# Thread: DC coupled C.F. question

1. ## DC coupled C.F. question

Hello, how to avoid grid current into a 12ax7 C.F. DC coupled ,please ?
The tube is supplied at 300V around and coupled at 75V through a 1M/1M voltage divider from the 150V plate of previous stage.Should be biased at -2V arround but it is not. I instaled 1K grid resistor. I don't have a precise milivoltmeter but on my fluke175 still recorded 0.1mV over it which is pretty evident as time I cannot proper bias my C.F. stage. The actual cathode resistor is 100k and draw almost same amount of current as recorded over 220k plate resistor of previous stage. With 47k seems the derive is better somehow...Please let me know which is the best way to limit the current to none through the CF grid. Thanks.

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2. Originally Posted by catalin gramada
The actual cathode resistor is 100k and draw almost same amount of current as recorded over 220k plate resistor of previous stage.
That sounds about right to me. (300-150)/220k = 0.68mA. 75/100k = 0.75mA.

The plate of the stage driving the CF is at 150V so the grid of the CF should be at 75V and its cathode at 77V. Is it not like that?

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3. Originally Posted by Dave H
That sounds about right to me. (300-150)/220k = 0.68mA. 75/100k = 0.75mA.

The plate of the stage driving the CF is at 150V so the grid of the CF should be at 75V and its cathode at 77V. Is it not like that?
Thanks for answering. No, the difference is 6V around. I wonder why?

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4. You are measuring with one probe on grid and your other probe on cathode? (neither probe to ground)

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5. No, I measured voltage of divider and voltage over cathode resistor and I did the difference.

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6. Can you post a schematic of the stage with voltages?
Did you try different tubes?
At high cathode current grid bias of the DCCF may vary considerably with individual tubes.
It is not uncommon to find positive grid bias in the DCCF of e.g. an AC30.

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7. Is a straight design, changed tube, measured resistors and here we go...

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8. Originally Posted by catalin gramada
Is a straight design, changed tube, measured resistors and here we go...
So problem solved?

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9. ...

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10. Originally Posted by catalin gramada
...
First of all your voltages indicate that the DCCF has negative grid bias.
Most DMMs have an input impedance of 10M, so voltages across an 1M resistor may measure up to 9% low depending on rest of circuit.
Do you still measure a positive grid-cathode voltage at the CF, attaching the red meter lead to grid and the black one to cathode with the changed tube?

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11. Originally Posted by catalin gramada
No, I measured voltage of divider and voltage over cathode resistor and I did the difference.
Measure the plate voltage of the driver stage and cathode voltage of the CF.
Do they measure as you would expect knowing the divider ratio and -2V CF bias voltage?

Edit: I see Helmholtz beat me to it

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12. Originally Posted by Helmholtz
Most DMMs have an input impedance of 10M, so voltages across an 1M resistor may measure up to 10% low depending on rest of circuit.
If I insert that parallel 10M into that voltage divider I calculate 71V at the grid.

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13. Yes, should be like that. The divider resistors are perfectly matched so the voltage should be half (except if grid C.F. do not start to conduct) In this circumstance the difference was -2.8V around with 100k in cathode. Changed to 47k and adjusted the supply as same as before (300v) get -2.25V

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14. It have no sense to me. If I want to do the grid less negative in respect with cathode I suppose to raise the cathode value for the same fixed bias in grid.....but is happen exactly opposite....

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15. Originally Posted by catalin gramada
It have no sense to me. If I want to do less negative I suppose to raise the cathode value for the same fixed bias in grid.....but is happen exactly opposite....
Why? A lower value cathode resistor requires more cathode current (as cathode voltage follows grid voltage within a couple of volts), thus more positive/less negative grid bias.

What is the actual problem?

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16. Originally Posted by catalin gramada
It have no sense to me. If I want to do the grid less negative in respect with cathode I suppose to raise the cathode value for the same fixed bias in grid.....but is happen exactly opposite....
Do you mean to make the bias voltage less negative the cathode resistor value should be increased?

Increasing the cathode resistor means less cathode current which means more negative bias voltage not less so it's not opposite.

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17. Are you still trying to measure from the junction of the two 1M to ground?
Like Helmholtz said, I don't understand what problem you are trying to correct.

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18. The actual problem is: changing the cathode resistor from 47k to 100k did not do almost any difference for the same plate voltage (150v) before divider. Considering 75v in c.f. grid I get 77.8v with 47k and 78.3v with 100k just now. I wonder how can I bias the tube say -1.5v around for the same supply voltage?
No measured over divider, the voltages was measured into the plate and cathode.

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19. Originally Posted by catalin gramada
The actual problem is: changing the cathode resistor from 47k to 100k did not do almost any difference for the same plate voltage (150v) before divider. Considering 75v in c.f. grid I get 77.8v with 47k and 78.3v with 100k just now. I wonder how can I bias the tube say -1.5v around for the same supply voltage?
I don't think there is a problem. You don't set the bias voltage. You have already set the cathode current by choosing the cathode resistor value and grid voltage. The tube sets the bias voltage at the value required to provide that cathode current at the supplied plate to cathode voltage. You can look it up on the data sheet.

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20. Originally Posted by catalin gramada
The actual problem is: changing the cathode resistor from 47k to 100k did not do almost any difference for the same plate voltage (150v) before divider. Considering 75v in c.f. grid I get 77.8v with 47k and 78.3v with 100k just now. I wonder how can I bias the tube say -1.5v around for the same supply voltage?
No measured over divider, the voltages was measured into the plate and cathode.
I am getting the impression that you want to avoid grid current in the DCCF. The question is: Why?

It is common to have up to several hundred µA grid current in typical DCCFs, meaning that bias voltage is close to or even above zero depending on individual tubes and component values. This is described in tube amp literature (Zollner, Blencowe). The result is that this circuit shows a nonlinear transfer characteristic, meaning that it distorts easily. From a (hifi) engineering POV it is a poor design.
But it tends to have a positive influence on sound in a guitar amp. The only drawback is that its distortion characteristics may vary with the individual tube. The tube in the DCCF is maybe the most influential in the amp regarding sound.

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21. Thank for You answers gents, it helps me a lot to understand how this circuit works.

Talking about C.F. still I have one more question please. I want to play a little bit, the mixer from the picture above is a reasonable mixer circuit,Please? Thanks

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22. Originally Posted by catalin gramada
No, I measured voltage of divider and voltage over cathode resistor and I did the difference.
Do what G1 said.

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23. Thx Nick, I don't measured in the grid, but over plate instead. The divider resistors are matched so it was simple divided by 2.Anyhow the result was little far than expected. Now I think I understand the operation is conditioned by grid voltage in respect with what? .... internal resistance of the tube perhaps...Thanks

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24. Originally Posted by catalin gramada
Thx Nick, I don't measured in the grid, but over plate instead. The divider resistors are matched so it was simple divided by 2.Anyhow the result was little far than expected. Now I think I understand the operation is conditioned by grid bias voltage in respect with what? .... internal resistance of the tube perhaps...Thanks
No, no, no...

1) When you measure at the 220k about 15uA will flow thru your meter == 3.3V error
2) With 1% resistors you get up to a 2% error and the mid point i.e about 1.5V

So you connect the meter where minimum current flows (i.e. grid to cathode) and so the error is minimized. Try it now and see what you get.

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25. So you connect the meter where minimum current flows (i.e. grid to cathode) and so the error is minimized.
..because voltage difference is small.

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26. So, for 148v on plate before divider get 2.25v between follower grid to cathode

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27. 1) When you measure at the 220k about 15uA will flow thru your meter == 3.3V error
I think that the internal DC tube resistance can be assumed to equal the plate resistor. This means that the measurement error would be only 3.3/2= 1.65V.

But if 2 equal 1M divider resistors are used, the grid voltage of the CF must equal half the plate voltage - provided there is no grid current. Most probably there is some grid current, though.

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28. Originally Posted by Helmholtz
I think that the internal DC tube resistance can be assumed to equal the plate resistor. This means that the measurement error would be only 3.3/2= 1.65V.

But if 2 equal 1M divider resistors are used, the grid voltage of the CF must equal half the plate voltage - provided there is no grid current. Most probably there is some grid current, though.
I was taking about the measurement at the 220k resistor so 3.3v is correct.

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29. Originally Posted by catalin gramada
So, for 148v on plate before divider get 2.25v between follower grid to cathode
See! Go back and give g1 a big hug!

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30. Originally Posted by nickb
I was taking about the measurement at the 220k resistor so 3.3v is correct.
I think measurement between plate and ground means a source impedance of 110K, resulting in a measurement error of 1.63V.

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31. Originally Posted by Helmholtz
I think measurement between plate and ground means a source impedance of 110K, resulting in a measurement error of 1.65V.
Indeed yes! I forgot that.

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32. This one it stinks. Is not usable as mixer btw. vs common anode mixer.

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33. Originally Posted by catalin gramada
This one it stinks. Is not usable as mixer btw. vs common anode mixer.
Stinks in what way, not enough output?
I know that with the other grid grounded the signal at the top of the tail is about half the input.

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34. I get a severe asymmetrical wave form at the follower output no matter by voltage applied.

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35. Originally Posted by catalin gramada
This one it stinks. Is not usable as mixer btw. vs common anode mixer.

I don't think this a useful circuit. It consists of 2 cathode followers having their outputs connected and sharing a common cathode resistor. As cathode followers have low internal impedance, the outputs are loading down each other. This means low gain <0.5 and probably non-linearity.

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