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Suggestions on a bleeder resistor

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avdrummerboy

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bleeder resistor

Hi I was wondering if for my power supply output I put on a 100k resistor 1/2 watt that would work? It is 16 V output.
 

Re: bleeder resistor

Actually, the bleeder resistance is connected across the capacitor to discharge it when the supply is turned off. It should carry up to 10% of the rated output current, and helps to stabilize the operation of the supply, eliminating any rises in voltage at very low currents that may occur. For high-voltage (> 50V) supplies, a bleeder resistor is essential to remove the hazard of an unexpected voltage in the filter capacitor when the supply is turned off. The bleeder resistor should be rated to dissipate the necessary power in steady operation.

Regards
 

bleeder resistor

Okay, I just need one that quickly drains the caps so the LEDs don't stay lit after the AC mains power is removed.
 

Re: bleeder resistor

i think that 100k(1/2 watt) would work or i can say it maybe overdesign, since the maximum current drawn to the resistor is just 0.16mA. thus, the power drawn to the resistor is 2.56mW and it is very small compared to the specs of your resistor.
 

Re: bleeder resistor

avdrummerboy said:
Okay, I just need one that quickly drains the caps so the LEDs don't stay lit after the AC mains power is removed.
Click to expand...

I think a 4.7K resistor 1/2W will be enough.

Regards
 

Re: bleeder resistor

why is it called bleeder anyway?
 

bleeder resistor

Usually used in High Voltage power supplies to drain the high voltage of the Capacitors to prevent shock when supply is off. In my case just to drain the caps so the LEDs won't turn on when power is off.
 

Re: bleeder resistor

avdrummerboy,
The time required to bleed off the capacitor charge depends on the RC time constant. T = RC, Where C is the filter capacitance, R is the resistance, T is the time constant which is the time required for the capacitor voltage to be reduced to approx 37% of its fully charged value. The resistor required to discharge the capacitor to a gvien voltag (V) in a specified time (t) is given by
R = -t/[C ln(V/Vi)]
Where
. Vi is the inital (fully charged) capacitor voltage.
. t is the time at which the voltage must be equal to V
. ln means natural logarithm
Regards,
Kral
 

bleeder resistor

Thanks guys, I am probably go with the 4700 1/2 watt idea. Nothing fancy. :)
 

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