# Resistor to discharge capacitor filter (And changin voltage in a power supply)

#### Rufinus

Joined Apr 29, 2020
234
Hello. I open a new thread because I think is a different topic. I have made a power supply with a resounded MOT to get up 12, 18, 24 and 38 volts making a conection every some turns and using a commutator to connect each one.

These voltages are after the filter. For filter I have used 20 X 2200 uf electrolitic capacitors. I'd like to put a resistor to discharge the capacitors once the power supply is off. I don't know if this is common, and how fast it should discharge. Any advice about resistance and power of the resistor?

And another question. When I use the conmutator with the power supply turned on, I suppose there will be no problem going up in the voltage, for example from 12 to 18, but Can I go down? I mean, lets say the conmutator is in 38 volts, and I turn it to 24. The voltage in the capacitors will be higher than the voltage in the brige rectifier. Is that a problem?

Thank you

Best regards

#### Dodgydave

Joined Jun 22, 2012
11,234
Any value from 4K7 to 47K will do to discharge the capacitor.

#### crutschow

Joined Mar 14, 2008
33,977
The voltage in the capacitors will be higher than the voltage in the brige rectifier. Is that a problem?
Not for the bridge.
But you may have to wait until the output settles to the new voltage before attaching a load.

#### MrChips

Joined Oct 2, 2009
30,433
It is not common practice to include a discharge resistor.

If you do want to do this, the calculation is simple.
Time Constant (seconds) = Resistance (ohms) x Capacitance (farads)

In other words,
τ = RC

Time Constant is the time it takes for the voltage on the capacitor to decay to 37% of the initial value.
As a quick reference,

1 Ω x 1 F = 1 s
1 Ω x 1μF = 1μs
1000Ω x 1μF = 1000μs = 1ms
1000Ω x 1000μF = 1 s
1000Ω x 2200μF = 2.2 s

It would take 5 time constants for the voltage to decay to 1% of the initial voltage. In other words, multiply the time constant by 5 to get an estimate of how long it would take to discharge the capacitor.

#### Rufinus

Joined Apr 29, 2020
234

Not for the bridge.
But you may have to wait until the output settles to the new voltage before attaching a load.
Ok, you mean, if I have 35 volts and I connect a load that is rated for 12 volts I have to wait to the voltage goes to 12 (with the help of the discharge resistor) to not damage the load, but just for the load right?

#### crutschow

Joined Mar 14, 2008
33,977
Ok, you mean, if I have 35 volts and I connect a load that is rated for 12 volts I have to wait to the voltage goes to 12 (with the help of the discharge resistor) to not damage the load, but just for the load right?
Right.

#### Pyrex

Joined Feb 16, 2022
236
Hi,
you can use a time delay relay to protect the load. The coil to be fed from bridge AC1 and AC2, and the NO contact inserted in series with the load. When you select voltage, the coil voltage is interrupted for a brief time and relay disconnects load till capacitor is discharged. Time delay is to be set for full or partial discharge.
It's easy to find a time delay relay for voltages 12 to 38VAC ( 80.01) :

https://www.farnell.com/datasheets/2237944.pdf

#### Rufinus

Joined Apr 29, 2020
234
Thank you for all your help!

#### MrAl

Joined Jun 17, 2014
11,250
Hi,

For low voltage stuff you usually don't use one, but at 38v maybe not such a bad idea.
How fast you want it to discharge is up to you i guess. The voltage will drop down to about 37 percent of the max in one time constant, and one time constant is R*C where R is your added resistor and C is the TOTAL capacitance of all the caps in parallel, if they are all in parallel that is. This makes the discharge time easy to calculate as you can get away with t=R*C and that should be safe enough. If you want it to discharge to less than 1 percent of the max voltage, then you use t=5*R*C.