cathode bias how many miliamps should I see on 1 ohm resistor?

[67polara]

Building a TOS and I only have 10 miliamps on the 1 ohm resistor at idle is that right?  If not what should it be and what do I need to do to fix it.  This amp doesn't seem to have much power.

Tony

[HotBluePlates]

From the schematic shown on this page, it appears the cathode bias resistor is 430 ohms, and has 39v across it.

39v / 430 ohms = 90.69mA, but this is the total cathode current of 2 tubes. 90.69mA / 2 = ~45.35mA.

I assume you already know that you should measure the mV across the 1 ohm resistor, and you would see 45.35mV, give or take. This also assumes you have approximately 420v on the plate and screen. Is that the case? Are proper voltages present elsewhere?

If voltages look good, check for wiring errors. Make sure your meter probes have good firm contact when taking a reading.

[67polara]

Thank you for your reply.  Yes you are correct on the resistor size I have 41.5 volts across that resistor.  I only show 9.7ma across the 1 ohm resistor though.  Plate and grid are at 474v ea.  So using your calculations it should be about 48 ma each but that is not what I get!  What could be wrong?

Tony

[woolly]

you can't read milli amps across a resistor. you must set your meter to milli Volts.

[Geezer]

you can't read milli amps across a resistor. you must set your meter to milli Volts.

Exactly (Woolly has it right)!

We use the 1ohm as a sorta "trick" to easily read the current, as (conveniently) the voltage reading across that resistor converts directly to mA (so we don't have to do the math).

G

[tubenit]

Simply as an FYI,  I have a 360R/10W & I read 41.3ma with lower plate voltages. I think anything from 37-48ma would probably be OK.


With respect, Tubenit

[67polara]

Thanks everyone.  I know that I just had a brain fart LOL.

Tony

[HotBluePlates]

you can't read milli amps across a resistor. you must set your meter to milli Volts.

Well, you probably could. But that defeats the point of having the 1 ohm resistor.

When you set you meter to measure current, it tries to act as a good-uality short-circuit between the meter leads. So in effect, you are using the meter to bypass the 1 ohm resistor, with the tube current flowing through the meter leads and on to the shared cathode resistor.

... Yes you are correct on the resistor size I have 41.5 volts across that resistor.  I only show 9.7ma across the 1 ohm resistor though.  Plate and grid are at 474v ea.  So using your calculations it should be about 48 ma each but that is not what I get!  What could be wrong?

Tony

And here is the clue. 474v sounds high, but given your measurements and that you're lookng for 48mA per tube, you appear to have a 430 ohm cathode resistor (.048 * 2 = .096A, 41.5v / .096A = ~432 ohms). So on the basis of the cathode resistor voltage, you have 474v - 41v = 433v, 433v * .048A = 20.78w. That's fine for 6L6GB/5581/6L6GC, etc., and is a high estimate because screen current was not subtracted from cathode current.

So here's the catch: We assumed that when the meter was set to measure current that its leads essentially were a short circuit. This is nearly-true for an external circuit resistance of hundreds-of-ohms or more, but maybe not true for an external circuit resistance of 1 ohm.

It's considered a little old now, but I have a Fluke 87III meter, which I consider to be fairly high-end for service work. Table 14 in the User's Manual lists the meter's specs when measuring current. The meter measures voltage well, and likely actually measures current by measuring the small voltage drop across an internal resistance, much as we use our 1 ohm resistor. The specs say that the "burden voltage" for the 400mA range is 1.8mV/mA; Note that Ohm's law is arranged this way when solving for resistance, and you see that the meter looks like a 1.8 ohm resistor when in the 400mA range.

So while the meter is bridging the existing 1 ohm resistor in the amp, the total circuit looks like a 1 ohm resistor in parallel with a 1.8 ohm resistor (the meter). Since the meter resistance is about twice the external circuit resistance, the current flowing through the meter will be about half that which flows through the external 1 ohm resistor; so of the total current flowing, only 1/3 is passing through the meter and registering on the readout.

Maybe your meter has a little higher "burden voltage" or your 1 ohm resistor is a shade under 1 ohm. Either way, if you have your meter set to read current, and bridge it across the 1 ohm resistor, it is believable that you will see 1/3 or 1/4 the total actual current.

This jives with what you're seeing. I think your amp is fine. Set your meter to measure millivolts across the 1 ohm resistor instead.