Bleeder Resistor

[davidhugo]

Hi

I have built a 30 v PSU for a chip amp I have built building it has 40,000uf of smoothing.
This bank of capacitors stays charged which is rather dangerous as I am still adjusting the amp.
What value ohms and watts would I need for a bleeder resistor? There’re 4 capacitors in the PSU.
Am I right in thinking I can just use one resistor across the leads going to the amp.

Any help would be very much appreciated

David

 

[richie00boy]

Depends how quickly you want it to decay the voltage. Alternatively string 3 car flasher lamps together and just apply across each rail every time you want to mess about with the internals. When the bulbs go out it's ready

 

[davidhugo]

Decay times

Hi

I would like it to decay within 5 minutes

Thanks

dh

 

[RJM1]

T=RC , T= 5 minutes * 60 seconds = 300 seconds, C= 40000uF = .04F.
300/.04 = R = 7500 ohms. Power dissipation = E^2/R = 30^2/7500 = .12W
If you use a 7.5K ½ W resistor it will discharge your 40000uF of capacitors in 5 minutes.
Yes, just one resistor across the leads going to the amp.

 

[davidhugo]

Values

Hi

Thanks for the formula and resistor values.

dh

 

[glennb]

T=RC , T= 5 minutes * 60 seconds = 300 seconds, C= 40000uF = .04F.
300/.04 = R = 7500 ohms. Power dissipation = E^2/R = 30^2/7500 = .12W
If you use a 7.5K ½ W resistor it will discharge your 40000uF of capacitors in 5 minutes.
Yes, just one resistor across the leads going to the amp.

The T=RC formula is for discharge to about 37% of full voltage. It takes 4 of these time constants to discharge to below 1%, so you could be waiting a while. Try two 5 Watt WW resistors (one across each +V and -ve bank) and run at about 2.5 Watts so they don't get really hot. Using R=E^2/P it gives R=360 Ohms. Using nearest E12 preferred value of 330 Ohms, T=RC=13.2 sec, giving 52.8 sec for discharge to 1%.

 

[AndrewT]

The T=RC formula is for discharge to about 37% of full voltage. It takes 4 of these time constants to discharge to below 1%, so you could be waiting a while.

Hi,
yes one RC time period charges up the capacitor to ~63% of the charging voltage and equally discharges the capacitor down by ~63% of the starting voltage.

It is common to use 5 RC to 7RC periods to approximate to reaching full charge or full discharge.
i.e. 5periods uses 0.37 * 0.37 * 0.37 * 0.37 * 0.37 =~0.7%, 7periods would be ~0.1%

 

[RJM1]

Yes, actually it is 2/3’s of the voltage(66.666%). After 5 minutes starting with 30V and 7.5K you would be at 10V, after 10 minutes you would be at 3.333V and after 15 minutes you would be at 1.111V. If you want it down to 1.111V after 4 minutes and 48 seconds use a 2.4K 1/2W resistor.
(dissipation at 30V=.375W)

 

[Javin5]

Yes, actually it is 2/3’s of the voltage(66.666%).

No. It discarges within RC time to 1/e = 36.78%. If you talk about charging the cap with a resisstor R, then it charges within RC time to 1 - 1/e, which is 63.21%. It is not a fractional number like 2/3.