Thread: Capacitor Calculation

Hi all,

Just wanted to run my numbers past another person, I am trying to calculate how many farads I need for a capacitor...

F = (A x S) / V

A = Amps, S = Second, V = Volt, F = Farad


My load runs a 15% duty of 1KHz cycles - this was calculated as 0.00015 seconds, so:


(0.3 x 0.00015) / 12 = 0.00000375 Farad (3.75 x 10e-6) = 3.75 microfarads


But then I thought that if the load is only running for 15% of 1KHz then I will need capacitance for 85% of the time, so:

85/15 = 5.667
5.667 x 3.75 = 21.25125 microfarads


Any of this look correct at all?

Originally Posted by fatman57

Hi all,

Just wanted to run my numbers past another person, I am trying to calculate how many farads I need for a capacitor...

F = (A x S) / V

A = Amps, S = Second, V = Volt, F = Farad


My load runs a 15% duty of 1KHz cycles - this was calculated as 0.00015 seconds, so:


(0.3 x 0.00015) / 12 = 0.00000375 Farad (3.75 x 10e-6) = 3.75 microfarads


But then I thought that if the load is only running for 15% of 1KHz then I will need capacitance for 85% of the time, so:

85/15 = 5.667
5.667 x 3.75 = 21.25125 microfarads


Any of this look correct at all?

You left out a key thing: What are you trying to accomplish? Without that important context, it's impossible to say whether or not your calculation is correct.

I have no idea what you are talking about. What is 15% duty of 1KHz cycles?

tk421 -

sorry, I am using PWM for an arduino project. I am powering a 12V 0.3a PC fan and it is making a PWM noise. I added a cap and that made it quieter, now I simply want to find a more accurate value to make my circuit a little more efficient.


Harold14370 -

See code below:

digitalWrite(fan1, HIGH);
delayMicroseconds(150); // Approximately 15% duty cycle @ 1KHz
digitalWrite(fan1, LOW);
delayMicroseconds(1000 - 150); //1000 = 1KHz

The signal is only sent for 150 microseconds. Then there is a pause for 1000-150 microseconds, this creates cycles of 1KHz. Any clearer?

What you are trying to do is to control the power to the fan by turning it on and off, with the power to the fan being on for 15 percent of the time. Naturally, this causes the fan to continually stop and start, hence the noise. When you put a capacitor in parallel with the fan, you are smoothing out the voltage, but you are also defeating the purpose. The fan will draw the same amount of power during the "on" cycle as it did before the capacitor, then it will draw additional power from the capacitor as the capacitor discharges during the "off" cycle. When the capacitor is charging up, the current could be rather high. It is only limited by the internal resistance of the power supply. By turning the capacitor on and off, you could be putting a lot of voltage stress on your power supply. I'd try to find a different way to do it.

Originally Posted by Harold14370

What you are trying to do is to control the power to the fan by turning it on and off, with the power to the fan being on for 15 percent of the time. Naturally, this causes the fan to continually stop and start, hence the noise. When you put a capacitor in parallel with the fan, you are smoothing out the voltage, but you are also defeating the purpose. The fan will draw the same amount of power during the "on" cycle as it did before the capacitor, then it will draw additional power from the capacitor as the capacitor discharges during the "off" cycle. When the capacitor is charging up, the current could be rather high. It is only limited by the internal resistance of the power supply. By turning the capacitor on and off, you could be putting a lot of voltage stress on your power supply. I'd try to find a different way to do it.

Thanks Harold. The fan does slow down or move faster according to how much time it is turned off or on. I see where you are coming from about it taking more power, and I will take heed of the current/voltage issue.

When I add the cap (1500 uF I had lying around) the PWM sound diminishes completely and fan speed works the same as before. I was thinking the cap would smooth the transition so the digital on and off is not as abrupt as before making it quieter. I'm not sure of other ways to do this apart from using a voltage regulator.

Alternatively - would a large cap like the one I am using cause less stress on the power supply? Or, is my best bet to measure what the cap is doing to consumption and build that into the system so the power supply can accommodate it? To do this I would get my observed measurements and simply add this to the total load to be expected from the system - would that work ok?

EDIT:

MY observed measurements are:

With Capacitor:

constant on = 0.28a
15% = 0.07a
50% = 0.2a
75% = 0.26

Without Capacitor:

constant on: 0.28a
15% = 0.02a
50% = 0.11a
75% = 0.18a

So I can see that you are right Harold. I am just not too sure how to factor the power supply into this. And if I choose this route I will still need to figure out how get the right values for my caps!