Series capacitors MATH ?

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gary350

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It has been 50 years since I was in school I forgot how to do this math. I need a capacitor for 180 VDC probably 250 VDC will be good but the largest voltage I have is 50 VDC. I have 50 caps brand new 3300 uf each 50 VDC.

Ct = 1 / 9900uf then 1 divided by 9900 = .000101 uf this has to be wrong ?????????

 
Your math is way wrong.

Example:

1/3 + 1/3 + 1/3 + 1/3 = 4/3.

Add the numerators. The denominator stays the same.
 
Right formula, wrong math:

You can do: =1/(1/0.0033F*3) capacitors) is also 1100 uF; answer is Farads

or 1/(1/(3300 uF*3) capacitors) = 1100 uF; This is a trick.

Kinda like using mA/kOhms and mV the answer is in Volts, ohms and amps.

Visitor beat me. You have three, soft of, different methods.
Even I had to go back and fix the parenthesis.
 
For equal value caps, it's even simpler - just divide the value of one cap by the number you have connected in series.
3300 / 3 = 1100.

For different values & a way to do it with simple mental arithmetic, pick a value that's near a common multiple of the values, then count each lower value of cap as (however many times it goes in to the reference value) in series, add all the number from different actual caps, and divide the reference value by that.

eg. 2200 and 3300; use 6600 as a reference, 2200 = 3x, 3300 = 2x, so result is 6600 / 5 = 1320

1 / (1/2200 + 1/3300) = 1320

It works with any number of series items, and the same principles work for resistors in parallel.
 
On a practical note, electrolytic capacitors have leakage current and they have poor tolerance. The leakage will differ between capacitors, so you should always have resistors in parallel with each capacitor to even out the voltages so that the voltages don't get too far different.

Also, when you first turn on, any capacitors that are in series that have a lower value will get a higher voltage across them, so you should have ample voltage in your capacitors. For instance, if the peak voltage is 250 V, use at least six 50 V capacitors.

Buying the correct capacitor is easier.
 
50 caps in series/parallel will be a lot of wasted space. I have about 40 caps 470uf 200VDC that will be a better choice.

The circuit is 120vac to bridge rectifier = 180VDC to filter caps, 20a circuit with current limiting to 10a. = 1800w

Diver300, I was thinking there needs to be a good safety factor of 250v or more. I don't have many high voltage caps.
 
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Kinda another way for resistors in parallel and capacitors in series.

the value is always less than the lowest value.

Two 8 ohm speakers in parallel is 8/2 or 4 ohms. So, two 1000 uF in series is 1000/2 uF.
 
The series capacitor value is the reciprocal of the sum of the reciprocals.
Thus, for your example, it would be 1 / (3/3300) = 1,100µF
 
Is there a formula to determine filter capacitor value for 180vdc 10a or is it better to trial & error test it with the scope or VOM ?

I found a few caps I can use to build a 2740uf test filter to see what value works. I have another 20,000.uf 200v caps that I can use too just for a test to get the best 180v flat line DC.

Not sure what target voltage to buy new cap maybe 300v or 400v.







 
gary350 said:
Is there a formula to determine filter capacitor value for 180vdc 10a or is it better to trial & error test it with the scope or VOM ?
Click to expand...
The value of capacitance needed is determined by how much voltage ripple you can tolerate in the DC output.
You show 2Vac (RMS or PP?).
Is that what you are aiming for?
 
The basic formula is that 1F will discharge at 1V per second at 1A

So with eg. 100Hz full wave rectified, it has 10mS to discharge per half cycle (or somewhat less due to being recharged at each voltage peak).

That means 10,000uF would have something under 1V ripple per amp load, in theory.

So 20,000 uF at 10A = 5V hypothetical ripple (or a bit less; and a bit less again for 60Hz power).

That will be increased by the series resistance (ESR) of the capacitors.
 
What is formula for bleeder resistor to discharges caps. I thought it was T=1/F but there is no F. 1.8W is .010ma at 180vdc

What are recommendation for bleeder resistor value? This thing worries me not being inside a case be careful not to touch it.

 
Capacitor discharge -



You solve the first equation (in red) for R for the final voltage you want in what time period.


Regards, Dana.
 
danadak posted what you need. Basically 5 time constants (R*C) and the cap is nearly discharged. If you want to say discharge it to 25V, a safe value look at the curve. If you would like it discharged in 5 seconds look at the curve/
 
you need some form of inrush current limiting in that circuit, otherwise you will short out the bridge rectifier the first time you plug it in... offline SMPSs use a NTC (negative temperature coefficient) thermistor in series with the incoming AC. also be aware that your "ground" will be floating at about 60-70VAC from the house ground.... you can't use that hot ground as a chassis ground, as it's a safety hazard.
 
I was not expecting 163 vdc. 120 x 1.414 = 169.68 vdc.

I wonder if voltage readings will change with a 10a load?

Tomorrow I check wall outlet voltage to see if it is still 120 volts.

Tomorrow I test circuit with a 12 ohm resistor = 10a load to see if 163v drops and if ripple voltage changes.



 
This is kinda like the basis of most 120/240 switchmode power supplies. 120V full wave rectifies the line voltage and 240 V half wave rectifies. They both then start out with aprox 160 VDC to derive all the other voltages. 120 has those pesky diode drops so GNDs are different by 1.4V.
 
New circuit with 5 caps 425 uf each in parallel. Load resistor drops 164v to 12v in 30 seconds. What size load resistor is typically used I seen to recall reading 10% but I don't remember 10% of what? I feel much safer with these smaller caps and AC ripple is only .04 vac that seems small.

 
My experience with smps with 120/240 range switches, is that the 120VAC would go through a voltage doubler, and the 240 was full wave rectified, each producing ~340 VDC.
 
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