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Hoffman Amps Forum image Author Topic: Another question on cathode bias     &nb  (Read 3498 times)

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Offline Geezer

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Another question on cathode bias     &nb
« on: June 18, 2005, 05:17:45 am »

  Hoffman Amplifiers
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j allen shaw
Junior tube assistant
Posts: 12
(3/2/04 10:24 am)
  Another question on cathode bias
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 I'm building a cathode biased amp using EL84's, and I got to thinking "How do you check the bias on cathode biased tubes?"
Since I normally use the "measure voltage across 1 ohm resistor from cathode to ground" method, I suppose this would not work with the cathodes already connected to ground thru the resistor/cap for bias.
Thanks! Jeff

Edited by: j allen shaw at: 3/13/04 5:17 am
 
Casey4s
Senior tube assistant
Posts: 142
(3/2/04 11:40 am)
  Re: How to check cathode bias?
--------------------------------------------------------------------------------
 Actually, you do the same thing. Measure voltage drop accross your cathode resistor, then divide that by the value of your resistor = Ma.

Casey4s
 
HotBluePlates
I only work on Fender's
Posts: 697
(3/2/04 5:04 pm)
  Checking Cathode Bias
--------------------------------------------------------------------------------
 Since you won't need to adjust a cathode biased tube, the reading will mainly be for your information.

Tou could insert a 1 ohm 1 watt resistor between the cathode of the tube (pin 8 for 6V6, 6L6) and the non- grounded side of the cathode resistor. That way, you can still read mv's directly to ma's.

The extra 1 ohm shouldn't cause any problems, as it's only 1/250th or less of the cathode resistor value.
 
LooseChange
Senior tube assistant
Posts: 210
(3/2/04 7:13 pm)
  Re: Checking Cathode Bias
--------------------------------------------------------------------------------
 Is wattage important on that 1ohm R?? Or does the cathode resistor (250ohm typical) take it all??
 
HotBluePlates
I only work on Fender's
Posts: 701
(3/2/04 7:54 pm)
  Re: Checking Cathode Bias
--------------------------------------------------------------------------------
 I would use a 1 watt resistor.

If you think that average tube current will be between 20-40mA, then that's 20-40mW of power dissipated.

The larger the resistor, the more voltage dropped across it. Voltage x current equals power, so most of the power will be dissipated in the larger cathode resistor.

If a tube were to fail and draw massive current, it would be nice to have that 1 watt resistor there to burn up like a fuse before there is enough current to damage the transformer.
 
j allen shaw
Junior tube assistant
Posts: 13
(3/2/04 7:54 pm)
  Re: Checking Cathode Bias
--------------------------------------------------------------------------------
 LooseChange brings up a good point: Since the cathode resistor is typically 5w to 10w, how will using a 1/2w to 3w 1ohm resistor work? Will the 1ohm also need to be 5w to 10w?

Another question: What value cathode resistor to use? Above it was mentioned that 250ohms is "typical", but I saw on 18watt.com that they are using 120ohm to 150ohm as cathode resistors.
This is the reason for my original question about how to read the cathode bias. Since this is a new build, I need to know what value resistor to start with, realizing I'll need to swap values up or down to get the tubes operating within their recommended parameters.

Jeff
Edit: I see some of this was answered while I was typing!!

Edited by: j allen shaw at: 3/2/04 7:56 pm
 
HotBluePlates
I only work on Fender's
Posts: 702
(3/2/04 8:18 pm)
  Re: Checking Cathode Bias
--------------------------------------------------------------------------------
 Good that you saw the answer to your question above.
I mentioned the fuse value because in a tube failure (for a cathode biased amp, this will probably be a short from plate the heater), the tube will draw likely up to 100 or more mA's and destroy the tube and possibly the OT if the fuse doesn't blow in time. The 1 ohm resistor is cheap compared to a burnt transformer.

For figuring the needed cathode resistor, simply follow the formula given in the first reply, and use algebra. So, if you know what bias voltage is needed and the idle current, rearrange the formula to solve for resistance.

You will often see 250-270 ohms as the cathode resistor in cathode biased 6V6/6L6/EL34 amps. 130 ohms to maybe 180 ohms are seen more in EL84 amps.

Lower plate voltages allow for biasing a tube into class A (if desired), which would take lower bias voltages and therefore a lower cathode resistor value.
Tweed amps (except the Champ), as well as Vox amps use a bit higher cathode resistor or wind up running the tubes in very hot class AB.

 
j allen shaw
Junior tube assistant
Posts: 15
(3/2/04 8:42 pm)
  Re: Checking Cathode Bias
--------------------------------------------------------------------------------
 Thanks HBP, that info is very helpful, as I have now decided I will be using 6V6's instead of EL84's in my project (as can be seen in this thread).
Jeff
 
j allen shaw
Junior tube assistant
Posts: 18
(3/12/04 5:38 pm)
  Re: Checking Cathode Bias
--------------------------------------------------------------------------------
 Since the (4) cathodes of the tubes are sharing (1) bias resistor, do I divide the bias current voltage result by 4?
Example: 145ma is read across the shared 1 ohm resistor, 145ma divided by 4 tubes = 36.25ma bias current per tube.
If plate voltage is 350: 350 x .03625 = 12.6875 watts dissipation per tube
Is this correct??
Also, what can happen if the 10w wire resistor is placed too close to the electrolytic bypass cap?
How close is too close?
Thanks!! Jeff
 
GroundhogKen
Forum Moderator
Posts: 2317
(3/13/04 8:59 am)
  Re: Checking Cathode Bias
--------------------------------------------------------------------------------
 The only reason to use 1 watt resistors is so you can put one on each tube. Your calculations above are correct, but you could do the same thing with the main big cathode resistor.

Say... 10v / 50 ohms equals 200mA through 4 tubes or 50mA per tube. Assuming all tube carry equal current.

If you put a 1 ohm resistor on each tube and then join the 1 ohmers at the main cathode resistor, then the 1 ohm resistors will tell you the exact current in each tube.

If you are just using a single 1 ohm resistor as a fuse, then why not use a real fuse?


Ken

EDIT: Also you don't want the hot cathode resistor to be too close to it's electrolytic bypass cap because electrolytics are usually not rated for temperature above 80 degrees C.

Edited by: GroundhogKen at: 3/13/04 9:02 am
 
HotBluePlates
I only work on Fender's
Posts: 705
(3/13/04 11:15 am)
  1 Ohm resistor
--------------------------------------------------------------------------------
 I second what Ken said. I intended to say use one 1 ohm resistor per tube.

The real virtue of the 1 ohm resistor is to measure actual tube current by "inference" when you measure the voltage drop in millivolts across he resistor.

So it's more handy to have one resistor per tube, which connects from the tube cathode to the top of the bias resistor (or cathode resistor, however you'd like to term it).

Ultimately, the fuse value of the resistor is a secondary effect of using the low wattage rating. The primary use is to measure each tube's current separately.
You could also run a wire from the cathode side of each resistor to its own meter-tip jack (a red one for instance), and a wire from the connection of all four to the top of the cathode resistor to its own tip jack (a black one for instance), and be able to measure tube current for each tube outside of the amp.
 
lowtech
Senior tube assistant
Posts: 183
(3/14/04 8:14 pm)
  Re: 1 Ohm resistor
--------------------------------------------------------------------------------
 As far as measuring plate disipation in a cathode biased amp, its a bit different. Measure your plate voltage. ( I'm assumng your referencing it to ground) Measure cathode voltage. ( again refenced to ground) Subtract the cathode voltage from the plate voltage and use that resultant for the voltage portion of the power equation for individual plate disipation. And yes like you stated earlier, if you're are sharing a cathode resistor for "n" number of tubes, divide your total current by "n" .
 
 
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