Hi Al, no, I don`t use a 14k resistor, but some 220 nF capacitor, and an extra
resistor, for the peak-currents.
And 2 W resistors are cheap enough, so no problem for the dissipation.
Heah? A decimal DOT ??? Indeed, we use a decimal COMMA, and a dot for
1000-separation.
ok, problem solved, thanks ecerfoglio !
ericgibbs said:
Its because we have no idea of your abilities and where the LED application is going to be used.
ok, indeed, but a small hint could have been my first line in my initial post:
"I am ashamed to ask, but just to check if I am correct with this."
A beginner would not be ashamed; I am already working for over 30 years in electronics,
designing and building. But until now, I haven`t used the seriescapacitor for this purpose
before, and this has to go into a children`s room. But NO problem abt safety, that will be ok!
A decent plastic box completely closed, with only the LEDs shining out, and a normal
mainscable coming out of the top, and will be hanged abt 8 inches from the ceiling.
The stock of this kind of capacitors here, is extreme poor, so hardly possible to try out.
And bcs they cost more than `a few cents` I thought of asking for backup, before I order
the capacitors. (also mind the ordering/shippingcosts)
And if it was for 100s of these things, it would have been worth to order a stock of these,
but sadly, it is for just about 1 (one) lamp... :-(
It must give more light, than a wallpluglamp, how is that called ? but less than a normal
lightbulb. So some twelve hi-bright LEDs, shining onto the ceiling would be an alternative.
If too much light, add an extra capacitor
Maybe some red LEDs shining down from out of the bottom ? To look a bit funny ?
BUT, and that is why the question came up: a seriescapacitor in a.f. situations is also
different calculated:
- 3 dB = 1 / ( 2 x pi x R x C )
If you use Xc and calculating it as voltagedivider you get -1,5 dB at the same frequency.
Kind regards!
Edwin