Resistors will dissipate power, how about a charge pump type circuit,
where a capacitor and rectifiers deliver the power. This will give an
equivalent resistance without dissipating power. Something on the order
of R = 1/(C*60Hz). So 200V/10mA = 20K ohms => C = 0.8uF capacitor.
This will limit current to 10mA at 200V drop. If more power is needed,
a larger cap is needed, which will become impractical if it gets too
large since high voltage caps are large. (or else a higher frequency
will be needed). The problem with resistive current limit is the low
current at the rated voltage. Next step, dc-dc converter that will provide
some current an voltage regulation.
Wayne
At 08:16 PM 9/27/2003, you wrote:
>If you put a 10mA fuse in series, I believe it will burn immediately, the
>initial charging current is much higher than that and may keep for a few
>seconds. If you want to use other transformers and limit the current you can
>add a power resistor in series, a good value may be around 2K ohms 20W.
>
>C.H.Ling
>
>----- Original Message -----
>From: "Piers Hemy" <piers@xxxxxxxx>
>
>> Noted, good point!
>>
>> The label actually* says "Output DC 200V 10mA", so how about we add "and a
>> 10mA fuse" to "with a bridge rectifier"?
<snip>
|