Re: Reason for my electronics 101 question

Posted by KEB on October 26, 2006 at 08:57:01 from (192.91.171.36):

In Reply to: Reason for my electronics 101 question posted by MMB on October 26, 2006 at 06:36:24:

Hi, while Bob's calculations are correct, assuming the full 6 volts is applied to the field resistor is not. With the resistor in the circuit, the field winding and the external resistor are in series. Assuming 4 amps at 7 volts (roughly the charging voltage) with the field grounded, the resistance of the field winding is about 1.75 ohms. Putting that in series with the 3.3 ohm external resistor gives a field resistance of a little over 5 ohms.

At 7 volts and 5 ohms, the current through the series combination of the external resistor and the field winding is about 1.4 amps. 1.4 amps squared x 3.3 ohms gives about 6.5 watts being dissipated in the series resistor.

6.5 watts is pretty consistent with the size of the orignal wire wound resistor. A 10 watt resistor can't dissipate the full 10 watts inside the switch housing (they're rated in free air at a particular temperature) without overheating, but should be able to dissipate 6 or 7 watts just fine.

I'd go ahead and use the 10 watt resistor. If you're concnered, go to a slightly higher value, say 5 ohms. Under these same conditions, the current through the resistor would be a shade over an amp, and the resistor would dissipate about 5 watts. Since the output of the generator is directly proportional to the current through the field winding, the output current from the generator would be reduced by about 40% compared to a 3.3 ohm resistor, but should still maintain the battery without any problems.

Keith
(Electrical Engineer, 30+ years experience)

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