Calculating resistor wattage

[mike1 ryan1]

When calculating the power rating of a resistor, do you use the voltage drop across it. So for the lamp shunt resistor in a dac90a that requires 7 volts for the two lamps divided by the resistance to get the current, then I2R to get watts. Comes to around 0.6W. Is this correct?

[Radio Wrangler]

I squared R or V squared over R or V times I will give you the wattage which will be dissipated.

So you need a resistor rated to at least this power. Going for one rated about twice that power is good for reliability.

Most modern resistor materials can be run at much higher temperatures than vintage ones, the exception being wirewound parts which haven't change. Resistor manufacturers have exploited this and resistors now are often appreciably smaller than they used to be. But a Watt is still a Watt and the same dissipation from the smaller body means the smaller part will run at a higher temperature.

In vintage equipment, where you use a modern resistor as a replacement, you may want to pick a resistor rated for higher power than the original just to get a similar body size so your change isn't obvious. This also has the advantage that the surface temperature of the new resistor won't be higher than that of tho old. This can be important if insulation touches the resistor.

David

[broadgage]

In the case of a lamp shunt resistor, I believe that the above calculation is incorrect.
One needs to calculate the wattage dissipated when the lamp is defective. This will be much greater than with a sound lamp.

[snowman_al]

Originally Posted by mike1 ryan1

When calculating the power rating of a resistor, do you use the voltage drop across it. So for the lamp shunt resistor in a dac90a that requires 7 volts for the two lamps divided by the resistance to get the current, then I2R to get watts. Comes to around 0.6W. Is this correct?

Broadgage, you beat me to it...

The whole mains current passes through the bulb/resistor in normal operation.
So 0.1A for the heaters plus 0.04A for the HT = 0.14A.
If as broadgage says a bulb goes open then the 75 ohm resistor takes the full current. And has to dissipate about 1.5 Watts. Therefore choose a 3 or 4 watt type.
Alan

[cathoderay57]

Hi Mike, to avoid confusion over the lamp shunt resistance value see the following thread https://www.vintage-radio.net/forum/...ad.php?t=95408 . This explains why the service sheets say 250R whereas in many sets a 75R is fitted. As stated in the previous post the recommendation is good. The max current rating of the resistor isn't critical so long as it is at least 3 Watts. Cheers, Jerry

[mike1 ryan1]

Brilliant thanks very much for that.
Mike.

[TrevorG3VLF]

If a valve rectifier is replaced by a semiconductor diode and resistor, then the voltage drop and dissipation will be much higher than that calculated from the average current.

Suppose the instantaneous current is five times the average for one fifth of the time, the voltage drop will be five times and the dissipation will be five times that calculated from the simple case.

[Craig Sawyers]

Resistor dissipation rating is in free air. So don't mount it hard onto a circuit board if it is dissipating a fair chunk of a watt - space it above the board by a few mm so that air can circulate via convection. If you don't, you get a hot spot where it contacts the board, with the possibility of resistor burn out or board damage.

Craig

[David G4EBT]

Originally Posted by mike1 ryan1

When calculating the power rating of a resistor, do you use the voltage drop across it. So for the lamp shunt resistor in a dac90a that requires 7 volts for the two lamps divided by the resistance to get the current, then I2R to get watts. Comes to around 0.6W. Is this correct?

No that isn't correct - it needs to be ten times that Wattage.

The shunt (R15, 250R on the Bush datasheet), is in parallel with the two lamps - not in series and it's there to ensure that should either or both of the dial lights fail, the set still has continuity with the incoming mains neutral (or line, as the case might be, given the manky reversible mains connector, which I won't call a 'plug' 'cos it's actually a socket - the plug in fixed and on the rear apron).

Bush state that R15 is rated as 6 Watts, which accounts for its large physical size.

A modern ceramic wirewound would be much smaller, so you might as well get a higher wattage - a 10 Watt would only be 48mm x 10mm x 10mm, but the modern value would be either 220R or 270R rather than 250, which I'd have thought would be fine, given the debate about the value of the resistor anyway, which seems to have changed over time.

You'd get either value resistor from ESR Electronics;

270R Stock No 926 -127
220R Stock No 926 - 122

https://www.esr.co.uk/components/pro...-resistors.htm

10 Watt 250R resistors are available on e-bay cheaply, but from Hong Kong and I imagine you'd rather press on than to stall the repair.

Hope that helps a bit Mike.

Good luck with it.

[ms660]

For an approximate figure multiply the DC load current (approx. 0.04 amps in the trader sheet) by 1.75 then add that result to the heater current (0.1 amps) then do the Isquared*R calc, whatever that turns out to be double it for the rating of a replacement resistor.

Unless yours is an unmodified receiver the shunt resistors value should be 75 ohms as per the Trader sheet.

Multiplying the DC load current by 1.75 is to allow for the dissipation in the resistor due to the current flowing through the rectifier.

Lawrence.