Q331 and the resistor are part of the overcurrent protection and trip the relay under high current conditions such as shorted speaker leads etc. Nothing there will impact the the main amp as such although a fault in the overcurrent part may stop the relay pulling in.
I meant to send that last night, something I stumbled across while inspecting the amp. Funny thing is with the resistor removed the relay engages. Im sure the original resistor i pulled had been there since the original accident occured
I would worry about the protection circuit once the amp is up and running. You can leave Q331 out for now and provided the DC offset is low on both channels then the relay should engage.
OK, so attach the black lead at the junction of the .22 ohm resistors and then check each base/emitter voltage of all the transistors in that circuit?
You are just working back with the measurements. Just reconfirm you still have the 6.5 volts or so across Q311 and then do the other measurements as outlined. Each circled measurement should only increase by around 0.6v. Any issue should stand out a mile.
This is a quick simulation of the amp.
The voltage across Q311 is 3.7 volts. Its important you measure across the transistor and you recorded 6.5 volts. The 1.8 and -1.9 are what is seen measuring from ground but I want you to measure across the transistor. In this example you would measure 1.8v+1.9v giving a total of 3.7 volts across the transistor.
More than that, each base/emitter junction drops approx 0.6 volts. Measuring from the junction of the resistors takes any DC offset out of the equation.
This is a quick simulation of the amp.
The voltage across Q311 is 3.7 volts. Its important you measure across the transistor and you recorded 6.5 volts. The 1.8 and -1.9 are what is seen measuring from ground but I want you to measure across the transistor. In this example you would measure 1.8v+1.9v giving a total of 3.7 volts across the transistor.
More than that, each base/emitter junction drops approx 0.6 volts. Measuring from the junction of the resistors takes any DC offset out of the equation.
I now have 2.32 vdc across Q311.
q315
E= .652 vdc
C= 44.3vdc
B= 1.2vdc
q317
E= -6.82vdc
C= -41.2vdc
B= -1.2vdc
Q319
E=6mv-2mv constantly moving
C=44.5vdc
B=490mv moving downward
Q321
E= -28.2vdc
C= -42vdc
B= -16.2vdc
It was getting very hot so I shut it down to cool off. The DBT light was slowly getting brighter
q315
E= .652 vdc
C= 44.3vdc
B= 1.2vdc
q317
E= -6.82vdc
C= -41.2vdc
B= -1.2vdc
Q319
E=6mv-2mv constantly moving
C=44.5vdc
B=490mv moving downward
Q321
E= -28.2vdc
C= -42vdc
B= -16.2vdc
It was getting very hot so I shut it down to cool off. The DBT light was slowly getting brighter
You have big problems with the DC conditions. These were all measured using the 0.22 ohm junctions as a reference point?
Q315 and Q319 readings are possible and so valid at this point.
Q317 and Q321 show problems and their emitter voltages put nearly 30 volts across R355 which will either be very hot or burning. If its not then its already open circuit.
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Lets concentrate on Q321 and you really need to understand the basic semiconductor theory here 🙂 Don't just take what I say as always correct all the time.
So measuring from that 0.22 ohm junction you see -28 volts on Q321 emitter. That puts the -28 volts across the B/E junctions of the output transistors and the 4.7 ohm base resistors which isn't possible unless there is an open circuit in that path.
You have this which if we look at just one current path tells us either the 4.7 ohm's are open circuit or the 0.22 ohms are open or the B/E junctions of the outputs are open (most unlikely). The base of the output transistors can not ever be more than around 0.6 volts more negative than the emitter for a PNP type (it behaves like a diode). So where is the -28 volts being dropped? It has to be either open print or some or all of the 4.7 ohm and 0.22 ohm are open.
Q315 and Q319 readings are possible and so valid at this point.
Q317 and Q321 show problems and their emitter voltages put nearly 30 volts across R355 which will either be very hot or burning. If its not then its already open circuit.
--------------------------------------
Lets concentrate on Q321 and you really need to understand the basic semiconductor theory here 🙂 Don't just take what I say as always correct all the time.
So measuring from that 0.22 ohm junction you see -28 volts on Q321 emitter. That puts the -28 volts across the B/E junctions of the output transistors and the 4.7 ohm base resistors which isn't possible unless there is an open circuit in that path.
You have this which if we look at just one current path tells us either the 4.7 ohm's are open circuit or the 0.22 ohms are open or the B/E junctions of the outputs are open (most unlikely). The base of the output transistors can not ever be more than around 0.6 volts more negative than the emitter for a PNP type (it behaves like a diode). So where is the -28 volts being dropped? It has to be either open print or some or all of the 4.7 ohm and 0.22 ohm are open.
I will check that @wg_ski . @Mooly I will pull and replace the .22 ohm resistors as I have spares. R355 is hot as a pepper. I checked with the amp off but in circuit and it ohmed out but to be sure I will replace it as well. Once I have all these done I will recheck my readings at Q321. I have to work tonight so it might be way later pl;us we have a winter storm rolling thru 😵
The 0.22 ohm resistors being so low in value can be checked in circuit. Once values are in the 100's of ohm and above is when you start to have to check them out of circuit. Same for those 4.7 ohms. In circuit check is fine. Remember to check not just across the leads themselves but also from the points they connect to as checking across the leads alone will not reveal any break in the print.
Based on your measurements the 4.7 ohm's are the most likely suspects.
Based on your measurements the 4.7 ohm's are the most likely suspects.
I'm curious as to what in the measurements points you to the 4.7 ohm resistors so I can enlighten myself for future troubleshooting
OK 🙂
So we start with the -28 volts that you measure between those two points. That area of the circuit reduces to this (left hand diagram) and the PNP transistor can be drawn as its diode equivalent.
So as shown you have the voltages you should be seeing if you have -28 volts as shown. The base/emitter junction behaves as a diode.
Work out the current flow in the 4.7 ohm and 0.22 ohm using ohms law and you will see why we have a problem. The circuit can not generate that level of current, and if it could there would be smoke and flames... so your 28 volts while real, is not appearing across those components values. If there is a break in the circuit or any of the three parts is open circuit then the -28 volts is 'lost' across that failure.
So work out the power dissipated in the 4.7 ohm under the conditions in this diagram. W=I*V Its all just ohms law.
So we start with the -28 volts that you measure between those two points. That area of the circuit reduces to this (left hand diagram) and the PNP transistor can be drawn as its diode equivalent.
So as shown you have the voltages you should be seeing if you have -28 volts as shown. The base/emitter junction behaves as a diode.
Work out the current flow in the 4.7 ohm and 0.22 ohm using ohms law and you will see why we have a problem. The circuit can not generate that level of current, and if it could there would be smoke and flames... so your 28 volts while real, is not appearing across those components values. If there is a break in the circuit or any of the three parts is open circuit then the -28 volts is 'lost' across that failure.
So work out the power dissipated in the 4.7 ohm under the conditions in this diagram. W=I*V Its all just ohms law.
Okay, I replaced R363 and R371 as well as Q321`.
Q321
E=13.6vdc
C=6.3mv
B=28.6vdc
I also get 1.9mv on the bias voltage but it wont move from that
Q321
E=13.6vdc
C=6.3mv
B=28.6vdc
I also get 1.9mv on the bias voltage but it wont move from that
I made a misrake. The first measurements of Q321 I got the emtter and baseflipped as I was looking at the wrong datasheet. I feel so dumb and very sorry for making that mistake.
No problem 🙂
For any normal silicon transistor (like the ones in this amp) you should never see more than around 0.6 to 0.8 volts across the base/emitter junction when forward biased (and they should always be in that state, forward biased). If a fault reverse the voltages across a base/emitter junction then they often tend to limit out at around 7 volts or so (behaving a bit like a Zener diode).
For an NPN the base is always the most positive and vice versa for PNP's like Q321
For any normal silicon transistor (like the ones in this amp) you should never see more than around 0.6 to 0.8 volts across the base/emitter junction when forward biased (and they should always be in that state, forward biased). If a fault reverse the voltages across a base/emitter junction then they often tend to limit out at around 7 volts or so (behaving a bit like a Zener diode).
For an NPN the base is always the most positive and vice versa for PNP's like Q321
They are low enough in value to measure in circuit at this point. Measure from the emitter of Q321 (on its actual lead) to the base of both PNP output transistors (again on the leads, not the board) and you should read 4.7 ohms for both.
Measuring TO the actual device leads also includes any possible breaks in the print, measuring directly across the 4.7 ohm doesn't take that possibility into account.
Measuring TO the actual device leads also includes any possible breaks in the print, measuring directly across the 4.7 ohm doesn't take that possibility into account.
So think it through... look at the circuit.
You have 28 volts across the points arrowed (that is your measurement)... so now measure from the emitter of Q321 to both legs on both of the 4.7 ohm resistors. You should read 0.00 volts. Now still measuring from the emitter of Q321 measure to the base of the output transistors. Again you should see 0.00 volts. Now move from the base to emitters of the outputs. Again you should see 0.00.
You are now up to the 0.22 ohm resistors and you must (by your own previous measurements) now see 28 volts on at least one leg of each 0.22 ohm. If you don't then the original 28 volt measurement is in error or there is a break in the print from the 0.22 ohm's to the point where you measured the 28 volts.
Do you you see 🙂
Either the measurements are in error or the 28v is not present... or there is a break somewhere.
You have 28 volts across the points arrowed (that is your measurement)... so now measure from the emitter of Q321 to both legs on both of the 4.7 ohm resistors. You should read 0.00 volts. Now still measuring from the emitter of Q321 measure to the base of the output transistors. Again you should see 0.00 volts. Now move from the base to emitters of the outputs. Again you should see 0.00.
You are now up to the 0.22 ohm resistors and you must (by your own previous measurements) now see 28 volts on at least one leg of each 0.22 ohm. If you don't then the original 28 volt measurement is in error or there is a break in the print from the 0.22 ohm's to the point where you measured the 28 volts.
Do you you see 🙂
Either the measurements are in error or the 28v is not present... or there is a break somewhere.