Am I on crack here, or do I see a spot above R3 on the PSU in Peter's LM3875 kit for a LED?
I'm using a Avel torodial transformer @ 22+22 V secondaries, so that puts me at what, about 34V rectified at the PSU?
I'm assuming that's what that is.. it'd be awful nice if it was. If that's the case if I were to go with a 3.3v LED with the standard 20-25ma current you'd need a mighty large resistor (about 1.5K/2W) - and from what the online calcuators are telling me, that resistor is going to get mighty warm.
Is there a solution for this? I suppose I could add a small transformer and wire it in parallel with the Avel just to drive the LED, but that seems silly. There's gotta be a better way.
Course, if my understanding is incorrect and the board does NOT accomidate for a LED, well.. where would I add one I wonder? between V+/V- on the output side?
I'm using a Avel torodial transformer @ 22+22 V secondaries, so that puts me at what, about 34V rectified at the PSU?
I'm assuming that's what that is.. it'd be awful nice if it was. If that's the case if I were to go with a 3.3v LED with the standard 20-25ma current you'd need a mighty large resistor (about 1.5K/2W) - and from what the online calcuators are telling me, that resistor is going to get mighty warm.
Is there a solution for this? I suppose I could add a small transformer and wire it in parallel with the Avel just to drive the LED, but that seems silly. There's gotta be a better way.
Course, if my understanding is incorrect and the board does NOT accomidate for a LED, well.. where would I add one I wonder? between V+/V- on the output side?
Keyoke said:
LEDs glow quite well on a tiny fraction of the "spec" current, which is quite often either a maximum or to give the quoted peak brightness (and a few hundred or more mCd is OTT for an indicator light).
5mA will increase the resistor (on 34V) to around 6.2k but reduce its wattage to under 1/4W.
http://led.linear1.org/how-is-led-brightness-related-to-current/
How does wrapping magnet wire around the torroid not generate AC current?
I mean, I understand how it would work - the magnetic field generated by the current passing through the toroid moving the electrons in the wire, but if the torroid is fed from mains.. isn't the current in the torroid fluxuating +/- 115V @ 60Hz? So wouldn't it stand to reason that the .. um.. parasitic(?) current picked up by the additional winding would need to be rectified somehow to produce DC current?
Sorry if I'm being overly dense on this one.
I mean, I understand how it would work - the magnetic field generated by the current passing through the toroid moving the electrons in the wire, but if the torroid is fed from mains.. isn't the current in the torroid fluxuating +/- 115V @ 60Hz? So wouldn't it stand to reason that the .. um.. parasitic(?) current picked up by the additional winding would need to be rectified somehow to produce DC current?
Sorry if I'm being overly dense on this one.
yep, there will be a 60Hz flicker, easily fixed with a diode..which is why I prefer the resitor and a connection to + and - rail of PSU --- this also drains the caps for you on power down.
Its realy fine to use a 1/4W 10k to 20k resistor...
lets say you have a 2V led and a 70V rail to rail transformer. You decide on about 5mA for the LED.
now some of the principles you want ot keep an eye open for is, the when you place resistors in series, they will all have the same amount of current through them, but have voltages = to the portion of the total resistance of the series chain.. only the current part is needed here...
So seeing the LED as resitor (due to the voltage drop over it) it stands to reason the other resistor would have to carry and dissipate the remaining energy. which would be 70-2V=68V, because we want only 3mA in the chain, it would stand to reason the resistor needs to disipate.
Lets say we want to not exceed 0.25W
0.25W / 68V = 0.0037A (3.7mA)
3.8mA is enough for our LED
R = V/I = 68 / .00378 =18.5K ohm
Its realy fine to use a 1/4W 10k to 20k resistor...
lets say you have a 2V led and a 70V rail to rail transformer. You decide on about 5mA for the LED.
now some of the principles you want ot keep an eye open for is, the when you place resistors in series, they will all have the same amount of current through them, but have voltages = to the portion of the total resistance of the series chain.. only the current part is needed here...
So seeing the LED as resitor (due to the voltage drop over it) it stands to reason the other resistor would have to carry and dissipate the remaining energy. which would be 70-2V=68V, because we want only 3mA in the chain, it would stand to reason the resistor needs to disipate.
Lets say we want to not exceed 0.25W
0.25W / 68V = 0.0037A (3.7mA)
3.8mA is enough for our LED
R = V/I = 68 / .00378 =18.5K ohm
Keyoke said:
Heat (in Watts) = Volts x Amps.
If you're aiming to drop 30V from the supply, running a LED at 25mA generates 0.75W in the series resistor (which needs to be 30/0.025= 1.2k); running it at 5mA generates 0.15W in the necessary 6k resistor (or 0.145W in the nearest stock 6.2k one which allows a near-enough 4.8mA to flow).
A 2W resistor doesn't produce 2W, it will stand 2W without damage.
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