RE: [CAD_CAM_EDM_DRO] Bleed resistor AND Led

>For the LED across the CAP what size resistor do I put in series and
>again the math to figure this out?

You want 10mA to 20mA for the LED. Assume the voltage across the LED is
1.5V

Then 33.6V - 1.5V is 31.1V.

R = V/I

33.1V / .015A = 2206 Ohm

The closest standard value is 2K2 ( K means 1000, and electronic
engineers being to
lasy to type 2.2K abbreviate it to 2K2 ).

Now calculate power dissipated in the resistor:

P = I * V

0.015A * 33.1V = 0.497W

So you need a resistor of at least half a watt.

Now you ditch the math and use some gut feel. Resistor dissipation you
give at
least 100% margin, so use a 1W resistor.

Instead of using one 1W resistor use two 0.5W carbon resistors in
parallel, that
way they don't get so hot.

When using two resistors of the same resistance in parallel their
combined resistance is
halved, and combined power handling is doubled.

The formula is

(1 / Rparallel) = ( 1 / R1 ) + ( 1 / R2 ) + .... for as many as you
want.

So you want to use two 4K4 half watt resistors. 4K4 is available but
that is not a standard
value. The closest standard values is 3K9 and 4K7. Both will give you
acceptable results,
or use one of each to get really close.

If you redo the math without taking the voltage drop across the led into
account you'll notice
that the difference is negligible at high voltages.

For a list of standard values and explanations of why we use them do a
search for "E12 resistor"
on www.google.com

Regards,
Nic