BrianGT LM3875 and power LED - how to?

Status
This old topic is closed. If you want to reopen this topic, contact a moderator using the "Report Post" button.
I have Rev C of BrianGT's LM3875 that I put together few years ago. I am now building a newer case for it and would like to have power LED. I have noticed that PSU board has space for LED and resistor. Now, since my electronic knowledge is very limited, how do I know what resistor value do I need to use? My trafo is 42VCT 21V-0- 21V 80VA which gives me about 29V after rectification.

I am also not sure if I can use the LED with PSU modified for CT transformer (I am only using 4 diodes and 1 PSU board).

Any help is appreciated.
 
I'm sure someone on these forums will say I'm full of it, but:

Method 1) Lift the existing resistor, either use it directly or measure it and use a resistor of similar value.

Method 2) Review the schematic and acquire a resistor of similar value.

Method 3) Assuming the LED drops 2V (which can vary a lot), and aiming for 0.020A (20mA) :
Resistance:
29V - 2V = 27V
27V / .020 = 1350 ohms
Power:
27V * 0.02A = ~0.5W

So : ~1350 ohm 1/2W resistor minimum (if you believe my assumptions). A 1W resistor would be mo-betta.
 
Using standard resistor values:

1.5K resistor gives about 18mA current for the LED
2.2K gives about 12mA
3.3K gives about 8mA
4.7K gives about 6mA

More current makes it brighter.

It should work fine with your single PSU board and CT transformer.

Most common LEDs drop a little under 2V and will handle at least 20 mA. I'd aim lower than that though so it's not too bright.

Resistors only cost a couple of cents anyway so it's cheap to experiment.
 
Last edited:
Status
This old topic is closed. If you want to reopen this topic, contact a moderator using the "Report Post" button.