# How does it affect power supply operation if filter caps are doubled in size?

Discussion in 'Electronic Basics' started by at, Aug 30, 2004.

1. ### atGuest

I've got a power supply that has 2 470uF 400V filter caps in series. The
power supply puts out 400VDC.

Now if I change the caps to 1000uF 400V, what will happen? Will the output
voltage of the supply change?

(this is just a basic supply that fullwave rectifies AC with diodes, then
comes the filter stage between the supply voltage and ground)

-at

2. ### Tom BiasiGuest

With the supply that you describe the result will most likely be a little
better filtering. This may raise the average output under load but you
mention no regulation anyway. Are you sure the caps are in series? Series
filter caps are not the norm.
Regards,
Tom

3. ### atGuest

Thanks. There are 2 caps in series, I guess it's to give better voltage
handling capacity.

I've got one other question.

Now if I change the diodes into a rectifier tube, that has a max capacitance
in the first filter as 60uF, and after that I place a resistor or a choke,
what is the max capacitance the second filter stage can have (it's after the
resistor/choke)? I've heard people tell me it can be bigger than the first
stage, and I've heard it cannot. So I'm somewhat confused what's the truth.
If I use a resistor, does it isolate the second filter stage from the first?
If I use a choke, does it do the trick?

If you've got a regulated power supply, does that make things different?

-at

4. ### CFoley1064Guest

Subject: How does it affect power supply operation if filter caps are doubled
Good morning. You didn't say if you've got 120VAC or 240 VAC input. Here's
one way you can get "400VDC" from 120VAC using two caps in series (view in

1N4006
o-----o--->|----o-----o+
| +|
120VAC | ---
| C ---
| |
o-----|---------o "400VDC"
| +|
| ---
| C ---
| |
'---|<----o-----o-
1N4006
created by Andy´s ASCII-Circuit v1.24.140803 Beta www.tech-chat.de

As was mentioned in another post, putting 2 470uF caps in series will give you
a net capacitance of 470/2 = 235uF. Doubling up the caps to 1000uF will get
you a net capacitance of 1000/2 = 500uF.

In an unregulated supply, your DC voltage will be dependent on ripple voltage
(each of the 2 caps is recharged 50 or 60 times a second). Doubling the
capacitance will cut the ripple voltage in half, which will result in a higher
DC voltage, and that could cause problems. Look at the load and evaluate
whether it can handle the higher voltage. Sometimes it makes a difference.

Oh, yes. Do be careful here. In this type of unregulated supply, both + and -
are at line voltage potential, which could be dangerous.

Good luck
Chris

5. ### Randy DayGuest

Having the 2 in series gives you an 800V cap.
Otherwise the OP would be running too close
to the limit and be risking a blowout...

6. ### atGuest

It's good to know how it works. If there's a problem with the higher DC
voltage, I'll increase the resistor size in the power supply. That will drop
the voltage some. Or is there a better way?

Don't the filter stages also drop voltage? Or do they increase it? Is it the
resistors between the filter stages that make the voltage drop, and not the
caps?
If I put resistors parallel with the caps (as is sometimes used), the
resistors will also drop some voltage, is this correct?

-at

7. ### Tom BiasiGuest

Hi at,
It would help if you told more about your application for this supply.
You sound as though you have little experience with high voltage supplies,
be careful.
You are describing very old design charistics. If you are working on and old
radio or other old device tell us what it is.
Regards,
Tom

8. ### atGuest

This is just an example supply that's like many used in old tube-based audio
equipments. I know the dangers with high voltages, I'm not going to get
hurt. It's true I don't know much about the theory details, that's why I'm
writing to this basics newsgroup. I would like to know how the capacitors
work so I could see if I could benefit from better filtering. Big cans are
expensive so it's better to ask questions before deciding. I just would like
to know how big capacitors you can put into an tube amp with a tube
rectifier. I would want to get good filtering, so I'm interested to know if
I could put bigger caps in the other filtering positions, than what the tube
rectifier allows for in the first filter position.

-at

9. ### CFoley1064Guest

Subject: Re: How does it affect power supply operation if filter caps are
I believe in your response to another answer you mentioned you've got one of
those "old tyme tube-type" setups. In days of yore, back when "men were men
and women were glad of it" (Three Stooges), they did something like this (view
in fixed font or M\$ Notepad):

1N4006 ___
o-----o--->|----o--|___|-o----------o
| +| R |
120VAC | --- ---
| C --- C ---
| | | "400VDC"
o-----|---------o--------o
| +| |
| --- ---
| C --- C ---
| | ___ |
'---|<----o--|___|-o----------o
1N4006 R
created by Andy´s ASCII-Circuit v1.24.140803 Beta www.tech-chat.de

except they used tubes as rectifiers instead of silicon. You've got an R-C
filter at the output, which reduces ripple voltage as well as lowering the
output voltage somewhat. A good rule of thumb commonly followed was to make
all four caps the same size, but nowhere is it written in stone one way or
another. Do the math, or just try it and see. Just make sure you've got high
enough wattage for your power resistors.

Using a series R as a filter element is much smarter than just using the
resistor to load down the cap. You're just wasting capacitance that way, and
at 400V, uF get pretty expensive. But, if you have an existing layout and
don't want to rewire, it's a safe way to get the voltage back to where it was.
Double the capacitance, double the load, and you have the same output voltage
as the original, just more heat.

Good luck, and thanks for a good question
Chris

10. ### Tom BiasiGuest

"> This is just an example supply that's like many used in old tube-based
audio
Back in the day, we liked the time constant discharging the caps to be at
least three times that of the charging time constant.
What do those terms mean? You should be able to find something and bring