# Disparate Resistors in Parallel. POWER HANDLING Question?

#### SONDEKNZ

I need to reduce the value of my (shared) Cathode Resistor, to increase the bias, per pair of KT88 output tubes.

Presently, the Cathode Resistor comprises two 750K/12W wirewound resistors in parallel, combined to deliver 375K/24W.

With 370V measured from Cathode to Anode - and 40V measured across the Cathode Resistor at idle - the KT88 tubes are currently operating well below 50% of maximum dissipation.

To increase the bias, I need to bring the Cathode Resistor down to at least 300K, which is very easy to achieve by adding a third resistor in parallel - at a value of 1.5-Megaohms.

QUESTION: What minimum WATTAGE does the third paralleled resistor need to be, in order to maintain around 24W of power handling at this Cathode Resistor?

Apologies that my extensive scouring of the Internet has not been able to unearth a definitive answer...

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#### grovergardner

First of all, that's an awfully large cathode resistor. Are you sure you have the values right? A typical cathode resistor would be 750 ohms, not 750K ohms. At any rate, adding a resistor in parallel follows Ohm's Law just like the other resistors.

750 ohms with 40 volts (bias voltage) across it is V/R = I (current) so

40/750 = .053 amps.

I x V = P (watts) so 40 x .053 = 2.133 watts.

Always uprate by at least 2x, preferably 3x to avoid overheating. So a 7 watt resistor would be safe here.

Now parallel with 5K resistor to reduce the resistance.

40/5000 = .008 amps.

.008 x 40 = .32 watts.

So a 1 watt resistor would be safe.

There are online calculators that make this easier. ;-)

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#### SONDEKNZ

@grovergardner

Many thanks for your quick answer - and for spotting my error above.

Yes. the present (shared) Cathode Resistance is a total of 375-Ohms 24W. [Not 375K/24W]

I assume the 1W paralleled resistor diminishes my (previous) 24W power handling, but I am unsure to what degree.

QUESTION: If I do add a third parallel resistor - of only 1W power handling - what total power handling does my new Cathode Resistor now have?

#### Galu

There's a wee bit of confusion in post #2 in that 40/750 does not equal 2.133

Better to use P = V^2/R = 40^2/750 = 1600/750 = 2.13 W

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#### grovergardner

Yes, I updated the comment.

#### grovergardner

@grovergardner

Many thanks for your quick answer - and for spotting my error above.

Yes. the present (shared) Cathode Resistance is a total of 375-Ohms 24W. [Not 375K/24W]

I assume the 1W paralleled resistor diminishes my (previous) 24W power handling, but I am unsure to what degree.

QUESTION: If I do add a third parallel resistor - of only 1W power handling - what total power handling does my new Cathode Resistor now have?

No, adding another resistor in parallel does NOT reduce the power handling capacity of the other resistors. Ohm's Law is inviolable. ;-) Each separate resistor has voltage across it, and that determines the amount of current it sees and the power generated.

Also, I am NOT recommending any resistor values. I just chose random values. To determine what resistor to add in parallel, you have to decide what you want the new cathode resistance to be and add an appropriate resistor in parallel.

Bear in mind that changing the cathode resistance will increase the current of the tubes, but also reduce the cathode voltage itself, so you have to balance those two things.

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#### SONDEKNZ

So, the 1W resistor definitely won't melt?

#### grovergardner

No, it's a 5K resistor with 40 volts (or whatever) across it. Period. Ohm's Law applies.

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#### Galu

In that case, P = V^2/R = 1600/5000 = 0.32 W

So yes, 1W would be fine.

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#### Tom Bavis

A lower value than 5K would have a higher dissipation - i.e 1K with 40V would be 1.6W, so you'd use 3W or higher. Assuming this resistor is properly rated, overall dissipation rating of the total cathode resistance can only go higher.

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