Check out this PS Schematic of Kenwood KR-9050

Endzone

Endzone

Active Member
Here is the power supply section for the tuner on the KR-9050. The voltage at the emitter of Q11 supplies a large part of the tuner via R123. D17 & D16 have been replaced because they were shorted. This P/S just seems odd to me as it has both rectified AC at pins 13 & 14 and also +30VDC coming in at pin 51 via R120.

There is 23VDC at Q11 collector and 0VDC on the base of Q11 which means Q11 is not turning on. I'm guessing Q12 is a voltage regulator via Q10 for Q11. I'm not sure what the function of Q13 is. Anyway my plan is to shotgun Q11, Q10, Q12, &Q13. It's a little tough to find these old transistor parts.


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How about a bit more of the schematic?

Seems to me that Q10/11 are a Darlington pair driving the pass xister, Q12. Then Q13 appears to be a constant current source with a zener in its emiter to set voltage regulation. In this case, the 2 diodes with the electrolytic caps do something else and the 3 lines labeled 50-52 would appear to carry the mains rails?

There's more here than meets the eye.

Cheers,

David
 
dshoaf said:
How about a bit more of the schematic?

Seems to me that Q10/11 are a Darlington pair driving the pass xister, Q12. Then Q13 appears to be a constant current source with a zener in its emiter to set voltage regulation. In this case, the 2 diodes with the electrolytic caps do something else and the 3 lines labeled 50-52 would appear to carry the mains rails?

There's more here than meets the eye.

Cheers,

David
Click to expand...

Thanks for your circuit description. I guess Q12 is in fact the series pass transistor being controlled by the voltage drop across R123. Q12 does in fact supply everthing on the front end and I.F. section of the FM receiver including the 1st L.O., 1st RF amp, I.F. amp and quadrature detector. Line 51 is B+ to another section of the receiver. The negative side of C101 & C105 are hard to follow because the schematic splits in 2 sections right there. I thought they were filler caps for the collector voltage of Q11, but it is hard to follow. Thanks for your analysis.
 
Q11 is the series pass transistor.

The main supply to Q11, and Q10, comes from D17. D16 is sending a voltage in a part of the circuit that is not on the schematic, and is not important here.

The voltage at pin 51 is the one that turns Q10 ON.

As the output voltage start to raise, capacitor C104 will charge until his voltage will reach the avalanche voltage of the zener D18 + 0.6V (the voltage drop in the base-emitter junction of Q13). At this moment, Q13 will start to conduct and will divert part of the current coming from R120, R121. This will regulate the amount of current going in the base of Q10. Thus, Q13 is the voltage regulator in this circuit.

If, for any reason, the current in the load is large enough to have a voltage drop of ~0.6V in R123 (600mV / 3.3 Ohm = 180mA), Q12 will turn ON and will divert part of the current going in the base of Q10. Thus, Q12 is a current limiter in this circuit.

If you have 0VDC at the base of Q10, the base-emitter junction of Q10 is open. This transistor is bad.

As for the replacement, the exact transistor is not critical.
 
ecluser thanks for taking the time to explain how that circuit works. I read your description a few times until I basically got it. That is a big help.

Regards,
 
Check resistors R120 and R121. If one of them is open you will have 0V at the base of Q10. This means that Q10 is possibly good...
 
This is simple. Q13 is the DC feedback transistor.

Say the voltage on the downside of R123 wants to drop. This turns off sharp cutoff transistor Q13, which shunts the current provided by R121 to ground. When Q13 cuts off, R121 acts as a pull-up resistor for the base of Q10. Since Q10 and Q11 are wired in Darlington configuration, this pulls up on the voltage on R123.

If the voltage on R123 wants to rise, the opposite sequence occurs.

The entire feedback matrix is stabalized by reference diode D18.

There you have it.

Fred
 
ecluser said:
Check resistors R120 and R121. If one of them is open you will have 0V at the base of Q10. This means that Q10 is possibly good...
Click to expand...

There is 30VDC (via pin 51) on the base of Q10, but there is 0VDC at the emitter of Q10 and base of Q11 which could mean that Q10 is open. I seem to remember this transistor getting very hot for some reason during the repair process. Thanks.
 
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Fred Longworth said:
This is simple. Q13 is the DC feedback transistor.

Say the voltage on the downside of R123 wants to drop. This turns off sharp cutoff transistor Q13, which shunts the current provided by R121 to ground. When Q13 cuts off, R121 acts as a pull-up resistor for the base of Q10. Since Q10 and Q11 are wired in Darlington configuration, this pulls up on the voltage on R123.

If the voltage on R123 wants to rise, the opposite sequence occurs.

The entire feedback matrix is stabalized by reference diode D18.

There you have it.

Fred
Click to expand...

Yes that's a good description of how the regulator works to hold 13.5VDC supply. I wonder though if D18 is open if that would cause Q10 base to remain at the voltage of pin 51 (+30VDC) and not drop sufficient voltage across R120 and R121 to turn on Q10.
 
Sometimes its hard to see because of the way they drew the schematic, but if you just rearrange things, you can see that you have a darlington totem-pole:

darlington-totem.png


Edit: Corrected a mistake.
Edit: Fixed again! :drool:
 
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Artie said:
Sometimes its hard to see because of the way they drew the schematic, but if you just rearrange things, you can see that you have a darlington totem-pole:

darlington-totem.png
Click to expand...

I see the Darlington pair on top. I wonder if you might have missed it a little bit on the bottom though. R122 is bussed to the emitter of Q12 and one side of R123 is also bussed to the same place. That drawing is a lot clearer though. This schematic that Kenwood had is tougher to follow.

I did an in-circuit diode check on Q12 and Q10. The base-emitter junction of Q10 is wide open and all junctions on Q12 are shorted. I think replacing the 4 transistors would be a good place to start.
 
Endzone said:
Yes that's a good description of how the regulator works to hold 13.5VDC supply. I wonder though if D18 is open if that would cause Q10 base to remain at the voltage of pin 51 (+30VDC) and not drop sufficient voltage across R120 and R121 to turn on Q10.
Click to expand...

NO, if D18 is open, you would have a high voltage at the base of Q10 (for this you are right) and Q10 would be in conduction.

The voltage drop in R120 and R121 resulting from the base current to Q10 is so small (~180 mA / 100 X 100) that you can neglect this. This is why Q13 is effective to regulate the voltage (he can sink current from R121 if the output voltage is too high, and this has the effect of lowering the voltage at the base of Q10), and Q12 is effective to limit the current (he can sink the current from R121 if the current in R123 becomes too high and this has the effect of lowering the voltage at the base of Q10).

From your last measurement, now I am sure Q10 is bad. The base-emitter junction of this transistor is open.

Artie, your schematic is wrong because R123 is between the base and emitter of Q12 in the original schematic... Don't change the schematic, it will fool Endzone. He doesn't need that at this moment, this is not a graduate exam...
 
Endzone said:
I did an in-circuit diode check on Q12 and Q10. The base-emitter junction of Q10 is wide open and all junctions on Q12 are shorted. I think replacing the 4 transistors would be a good place to start.
Click to expand...

In circuit... don't forgert that you have a 3.3 Ohm resistor between the base and emitter of Q12 in the circuit, it is normal that your meter gives you a dead short for the B-E junction of Q12

But if the B-C junction of Q12 was shorted, you would have a high voltage at the output. The same voltage that you measured at the base of Q10.
 
I would also look for a shorted bypass cap in the tuner itself, that could be shutting down the supply...
 
bharper said:
I would also look for a shorted bypass cap in the tuner itself, that could be shutting down the supply...
Click to expand...

I have about 100K from R123 to ground, and I can tell on the meter that I am charging a cap when I measure. At least without power applied there doesn't seem to be a short. R123 measures 4.4 ohms in-circuit. I've ordered the transistors from MCM Electronics.
 
Of Q10, Q12, & Q13 only one transistor was bad, and the others 2 were just fine. The one transistor that was bad was either Q10 or Q13, but I'm not sure which one it was. The bad transitor was open at both junctions.

If you just look at the traces wrong on this tuner PCB, they will pull off the board. They are weak, and also Kenwood never bothered to clean the flux off the back of the board. For a unit that cost $1150 in 1979, I'm a little surprised at that. But I've had Kenwood amateur radio transceivers that had a lot of mods glued onto the boards and what not. Maybe I shouldn't be so surprised. Yes I did lift one pad on one of the transistors, but I will just have to use the lead of the transistor to solder to the trace. Oh well...


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