Quick LED Question

[surfsup]

Just want to make sure this is right. To use the LED as an indicator, to tap off the 6.3VAC heater secondaries, and use a diode, LED, R in series. red LED is 2V so:

6.3 rectified here would be 6.3*0.7 = 4.4VDC
4.4-2V = 2.4V
2.4V/0.02A = 120R resistor

[img:720:366]http://chicagocadcam.com/ChrisHahn/sche ... ircuit.jpg[/img]

[martin manning]

The peak rectified voltage is 6.3 x 1.414 - 0.6, so about 8.3V. Assuming 2V drop for the LED leaves 6.3V. You want to limit current to maybe 10mA, so 6.4/0.010 = 630R

[surfsup]

martin, thanks. I know you are correct, but I thought I was only rectifying half the cycle with the one diode? (which is why I thought I needed 20mA - though I'm probably confusing myself trying to be smart about it)

[martin manning]

It's the peak current that you need to control, though, so half wave or full it's the same. Personally I like to keep the brightness down, so I'd probably go to 1k or more.

[surfsup]

Ok thank you martin.

[surfsup]

One more question, if I ran the resistor to ground (the two 100R ground), Would my calculation be correct then?

3.15*1.4 = 4.4VDC, etc...

[ToneMerc]

It's the peak current that you need to control, though, so half wave or full it's the same. Personally I like to keep the brightness down, so I'd probably go to 1k or more.

I like use a 330-390 ohm resistor in conjunction with a diffused LED that has an mcd rating of 15 or lower.

TM

[ToneMerc]

One more question, if I ran the resistor to ground (the two 100R ground), Would my calculation be correct then?

3.15*1.4 = 4.4VDC, etc...

Yes good enough for gov't work. At an assumed 3.15VAC on each filament side, each 100 ohm resistor will load the each half an additional 32mA(64mA total) which is nil in the whole scheme of total current drawn upon the filament supply winding.

TM