Re: [CAD_CAM_EDM_DRO] Lightbulbs and Power Resistors?

Chris and Dee wrote:

>So I built my power supply and it outputs 63v. Great! Except that
>once its charged, even if I shut down the p/s, I can't hookup
>anything to the connector I have coming off of the capacitor unless I
>creatively and safely discharge the capacitor (we all know how much
>DC voltage likes to ARC). I also don't have anything but my
>multimeter to tell me if the capacitor still has a charge. I tried
>using a power resistor from Rad shack on the capacitor so that when
>the AC power was removed the cap would discharge somewhat quicker.
>However, they don't have any that are rated high enough (or I'm too
>electronically challenged to know what one to buy or how to hook it
>up). I also wanted to hookup 4 lightbulbs (or LEDS, I don't care) to
>show me when the cap has a charge and for each of my 3 drivers to
>show when power is going to the driver. The only problem with this
>is that I can't find any bulbs rated for that voltage does rad shack
>carry any voltage regulators for that voltage (not to mention that I
>would prefer not to have to solder voltage regulators to each bulb).
>
>Can anybody tell me what resistor to buy and how to hook it up safely
>(I thought it just went across the leads of the cap, but maybe this
>is wrong). Also, any ideas on how to install bulbs or LEDs to tell
>me when the cap and drivers have voltage?
>
>

If you can find a modest light bulb with a 24 V rating (48 would be even
better) and can get the
design current rating for that bulb, you can then figure out a resistor
to use for it.

Let's say you have a 24 V, .020 Amp bulb. Power supply voltage is 63.
So, the series resistor
needs to drop 63-24=39 V at 20 mA. R=V/I = 39/0.020 = 1950 Ohms. So,
you want a 2 K Ohm
resistor. now, for the power, P=I^2 * R. So, P = (.020)^2 * 2000 = .8
Watts. So, a 1 W
resistor will work, but will run somewhat hot. A 2 W resistor would be
good, if you can find it.
Otherwise, four 510 Ohm, 1/2 W resistors in series will do as well.

If you use an LED, the red ones drop 1.6 V at all reasonable currents,
the green ones are about 2 V.
You can run just about all LEDs on 20 mA, and some of the bigger ones
will tolerate 50 mA
pretty well. With the LEDs, the resistors therefore have to dissipate
practically all the power.

If your cap bank is really large, you may find it takes many minutes to
discharge the caps this
way. You could rig a 120 V relay with a normally closed contact to
connect a larger power
resistor across the cap bank when the AC power goes off. You could make
a bigger power
resistor from a string of smaller ones. You can calculate current with
I = V/R.

Jon