Hi
I have a tube amp on my bench that blew the bias adjustment pot. The manufacturer is on vacation till the end of August, it would be a month or so till we got the correct replacement part.
I dialled the bias pots back all the way and basically have to run the entire adjustment procedure from scratch.
I've never worked with tube amps at this level, so was hoping I'd get a few pointers.
This is a KT88 PP amp with fixed bias, and two 12AU7s per channel. One for preamp duties, the second for phase splitting.
The bias for the output tubes are derived from a small negative supply generated by a series-pass transistor, and there is a pot that adjusts the level of bias. Each tube has a single pot, and one of the pots is blown (measures dead short) and the bias is set to minimum level, refusing to adjust upward.
User has already switched the tubes and the problem stays where it is, so the tube is probably fine. Amp has not been powered up since it came in, still on the workbench. The diagnosis was made by looking at the schematic and testing a few parts in-circuit and on their own. Relevant schematic section is attached:
Part values:
R19, R510: 100K
R20: 1K
R505: 10K
R26: 100E
VR2: 22K (test shows 0ohm across the resistive element, all other pots test fine)
Do let me know if you need any more info/pictures.
The pot is not available locally. It's 22K multi-turn 1 watt wirewound with a 4mm shaft, and the only alternative is a 10K part or the right value with a 6mm shaft (which won't fit the panel hole). I'm guessing a 10K part won't work?
Here was the plan:
1. I was hoping to be able to power up the amp without any tubes, and check which pot position yields most negative grid voltage (to shut the tubes down) to prevent tube damage during the first power up. Is this at all possible? The tubes are Gold Lions, so was keen to prevent any wear and tear if I could help it 🙂
2. Once I get minimum grid voltage from pin 5 of KT88 to ground, I would insert tubes in the working channel and adjust bias to the manufacturer-recommended level of 300mV (test points E/F in the schematic, range recommended is 250-450mV, lower bias for first 100 hours).
3. Once this is done I would measure the resistance of VR3/4 (working channel) and insert a fixed resistor of approximately the same value in place of the fused part VR2. This would enable the owner to use the amp till I got the right part.
4. When the manufacturer returns from vacation in September, I would insert the correct part and remove the fixed resistor, and dial the bias to the long-term level of 450mV.
Awaiting your comments, feedback and suggestions. Help is much appreciated 🙂
Edit: here's the thread with the pictures of the part: http://www.diyaudio.com/forums/parts/171427-help-identify-multiturn-pot.html
I have a tube amp on my bench that blew the bias adjustment pot. The manufacturer is on vacation till the end of August, it would be a month or so till we got the correct replacement part.
I dialled the bias pots back all the way and basically have to run the entire adjustment procedure from scratch.
I've never worked with tube amps at this level, so was hoping I'd get a few pointers.
This is a KT88 PP amp with fixed bias, and two 12AU7s per channel. One for preamp duties, the second for phase splitting.
The bias for the output tubes are derived from a small negative supply generated by a series-pass transistor, and there is a pot that adjusts the level of bias. Each tube has a single pot, and one of the pots is blown (measures dead short) and the bias is set to minimum level, refusing to adjust upward.
User has already switched the tubes and the problem stays where it is, so the tube is probably fine. Amp has not been powered up since it came in, still on the workbench. The diagnosis was made by looking at the schematic and testing a few parts in-circuit and on their own. Relevant schematic section is attached:
An externally hosted image should be here but it was not working when we last tested it.
Part values:
R19, R510: 100K
R20: 1K
R505: 10K
R26: 100E
VR2: 22K (test shows 0ohm across the resistive element, all other pots test fine)
Do let me know if you need any more info/pictures.
The pot is not available locally. It's 22K multi-turn 1 watt wirewound with a 4mm shaft, and the only alternative is a 10K part or the right value with a 6mm shaft (which won't fit the panel hole). I'm guessing a 10K part won't work?
Here was the plan:
1. I was hoping to be able to power up the amp without any tubes, and check which pot position yields most negative grid voltage (to shut the tubes down) to prevent tube damage during the first power up. Is this at all possible? The tubes are Gold Lions, so was keen to prevent any wear and tear if I could help it 🙂
2. Once I get minimum grid voltage from pin 5 of KT88 to ground, I would insert tubes in the working channel and adjust bias to the manufacturer-recommended level of 300mV (test points E/F in the schematic, range recommended is 250-450mV, lower bias for first 100 hours).
3. Once this is done I would measure the resistance of VR3/4 (working channel) and insert a fixed resistor of approximately the same value in place of the fused part VR2. This would enable the owner to use the amp till I got the right part.
4. When the manufacturer returns from vacation in September, I would insert the correct part and remove the fixed resistor, and dial the bias to the long-term level of 450mV.
Awaiting your comments, feedback and suggestions. Help is much appreciated 🙂
Edit: here's the thread with the pictures of the part: http://www.diyaudio.com/forums/parts/171427-help-identify-multiturn-pot.html
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I would use a VOM on the ohms scale on the good pots to determine which pot position will yield the max negative grid voltage. It's the direction that puts the B closest to the -62v buss. No need to power up.
But, may I suggest that you use the 10 K pot that you have on hand in series with a 12K resistor to make up a quick fix. The only question is which end of the pot to install the resistor. I'd place the pot closest to the -62 v bus (higher on the drawing) first as this would be most conservative. If that wouldn't allow for proper adjustment (high enough plate current), I'd swap them.
BTW- I totally approve of multi turn pots in this application. I'm just astounded to see how often manufactures use 5 cent single-turn pots in this application. It makes bias adjustment tedious and is a major pet peeve of mine.
But, may I suggest that you use the 10 K pot that you have on hand in series with a 12K resistor to make up a quick fix. The only question is which end of the pot to install the resistor. I'd place the pot closest to the -62 v bus (higher on the drawing) first as this would be most conservative. If that wouldn't allow for proper adjustment (high enough plate current), I'd swap them.
BTW- I totally approve of multi turn pots in this application. I'm just astounded to see how often manufactures use 5 cent single-turn pots in this application. It makes bias adjustment tedious and is a major pet peeve of mine.
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VR2 is a potentiometer, not simply a resistor, so it can't be replaced by a fixed resistor. It would have to be replaced by two resistors. You could use a 10K pot instead of 22K, if you change R505 from 10K to 3.9K or 4.7K. You can't assume that each valve will have the same setting - this is why there are separate pots.
I am puzzled why the pot has shorted - this is unusual. Is there a short in the wiring? Are you sure the pot has shorted? The symptoms you describe would happen if the pot slider has simply lifted off the track. Then R510 would apply the maximum negative bias - that is why R510 is present as a safety feature.
I am puzzled why the pot has shorted - this is unusual. Is there a short in the wiring? Are you sure the pot has shorted? The symptoms you describe would happen if the pot slider has simply lifted off the track. Then R510 would apply the maximum negative bias - that is why R510 is present as a safety feature.
Yeah, I like the idea of replacing R505 with a higher value. It's much cleaner. But that will only work if you are lucky enough to need the pot at the "top" of the circuit. But there is a way to improve your luck. You can rotate the tubes until you find the "hottest" tube (the one that needs the highest negative bias voltage) and place that tube in the "B" socket.
Yes, you can have the KT88 tubes removed and set the negative bias voltage at the grid.
Once this is done, power off the amp and put the tubes back.
Power it up and check the negative bias voltage is present and further adjust the KT88 idle current as required.
Johnny
Once this is done, power off the amp and put the tubes back.
Power it up and check the negative bias voltage is present and further adjust the KT88 idle current as required.
Johnny
It all depends whether you want to do a proper repair, or a temporary lash-up. To do a proper repair you need either the correct component, or a simple compensation for the 'wrong' component (i.e. 10K pot instead of 22K pot, with corresponding reduction in the series resistor R505). If you are charging a customer then you need to do a proper job.
Wow, thanks for all the replies.
@DF96: The amp is within warranty, so no, we're not charging for the repair but keen to get it back to him so he can enjoy it till the proper part arrives, which is why the lash-up is being planned. We would like to get the amp to factory condition (no authorised tech in my country, I'm hopefully being compensated for the repair but by the distributor. Client is not paying a dime.) To your question, the pot is measuring a short across the track. Even after disassembly, with the wiper out of the circuit. I agree it is most unusual for that to have happened as the current through the pot is miniscule.
@Johnny, thanks so much - that's very helpful and probably answers my questions extremely well. The schematic indicates a desired grid voltage of -55V, which I guess I can measure on the tube sockets. I do understand that each tube will have slightly different cathode voltage (which is really the exposed test point). The manufacturer has numbered the tubes from 1 through 4 already, I assume this is done as part of initial factory setup. This is very helpful, thanks so much. I was worried of blowing something if the tubes were not in the circuit.
@Capt Dave: Thanks for the very helpful reply 🙂 The pot is in parallel with one of the resistors, I assume the top of R510 is where I would get maximum grid voltage, so would a reading close to zero across R510 indicate I have my target? Which would also probably mean that the added 12K would have to be toward the ground (more positive voltage) to keep the bias on the lower side?
Appreciate the great help, all 🙂
I do have another possibility, using a single-turn, small sized (1/8W) volume control pot. If I get a dual, say 47K, and parallel the two gangs I could get about 1/4W of power handling and a value close enough to the original, would this work? The only problem is that these usually have carbon tracks and will fail to open circuit, putting the tube at full conduction or close enough 🙁 The original was rated at 1 watt (I think) but I don't think the grid really draws too much current - would this work? Too risky?
@DF96: The amp is within warranty, so no, we're not charging for the repair but keen to get it back to him so he can enjoy it till the proper part arrives, which is why the lash-up is being planned. We would like to get the amp to factory condition (no authorised tech in my country, I'm hopefully being compensated for the repair but by the distributor. Client is not paying a dime.) To your question, the pot is measuring a short across the track. Even after disassembly, with the wiper out of the circuit. I agree it is most unusual for that to have happened as the current through the pot is miniscule.
@Johnny, thanks so much - that's very helpful and probably answers my questions extremely well. The schematic indicates a desired grid voltage of -55V, which I guess I can measure on the tube sockets. I do understand that each tube will have slightly different cathode voltage (which is really the exposed test point). The manufacturer has numbered the tubes from 1 through 4 already, I assume this is done as part of initial factory setup. This is very helpful, thanks so much. I was worried of blowing something if the tubes were not in the circuit.
@Capt Dave: Thanks for the very helpful reply 🙂 The pot is in parallel with one of the resistors, I assume the top of R510 is where I would get maximum grid voltage, so would a reading close to zero across R510 indicate I have my target? Which would also probably mean that the added 12K would have to be toward the ground (more positive voltage) to keep the bias on the lower side?
Appreciate the great help, all 🙂
I do have another possibility, using a single-turn, small sized (1/8W) volume control pot. If I get a dual, say 47K, and parallel the two gangs I could get about 1/4W of power handling and a value close enough to the original, would this work? The only problem is that these usually have carbon tracks and will fail to open circuit, putting the tube at full conduction or close enough 🙁 The original was rated at 1 watt (I think) but I don't think the grid really draws too much current - would this work? Too risky?
The pot carries about 2mA, and the grid should draw almost no current at all, so power rating should not be an issue. The main risk, wiper losing contact, is already taken care of.
The exposed test point, "B" for example, is the grid voltage. I think you understand that changing this grid voltage changes the idle current which you measure across the cathode resistors as cathode current. I just want to be sure we're clear on that.
V505 and your fouled up pot in series form a voltage divider. Your pot is at the top of the voltage divider so the wiper picks up voltage at the top of the range. R510 has a very important function: it prevents the destruction of the very expensive gold lion KT88 if the wiper in VR2 should lift. If it should, the grid would see -62 volts which would put the tube at cut-off. Without that resistor, the grid would see no bias voltage and the tube would be lost. R510 does not have another function in the circuit per se.
The added 12 K resistor could be in either position. Remember the circuit is a voltage divider. You can only access the voltage through the wiper so the pot needs to be installed in a range of the divider where you expect to find the required voltage. Perhaps a better suggestion would have been to place a 6K at each end of the pot. Do you follow that? That's why I mentioned rolling the tubes to find the "lucky" tube. Each one will need a different grid voltage to produce the spec cathode current - that's just the nature of tubes as imprecise devices.
You could use a single 47K pot and get away with that too. Leave the R505 in the circuit. Be careful, the single turn pots can be touchy.
Having said that, you should be careful not to try anything you do not fully understand because those Gold Lion tubes are very expensive and it doesn't sound like your customer or distributer are fully committing to paying for the repair let alone a replacement tube. You don't want to be caught holding the bag.
V505 and your fouled up pot in series form a voltage divider. Your pot is at the top of the voltage divider so the wiper picks up voltage at the top of the range. R510 has a very important function: it prevents the destruction of the very expensive gold lion KT88 if the wiper in VR2 should lift. If it should, the grid would see -62 volts which would put the tube at cut-off. Without that resistor, the grid would see no bias voltage and the tube would be lost. R510 does not have another function in the circuit per se.
The added 12 K resistor could be in either position. Remember the circuit is a voltage divider. You can only access the voltage through the wiper so the pot needs to be installed in a range of the divider where you expect to find the required voltage. Perhaps a better suggestion would have been to place a 6K at each end of the pot. Do you follow that? That's why I mentioned rolling the tubes to find the "lucky" tube. Each one will need a different grid voltage to produce the spec cathode current - that's just the nature of tubes as imprecise devices.
You could use a single 47K pot and get away with that too. Leave the R505 in the circuit. Be careful, the single turn pots can be touchy.
Having said that, you should be careful not to try anything you do not fully understand because those Gold Lion tubes are very expensive and it doesn't sound like your customer or distributer are fully committing to paying for the repair let alone a replacement tube. You don't want to be caught holding the bag.
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@Captn Dave: Actually, the 4 test points are EFGH, which is the cathode voltage. ABCD are internal connection points, at the grid - they basically are marked on the PCB for connection to the pots (which are panel mounted). Sorry if that wasn't clear in the schematic.
Thanks for the advice on the tube. I've actually warned the distributor I may yet require one Gold Lion at the end. We're friends, and he fully understands the risks of letting me loose on a tube amp with my lack of experience with tube amps.
I understand the bias section fully now after your explanation of R510. Would it also hold true if, for example, the carbon track on a 47K pot (with gangs paralleled for 23.5K) went open-circuit? After listening to your explanation, it would seem that even that condition would put full negative voltage on the grid, placing it in cut-off condition?
I'm not keen to replace or otherwise modify much of the circuit. The pot is wired to the bias section on a leadout, and the wiper of the pot is connected to the grid with another leadout. I just was trying to get a working amp till the replacement part arrived, so a dual-gang pot seemed to be the simplest way to achieve it *if the power rating was sufficient*.
Thanks for the advice on the tube. I've actually warned the distributor I may yet require one Gold Lion at the end. We're friends, and he fully understands the risks of letting me loose on a tube amp with my lack of experience with tube amps.
I understand the bias section fully now after your explanation of R510. Would it also hold true if, for example, the carbon track on a 47K pot (with gangs paralleled for 23.5K) went open-circuit? After listening to your explanation, it would seem that even that condition would put full negative voltage on the grid, placing it in cut-off condition?
I'm not keen to replace or otherwise modify much of the circuit. The pot is wired to the bias section on a leadout, and the wiper of the pot is connected to the grid with another leadout. I just was trying to get a working amp till the replacement part arrived, so a dual-gang pot seemed to be the simplest way to achieve it *if the power rating was sufficient*.
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It took me so long to write that I missed the other posts. Either idea will work just fine: The 47K single turn or the multi turn with series resistors. You don't need the dual gang, one is fine. As F96 points out, the current is very small in a bias circuit because the grid impedance is very very high. The multi turn are a lot less hassle to adjust.
Yes, the same fail safe applies. The "keep it simple" solution is to sub in a single 47K pot. That will widen the "window" in the voltage divider from which you may select your voltage, rather than narrow it so that is the obvious choice. Again, a single 47K pot is fine.
It's funny in a way, I'm working right now on a hybrid (tube/solid state), but while I understand the tube aspect well enough, I struggle with the solid state end where the problem seem to lie. We should trade. LOL
😆 I'm equally terrible at solid state, but don't tell anyone!!!
Thanks, I guess I'll go for it then. Will update the thread tonight with the results, hopefully with the amp playing in the background 🙂
Thanks, I guess I'll go for it then. Will update the thread tonight with the results, hopefully with the amp playing in the background 🙂
Right.
Put in a 47K dual-gang pot with the two halves in parallel. Spent an hour or so checking my connections, also checked the voltages at the sockets after firing up the amp without the tubes. Repeated for all tubes and pots till I got a feel of it, then fitted the tubes (rather nervously at first).
Fired the amp up, set cathode voltage to 250mV and the amp came to life. Sounds fantastic, should be with me for the next few days as the tubes settle down. Will be temporarily gone and then will be back in September for the final replacement to be fitted. The adjustment is very jumpy with the single-turn pot, as expected, but I guess I can just about handle it...
Thanks for the help, wouldn't have been able to do it without you 🙂
Put in a 47K dual-gang pot with the two halves in parallel. Spent an hour or so checking my connections, also checked the voltages at the sockets after firing up the amp without the tubes. Repeated for all tubes and pots till I got a feel of it, then fitted the tubes (rather nervously at first).
Fired the amp up, set cathode voltage to 250mV and the amp came to life. Sounds fantastic, should be with me for the next few days as the tubes settle down. Will be temporarily gone and then will be back in September for the final replacement to be fitted. The adjustment is very jumpy with the single-turn pot, as expected, but I guess I can just about handle it...
Thanks for the help, wouldn't have been able to do it without you 🙂
Good Work!
The bias supply circuit looks interesting. Could you tell us the part number for that series pass transistor (2n3904??) and the resistor and capacitor values. I'd like to breadboard that circuit. I'm also curious about IC1.
The bias supply circuit looks interesting. Could you tell us the part number for that series pass transistor (2n3904??) and the resistor and capacitor values. I'd like to breadboard that circuit. I'm also curious about IC1.
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3904 won't work at 92V, for long any way.
Try a MPSA92 (PNP).
Good resistor values would be 10K (R502) and 20K (R503) with the filter cap at the base connection around 330uF C512 (tau = 0.22seconds).
The input filter cap (C503) should be fair sized for minimum ripple, say 100uF at 150V since it is a low current supply.
The output filter cap (C504) is probably ok with a small value for transient suppression, so maybe 1uF metal film 100V or greater.
Power dissipation of Q101 is about 0.7mA*4*30V = 84mW so dissipation is not an issue. Make R501 3K3Ohms as this will drop less than 10V. It is probably an inrush limiting resistor.
Since we know VRn is 20K and ideally it should be in the center of adjustment at -55V, we can infer it will drop 62-55= 7V per half or 14V for the entire pot. so current is about 14V/20K=0.7mA. The remaining supply (62-14=48V) is dropped by R504,5,6,7 at 0.7mA for an approximate value of 48/0.0007=68571 or 68K.
I would make R508,9,10,11 270K each for starters but should consult the KT88 data sheet for maximum value.
Try a MPSA92 (PNP).
Good resistor values would be 10K (R502) and 20K (R503) with the filter cap at the base connection around 330uF C512 (tau = 0.22seconds).
The input filter cap (C503) should be fair sized for minimum ripple, say 100uF at 150V since it is a low current supply.
The output filter cap (C504) is probably ok with a small value for transient suppression, so maybe 1uF metal film 100V or greater.
Power dissipation of Q101 is about 0.7mA*4*30V = 84mW so dissipation is not an issue. Make R501 3K3Ohms as this will drop less than 10V. It is probably an inrush limiting resistor.
Since we know VRn is 20K and ideally it should be in the center of adjustment at -55V, we can infer it will drop 62-55= 7V per half or 14V for the entire pot. so current is about 14V/20K=0.7mA. The remaining supply (62-14=48V) is dropped by R504,5,6,7 at 0.7mA for an approximate value of 48/0.0007=68571 or 68K.
I would make R508,9,10,11 270K each for starters but should consult the KT88 data sheet for maximum value.
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Yes, it is a simple pass regulator with the output set as a fixed portion of the input. No fixed value regulation per-se.
R510 is up in the air, as we don't know R18s value either.
I've not done fixed bias so my question is can the grid resistor in fixed bias be much greater than the max for cathode bias?
Penta Labs shows Rg1 220K for <40W and 100K for Po>40W this gives us the R18,9.. value. With less than 400V plate supply I would infer less than 40W. So I expect a low value for R510 to insure cutoff in the event the wiper went open. My original guestimate of 270K is probably safe as it is large enough not to upset the potentiometer setting, but small enough that it should shut off the tube if the wiper goes open.
R510 is up in the air, as we don't know R18s value either.
I've not done fixed bias so my question is can the grid resistor in fixed bias be much greater than the max for cathode bias?
Penta Labs shows Rg1 220K for <40W and 100K for Po>40W this gives us the R18,9.. value. With less than 400V plate supply I would infer less than 40W. So I expect a low value for R510 to insure cutoff in the event the wiper went open. My original guestimate of 270K is probably safe as it is large enough not to upset the potentiometer setting, but small enough that it should shut off the tube if the wiper goes open.
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