LED question, SA-9800 project

I am working on this:

http://audiokarma.org/forums/showthread.php?t=73537&highlight=sa-9800

I am a bit of novice when it comes to AC, more of a computer guy by trade The power lamp supply voltage I am taking the AC power from registers at 25v AC. I am then running, in series a 4004 (1.6v drop) rectifier, a white led (3.2v) for the power lamp, three blue leds (3.2v each), the power limiting resistor (can't remember this off the top of my head) and another rectifier (1.6v). They work fine, but I am more concerned with the rest of the circuit.

First off, do I need two rectifier diodes, or just one to prevent reverse voltage? And secondly, am I affecting the power supply in any noticable way that will harm the rest of the functionality of the amp? The original specs are designed just for a 8v drop from a power lamp.

Please advise, and thanks.

dew.

One diode will do it. The small current pulled by the LED's isn't enough to affect the PS in any meaningful way, but there is likely a couple of power resistors that were in series to the original lamps that will have to be taken into consideration when calculating the current limiting resistor for the LED's.

EchoWars said:

... but there is likely a couple of power resistors that were in series to the original lamps that will have to be taken into consideration when calculating the current limiting resistor for the LED's.

Click to expand...

Do I need a diagram to find that out, or is that something I can track down inside? Why would the resistors in series be an issue? I assume it'll throw off the measurements somehow?

Is there an easier way to pull voltage from somewhere to run LED arrays?

I appreciate the hand holding here

dew.

With LEDs all you have to worry about is current. Just measure the voltage drop across the current limiting resistor and figure the current (V=I*R). Most LEDs are good for about 30 mA, though 15-20 mA is better. If it's too high, use a higher value resistor.

Tom Bavis said:

With LEDs all you have to worry about is current. Just measure the voltage drop across the current limiting resistor and figure the current (V=I*R). Most LEDs are good for about 30 mA, though 15-20 mA is better. If it's too high, use a higher value resistor.

Click to expand...

I design LED lamps for 10 mA. Use Ohms law to get R, assuming a 2 volt forward drop across the LED. On AC, avoid PIV greater than their 5 volt limit - it is very low compared to most diodes. On DC don't ever connect the LED the wrong way around! In fact, keep PIV down to 0.6 volts max. with a reverse diode across the LED - you can use virtually any small Si diode, I get protection diodes from dead PC power supplies (along with HV caps and power rectifiers for other projects.)
Cheers,
Roger

"Reclaim, repair, refurbish, rebuild, reuse, recycle"
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