Hi all. I have a question and have not been able to find an answer searching. A little help would be greatly appreciated.
I have an SE amp running 6sn7, to 6sn7, to 3 parallel 6550eh. The 6550's are biased at -47 (fixed bias), running 71ma each. Looking at the chart for 6550 it shows possible grid voltages from 0 to -70 volts before you reach cutoff. So if my quiescent point is already at -48, I wouldn't think I have much wiggle room as far as voltage. But looking at an oscilloscope probe on the grid I see an input of well over 200v peak to peak before I see the output waveform of the 6550 start to distort (yes it distorts on top first, it cuts off first)
So my question is this, how do you figure how much of a peak to peak signal you need to apply to the grid of an output tube in a SE amp for maximum output before cutoff. It can't be as simple as look at the chart because the chart shows that from a starting point of -48 I would only have to go 22v more negative to reach cutoff. Again, I'm looking at the sine wave (1khz) being over 100v positive peaks and 100v negative peaks (200v p-p) going onto the grid of the 6550 before the output taken at the speaker tap starts to distort?
Thank you all so much, and I'm open to any reading. It's just that so far everything I've read doesn't go into detail about the voltage of the signal. I need to figure out the formula for grid voltage change on a plate chart, vs grid voltage change on an oscilloscope measuring peak to peak.
Thanks again all
I have an SE amp running 6sn7, to 6sn7, to 3 parallel 6550eh. The 6550's are biased at -47 (fixed bias), running 71ma each. Looking at the chart for 6550 it shows possible grid voltages from 0 to -70 volts before you reach cutoff. So if my quiescent point is already at -48, I wouldn't think I have much wiggle room as far as voltage. But looking at an oscilloscope probe on the grid I see an input of well over 200v peak to peak before I see the output waveform of the 6550 start to distort (yes it distorts on top first, it cuts off first)
So my question is this, how do you figure how much of a peak to peak signal you need to apply to the grid of an output tube in a SE amp for maximum output before cutoff. It can't be as simple as look at the chart because the chart shows that from a starting point of -48 I would only have to go 22v more negative to reach cutoff. Again, I'm looking at the sine wave (1khz) being over 100v positive peaks and 100v negative peaks (200v p-p) going onto the grid of the 6550 before the output taken at the speaker tap starts to distort?
Thank you all so much, and I'm open to any reading. It's just that so far everything I've read doesn't go into detail about the voltage of the signal. I need to figure out the formula for grid voltage change on a plate chart, vs grid voltage change on an oscilloscope measuring peak to peak.
Thanks again all
Volts is volts, so there's no translation needed between a 'scope reading and a plate chart prediction. The 100 peak volt number seems like an error because it would mean very considerable positive grid voltages. If the grid gets positive WRT cathode it conducts current, loading the driver stage notably. If it were me making the measurement, I'd blame a scaling error with the 'scope.
All good fortune,
Chris
All good fortune,
Chris
For any amp, we try to avoid getting grids positive, so your -48V fixed bias means positive peak can't go beyond 48V , period.
If you "see" 100V peaks can only mean that your scope control, attenuation or probe settings are not correct, chbeck that.
Besides that, in SE amps which by definition must be run Class A, rather than thinking "voltage" first, think needed *current* and only after that set bias to the needed value to get such current.
If you "see" 100V peaks can only mean that your scope control, attenuation or probe settings are not correct, chbeck that.
Besides that, in SE amps which by definition must be run Class A, rather than thinking "voltage" first, think needed *current* and only after that set bias to the needed value to get such current.
The scope is set correctly
The scope is set correctly. I have verified this with multiple known sources. I agree, with the 6550 set at -48, it can't go above 0, so can't go more than 48 volts positive. Heck, with an even wave (same positive and negative) it can't even go that far, because cutoff occurs around -70 with a 6550. Yet, I am seeing a sine wave of 200V peak to peak on the grid of the 6550. It's not a mistake. The question is, should I be doing an RMS equation to that 200v p-p? Or is it that some of the voltage is wasted on feedback from the tertiary winding in the OPT that the cathode hooks through to ground? Or? There is, definitely, 200V peak to peak coming out of the coupling cap from the preceding stage. Maybe it's not all going on to the grid, but then where?
As to the current post. Could you please explain a little, I'd really appreciate it. I've always figured gain by voltage requirements.
Thank you all so much,
Loren
The scope is set correctly. I have verified this with multiple known sources. I agree, with the 6550 set at -48, it can't go above 0, so can't go more than 48 volts positive. Heck, with an even wave (same positive and negative) it can't even go that far, because cutoff occurs around -70 with a 6550. Yet, I am seeing a sine wave of 200V peak to peak on the grid of the 6550. It's not a mistake. The question is, should I be doing an RMS equation to that 200v p-p? Or is it that some of the voltage is wasted on feedback from the tertiary winding in the OPT that the cathode hooks through to ground? Or? There is, definitely, 200V peak to peak coming out of the coupling cap from the preceding stage. Maybe it's not all going on to the grid, but then where?
As to the current post. Could you please explain a little, I'd really appreciate it. I've always figured gain by voltage requirements.
Thank you all so much,
Loren
You hadn't measured cathode feedback before - yes, you will need more drive voltage - it's increased by the amount of cathode voltage. The cathode to grid voltage is considerably less.
Thank you Tom
Thank you Tom. So is it a direct increase, the cathode voltage increases the drive requirement by the amount of cathode voltage exactly, or close to exactly that much? And, maybe more importantly to me trying to understand what I'm seeing on the scope here... Is the drive requirement as a whole elevated by the cathode voltage, or is each peak of a whole wave elevated....What I mean is, lets say there's 20V on the cathode. Does that mean I would have a requirement raised by 20v on the peaks, and 20v on the troughs? Or by 10 and 10?
Thanks so much, I really appreciate it.
Loren C.
Thank you Tom. So is it a direct increase, the cathode voltage increases the drive requirement by the amount of cathode voltage exactly, or close to exactly that much? And, maybe more importantly to me trying to understand what I'm seeing on the scope here... Is the drive requirement as a whole elevated by the cathode voltage, or is each peak of a whole wave elevated....What I mean is, lets say there's 20V on the cathode. Does that mean I would have a requirement raised by 20v on the peaks, and 20v on the troughs? Or by 10 and 10?
Thanks so much, I really appreciate it.
Loren C.
So my question is this, how do you figure how much of a peak to peak signal you need to apply to the grid of an output tube in a SE amp for maximum output before cutoff. It can't be as simple as look at the chart because the chart shows that from a starting point of -48 I would only have to go 22v more negative to reach cutoff. Again, I'm looking at the sine wave (1khz) being over 100v positive peaks and 100v negative peaks (200v p-p) going onto the grid of the 6550 before the output taken at the speaker tap starts to distort?
Thanks again all
It's really quite simple. If you already know that cutoff is -70v then the peak (+) signal on the grid can only be 35v. And so you would set your biasing to -35v. And you would swing the grid with a signal from +35v to -35v for the max (-) input of -70v on the grid.
?
You can't just set your bias to whatever you want, the tubes operating points are fixed based on your load lne. You can pick one, but the 6550 pulls a LOT of current in a quiescent state at a -35 Bias, with a B+ of 523v and a 5K primary.
I've asked this question all over, and it seems a lot of people are confused, or think that I am. The grids of the 6550's are indeed seeing 200v peak to peak AC signal, coming from the coupling cap of the previous stage. The question is, why can this be happening. Some of that gets burned up in the cathode to grid voltage as Tom said, but seeing as though the 6550 can only possible have between 0 and -70 volts on the grid based on it's chart...When I see 200v peak to peak on the scope, at the junction of coupling cap and 6550 grid, is there an RMS equation that needs to be done to that 200V before I see how it lays out on a load line?
It's really quite simple. If you already know that cutoff is -70v then the peak (+) signal on the grid can only be 35v. And so you would set your biasing to -35v. And you would swing the grid with a signal from +35v to -35v for the max (-) input of -70v on the grid.
You can't just set your bias to whatever you want, the tubes operating points are fixed based on your load lne. You can pick one, but the 6550 pulls a LOT of current in a quiescent state at a -35 Bias, with a B+ of 523v and a 5K primary.
I've asked this question all over, and it seems a lot of people are confused, or think that I am. The grids of the 6550's are indeed seeing 200v peak to peak AC signal, coming from the coupling cap of the previous stage. The question is, why can this be happening. Some of that gets burned up in the cathode to grid voltage as Tom said, but seeing as though the 6550 can only possible have between 0 and -70 volts on the grid based on it's chart...When I see 200v peak to peak on the scope, at the junction of coupling cap and 6550 grid, is there an RMS equation that needs to be done to that 200V before I see how it lays out on a load line?
You can't just set your bias to whatever you want, the tubes operating points are fixed based on your load lne. You can pick one, but the 6550 pulls a LOT of current in a quiescent state at a -35 Bias, with a B+ of 523v and a 5K primary.
I've asked this question all over, and it seems a lot of people are confused, or think that I am. The grids of the 6550's are indeed seeing 200v peak to peak AC signal, coming from the coupling cap of the previous stage. The question is, why can this be happening. Some of that gets burned up in the cathode to grid voltage as Tom said, but seeing as though the 6550 can only possible have between 0 and -70 volts on the grid based on it's chart...When I see 200v peak to peak on the scope, at the junction of coupling cap and 6550 grid, is there an RMS equation that needs to be done to that 200V before I see how it lays out on a load line?
You use peak voltage. If your input signal to the grid is really (+)100v then you are overdriving the grid when you have a bias voltage of -47v. Turn down the volume control. Next, I didn't say that (-) 35v. was going to be the proper bias point for your amp. But you wanted to know how to get the input signal and your selected cutoff voltage of (-) 70v. to be centered over the bias point. If you don't want your drive signal to push the output into cutoff then you have to keep the drive signal low enough not to go that far (-). If you are the amp designer, then you choose the bias voltage for an SE amp based on your B+ (the anode V. to be more precise) and the quiescent current which runs your plates at 90% - 95% of the max power dissipation. But if you really have as large of a drive signal as you say, then you have a driver circuit problem and need to get that under control, first.
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Measure the voltage swing at the cathodes. Subtract it from the measured voltage swing at the grids. This difference is the voltage swing that the tube sees, and will conform to plate curve predictions.
All good fortune,
Chris
All good fortune,
Chris
Or is it that some of the voltage is wasted on feedback from the tertiary winding in the OPT that the cathode hooks through to ground
You have a feedback voltage applied to the cathode AND drive voltage applied to the grid. The 6550 only understands the DIFFERENCE between the cathode and grid. That is the true instantaneous "bias" voltage.
You stated that there is a tertiary winding on the OPT that is in series with the cathode. This also has some DC resistance. There will be a dc bias voltage on the cathode at idle. The tube's bias point will be the DC difference between the grid and cathode at idle with no signal applied. Measure the cathode voltage at idle. Measure the grid voltage at idle, subtract. That is your true idle bias voltage.
With drive applied you will see an AC voltage at the grid, AND an AC signal on the cathode. The actual drive voltage seen by the tube is again the difference between these two AC signals.
If you have a dual trace scope, put one trace on the cathode, and the other on the grid. The difference between these two signals is the true instantaneous drive voltage. The two traces should be in phase...or close.
Some scopes like my old Tek 2232 have the ability to display this directly. There is a button to invert one channel. Push it. The two traces should now be out of phase. There is a switch to select, "alternate, chop, or add" select add. The scope will add one trace to the inverted other trace, which subtracts the two, displaying the true drive voltage.
73's KB4LRE
Thank you so much! So...
Thanks so much KB4LRE. So funny another Amateur would answer, I love being a ham. Between You and Chris, who also said it (thanks), okay I get it. And I will measure it all to see it. So...
If I were starting with this output circuit and wanted to redesign the driver stage, how would I figure out how much voltage was needed to drive it? And I'm up for reading too, but haven't been able to find anything. All the literature says "look at the chart, see your quiescent point, and you need so much to the negative and positive from there". And it is what you "need", but being that it's the difference between two voltages, how would I calculate it?
Again, I can't thank you guys enough, this is really helping me. I'm a member of a few forums and you guys are the ones who came out with answers instead of flames 🙂
You have a feedback voltage applied to the cathode AND drive voltage applied to the grid. The 6550 only understands the DIFFERENCE between the cathode and grid. That is the true instantaneous "bias" voltage.
You stated that there is a tertiary winding on the OPT that is in series with the cathode. This also has some DC resistance. There will be a dc bias voltage on the cathode at idle. The tube's bias point will be the DC difference between the grid and cathode at idle with no signal applied. Measure the cathode voltage at idle. Measure the grid voltage at idle, subtract. That is your true idle bias voltage.
With drive applied you will see an AC voltage at the grid, AND an AC signal on the cathode. The actual drive voltage seen by the tube is again the difference between these two AC signals.
If you have a dual trace scope, put one trace on the cathode, and the other on the grid. The difference between these two signals is the true instantaneous drive voltage. The two traces should be in phase...or close.
Some scopes like my old Tek 2232 have the ability to display this directly. There is a button to invert one channel. Push it. The two traces should now be out of phase. There is a switch to select, "alternate, chop, or add" select add. The scope will add one trace to the inverted other trace, which subtracts the two, displaying the true drive voltage.
73's KB4LRE
Thanks so much KB4LRE. So funny another Amateur would answer, I love being a ham. Between You and Chris, who also said it (thanks), okay I get it. And I will measure it all to see it. So...
If I were starting with this output circuit and wanted to redesign the driver stage, how would I figure out how much voltage was needed to drive it? And I'm up for reading too, but haven't been able to find anything. All the literature says "look at the chart, see your quiescent point, and you need so much to the negative and positive from there". And it is what you "need", but being that it's the difference between two voltages, how would I calculate it?
Again, I can't thank you guys enough, this is really helping me. I'm a member of a few forums and you guys are the ones who came out with answers instead of flames 🙂
Thanks all, still would like the drive answer though
Thanks all so much. With the help, changed a few things, 44W out of 3 6550 PSE! Would still like to know the math though, on drive requirements 🙂
Loren
Thanks all so much. With the help, changed a few things, 44W out of 3 6550 PSE! Would still like to know the math though, on drive requirements 🙂
Loren
Thanks all so much. With the help, changed a few things, 44W out of 3 6550 PSE! Would still like to know the math though, on drive requirements 🙂
Loren
(+) signal peak swing voltage (+)(-) bias voltage = < 0v. Keep the total to slightly under 0v. That is the fixed bias way. For cathode biasing with a (+) cathode voltage and 0v on the grid keep the (+) swing less than the bias voltage. Just figure out what the true bias voltage is and keep your peak signal on the grid smaller. This is for a SE amp. Class A. If your amp has special current restrictions then you may have to run it colder with a more (-) biasing and so you'd have to keep your signal level smaller to stay out of the cutoff zone.
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I (all of us?) feel cheated, you should have mentioned this on the *first* post.Or is it that some of the voltage is wasted on feedback from the tertiary winding in the OPT that the cathode hooks through to ground? Or? There is, definitely, 200V peak to peak coming out of the coupling cap from the preceding stage. Maybe it's not all going on to the grid, but then where?
What Chris mentions on post #10 is right, of course.
Again.As to the current post. Could you please explain a little, I'd really appreciate it. I've always figured gain by voltage requirements.
I'm not talking *gain* but proper idle current which in a Class A design sits at exactly half peak value, because it swings from cutoff to peak extremes, by definition.
That, if you want a reasonably efficient design.
Of course it will still work if you "waste" range one or both ways, just will lose possible power and be even less efficient, so standard practice is to set it in the middle of the range.
The feedback signal level creates the bias voltage computation problem. What is the ratio of plate signal to feedback signal applied to the cathode? ....? Whatever signal gets applied to the cathode is calculated into the instantaneous bias voltage. Does the OPT have specs for the tertiary winding ratio? Can you juggle the grid signal against the cathode signal against the quiescent bias voltage?
I went back to carefully reread all of this and find we are assuming the FB goes to the cathode of the 6550's. Is that where it's going? Or the drivers?
"Tertiary windings" usually refers to NFB ones which go to the cathodes.
Of course, being "negative" means that they substract from drive signal so you need more of it to compensate.
In that case, the 200Vpp (to ground, NOT to cathode) drive signal makes sense, and not otherwise ... but we should have been warned of it in the first post, instead of being sent on a wild goose chase .
Of course, being "negative" means that they substract from drive signal so you need more of it to compensate.
In that case, the 200Vpp (to ground, NOT to cathode) drive signal makes sense, and not otherwise ... but we should have been warned of it in the first post, instead of being sent on a wild goose chase .
I would seriously suggest that the OP posts a schematic of what is going on, so that one can have an exact picture of the whole affair before continuing.
I also suggest the OP get a differential probe for their scope...to properly measure floating signals..
I use the Tektronix P5205 ...
You can find used ones on Ebay for fairly cheap.....
I use the Tektronix P5205 ...
You can find used ones on Ebay for fairly cheap.....
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