Take it R156 and R157 are still present ? Are they the same value as on the schematic ?
I would love to have this under my hands to poke around with a DVM checking for weird DC voltages.
I understand that VR6 is for setting bias, but if the wiper is attached to both of the pot, the whole thing is like a wire. I am now aware that your CD playes does not have one, but take a look at the schematic and imagine taking out VR6, would it make a difference for the function?
You mentioned that on the good channel, there is 12V over the plate resistors. That is 9mA per leg, or 18mA total. The cathodes should be at about 6V and the B- is -15V. If you take a 1k2 resistor, the 21V over it will cause a current of about 18mA. So you could take out the IRF610 and put the 1k2 resistor from the connection of R152 and R153 (that was connected to the drain of IRF610) to -15V to emulate the CCS, and see if the stage behaves better.
Best regards,
Erik
- are all grids at 0V (WRT gnd)?
- are the cathodes positive WRT gnd?
- what is the voltage between the source of Q14 to the -15V?
I understand that VR6 is for setting bias, but if the wiper is attached to both of the pot, the whole thing is like a wire. I am now aware that your CD playes does not have one, but take a look at the schematic and imagine taking out VR6, would it make a difference for the function?
You mentioned that on the good channel, there is 12V over the plate resistors. That is 9mA per leg, or 18mA total. The cathodes should be at about 6V and the B- is -15V. If you take a 1k2 resistor, the 21V over it will cause a current of about 18mA. So you could take out the IRF610 and put the 1k2 resistor from the connection of R152 and R153 (that was connected to the drain of IRF610) to -15V to emulate the CCS, and see if the stage behaves better.
Best regards,
Erik
Only VR5 and VR2 are actually present. VR5 for bad right side, VR2 for left. Still reading the rest of your post, but wanted to throw that out
I just measured the voltage across the plate resistors. So 12v is the good side. 90v is the bad side.
My suggestion is then to take out Q14 and from pin 2 (drain) of it connect a 1k2 resistor to -15V, and see what happens. Q15 and surrounding passive parts will be left floating, but that should not be a problem.
It is a constant current source. Given some limitations, it pulls a predefined current (I calculated 18mA, see post 44), independent of the voltage over it,
I am not claiming to be smarter than the BAT guys, but it is not really necessary here, though it may improve the performance. Using the 1k2 resistor is also not proposed as a definitive solution, but to isolate the problem - a single resistor can cause much less problems than two active devices.
I am not claiming to be smarter than the BAT guys, but it is not really necessary here, though it may improve the performance. Using the 1k2 resistor is also not proposed as a definitive solution, but to isolate the problem - a single resistor can cause much less problems than two active devices.
So to try this I need to lift pin 2 and tie it to the resistor, and where does the other end of the resistor go? Can I remove Q14?
Best is to indeed remove Q14. Connect the 1k2 between the middle hole (Pin 2 indeed) and -15V supply, so the lower part of R157.
Looking at it again, to make it easier: Remove Q14 and put the resistor between pin 2 (drain) and pin 3 (source). You can also use a 1k there, as you have some series resistance (R156 and R157)
Looking at it again, to make it easier: Remove Q14 and put the resistor between pin 2 (drain) and pin 3 (source). You can also use a 1k there, as you have some series resistance (R156 and R157)
Well, I think you're on the right track. First time something has changed. Voltage on the plate resistors changed to 162V on each side with only .5 volts across the resistor. But, the working side changed to 150v on one leg and 90v on the other, or about 55v across the resistor.
So somehow there was an interaction with the working side as there was only a small voltage across it and now there is 55v which among other things is a heat issue. And instead of 150v on the bad side we have 162 on each resistor leg.
So somehow there was an interaction with the working side as there was only a small voltage across it and now there is 55v which among other things is a heat issue. And instead of 150v on the bad side we have 162 on each resistor leg.
OK. Moved the resistor to pins 2&3. Similar to pins 1&2 but now, I get 150v on one leg and 162V on the other leg of the bad side. So 12v across the resistor. But back to the working side is the same as above with 150v on one leg and 90v on the other side. So no change from pins 1&2 to pins 2&3. Still have an interaction that has caused the working side to have 55v across the resistor.
You are welcome. After you did the above, did the CD player work and produce sound from both channels? If yes, how the did the output levels compare?
It sounds as if the bad channel was pulling to much current, interacting with the supply to the good channel. 55V over the resistors on the good channel sounds a bit too high, but more reasonable then 12V. Can you check if on the good channel all plate resistors are still good, and that the effective resistance there is around 1k275?
Today I had a look again at the working of the Q15 and Q14, which I incorrectly described as a constant current source, it is actually a constant current sink. Anyway, the Q15 has a constant voltage between base and emitter of ~0.7V. A fixed voltage over a fixed resistor (the 10R of R157) defines a constant current of circa 70mA. This 70mA is split over the two legs, so 35mA per leg. As each leg has an effective 1k275 resistor in the plate, the voltage drop is indeed about 45V, which is closer to your 55V than the previous 12V. Can you measure the vlaue over R121 (110 ohm resistor in the left channel)? This should give a good indication of the target current BAT had in mind for this circuit.
So you measure 12V over the resistors in the bad channel. This was the expected result with the 1k2 resistor between pin 2&3 of Q14. Lets get the bad channel to pull around 70mA through the tubes now. Doing the same calculation as before: cathodes at circa 3V, negative supply of -15V, give a span of 18V. To get 70mA, we need 18V/0,07A = 257R. You already have 120R, so you need another 137R. So replace the 1k2 resistor between pin 2&3 with something around 140R, power rating shoud be around 2W.
It sounds as if the bad channel was pulling to much current, interacting with the supply to the good channel. 55V over the resistors on the good channel sounds a bit too high, but more reasonable then 12V. Can you check if on the good channel all plate resistors are still good, and that the effective resistance there is around 1k275?
Today I had a look again at the working of the Q15 and Q14, which I incorrectly described as a constant current source, it is actually a constant current sink. Anyway, the Q15 has a constant voltage between base and emitter of ~0.7V. A fixed voltage over a fixed resistor (the 10R of R157) defines a constant current of circa 70mA. This 70mA is split over the two legs, so 35mA per leg. As each leg has an effective 1k275 resistor in the plate, the voltage drop is indeed about 45V, which is closer to your 55V than the previous 12V. Can you measure the vlaue over R121 (110 ohm resistor in the left channel)? This should give a good indication of the target current BAT had in mind for this circuit.
So you measure 12V over the resistors in the bad channel. This was the expected result with the 1k2 resistor between pin 2&3 of Q14. Lets get the bad channel to pull around 70mA through the tubes now. Doing the same calculation as before: cathodes at circa 3V, negative supply of -15V, give a span of 18V. To get 70mA, we need 18V/0,07A = 257R. You already have 120R, so you need another 137R. So replace the 1k2 resistor between pin 2&3 with something around 140R, power rating shoud be around 2W.
"It sounds as if the bad channel was pulling to much current, interacting with the supply to the good channel."
With the above in mind in might a good idea to get a set of measurements off the good channel without the bad channel interfering by removing the valves from the bad channel.
BTW, thought it would interesting to build the circuit and then I saw the price of a 6H30.
With the above in mind in might a good idea to get a set of measurements off the good channel without the bad channel interfering by removing the valves from the bad channel.
BTW, thought it would interesting to build the circuit and then I saw the price of a 6H30.
If there is a correctly working CCS in the cathode circuit of the good channel, the voltage drop over its respective plate resistors should be constant, even if the supply voltage is off due to problems at the other channel. So I do not really get how the voltage over the plate resistors in the good channel could be at 12V?? I assumed this was the designed voltage, even though I found it strange to have so little voltage drop over the plate resistors (less gain, more distortion), therefore my first recommendation was to target 12V over the resistor plates of the bad channel, which apparently worked with the 1k2 resistor instead of the CCS. Now get more current through the tubes, if that works stably, replace (again) all the components in the CCS circuit of the bad channels?!
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