discharging a Psu capacitor ?

[harps]

hello.

i made a 24v regulator circuit using T0220 regulator, a .1uf and a 470 uf cap. I am taking the power from a 40v 400ma dc psu . it all works fine .However when i turn the power off at the wall. the little led I soldered just after the regulator stays on for ages ...and ages...

i tested the psu in the wall by itself . and found when i turn it off the capacitors inside stay charged for many minutes ( maybe hours) until i short the connector and get a very powerful 40v spark on my screw driver.

i wonder can i solder a resistor across the + - of this psu so that it will discharge when I turn it off , would i need a certain type of resistor ( of a certain wattage?)


thanks for any help.

 

[audioguru]

Is the 470uf capacitor at the output of the LM317? Its charge is powering the LED when the power is turned off.Why is its value so high?
The datasheet for the LM317 shows a 1uF output capacitor. The 1uf capacitor will power the LED for only 1/470th as much time.

 

[harps]

hi : )
the 470 uf is before the regulator, but the psu (by itself) that powers this regulator circuit remains charged until I short it.

the 240ac>40dc psu must have some caps inside it too..but i cannot get inside the plastic box as it is sealed. so i thought maybe a resistor across the + - of this psu would help discharge it?

: )

 

[Reloadron]

If you want to you could place a bleeder resistor across the output of the 40 volt supply. However, you will still have your 470 uF cap charged in your 24 volt regulatior circuit illuminating the LED since the 24 volt regulated voltage is apparently unloaded. Just remember that the resistor will always be powered and thus always using some of your power. Calculate the RC time and select a resistor of proper wattage. Also, don't directly short the PSU to discharge that cap as it is really unwise to do so. Caps hate that.

Ron

 

[harps]

thanks
: )
how about a bleeder resistor after the 470uf ? i wonder, how would I calculate select the resistor?

 

[audioguru]

The main filter capacitor at the input of the regulator and the capacitor at the output of the regulator both power the LED when the power is turned off. So of course the LED is still powered for a while when nothing else discharges the capacitors when the power is turned off.
Who cares and why??

 

[harps]

its not just the 470uf cap, there are big caps inside the main psu also. which keep the device on for a long time( not just the led) the whole circuit. if you turn a device off you want it to go off , not stay on for 10 seconds.

 

[audioguru]

its not just the 470uf cap, there are big caps inside the main psu also. which keep the device on for a long time( not just the led) the whole circuit. if you turn a device off you want it to go off , not stay on for 10 seconds.

Then you switch off the voltage to the powered circuit, not to its power supply. But then the power supply is always powered.
My Hi-Fi stereo receiver turns off in about half a second when its power supply input is turned off.

 

[Reloadron]

Since you really don't know the value of the output capacitor(s) on the 40 volt side of your regulator it gets a little hard to calculate a RC time constant. However, if you want to chase this I guess you could get several 1K 2 Watt resistors and start with placing 1 across the 40 volt side and see what happens. Then parallel another. The big downside to this is you have a 40 volt 400 mA supply and whatever you shove in there will load the supply as in wasted power. The same holds true if you start placing bleeders across your 470 uF output filter cap.

Since the thing is unloaded the LED will glow. I agree with audioguru in that I wouldn't worry about it in your case as I have no clue how much power during use you can afford to lose.

Ron

 

[RCinFLA]

The 40 v switching power supply has filter caps on the rectified AC input. With light loads the charge on the high voltage cap will run the switcher for quite a while until the high voltage drops below a level that the switcher can maintain the regulated 40 vdc output.

 

[ronv]

You should be able to get rid of the 470 Ufd. cap on the input to the regulator since the 40 volt supply is aready regulated.. That will help a little. Since you don't know the capacitors inside the 40 volt regulator you can't calculate the time constant to discharge them, so do as Ron suggested and add some load across the 40 volt supply untill you get the desired results. Don't forget when you add more "stuff" besides the LED the load will go up and the time it stays on will go down. Also for every 1K you add across the 40 volt supply will reduce the available current by 40 ma. So your 40 volt 400ma supply is now a 360 ma supply and a 1.6 watt heater.

 

[harps]

thanks : )

probably a silly question.

Am i right in thinking that as this load resistor will be in parallel, the rated wattage is not too important? : ?

(ducks down for cover)

 

[audioguru]

thanks : )

probably a silly question.

Am i right in thinking that as this load resistor will be in parallel, the rated wattage is not too important? : ?

The resistor has 40V across it for the entire time the power supply is turned on so of course its power rating must be calculated and used.
The resistor gets as hot as the voltage times the current or the voltage squared divided by the resistance.

 

[ronv]

Harps,
I'm not sure how you turn the big supply off, but you might consider using a switch that lets you turn off the dc at the same time you turn off the ac.