[Boatanchors] Ohms Law

[Singley, Rodger]

Barrie,

Make sure that you are making the measurement with the tubes in place and at the tube pins with an ACCURATE AC meter.  Most meters are least accurate on the AC range and a lot of the older analog types are woefully inaccurate due to calibration drift over the years.

I homebrewed an amp using three Russian 4CX800 tubes and I used some additional wire (loosely rolled up) between the transformer and tube heaters to provide the additional half volt of drop I needed.

Back to your resistor question, 15 watts generates a fair amount of heat so your resistor is going to get hot.  The only difference is your 30 watt unit is big enough to safely dissipate 15 without burning up; a 5 watt rated resistor would develop the same amount of heat over a smaller surface for a short time before it went up in smoke while a 100 watt unit would develop the same amount of heat but over a larger surface.  Think of the amount of heat produced by a little #47 pilot light which is only drawing about 1 watt total.  The wattage dissipated by the resistor is easy to find through Ohm's law as you have found.  With a known resistance and accurately measured voltage drop you can determine the current and watts dissipated.  The only important voltage measurement is that developed across the resistance-the source voltage doesn't matter.

Rodger WQ9E


-----Original Message-----
From: boatanchors-bounces at mailman.qth.net on behalf of Barrie Smith
Sent: Sun 4/18/2010 5:40 PM
To: Boatanchors Mail List
Subject: [Boatanchors] Ohms Law
 
I've been juggling 5 volt transformers all day, trying to find one that would actually produce close to 5 volts for a pair of tubes.

They were all either too high in voltage, or too low.  One did give me almost exactly 5 volts, but it ran way too hot.

So I've selected one with plenty of current capacity and am playing with resistance in the primary to get the secondary voltage down to 5 volts.

Using a 25 ohm, 150 watt rheostat, I find that 11 ohms gets me real close.  The rheostat runs quite warm, which leads me to my question.

The voltage drop across the rheostat is a measured 13 volts.  Knowing that and the resistance, Ohm's law tells me that I have 1.18 amps running through, which equals 15 watts.

I don't want to install the rheostat in the transmitter because it is quite large, so I thought I'd substitute a fixed resistor.

I tried a 15 ohm, 30 watt fixed resistor.  It got very, very hot real quick!

I must be making a mistake.  I thought I should use the "dropped voltage" in Ohm's law.  Perhaps I must use the full 115 volts, instead.

The amount of heat I'm producing indicates that is true.

Thoughts?

73, Barrie, W7ALW
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