Hifi amplifier biasing issues and output stage shorting?

[Yamin]

Hi guys,
Just wanted to clarify some doubts I had with biasing issues on SS amplifiers. I am really interested on working on amplifiers and anything related to audio. I have been reading on amplifiers.

So here is the series of events on the project I am working on:

• Output transistor Q6050 was shorted and one half of the emitter resistor R6100 was open (left channel) and the B+ fuses were open.
• So I borrowed the two output transistors and the emitter resistor from a working channel which was not used. Replaced the fuses however I didn’t replace the driver transistors(I should have!).
• Powered on the unit and measured the biasing voltage at the test points and it was within specs given on the SM. However I should add that I didn’t monitor it long enough.
• Played some signal through it and it was working perfectly fine.
• So assembled the unit and returned it to the owner to be used until the replacement parts arrived.
• Checked it at his place, was working fine and after I got back he called me back saying that the amp had failed and that there was no output.
• I checked the amplifier again the same left channel had shorted. The two output transistors were shorted and the B+ voltage fuses were blown. The emitter resistor was intact this time though.

As per what I have read online regarding what could blow the output stage (some reasons):

• The speaker being shorted - I checked the speakers and I didn’t measure a short across the terminals. However it had a crossover circuit and only what I could read was a capacitor charging and both speakers showed similar behaviour. The L and R speaker sounded same to me when it was working.
• Cold/broken solder joints - I have checked for these and didn’t come across any obvious points.
• Other components failed - I have checked almost all the components on the left channel and it measures ok.
• Biasing issues.

I would like to clarify about the biasing circuit used on this amp. I have done some research on Vbe multiplier circuits but most of the circuits I came across the variable resistor is connected to the base of the Vbe multiplier. However in this unit the VR is connected to another transistor’s base Q6010. I am finding it a bit hard to grasp how this works. Could someone kindly explain.

Q6010 is placed next to one of the drive transistor and it has got what looks like thermal compound applied to it. What is the reason for this?

If it was a biasing issue and somehow it caused the output transistors to short out, note B-C-E were shorted - why didn’t it take out the drive transistors? The drive transistors still checks out ok.

All of the components related to the biasing part of the channel checks out ok but hypothetically speaking what component failure for e.g.: the potentiometer’s contacts loosing connection, some kind of thermal effect messing up with the biasing circuitry’s transistors?.

If its some kind of an issue which only causes the fault when power is applied what further troubleshooting should I be doing on this circuit? At the moment I am waiting for the replacement parts to arrive but meanwhile I would like to figure out what exactly caused the fault - could I power up the amp without the output transistors and see if the proper voltages are preset which is shown on the SM and lets say if the drive transistors are faulty, ‘leaky’ for instance would I be able to measure that?

Based on what I have detailed above, what’s your opinion for the reason for the fault to occur?
Thanks in advance for the help.

P.S - I couldn’t find replacement for some of the transistors and resistor so here are the equivalents that I have ordered
2SC1740S-S ——— 2SC2240
2SA1930 ——— A1837
0.22 Ohm 2W emitter resistors —— 0.22 Ohm 5W 
Also note that there are some film capacitors which looks a bit black, I am not sure it was like it to start with. It measures ok especially when compared to a working circuit. Hope you can see it in the photo

[magic]

If we replace Q6000, R6000 and R6010 with a short, the Q6010 circuit is a Vbe multiplier.

The Q6000 circuit also is a Vbe multiplier and as such it behaves like a zener. It forces Q6010 to increase its collector by Q6000 circuit's "breakdown voltage".

Total bias is the sum of two Vbe multipliers which operate at different temperatures. I suppose there is some chance that if the inner multiplier doesn't work correctly, you can compensate its error by readjusting the outer multiplier, but then things get bad when the amp warms up.

Or it could be some other problem altogether causing it to blow up, dunno.

[coromonadalix]

q6000 could be a constant current reulator too, you need to be stable in voltage and current for the biais adjustment

Is it a sony amplifier / tuner ?    it's a direct coupled amplifier model,  meaning if something goes wrong  lots of parts may die.

You have to remove all transistors, the biais one too,   check them carefully on diode and ohms mode each pins and reverse polarity too, not in the circuit.

Sometimes in the past   the only solution i had  was to replace all transistors ... check all resistors, and check for cold joints

[SpecialK]

i believe it's an Onkyo HT-RC440 or similar model.  Googled one of the transistors model number and it's silkscreen ID.

[xavier60]

Two main possible reasons for that kind of failure. Even a brief open circuit in the idle current biasing circuit such as a fractured joint or bad contact of the trimpot's wiper will cause destructively high rail to rail current through the output transistors.
There could be a problem at the owners place. Many people just don't realize the harm that short circuiting stray speaker lead strands can cause.

When an output stage blow up occurs, because the VAS transistors are configured as current sourcing, damage rarely occurs to them, this also applies to earlier stage transistor.
Transistors and other components in the bias circuit do get damaged.

[Yamin]

Thanks everyone for the input really do appreciate it.

q6000 could be a constant current reulator too, you need to be stable in voltage and current for the biais adjustment

Is it a sony amplifier / tuner ?    it's a direct coupled amplifier model,  meaning if something goes wrong  lots of parts may die.

You have to remove all transistors, the biais one too,   check them carefully on diode and ohms mode each pins and reverse polarity too, not in the circuit.

Sometimes in the past   the only solution i had  was to replace all transistors ... check all resistors, and check for cold joints

Its an Onkyo SR508, sorry I should've mentioned that in my first post. Yes I was also planning on taking the small transistors out of circuit and testing them. I have tested them in circuit and didn't come up with an obvious fault. I also measured them in ohms mode and compared the results with the working channel's transistors they seem to match up   

Two main possible reasons for that kind of failure. Even a brief open circuit in the idle current biasing circuit such as a fractured joint or bad contact of the trimpot's wiper will cause destructively high rail to rail current through the output transistors.
There could be a problem at the owners place. Many people just don't realize the harm that short circuiting stray speaker lead strands can cause.

When an output stage blow up occurs, because the VAS transistors are configured as current sourcing, damage rarely occurs to them, this also applies to earlier stage transistor.
Transistors and other components in the bias circuit do get damaged.

Can I ask please do correct me  - for instance if the VR was open that should 'turn off'  Q6010 right? Could that take out the Output transistors? If so why does it happen? I was under the impression that it'll just make the output sound distorted with cross over distortion.
Sorry for the noob question - could you please elaborate on VAS stage of this amplifier, I know it stands for the voltage amplification stage  .
What is the purpose of Q5030 & Q5040?

Why doesn't biasing issue take out the drive stage transistors like the output transistors? I haven't had much experience working on amps but from most of the stuff that I have read and seen the driver transistor seems to be intact even when the outputs short out, but everyone recommends to change them anyway.

'IPRO' is current protection, shouldn't this protection be engaged when there is a bias issue? There are other two protection signals but I'm not sure what it stands for 'FLVPRO' and 'FLVOLH' the service manual is not really descriptive on this.

Nice pointer about the strands causing shorts, I'll have a proper look at the speakers terminal at the owners place.

Once again thanks so much for the help everyone, really do appreciate it.

[xavier60]

Thanks everyone for the input really do appreciate it.

q6000 could be a constant current reulator too, you need to be stable in voltage and current for the biais adjustment

Is it a sony amplifier / tuner ?    it's a direct coupled amplifier model,  meaning if something goes wrong  lots of parts may die.

You have to remove all transistors, the biais one too,   check them carefully on diode and ohms mode each pins and reverse polarity too, not in the circuit.

Sometimes in the past   the only solution i had  was to replace all transistors ... check all resistors, and check for cold joints

Its an Onkyo SR508, sorry I should've mentioned that in my first post. Yes I was also planning on taking the small transistors out of circuit and testing them. I have tested them in circuit and didn't come up with an obvious fault. I also measured them in ohms mode and compared the results with the working channel's transistors they seem to match up   

Two main possible reasons for that kind of failure. Even a brief open circuit in the idle current biasing circuit such as a fractured joint or bad contact of the trimpot's wiper will cause destructively high rail to rail current through the output transistors.
There could be a problem at the owners place. Many people just don't realize the harm that short circuiting stray speaker lead strands can cause.

When an output stage blow up occurs, because the VAS transistors are configured as current sourcing, damage rarely occurs to them, this also applies to earlier stage transistor.
Transistors and other components in the bias circuit do get damaged.

Can I ask please do correct me  - for instance if the VR was open that should 'turn off'  Q6010 right? Could that take out the Output transistors? If so why does it happen? I was under the impression that it'll just make the output sound distorted with cross over distortion.
Sorry for the noob question - could you please elaborate on VAS stage of this amplifier, I know it stands for the voltage amplification stage  .
What is the purpose of Q5030 & Q5040?

Why doesn't biasing issue take out the drive stage transistors like the output transistors? I haven't had much experience working on amps but from most of the stuff that I have read and seen the driver transistor seems to be intact even when the outputs short out, but everyone recommends to change them anyway.

'IPRO' is current protection, shouldn't this protection be engaged when there is a bias issue? There are other two protection signals but I'm not sure what it stands for 'FLVPRO' and 'FLVOLH' the service manual is not really descriptive on this.

Nice pointer about the strands causing shorts, I'll have a proper look at the speakers terminal at the owners place.

Once again thanks so much for the help everyone, really do appreciate it.

Most designs have one VBE multiplier so forget about Q6000 for now. The VBE multiplier's job is to clamp the voltage difference between the Bases of the driver transistors at something that will cause the correct idle current and to provide temperature compensation.
The C-E clamping voltage becomes roughly the B-E turn on voltage, 0.6V , multiplied by the divider ratio. The 2mV/degree c drop in B-E turn on voltage also gets multiplied which conveniently provides temperature compensation for all of the series-ed B-E  junctions in the output stage.

The top VAS transistor is sourcing current. The bottom VAS transistor is sinking current. Without the voltage clamping action of the VBE multiplier, the Base of Q6030 will go to a high positive voltage and the Base of Q6040 will go to a high negative voltage. This is what will happen if the VR's wiper goes open.

Because the VAS transistors are current source/sinking, they are inherently self protecting. You can't overload a current source by short circuiting it.

I'm not to sure about the other stuff for now,

[Yamin]

Thanks so much for the detailed explanation.

[xavier60]

The current at witch a BJT fails is affected by temperature and C-E voltage. Look at the SOA curve, the maximum allow current drops steeply above a certain voltage.
So it's not likely that an over-current protection circuit will always be effective.
Most over-current protection circuits are voltage compensated, so that less current is needed to cause a trip as C-E voltage increases.
This allows discrimination between high currents caused by normal speaker load and that caused by a short circuit on the output.

The over-current protection in this amplifier does not appear to have voltage compensation.

[Yamin]

Two main possible reasons for that kind of failure. Even a brief open circuit in the idle current biasing circuit such as a fractured joint or bad contact of the trimpot's wiper will cause destructively high rail to rail current through the output transistors.

Hi @xavier60, I was hoping to run something by you regarding the trimpot failure - this is quoted from Bob Cordell's book Designing Audio Power Amplifiers. This is under the Vbe circuit explanation -

"When the Vbe multiplier is used as a bias spreader, R2 will be made adjustable with a trim pot. As R2 is made smaller the amount of bias voltage is increased. Notice that if for some reason R2 fails open, the voltage across the Vbe multiplier falls to about one Vbe, failing in the safe direction."

Would that also be applicable in this unit I am working on?

[xavier60]

R6040 is configured as a potentiometer rather than a Variable Resistor. If the wiper loses contact with the resistive track, Q6010 will turn off.
This problem mainly happens when a Trimpot is first moved after sitting in the same spot for all its life.
As a precaution, I oil them and work them with the power off.
The original position can often be felt because the moving contact usually makes a small dimple in track

[Yamin]

R6040 is configured as a potentiometer rather than a Variable Resistor. If the wiper loses contact with the resistive track, Q6010 will turn off.
This problem mainly happens when a Trimpot is first moved after sitting in the same spot for all its life.
As a precaution, I oil them and work them with the power off.
The original position can often be felt because the moving contact usually makes a small dimple in track

That's an interesting note, thanks for the tip. So would a momentary loss of contact cause the output transistor to short? I am sort of scared to move it now

[xavier60]

R6040 is configured as a potentiometer rather than a Variable Resistor. If the wiper loses contact with the resistive track, Q6010 will turn off.
This problem mainly happens when a Trimpot is first moved after sitting in the same spot for all its life.
As a precaution, I oil them and work them with the power off.
The original position can often be felt because the moving contact usually makes a small dimple in track

That's an interesting note, thanks for the tip. So would a momentary loss of contact cause the output transistor to short? I am sort of scared to move it now

It's possible but not likely since it has been worked now. If I suspect  bias problems, I temporarily replace the Emitter resistors with something higher, 10 to 100 ohms.

Edit: If you haven't moved it at all yet, then don't while powered up.

[Yamin]

R6040 is configured as a potentiometer rather than a Variable Resistor. If the wiper loses contact with the resistive track, Q6010 will turn off.
This problem mainly happens when a Trimpot is first moved after sitting in the same spot for all its life.
As a precaution, I oil them and work them with the power off.
The original position can often be felt because the moving contact usually makes a small dimple in track

That's an interesting note, thanks for the tip. So would a momentary loss of contact cause the output transistor to short? I am sort of scared to move it now

It's possible but not likely since it has been worked now. If I suspect  bias problems, I temporarily replace the Emitter resistors with something higher, 10 to 100 ohms.

Edit: If you haven't moved it at all yet, then don't while powered up.

Nice tip about the emitter resistor, when you temporarily use a higher resistor and if the amp functions then you can be convinced that the issue is in the biasing right? Or is there any other thing I am missing? Could I know more about this procedure. Thanks so much

[xavier60]

The temporary Emitter resistors allow amplifier function checks to be made more safely by limiting the fault current if something does goes  wrong. It can be run into speakers only at very low level, it should sound normal if nothing is wrong.

[Yamin]

The temporary Emitter resistors allow amplifier function checks to be made more safely by limiting the fault current if something does goes  wrong. It can be run into speakers only at very low level, it should sound normal if nothing is wrong.

Thanks so much and thanks everyone really do appreciate it.

I'd like to add one more point for completion sake, when I was first testing out the amplifier on my bench after installing the borrowed transistors and resistor from the unused channel I heard a relay click while I was playing music, nothing ill happened the amp was working fine. I turned off the amp and back on again was playing music and after couple of seconds I heard the relay click. I didn't figure out what it was. Its not the speaker relay which clicks when the amp is first turned on. What could this be?

[xavier60]

Some amplifiers have some sort of switching tied in with the speaker A/B switch to change the rail voltage to the power amp stage.

[Yamin]

Some amplifiers have some sort of switching tied in with the speaker A/B switch to change the rail voltage to the power amp stage.

I also have a suspicion that it was the 'voltage shifting' relays. RL6091 and RL6902. Whats the reason for this B+ voltage change? and when is it suppose to happen maybe at certain temp or volume? When I heard the relay click I wasn't changing the volume. The SM shows that the control signal for this is 'SEC1H' I can't figure what this is. Also one more important thing I just realized is that the two fuses which were blown were F6901D and F6902D those are on the low voltage path  .

[xavier60]

Some amplifiers have some sort of switching tied in with the speaker A/B switch to change the rail voltage to the power amp stage.

I also have a suspicion that it was the 'voltage shifting' relays. RL6091 and RL6902. Whats the reason for this B+ voltage change? and when is it suppose to happen maybe at certain temp or volume? When I heard the relay click I wasn't changing the volume. The SM shows that the control signal for this is 'SEC1H' I can't figure what this is. Also one more important thing I just realized is that the two fuses which were blown were F6901D and F6902D those are on the low voltage path  .

When going from 8 to a 4 ohm load, the output power will potentially almost double, putting too much stress on the amplifier as a whole.
The idea is to reduce the rail voltages to what will cause the amplifier to develop approximately the same safe power level.

'SEC1H' could relate to "Secondary", as in switching of the transformer's secondary windings.

[Yamin]

One more question:)I am going through Bob Cordell's book - Designing Audio Power Amplifiers. I am trying to figure out stages of the onkyo amplifier and trying to find similarities from the examples given in the book.(I haven't gotten far into the book yet)
Please see the attach photo, the pink dotted line I have drawn is the negative feedback path right? R5030,R5230 (why have two resistors in parallel would a lot of current be flowing in this path)&R5200 forms the voltage divider for this path right? I read from somewhere that issues on this path could also take out the output transistors. I have tested the components on this path but would an issue on this path cause the output transistors to blow - or an issue this path would only introduce DC offset?

Thanks for the help

[Renate]

No, the gain is set by R5030, R5230 and R5040.
R5030 || R 5230 = 60 k
R5040 = 2.2 k
Gain = 60 k / 2.2 k + 1 = 28.27
20 * log10(28.27) = 29 dB (just like the upper left hand corner of the schematic says)
C5040 makes it that the DC gain is only 1 (we don't need or want DC gain)

Have you tested this without speakers?
If you want to run this without speakers you can replace the final output transistors (if they are pulled already or blown) with just a diode for the base-emitter part of the transistors.
You can't drive a speaker like this, but it will work fine if you want to check voltages.
Without the diodes the circuit won't balance.

Have you checked the bias? "+BL" to "-BL" is critical.
You know that there is a typo in the schematic? The collector of Q5040 ("-BL") should be negative 1.1 V

[Yamin]

No, the gain is set by R5030, R5230 and R5040.
R5030 || R 5230 = 60 k
R5040 = 2.2 k
Gain = 60 k / 2.2 k + 1 = 28.27
20 * log10(28.27) = 29 dB (just like the upper left hand corner of the schematic says)
C5040 makes it that the DC gain is only 1 (we don't need or want DC gain)

Have you tested this without speakers?
If you want to run this without speakers you can replace the final output transistors (if they are pulled already or blown) with just a diode for the base-emitter part of the transistors.
You can't drive a speaker like this, but it will work fine if you want to check voltages.
Without the diodes the circuit won't balance.

Have you checked the bias? "+BL" to "-BL" is critical.
You know that there is a typo in the schematic? The collector of Q5040 ("-BL") should be negative 1.1 V

Thanks so much, I am going to try that and thanks for pointing out the typo. I still haven't got the spares in yet. Last time I checked (the first time I 'repaired' and assembled the unit) I checked the +BL voltage and it measured around 54V. Same for the -BL.

[Renate]

Well, you can't just throw more transistors at it, especially since I believe that you are waiting for them.

As I said, replace the output transistors with two diodes, one in each position.
Just tack solder them from base to emitter of where the output transistors aren't.
Put them so that the diode symbol matches the little arrow on the transistor symbol,
Q6050: base = anode, emitter = cathode
Q6040: base = cathode, emitter = anode
This is like replacing the output transistors with transistors with a Beta of 1.
It makes the drivers do all the work.
Which is why you should not connect a speaker directly.
But it will close the loop and the voltages should be sane.
Because there will be less drop over the output emitter resistors the idle current through the drivers will be a bit higher, but not bad.
You should also be able to see that the bias adjustment works by looking at the voltage across R6080 & R6090 when adjusting the bias.
(Does the manual tell you how to adjust the bias? Probably monitoring P6080? Do not follow their procedure without the real output transistors!)

Note: Many measurements need precision and measuring to ground and subtracting is not accurate. You have to measure across components.

If the output voltage balances to zero now you could hook the output to a 500 ohm resistor and connect that to a speaker.
You should be able to play music softly using just the drivers.
There's no chance that big things will burn because the output transistors are gone.
In worst case you will burn up R6080 & R6090

If everything appears (and sounds) normal, put your voltmeter between +BL and -BL and put on a long playing record.
It could be that your problem is thermal though and it may not show unless the heatsink is warmed up, (maybe gently with a hot air gun?)

[Yamin]

Well, you can't just throw more transistors at it, especially since I believe that you are waiting for them.

As I said, replace the output transistors with two diodes, one in each position.
Just tack solder them from base to emitter of where the output transistors aren't.
Put them so that the diode symbol matches the little arrow on the transistor symbol,
Q6050: base = anode, emitter = cathode
Q6040: base = cathode, emitter = anode
This is like replacing the output transistors with transistors with a Beta of 1.
It makes the drivers do all the work.
Which is why you should not connect a speaker directly.
But it will close the loop and the voltages should be sane.
Because there will be less drop over the output emitter resistors the idle current through the drivers will be a bit higher, but not bad.
You should also be able to see that the bias adjustment works by looking at the voltage across R6080 & R6090 when adjusting the bias.
(Does the manual tell you how to adjust the bias? Probably monitoring P6080? Do not follow their procedure without the real output transistors!)

Note: Many measurements need precision and measuring to ground and subtracting is not accurate. You have to measure across components.

If the output voltage balances to zero now you could hook the output to a 500 ohm resistor and connect that to a speaker.
You should be able to play music softly using just the drivers.
There's no chance that big things will burn because the output transistors are gone.
In worst case you will burn up R6080 & R6090

If everything appears (and sounds) normal, put your voltmeter between +BL and -BL and put on a long playing record.
It could be that your problem is thermal though and it may not show unless the heatsink is warmed up, (maybe gently with a hot air gun?)

Yep I am going to try the procedure while I am waiting for the parts to arrive. Have to figure out a way to connect everything up and orient the main board in a way so that I would be able to take the voltage measurements. Such a nightmare.
Thanks

[Renate]

This is like replacing the output transistors with transistors with a Beta of 1.

Mmm, actually it means a Beta of zero.
Beta = Ic / Ib
In this case Ic = 0, therefore Beta = 0
I was thinking Ie / Ib which is not used for anything.