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Old 13th September 2004, 06:41 AM   #1
Prune is offline Prune  Canada
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Default Load sense

I need a circuit that will sense whether a sufficient load is connected to an HV, high power supply (brute force filtering), so it can turn power relay off if load is too low (both to avoid the rising voltage, and as a form of protection so there's no HV on the output power cable when it's disconnected). Any suggestions how to do this?
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Old 13th September 2004, 06:50 AM   #2
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That's what a "bleeder resistor" takes care of.
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Old 13th September 2004, 06:53 AM   #3
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That doesn't help with my second problem: when the HV power cable is unplugged from the amplifier, I want power off -- like a safety interlock. For the overvoltage issue, I'd have to bleed something like 10%, or about 100 W -- that's nasty.
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Old 13th September 2004, 07:19 AM   #4
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What kind of HV and current we talking about? A HAM radio 2KW amp supply with 3,300V @ 1.5 amps usually require that much [100W].

I'll do the math for ya
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Old 13th September 2004, 07:59 AM   #5
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Less than that. The microwave oven transformer I have is 2100 VAC RMS and about 700 mA. I don't need all of it. When I rectify and use a what I've got for filtering, CRCLC (82.5 uF, 300 Ohm, 80 uF, about 4 H, 80 uF), the simulator gives me about 2700 V, in my estimated 400 mA load (in practice it's a bit less, about 2600 V). So about 1 KW.
What if I just wind a current sensing coil around the output cable and use it to drive the relay transistors? How do I calculate such a setup?
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Old 13th September 2004, 09:06 AM   #6
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Bah. You could use a crowbar relay. When turned on, relay pulls in and opens circuit to a low-ohms resistor. When turned off, relay shorts HV output across resistor.

Running an idle load of 10% is typical. 100W seems like a lot, but then, 1kW is a lot too.

Tim
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