How to reduce DC a bit without resistor or capacitor? Is there a simple solution?

[stevenmahoney]

I have a simple schematics with a number of LED drivers. LED strings have a certain number of LEDs that I cannot change. In order for LEDs to operate properly, I have to feed them with 108V and 1.3A current (combined).
If I rectify 120V after diode bridge voltage increases to approximately 168V. I have excess of 32 volts (28V drops on LED drivers).

I can use a resistor or a capacitor (on AC side) to get rid off excess of 32 volts, but both are bulky (and in case of the resistor hot) solutions.

Is there any other smarter way to solve this problem? Any ideas will be appreciated.

Steven

 

[(*steve*)]

It's a terrible idea, but you could use a capacitor in the live lead that has a reactance of 25 ohms at 60Hz.

C = 1/(2*pi*f*Z) = 1/9425
= 106uF

So a mains rated 106uF (non-polarised) capacitor would work to replace the 25 ohm 50 watt resistor.

Get someone to double-check my math, and choose a LOWER value capacitor to increase the reactance.

HOWEVER: A series resistor is still a good idea, as is a resistor across the cap. Switching and transients can cause higher currents/voltages to be seen by the LED/regulator string. And this is partially why it's a horrible solution. (also the cap will be large and potentially not cheap)

However there are a lot of drawbacks...

 

[stevenmahoney]

*steve*,

I tried that, but capacitors turn out to be bulky. That is why I was looking for a different solution.

And your math is correct.

 

[(*steve*)]

OK, find a transformer with a mains primary and a secondary rated at around 32V 1.5A (or greater current).

place the mains winding across the mains, and wire the secondary in series with the live feed to your circuit with the transformer's winding arranged to be out of phase with the mains.

 

[duke37]

If flicker is not a problem, then C1 could be disconnected to give a lower average voltage

 

[stevenmahoney]

Thanks, tried that. It did not work...

 

[CDRIVE]

OK, find a transformer with a mains primary and a secondary rated at around 32V 1.5A (or greater current).

place the mains winding across the mains, and wire the secondary in series with the live feed to your circuit with the transformer's winding arranged to be out of phase with the mains.

Steve, that's just toooooo clever!! It brought a smile to my face. It's been a crappy month so I needed to smile.

Chris

 

[CDRIVE]

[/QUOTE]

Steven, I forgot to mention that a bridge connected to the mains is not a good or safe idea. Your schematic indicates that the negative end of your bridge is grounded. If that's earth ground you can't do that. If you do the diode at 10:00 will pop like a fire cracker! This is because circuit ground in your circuit will swing to -168V every negative half cycle with respect to earth ground.

Chris