6BQ5 Cathode Bias

[bobloblaws]

I have a Kent Model A-60 amp here, I'm guessing built sometime in the '60s. Unfortunately, no schematic. My tube tester showed the 5Y3 rectifier as very weak so I popped in a brand new one. Indeed the plate voltage on the 6BQ5/EL84 power tubes went up by about 50 volts. That put the plate current draw over spec so I want to bring the bias down. I'm used to seeing the cathodes connected either directly to ground in a fixed bias amp or via a resistor in a cathode biased design. In the case of this Kent amp I see neither, in fact the cathodes are commonly connected to the one of the heater pins. The heater supply is 6.3V AC and the DC voltage on the cathodes is 9VDC. I'm curious where that 9VDC comes from. But more importantly, if I wanted to reduce the bias level on those tubes would I have to get rid of that cathode/heater connection and connect the cathodes to ground via an appropriately valued resistor to get a more positive voltage on the cathode?

B.

[52 Bill]

There should still be a cathode resistor and maybe a cathode bypass cap in the circuit somewhere. The heaters are connected to the cathodes in order to elevate the heaters and reduce hum.

The 9 volts dc is a normal cathode voltage and is indicative of there being a cathode resistor in the circuit. If you read the resistance from the cathodes to ground do you get a resistance reading?

[eschertron]

+1
The heaters being present on the cathodes is immaterial to the operation of the EL84s. You will find a resistor somewhere. What do you calculate the dissipation at?

[bobloblaws]

There should still be a cathode resistor and maybe a cathode bypass cap in the circuit somewhere. The heaters are connected to the cathodes in order to elevate the heaters and reduce hum.

The 9 volts dc is a normal cathode voltage and is indicative of there being a cathode resistor in the circuit. If you read the resistance from the cathodes to ground do you get a resistance reading?

Ahh, OK, it is making sense now. There is a 130/5W ceramic resistor between the cathode and ground and bypassed by a 20 uF cap. The cap is in the multi-section can and the resistor is soldered directly to one of the lugs on the can. That and the fact I didn't expect a cathode resistor to be of a ceramic type threw me off, in addition to the heater elevation business which happens to be on my to-do list of things to get familiar with. Thanks Bill and other responders, much appreciated!

[bobloblaws]

+1
The heaters being present on the cathodes is immaterial to the operation of the EL84s. You will find a resistor somewhere. What do you calculate the dissipation at?

40 mA on one tube and 32mA on the other. Plate to cathode voltage is 320VDC so the 70% target is 26 mA. The difference presumably because the OT resistance is 187 ohm on one side and 207 ohm on the other, respectively.

[bobloblaws]

Does the resistor in question really need to be a 5W ceramic in this scenario? The closest I have is 2W metal film.

[Enzo]

You tell us. What voltage is dropping across that 130 ohm resistor? Now apply Ohm's Law.

Do the two plates have different voltages?

[Dave H]

40 mA on one tube and 32mA on the other. Plate to cathode voltage is 320VDC so the 70% target is 26 mA. The difference presumably because the OT resistance is 187 ohm on one side and 207 ohm on the other, respectively.

The bias current difference won't be because of the OT resistance imbalance. How are you measuring the plate currents? Swap the tubes over to see if the current readings follow the tubes.

[bobloblaws]

Swap the tubes over to see if the current readings follow the tubes.

I get the same plate voltage readings on either tube. The only variable is the OT resistance.

How are you measuring the plate currents?

I got identical readings using OT resistance method and OT shunt method.

[bobloblaws]

You tell us. What voltage is dropping across that 130 ohm resistor? Now apply Ohm's Law.

9V / 130 ohms x 9V = 0.62 W

I get it :-)

That number only applies to quiescent scenario though, no?

Do the two plates have different voltages?

Virtually the same.

[nickb]

...the 70% target is 26 mA.

This is just Internet lore. It's only a guideline for amps that are operating in class AB. For sure it's excessive for many amps and especially those that use tiny power transformers for cost reduction. This amp could be even class A in which case the measured current is low, but I'm pretty sure it's not unless the cathode resistor has been changed.

For a cathode biased amp, it's quite common to bias them on the hot side. The reason is because during higher drive each tube is cut off for part of the time so raising the average current through the cathode resistor. This raises the cathode voltage and lowers the bias current leading to crossover distortion. I suggest you leave things as they are.

[eschertron]

9V / 130 ohms x 9V = 0.62 W

I get it :-)

That number only applies to quiescent scenario though, no?



Virtually the same.

Do you have a test sig generator? drive the output to clipping and measure the voltage drop again. If the amp is class A, there won't be much difference. And the plate current will be roughly proportional to the voltage drop across the resistor. Actually a little less, since the plate voltage drops and the cathode voltage rises.

edit: I see Nick posted while I was writing. I agree, the readings you get for idle current aren't all that surprising.

[Dave H]

The only variable is the OT resistance.

Have you checked that the screen and control grid voltages (and screen resistor values) are they the same for both tubes? If the imbalance follows the tubes then it's most likely the tubes. They will balance better if individual 270R cathode resistors (and capacitors) are used. That 8mA difference will come down to around 2mA if I remember correctly and it's cathode biassed so you could run the tubes at more than 70%

Edit: I see nick has already explained the 70% "rule"

[bobloblaws]

Hey guys, thanks for setting me straight re: 70% etc. I was still thinking in terms of class AB since it has two power tubes. There's a ton of information out there on this subject so I'll study up. It stands to reason that everything is probably pretty close to spec as is with the new rectifier.

[bobloblaws]

Do you have a test sig generator? drive the output to clipping and measure the voltage drop again. If the amp is class A, there won't be much difference. And the plate current will be roughly proportional to the voltage drop across the resistor. Actually a little less, since the plate voltage drops and the cathode voltage rises.

edit: I see Nick posted while I was writing. I agree, the readings you get for idle current aren't all that surprising.

Great tip, thanks.