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Old 30th Oct 2019, 12:44 pm   #113
ms660
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Join Date: Apr 2011
Location: Cornwall, UK.
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Default Re: Kolster-Brandes MR10

Quote:
I'm conscious of the simplistic power calculations in my Post which gave 0.77W on FM and 0.74W on AM but I have fitted a 3W resistor which should suffice?
It's all to do with the conduction period of the rectifier and the current during that conduction period that's demanded by the load as well as that demanded by the partially discharged reservoir capacitor which supplied the load when the rectifier wasn't conducting, remember that the conduction period of a rectifier into a capacitor filter (the reservoir capacitor) is much less than the +ve half cycle period.

From that it can be deduced that the current flowing through the rectifier (and its series resistor) will be greater than that of the DC load current and therefore the power dissipated in the series resistor will be much greater than the figure calculated using the value of the DC load current alone.

Rough and ready guide for calculation purposes is to assume that the current flowing through the rectifier is approx. 2.2 times the DC load current.

Eg: DC load current = 0.07 amps, 0.07*2.2 = 0.154 amps, R = 150 ohms therefore Pdiss = 3.5 watts, if it were me I would fit a 5 watt job.

Lawrence.
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