Hi,

I'm attempting to build a bench power supply for myself since I do not have one and I thought I'd build one of those before I try some more complicated projects. I'll be using an LM350T with a DP rotary switch to change the voltages (3.3, 5V, 9V, and 12V).

On the power supply end of things. I hear it is good common practice to use bleeder resistors after the main filter caps. I have never actually done so...but for something like this, I thought it might be a good place to try to implement them.

I've tried some calculators that have the time constant and voltage change etc. BUT, I'm not quite sure what these calculators accomplish. Is the value critical? My low electronics knowledge tells me that I should get a resistor with the least resistance, to avoid wasting too much power, but to get one with a high enough wattage rating so it can dissipate lots of stored power.

I believe I am going to use a 4700uF 25V bulk cap. Would a 0.22Ohm 5W work? I'm sorry I don't know how else to explain it, but I would appreciate any input.

Thanks.

 

My understanding is that a bleeder resistor 'bleeds' off the charge on a filter cap after power is removed. Something between 470K and 2 Meg wired ACROSS the filter cap would do the job, removing charge completely in about 1-5minutes.

 

Hi TheLaw,

Place the bleeder resistor in parallel with the 4700uF capactitor. When the mains are turned off then the bleeder will ensure the power supply voltage goes to zero more quickly (few seconds) when the power supply is not loaded. The quesient current of the power supply circuits will also help to bleed to filter cap, but can be slow and unpredictable.

Time to bleed down to 37% of the voltage is: R(bleeder) x C(4700uf).
Time to bleed down to <1% of the voltage is: R(bleeder) x C(4700uf) x 5.

e.g. 1K(1 Watt) x 4700e-6 = 4.7 seconds.

Regards,
Ifixit

 

Hi TheLaw,

Place the bleeder resistor in parallel with the 4700uF capactitor. When the mains are turned off then the bleeder will ensure the power supply voltage goes to zero more quickly (few seconds) when the power supply is not loaded. The quesient current of the power supply circuits will also help to bleed to filter cap, but can be slow and unpredictable.

Time to bleed down to 37% of the voltage is: R(bleeder) x C(4700uf).
Time to bleed down to <1% of the voltage is: R(bleeder) x C(4700uf) x 5.

e.g. 1K(1 Watt) x 4700e-6 = 4.7 seconds.

Regards,
Ifixit

Thanks. Is 1K a good value? It seems a bit high to be constantly resisting.

And if it is a 2W resistor you would do 1,000ohms x 2? Just wondering if that is what you meant by your equation.

Thank you.

 

My understanding is that a bleeder resistor 'bleeds' off the charge on a filter cap after power is removed. Something between 470K and 2 Meg wired ACROSS the filter cap would do the job, removing charge completely in about 1-5minutes.

Yeah I knew that much. Just confused about the actual calculations.

 

The down side of a bleed resistor is that it uses power all the time. In the example I gave (1000Ω) P=E^2 / R = 25V x 25V / 1000 = .625 Watts. A 1 or 2 Watt rated resistor would do.

The bleed resistor is optional. Do you need it? I use them in my bench supplies because I'm fussy and like to see the panel meter go to zero shortly after I turn the supply off. Othewise it could take minutes.

There are more sophisticated ways to do this. Such as using a transistor for switching in the bleed resistor only when mains power is off.

Regards,
Ifixit

 

The down side of a bleed resistor is that it uses power all the time. In the example I gave (1000Ω) P=E^2 / R = 25V x 25V / 1000 = .625 Watts. A 1 or 2 Watt rated resistor would do.

The bleed resistor is optional. Do you need it? I use them in my bench supplies because I'm fussy and like to see the panel meter go to zero shortly after I turn the supply off. Othewise it could take minutes.

There are more sophisticated ways to do this. Such as using a transistor for switching in the bleed resistor only when mains power is off.

Regards,
Ifixit

Ahh...I think I have an idea on how that would work, but not too sure.

I'm willing to try the transistor method as long as it's not too complicated. Do you have a schematic on hand or a link to what you are refering to?

Thanks.

 

Wish I could have been of more help to you, but large value resistors are not what you had in mind.

Why, if I might ask, do you need the voltage to drop so quickly after power off? The load(if still attached) along with a large value resistor would have you down below 5 volts very quickly. True, those last few volts would take quite a bit longer with my suggested value vs. the 1K resistor, and a standard 1/4 watt would suffice with the large value. IF you needed the voltage to drop faster for some reason an externally connected resistor(alligator clipped) could always be applied.

Complications and details, details, details. I still persist despite my shortcomings in the game of life.

 

TheLaw:

A simple solution might be to use a DPDT power switch. Use one half to switch the mains and the other half to switch in a bleed resistor when the power sw is in the off position.

When I'm using my bench power supply to power an experimental circuit and a component starts to let the smoke out its nice to have a power supply that goes off quickly when frantic fingers hit the off switch..

Have fun,
Ifixit

 

Thanks guys. I guess the switch idea would work well.

I don't need the voltage to drop that quickly, but I just wanted to try it...AND, as you said, if I have something misbehaving and I need to cut the power quickly, the bleeder would be helpful. This is a bench supply. Something is bound to go wrong.

So I'm still interesting in the transistorized method, if you can remember how you'd do that.

Thanks again.