DC coupled C.F. question

[Dave H]

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.

[g1]

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.

[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.

[Dave H]

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.

[Helmholtz]

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.

[catalin gramada]

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

[nickb]

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

Do what G1 said.

Your meter is loading the grid and giving you an incorrect reading.

[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 voltage in respect with what? .... internal resistance of the tube perhaps...Thanks

[nickb]

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...

Simple divide by two? Not so. Think about your error sources
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.

[Helmholtz]

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.

[catalin gramada]

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

[Helmholtz]

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.

[nickb]

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.

[nickb]

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

See! Go back and give g1 a big hug!

[Helmholtz]

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.