Pioneer qx-8000 : capacitor or transistor problem?

[PJL]

This looks like a very odd design as there appears to be no reservoir capacitor for the bridge rectifier but I expect it is wired off board between B1 and E2. What value is it and are the connections good? If you have a scope I would take a look at the ripple on B1 as too much ripple will drastically impact the regulation.

[Chris55000]

Hi!

There is one main reservoir capacitor for lines B1 to B5, this is a large can–type electrolytic, 4700μ 63 V, reference C13 mounted on the main chassis metalwork!

Note the main circuit/interconnection diagram on page 22 (maker's printed page no) has got an extra decimal point on there, it says "6.3 V" – this is obviously erroneous, if you put in a 6.3 V component you'll get a massive explosion of paper, metal and borax everywhere!

This capacitor is 63 V working, ripple current rating 4.5 A minimum !

Looking at circuit Diagram AWR–004 for the power supply, tthe collector of Q1 is connected to the rectified +54 V line via R2 (220 Ω), so at a nominal current draw of 33 mA (from the diagram on p 22) the voltage at Q1 collector is (54 –(0.033 × 220)) which is 46.5 V nominal, if the other boards aren't drawing excessive current !

The voltage at Q1 base, by the potential–divider rule, works out at 34.6 V when the circuit is operating correctly, so you have :

Q1:– C 46.5 V, B 34.6 V, E 34.0 V
Q2 :– C 34.6 V, B 13.6 V, E 13 V

The voltage at the junction R1/R3/C1 is approx. 43.5 V when the circuit is working correctly, if this is too low, check R1/R3 for value change, C1 for leakage and Q2 for a base/emitter leak or s/c and ensure the correct 13 V nominal is present across zener diode D6 – this is the reference for both Q2 and the 12 V series stabiliser Q3 !

It's not that hard to follow,
really !

Q1 and Q3 can be replaced with BC337s, but watch the pinout, Q2 a BC547 & a BZY88C13 will do for D6 !

Chris Williams

[PJL]

We already have measurements for Q2 b=12.7, c=23.7, e=12.5.

Based on B6=23.4 and assuming low base current, Q2b=23.4*R6/(R6+R5) =23.4*33/80 = 9.65V but actual=12.7V. So suspects are R5, R6 or Q2.

[Chris55000]

Hi!

We already have measurements for Q2 b=12.7, c=23.7, e=12.5.

Based on B6=23.4 and assuming low base current, Q2b=23.4*R6/(R6+R5) =23.4*33/80 = 9.65V but actual=12.7V. So suspects are R5, R6 or Q2.

You've fallen into the common trap of trying to analyse a circuit not performing to the maker's original operating specifications!

The collector voltage of Q2, and therefore the base of Q1, is designed to be about 34.6 V as originally designed by Pioneer, so recalculating Q2 open–circuit base voltage :–

Vb(o) = 34 × (33/80) or 34 × 0.41 which is 14 V, therefore allowing for the base current of Q2 and the Vbe drop in Q2, the values of R5/R6 were correctly worked out for the designed open–circuit base voltage!

Therefore, Q2 should not be taking more collector current than is needed to drop approximately 20 V from the + 54 V line.

What the OP can try next, is to lift one end of R5 and R6, replace these with a 100 k preset with the track ends connected between Q1 emitter and chassis, with a 100k resistor in series from the preset wiper to Q2 base, and see if it is still possible to obtain the specified 35 V at terminal B6 by adjustment of this temporary preset – if the specified voltage still cannot be obtained, either there is a fault elsewhere that is drawing excessive current from the B2 or B3 lines or transistors Q1/Q2 are defective!

This type of circuit relies both on the correct hFE of Q1 and Q2 to provide the correct stabilised voltage, along with the specified design currents being drawn from the circuit – even allowing for a ± 10% tolerance in the maker's figures, the voltage at Q1 emitter should be no less than 30.5 V or more than 37.5 V!

Chris Williams

[Chris55000]

PS!

If the OP disconnects the leads from B2 and B3 of the power supply PCB AWR–004, and connects a 1 k 4 W resistor between B2 and earth, and a 15 k resistor between B3 and earth, this will equate to approximately the designed loading the regulator circuit was designed to give !

Chris Williams

[snowman_al]

Sorry everyone, but I'm a little confused still.
From the table below, B2 is 20 or so volts (nominally 25 volts) and B3 is 14 volts (again nominally 25 volts). They both come off the emitter of Q1 (?) so is the problem something on the B3 line pulling the circuit 'down'?

Originally Posted by jmcilkley

OK, more info. I measured voltages with no load and then again with one half of the control amp connected and finally with both halves of the control amp. All the outputs and also b, c & e of Q2
B1 B2 B3 B4 B5 B6 Q2b Q2c Q2e
no load 54.9 23.4 14.4 22.6 11.6 23.4 12.7 23.7 12.5
half c.amp 55.2 21.5 14.4 22.7 11.5 23.2 12.6 23.8 12.6
all c.amp 54.9 20.2 14.3 22.4 11.5 23.3 12.6 23.7 12.4

[PJL]

The manual shows B6=35, B2=25, B3=25 so there is a 10V drop on R7 and R8+R9 by design.

How it works. Assume the steady correct state:
Q1c=45, Q1e=34, Q1b=34.7
The potential divider of R5/R6 on the base of Q2 is the control mechanism and the Zener on the emitter the reference.
Q2e=13 (Zener), Q2c=Q1b=34.7, Q2b=13.7

Assume Q1e voltage drops due to loading
Q2b falls via R5/R6 divider
Q2c current reduces so Q1b rises with current from R1+R3
Q1e rises to stabilise Q1e

But, the actual measurements for Q2 don't add up: Q2b=12.7, Q2c=23.7, Q2e=12.5
If you calculate the expected Q2b based on the divider chain R5/R6 and Q1e (23.4) we get 9.65V.

Components involved include R5, R6, Q2 and C14 but the fact the Q2b is too high when it should be reverse biased and Vbc=0 to 0.2V suggests Q2 is at fault. Q2 (2SC368) is a high gain (hfe=250) 25V NPN transistor but it might be a good idea to replace it with a higher voltage device.

[jmcilkley]

Thanks Chris & Snowman. I did some more checking and the voltages on Q1 with no load are B23.6 C54.3 and E23.0 so a lot low.
As I said before the Q2 voltages are B12.6 C23.7 E12.4

I'll take out R5 and R6 again but I did check them before. I've put back the original Q1 as it tested ok. I still have the 2N2222 in for Q2 which is very similar to the one you suggested but with hFE of 70+
I'm going to remove all the resistors and retest them next - not easy as they have sort leads bent over at right angles and then soldered!
I have already replaced C1.

[PJL]

Just test R5/R6 and if they are good, assume the rest are OK.

Have you already replaced Q2 as it is marked as 2SC368 in the service data? Do you have any other NPN transistors at hand, a BC107 would work?

[jmcilkley]

A few days off with COVID! Getting over it now. I removed R5 & R6 and both measure ok. Interesting that in circuit the 22k resistor measured 13k so there must be some resistance through Q2? Seems odd to me. I looked at another Pioneer power supply from the same period that has some voltages marked and on there the resistance divider seems to do as expected and divide the Q1 emitter in the expected ratio onto the base of Q2 - I'm not getting that. (see attached screenshot).
Something is messing up the divider R%/R6. The only thing I don't have a part to swap is the zener diode - could that be the problem? I can't see any problem on the pcb tracks that could be a problem.

[jmcilkley]

I don't have another suitable transistor for Q2 but I could try putting the original back in - it tests ok.

[PJL]

When you are measuring a resistor in circuit, the rest of the circuit is effectively in parallel and can be difficult to interpret.

The Zener appears to be OK as Q2 emitter is running at 12.4V.

It's worth trying the old Q2. If it does need replacing then I would go for a transistor with at least 40V Vce.

[frsimen]

The voltage readings are very odd around Q2.

This type of feedback regulator can oscillate and if it is doing that it could explain the odd voltage readings. The C7 will keep the voltage across the Zener diode relatively level but the output voltage could be swinging wildly. A multimeter will end up averaging what it sees, resulting in the situation that you have where the voltage on the base of Q2 appears to be lower than the voltage on the emitter, yet that voltage is higher than would be expected from the measured output voltage.

Use your oscilloscope to examine the output at the emitter of Q1. If there is a high frequency oscillation, check C14. You may need to increase its value to restore stability.

If instead you find a 100Hz waveform, check the connections to the reservoir capacitor.

Paula

[PJL]

I agree with Paula. Connection to chassis is also worth checking.

[jmcilkley]

OK, scope out next then. I already changed C14 but for the same value as it was.

[PJL]

Did you try replacing the old Q2?
The suggestion at the moment are one of 3 scenarios:
1. Q2 and Q1 are merrily oscillating so that Q2 is only conducting part of the time hence the lower than expected average voltages.
2. The PSU reservoir capacitor C13 (4700uF 63V) is either faulty or not making good connection to the PSU board B1/E2 so that the input to the regulator has significant ripple again affecting average voltage measurements.
3. The regulator load is heavily fluctuating causing the regulator to misbehave resulting in low average voltages.

[jmcilkley]

OK, I checked the 25v supply with my scope and it seems ok, 20mA non-linearity. I had changed Q2 as well as Q1 but it had no effect so now I have replaced both with the originals as they both tested fine.
The problem still remains. Here are the voltages at Q1 and Q2 :
Q1 B:20.9 C:54 E:19.8 Q2 B:13 C:21.3 E:12.4
On an earlier post I attached the circuit from another Pioneer amp of the same period which had a similar power supply but it showed the voltages at various points. On it the base voltage of Q2 was exactly what would be expected by the voltage divider resistors but on mine it is not. If the emitter of Q1 is 19.8v then R5/R6 should make the base of Q2 2/5 of 19.8, i.e. about 8v but it measures 12.4. Clearly the voltage divider is being affected y something else in the circuit but for the life of me I can't see it! The other one (I'll post it again) seems to follow this perfectly. 30x15/31= 14.5 and the circuit shows 14.6.

[PJL]

Might be worth putting your scope on Q2 base.
The output voltage on Q1e has dropped further from 23.0 to 19.8. Vbe=0.6 which sounds low but possible.

It's going to be something stupid! Check the PCB carefully for dry joints, shorting or cracked tracks.
Where are you referencing your voltage measurements from (-ve probe)?
You could try tacking a resistor in parallel with R5 to see if Q1e rises. The circuit I am looking at has R5=47K, R6=33K. Something like a 150K would be OK.

[frsimen]

The voltages are slightly more sensible with the old transistors back in place. Whatever is wrong now appears to be around the base circuit of Q2.

To add to PJL’s list of possible causes, check that there is no contamination or discolouration on the PCB in the area around Q2/R5/R6, as that could cause leakage leading to the problem you are seeing. If your circuit has the equivalent of C12 from the circuit clip you posted, check that it hasn’t got high leakage.

I would also consider replacing the potential divider resistors. They may measure well out of circuit but one could be failing at the higher voltage present in the power supply.

Paula

[jmcilkley]

Thanks for those suggestions. I had already replaced the divider resistors even though the old ones tested ok. I'm tending to agree that there must be a track problem somewhere around the base of Q2. I'll try changing R5/6 for different values to see if that makes a difference as well as re-checking the pcb tracks.
I've just been using chassis as ground for the multimeter but I did check that it was a direct link to both of the ground pins on the power supply.