The bulb should be dim because there is no difference in voltage between the driver transistor base with the link fitted.
You need to determine where the current flow is occurring in this state by voltage measurement. Measure across each 0.22 ohm while the bulb is lit and see if any voltage is present across any of them.
You need to determine where the current flow is occurring in this state by voltage measurement. Measure across each 0.22 ohm while the bulb is lit and see if any voltage is present across any of them.
That seems to confirm an output stage fault. 0.13 volts over 0.22 ohm is 0.6 amp. It should be zero in this condition.
Another oddity is R375.
If R365 and R373 which are the upper 0.223 ohm have 0.13 volts across them then the total current is double the 0.6 amp in each. Where is that current going? R367 should see twice the volt drop if there is nothing over R375. That might be worth a recheck.
You are going to have to check very carefully all the output and driver circuitry. Voltage checks under the fault condition should help locate this.
This is what you have. There should be no significant voltage across any of the base/emitter junctions no matter what voltage you apply to the link at the left... such as the 10k we fitted. For current to flow in the 0.22 ohms means both an upper NPN and a lower PNP must be on together and that should not be possible with the driver base's linked.
It is worth checking for completeness that you have no 'short' from the main output line to ground or a supply.
In this diagram it is not possible to turn on both upper and lower halves at the same time.
Another oddity is R375.
If R365 and R373 which are the upper 0.223 ohm have 0.13 volts across them then the total current is double the 0.6 amp in each. Where is that current going? R367 should see twice the volt drop if there is nothing over R375. That might be worth a recheck.
You are going to have to check very carefully all the output and driver circuitry. Voltage checks under the fault condition should help locate this.
This is what you have. There should be no significant voltage across any of the base/emitter junctions no matter what voltage you apply to the link at the left... such as the 10k we fitted. For current to flow in the 0.22 ohms means both an upper NPN and a lower PNP must be on together and that should not be possible with the driver base's linked.
It is worth checking for completeness that you have no 'short' from the main output line to ground or a supply.
In this diagram it is not possible to turn on both upper and lower halves at the same time.
Base Emitter junctions of the NPN transistors (2sc3280) have 420mv across them. Rechecked R375 and again nothing significant (.2mv) across it.
Base Enitter junctions of the PNP transistors (2sa1301) have nothing significant across them (1.8mv).
Base Emitter junction of driver Q321 has .9vdc across it
Base Enitter junction of driver Q319 has 571mv across it
Base Enitter junctions of the PNP transistors (2sa1301) have nothing significant across them (1.8mv).
Base Emitter junction of driver Q321 has .9vdc across it
Base Enitter junction of driver Q319 has 571mv across it
420mv is just getting to the point where they come into conduction. That might explain why R373 has no voltage (current flow) and yet R365 does. Transistors replaced and from different batches... it not a problem just a possible reason.
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This is not expected simply because you measured 130 to 140mv across their emitter resistors. No voltage across the B/E junction means they should be 100% off and non conducting. So why are the resistors developing that 130 to 140mv drop.
So this is a very definite problem that needs resolving. Either the PNP output transistors are 'shorted' or perhaps shorted to the heatsink in some way. Or the emitter lead is shorted to something. It needs checking closely. No B/E voltage means no conduction and the only route for current in those resistors is via the emitter and either a a short or a shorted transistor.
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Suspiciously high. It is achievable in extreme case but there should essentially be very little current flowing in the B/E junction. It needs a lot of current to generate that sort of voltage across the junction... maybe destructively high. The transistor could also be defective.
Is this transistor getting hot when the bulb is lit? Check that the emitter isn't shorted to ground or the output line.
That sounds reasonable and would be expected with the 10k test resistor we have biasing the stage.
So a few things to resolve there one way or another.
Yes I will investigate. Robbing the old amp probably wasnt such a good idea since it was crapped out, that's why I obtained it. I did do a quick check of the transistors before installing them in the good amp but that doesn't mean they are good replacements.
I will check for shorts to ground, solder blobs, bad solder joints etc this evening after work. I think I will go ahead and obtain new outputs and drivers.
2SA1301----->MJL1302AG
2sc3280------>MJL3281AG
2SA1306------> 2SA1943
2SC3298------>2SC5200N
These are what I have found as proper replacements
I will check for shorts to ground, solder blobs, bad solder joints etc this evening after work. I think I will go ahead and obtain new outputs and drivers.
2SA1301----->MJL1302AG
2sc3280------>MJL3281AG
2SA1306------> 2SA1943
2SC3298------>2SC5200N
These are what I have found as proper replacements
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I wouldn't worry at this point over unmatched transistors and so on.
You have to understand how that stage works and find why it draws current. This is what you have. It is a text book emitter follower output stage. If the base of the drivers are linked then the output follows the input of that stage and no current flows in these transistors (unless there is a load on the output).
This is what you have.
Look at the input voltage rise and fall from -55 to +55 volts. Now look at the output voltage. It is superimposed on the input because the two follow each other.
Now look at the current in any 0.22 ohm. Each one would look like this. Just a couple of milliamps here because of my test load resistor. Without that load there would be nothing to see.
The 10k resistor at the base of the drivers in your case simply applies rail voltage to the stage and the output should swing to that voltage. I added the voltage generator as a source to show how it behaves under any possible input voltage. There is no current flow anywhere... unless you have a fault or a short somewhere.
You have to understand how that stage works and find why it draws current. This is what you have. It is a text book emitter follower output stage. If the base of the drivers are linked then the output follows the input of that stage and no current flows in these transistors (unless there is a load on the output).
This is what you have.
Look at the input voltage rise and fall from -55 to +55 volts. Now look at the output voltage. It is superimposed on the input because the two follow each other.
Now look at the current in any 0.22 ohm. Each one would look like this. Just a couple of milliamps here because of my test load resistor. Without that load there would be nothing to see.
The 10k resistor at the base of the drivers in your case simply applies rail voltage to the stage and the output should swing to that voltage. I added the voltage generator as a source to show how it behaves under any possible input voltage. There is no current flow anywhere... unless you have a fault or a short somewhere.
So in my case we have created a zero bias. The 10k resistor you had me install ensures that we have a positive voltage at the base which should result in positiver voltage on the emitters and the output.Since we do not have positive voltage on the emitters or the output and with current flowing thru the .22ohm resistors there has to be a short or a faulty device in the emitters follower output stage. I need to find where that fault is.
Cleaning up all the solder joints around the output transistors as well as the .22 ohm resistors. Using a magnifying glass looking for any little issue
One thing you can do is isolate each 4.7 ohm base resistor. For each resistor you isolate the current in that corresponding transistor and its 0.22 ohm should fall to zero. If it doesn't the transistor is faulty or there is a short in its mounting.
If you do that and then find that R355 or R347 or R349 then overheat (and they should not) then its still a valid test because because it tells us there is a problem around the driver and pre driver transistors,
If you do that and then find that R355 or R347 or R349 then overheat (and they should not) then its still a valid test because because it tells us there is a problem around the driver and pre driver transistors,
When you say isolate that means pull the other 4.7 ohm resistors except for the one I am working with?
It just means to remove one from the circuit such as pulling one of its lead out of the board. The transistor it feeds and the relevant 0.22 ohm should then pass zero current.
Isolated R369. R373 has .9mv and Q327 has 17.2vdc across collector to emitter. -9vdc from chassis ground to base
Those results are good and show the transistor is switching off and current flow in the 0.22 ohms stops.
Leave it in the that state and carry on looking for where the current is going and why the bulb is lit.
Is there still voltage across either R367 and R375 which are the lower two 0.22 ohm's ?
If there is then isolate R363 and R371 in the same way.
Leave it in the that state and carry on looking for where the current is going and why the bulb is lit.
Is there still voltage across either R367 and R375 which are the lower two 0.22 ohm's ?
If there is then isolate R363 and R371 in the same way.
Isolated R363, bulb goes out. 3.0mv across R367 . Q325 1.2mv collector to emitter voltage. -51.6vdc to base
Lets just be clear on this. Is R371 and its transistor and 0.22 ohm in place at this time? If they are fitted then there seems to be a problem with Q325 and surrounding parts.
If isolating R363 dims the bulb and refitting it brightens the bulb then you should see the same happen with R371.
The two transistors are effectively in parallel and so if both are in place then removing or isolating one of the two should not change the fault. If the circuit behaves differently with one vs the other in place then there is an issue.
There is also another problem showing.
With it left in that state with the upper NPN's still isolated (by removing the base drive to both) we have to ask where the high current needed for the bulb to light is coming from. The lower PNP can only draw that current from the output line that the 0.22 ohm's connect to and there is no obvious high current path available from the circuitry alone.
Been fighting the flu so havent spent as much time as needed on this.
R371 is still in place and the only base resistor left in.
R371 is still in place and the only base resistor left in.
I could reinstall one of the upper NPN's base resistor and see if the bulb lights.