks like you could be within specifications.

You do need to look at quiescent current, no load, no signal over time. Look at the voltage across say R64 or R73 over time for about 20-30 minutes. It should stay relatively stable over time, 0V isn't good.

I do think you should get a thermal adhesive for Q29. Not thermal paste. There might be some adhesives that will break ther bond when twisted. Not sure.

Remember, that you use only a really thin film of grease for insulators that require grease. Sill-pads do not require grease.
They are usually a soft grey, compressable and not smoothe. Kapton/Polyimid is orange and does require grease. Mica is an old material.

If you have a way to measure the temperature of the outputs, use it. Don't use the finger test unless you have say a thero-pad between you and your finger. Your looking at 50V.

You could even use the diode mode on your meter and the B-E junction of the transistor. Your interested in relative differences and not calibrated differences. It should be -10mV per deg C. So, you can say, thansistor A is hotter than transistor B by 10 degrees C.

Then do the test again if you can't what you need.

I think the amp has a clip indicator, but don't test it using speakers.

You have to get the repair right before you can push it hard.

Hopefully, you can see the importance of initial troubleshooting without a load and a signal applied.
No load, no signal is a bad place to be.

I do have a fluke thermomiter i could use if that would be any good for measuring the temperatures?

 

I have ordered the 200w 8Ω resister i posted earlier, should be here saturday.
I will try find a heatsynk to mount it to, this way i wont be deafening the whole house when it comes to checking max output

 

@KeepItSimpleStupid
In post #346 you said i had 250w output, how did you work that out, i though with the scope lead being on x10 i just needed to x10 the volts peak to peak to give me the total output but that gives me 107.9 ?

 

The power = V^2/R (Note with AC it is the RMS voltage that is used.)
So 38,9 x 38.9 /6 = 1513/6 =252 watts.
The impedance of loudspeaker is only a nominal value and varies over the frequency range and also with the characteristics of the enclosure that it is mounted in. So it is not certain that the output power is 252 watts.
I had a thought about there being no voltage across R64 and R73. When we had the bias regulator problem I wondered why the bias voltage was set to 6 base emitter voltage junction drops when there are 8 base emitter junctions in the emitter follower stages.
I am now thinking that at very low signal levels the last two transistors do not conduct. The quiescent current flows through R62 and R71. Once the voltage across these two resistors exceeds about 0.6 volts the output transistors will provide the extra current for the output. This might explain why the bias regulator only provides 6 BE junction drops.
I wonder what your thoughts are on this theory.
It would be interesting to see if the left hand channel behaves the same way. (When it has been repaired.)

Les.

 

So when you say 6 base emitter junction drops your refering to the 4v bias collector and emitter voltave on Q29, this ÷6 gives o.6v forward bias for 6 transistors, but there is 4 transistors on the output tails either side, 4+ and 4 --, so the bias should be closer to 5v ? Make any sence at all ?

 

The voltage across the emitter resistor will be in mV. Probably less than 10mV. Remember idle, no signal, no load for this one.

https://www.tnt-audio.com/clinica/bias_e.html

So when you say 6 base emitter junction drops your refering to the 4v bias collector and emitter voltave on Q29, this ÷6 gives o.6v forward bias for 6 transistors, but there is 4 transistors on the output tails either side, 4+ and 4 --, so the bias should be closer to 5v ? Make any sence at all ?

I explained that earlier. Yes and no.

There are two transistors that essentially have 2 B-E junctions in parallel and a couple in series × 3

So, 0.6*4 = 2.4 plus some number which is < (0.6 * 2).

 

I do have a fluke thermomiter i could use if that would be any good for measuring the temperatures?

Look at the manual and pay attention to the focal distance and spot size.

 

I know you didn't expect mV levels across a resistor, but it's a 0.22 Ohm resistor. There are 2 such resistors essentially in parralel in this Amp, so the quiesent current measured using the voltage across one resistor is really × 2.

 

Quiescent current. Remember 90V. If it was 100 mA that would be 9W which isn't unreasonable, then you would measure 50mA across 1 emitter resistor. There are 2 pairs across the +-45V supplies.

So V=IR, 0.050A × 0.22 Ohms =

A small number. 11mV

 

Hi kiss, I do not agree with some of your comments in post #354.
There are not two base emitter junctions in parallel. Remember Q27, Q35 R63 and R72 are not fitted.
Just looking at the negative side there are 4 emitter follower stages Q40, Q37,Q24 and Q36 so there are 4 B-E junctions in series.
The same applies to the positive side so we have 8 base emitter junctions.
If you look at the table of readings in post #306 you will see that there is only about 0.3 volts across the B-E junctions of Q20 and Q36 so they will not be conducting, So these 0.3 volt drops are developed across resistors R62 and R71 (10 ohms)
So the current flowing though these resistors will be 30 mA. This will be the quiescent current.
So between the input of the emitter follower stages and the emitters of Q26 and Q34 we have 6 B-E junctions which fits with the bias regulator producing 6 B-E junction drops.
When there is enough signal to produce a voltage of more than about 0.6 volts across R62 or R71 the B-E junctions of Q28 or Q36 will start conducting and providing the main part of the output current.

Rookieme, I am pleased to see from your post #353 that you seem to be understanding how the output stages of the amplifier works.

Les.

 

The not fitted makes sense. I should try to print the schematics better. I have a hard time reading them even with high power magnifiers.

 

Are we calling the right chanel done, or is there anything else you would like me to test,

Pete