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Old 9th October 2010, 02:12 PM   #1
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Default Bleeder Resistor

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
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Old 9th October 2010, 02:25 PM   #2
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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
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Old 9th October 2010, 04:47 PM   #3
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Default Decay times

Hi

I would like it to decay within 5 minutes

Thanks

dh
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Old 10th October 2010, 04:39 AM   #4
RJM1 is offline RJM1  United States
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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.
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Old 11th October 2010, 08:46 AM   #5
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Default Values

Hi

Thanks for the formula and resistor values.

dh
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Old 11th October 2010, 11:43 AM   #6
glennb is offline glennb  Australia
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Quote:
Originally Posted by RJM1 View Post
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%.
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Last edited by glennb; 11th October 2010 at 11:57 AM.
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Old 11th October 2010, 12:24 PM   #7
AndrewT is online now AndrewT  Scotland
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Quote:
Originally Posted by glennb View Post
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%
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Old 11th October 2010, 01:06 PM   #8
RJM1 is offline RJM1  United States
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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)

Last edited by RJM1; 11th October 2010 at 01:21 PM.
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Old 11th October 2010, 11:31 PM   #9
Javin5 is offline Javin5  Switzerland
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Quote:
Originally Posted by RJM1 View Post
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.
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