06-28-2010, 04:44 PM
You can closely estimate the wattage of the resistors if you know the voltage drop across the field coil and bias resistors. Unfortunately, the Rider's schematic for the 41-295 on Nostalgia Air doesn't include enough information to make the calculation. Maybe the Philco schematic does.
Subtract the voltage across the second filter capacitor from the voltage across the first filter capacitor to get the voltage drop, E. (The voltage across the first filter cap is missing from the Rider's schematic.) Then add the resistance in ohms of the field coil to the resistance in ohms of the three bias resistors for the total resistance, R. Use Ohm's law, E=IR, to get the approximate current, I, in amps. The bias current is so small that you can ignore it for the purposes of this calculation, so you can assume that the current across all three bias resistors is the same. In my experience, this calculation gives the B current within a very few milliamps, and any error is on the safe side.
Power in watts = I squared times R, so plug the value in ohms of each bias resistor into this formula to get the power each resistor must dissipate. I doubt that you'll need more than 2 watt resistors in that circuit, but since I don't know the voltage drop I can't say for sure.
It is hard to find power resistors in the exact values needed, so when replacing a candohm resistor, I'll usually put two or even three 1 watt resistors in series to get as close as I can to the right value and to be sure that each is dissipating no more than about 1/2 or 3/4 watt. Even then they can get pretty hot, so I make sure they are well separated and sticking up on long leads.
The only time I've had a problem doing this is when the resistors are squeezed in tight, as in the 46-350. The candohm resistors use the steel chassis as a heat sink, and I don't know any way to do that when replacing them with modern resistors. I know candohm resistors have a bad reputation, but now I usually won't replace them if they are still OK.
Another advantage to the candohms is that the wire-wound construction provides a little bit of needed inductance to the circuit. Of course if you can find modern power resistors of close to the right value, that problem goes away.
Subtract the voltage across the second filter capacitor from the voltage across the first filter capacitor to get the voltage drop, E. (The voltage across the first filter cap is missing from the Rider's schematic.) Then add the resistance in ohms of the field coil to the resistance in ohms of the three bias resistors for the total resistance, R. Use Ohm's law, E=IR, to get the approximate current, I, in amps. The bias current is so small that you can ignore it for the purposes of this calculation, so you can assume that the current across all three bias resistors is the same. In my experience, this calculation gives the B current within a very few milliamps, and any error is on the safe side.
Power in watts = I squared times R, so plug the value in ohms of each bias resistor into this formula to get the power each resistor must dissipate. I doubt that you'll need more than 2 watt resistors in that circuit, but since I don't know the voltage drop I can't say for sure.
It is hard to find power resistors in the exact values needed, so when replacing a candohm resistor, I'll usually put two or even three 1 watt resistors in series to get as close as I can to the right value and to be sure that each is dissipating no more than about 1/2 or 3/4 watt. Even then they can get pretty hot, so I make sure they are well separated and sticking up on long leads.
The only time I've had a problem doing this is when the resistors are squeezed in tight, as in the 46-350. The candohm resistors use the steel chassis as a heat sink, and I don't know any way to do that when replacing them with modern resistors. I know candohm resistors have a bad reputation, but now I usually won't replace them if they are still OK.
Another advantage to the candohms is that the wire-wound construction provides a little bit of needed inductance to the circuit. Of course if you can find modern power resistors of close to the right value, that problem goes away.
John Honeycutt
