28-10-2012, 09:46 AM
(28-10-2012, 09:29 AM)Yorkie Wrote: If I’ve done the sums right, with a 3,000 Ohm field coil, if 45 mA was to be drawn, then the voltage dropped across the coil would be .045 A x 3000 Ohms = 135 Volts. That Wattage dissipated in the coil would be .045 x 135 = 6.075 Watts.
With a 450 Ohm field coil, if 45 mA was being drawn, then the voltage dropped would be .045 x 450 = 20.25 Volts. The Wattage dissipated in the coil would be .045 x 20.25 = 0.91 Watts. (Call it one Watt).
Presumably a resistor in series with the 450 Ohm field coil to increase the total resistance to 3,000 Ohms would be in order? IE, 2550 Ohms. No such thing, so a 2K2 resistor would I assume be OK. The voltage dropped across that resistor would be 135V – 20.25V = 115 Volts, so the Wattage of that series resistor would be 135 x 0.045 mA = 5.15 Watts.
I appreciate that the field coil is of course an inductor – not a resistor, which has the dual role of energising the speaker and acting as a smoothing choke, but it terms of voltage dropped, 3000 Ohms is 3,000 Ohms. If for example - as often happens - someone substituted the speaker for a modern permanent magnet one, they'd dispense with the field coil (the usual reason being that it was OC anyway), and substitute a resistor of a similar value to the field coil across the smoothing/reservoir caps in the power supply.
Don’t know if that makes sense – any thoughts anyone?
Is the radio a 110v or 240v?
Lawrence.








