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Old 26th April 2009, 02:19 AM   #1
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Question Determining cathode resistor

Hi again, everyone !! I have a single-ended amp with a 6V6 a 6X5 and a 6SJ7. The 6SJ7 is grid leak biased and I would like to switch to cathode bias. How do I determine the value of the cathode resistor ???
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Old 27th April 2009, 03:35 PM   #2
Arnulf is offline Arnulf  Europe
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1: Measure grid voltage (that's the tricky part, especially is Rg is very high already) and cathode current, then calculate appropriate cathode resistor (cathode resistor = grid voltage / cathode current).

2: Measure plate voltage and cathode current, refer to datasheet for approximate grid voltage at that particular operating point, calculate cathode resistor and fiddle with its value until you get to same operating point where your amplifier was operating before the change.

A suggestion for option #2: take smaller resistor (smaller value than calculated, say 20% smaller) and wire a trimmer (its value at least twice the 20% you subtracted from the resistor) as rheostat in series with it so you can adjust cathode resistance without having to rewire the amplifier. I.e. if your calculation tells you to use 1K resistor, use ~800R + 470R trimmer (rheostat !) in series.

Since this is a signal pentode you are using you probably don't need to concern yourself with wattage ratings of the resistor and potentiometer/trimmer.
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Old 27th April 2009, 10:32 PM   #3
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I can guess that measuring the grid voltage will be really tricky, even with a VTVM! I would not trust any meter to do this, due to their non-infinite input resistance. Maybe a scope with a 100x probe?

Conclusion: Just look up the values given in the datasheet, and make an educated guess. Then adjust accordingly.
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Old 27th April 2009, 11:32 PM   #4
Arnulf is offline Arnulf  Europe
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Originally posted by costis_n
[B]I can guess that measuring the grid voltage will be really tricky, even with a VTVM! I would not trust any meter to do this, due to their non-infinite input resistance.
That's the tricky bit

However if meter probe impedance is relatively high (comparable to the value of Rg) one could simply remove Rg from circuit and let the meter sit there in its place.
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Old 27th April 2009, 11:56 PM   #5
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Angry I'm a dumba**

Well...I am definitely a dumba**. Of course using a pot in place of the cathode resistor and dialing it in until the plate voltage is the same as it was before will give me the proper value. Sorry to waste everyone's time...I KNEW that, my brain just passed a bit of flatus on me....
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Old 28th April 2009, 12:33 AM   #6
tomchr is offline tomchr  Canada
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Do yourself a favor and measure the grid, collector, and anode voltages first. Any digital voltmeter should have enough input impedance to do this without issues. Then do the math and figure out the appropriate cathode resistor. You can then use a fixed resistor plus a pot to find the final value if desired.

If you just insert any random pot without a fixed series resistor you run the risk of biasing the tube *way* hot (much too high anode current), thereby, wrecking the tube. If you decide to go this route anyway, at least turn the pot to the max resistance (lowest current).

Assuming the grid and cathode are referenced to the same supply rail (typ ground), you should be able to calculate the cathode resistor like this: Rk = Vgk/Ia, where Rk is the cathode resistor in ohms, Vgk is the grid-to-cathode voltage in volts, and Ia is the anode current (assuming zero grid current) in amperes.

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