Re: [CAD_CAM_EDM_DRO] Bleed resistor AND Led

Jay: First, MEASURE the actual voltage you have across your cap with the
power ON. Then shut down, and install a resistor-in-series-with-a-LED across
the DC lines (same as cap terminals). THAT will be your "bleed resistor".
You do NOT need another resistor, as the LED will bleed the charge in
seconds.

A red LED will have nearly 2 VDC across it, and figure on a max. of 20 mA
through it. So, say you measure 34 VDC with power-on, then your resistor
will have (34-2) = 32 volts across it. Using Ohm's law, R=E/I = 32/0.02
=1600 ohm. Use an 1800 ohm R. A 1-watt, since there will be 32 volts across
it. A ½=W would get quite warm, and may-well have a shortened life.

Watch the POLARITY! The notch or flat side of the LED is (-). You can
confirm this with a SINGLE flashlight-cell, no resistor: A LED will glow,
maybe not too brightly, if the correct lead is connected to the + end of the
cell, and the other lead to the (-) end (the "flat end"). (You'd need a bit
of wire, or a "jumper" for this, of course!)

This is for "everyday" LED's! I have seen "new versions" which are called "5
volt LED's". They have built-in current-regulators so they work "just right"
on 5 VDC, with no resistor. BUT, you must make SURE you have EXACTLY that
kind, as 5 VDC will SMOKE an "everyday" LED! And, for your "bleeder"
application, you will STILL need 1500 to 1800 ohm resistor in series, for the
30+ volts!

Lotsa luck! Jan Rowland