My Pearl One. Some questions

[nar]

Originally Posted by blackdod

Thanks Nar

Ok I'm clear that I should not use the Pad 4.

I take your advice and bring out a led + resistance from the V+ and Ground pads on the PSU.

For about 45 volts. Should I use a very large resistance?

Thanks

Depends if you use a high efficency one or standard.
Range between 10k and 68k should be ok. For my high eff white ones I use about 137k 0,25W on 48V

Best,

nAr

[blackdod]

Nice Nar !!!.

OK, I have some LED High Efficiency. I will test with that value of resistance.

Regards

[blackdod]

Hello guys!

I added a 10K resistor to the LED and the result is very good, but when I turn off the LED takes a long time off. I guess adding greater resistance than take off.

I've been testing the masses and how best to go is as I have now (safety Earth).

On the other hand, wanted to ask if the voltage on Q8-12 zone is crucial. I have a voltage of 0.1 V in this area.

Regards

[Zen Mod]

nope

when LED is lit for long time , that means its connected across big cap
if you put bigger resistor , it will stay lit for even longer time
either live with it , or feed LED from AC - put diode and resistor in series

[blackdod]

Thanks Zen,

Yes, in fact I'll have to live with it or use a diode more resistance (AC).
I am seriously thinking to change it to AC, and add a small capacitor



Regards

[Zen Mod]

you don't need capacitor

tell me AC voltage you have , which LED you want to use and I'll do calc for you

[blackdod]

Originally Posted by Zen Mod

you don't need capacitor

tell me AC voltage you have , which LED you want to use and I'll do calc for you

Thanks Zen

I can use +-37 or +-18 volt DC and the LED is a WP7104QBC/G (3.3 volt/20mA)

I would like the LED turns off slowly fod this I was think on a cap

Regards

[Zen Mod]

say that you're going to use +/-18 ; that's 36Vac ;

say that you're going for max 3mA (plenty for any signal LED)

say that you're going for just one diode in series - that's half rectifier , so you'll have (roughly , but precise enough for purpose ) 36/2 x 1.41 volts DC after diode , with modest cap ; say that 10uF/50V is sufficient

so - we have around 25Vdc

R=U/I
R=(25-3,3)/0,003
R=21,7/0,003
R=7233 Ohms

use first standard value - say 7K5

without recalc of current with altered value , we assume that we still have near 3mA

so - dissipation in resistor is :
P=I^2 x R = 0,003^2 x 7500 = 67mW , so any small metal film resistor is good enough ( factor 3 is always min in my book)

so - secondary - just one diode - cap for smoothing - resistor - LED

easypeasy

if LED is still blinding ya from listening position , you can solve that in 3 ways :

1- turn case for 90 degs , so you can't see LED
2- glue something diffuse across LED (foil or something)
3 - increase value of resitor

[blackdod]

Thanks for your good reply Zen.

I keep your post, since it is very interesting and helpful.

I used the 18 volt (DC) and added thanks to your recommendation a cap 10 uF but I used a 10K resistor.

According to my calculations, the result would be: (18-3.3)/0.003= 4900 Ohm; 5K

The result is very good, turning off with a slow.