Capacitor smoothing clarification

[MrAl]

Just wanted to know if my calculations were correct for voltage smoothing!

Example: If i wanted to smooth out a 1000Hz,20mA signal from a 12V DC power supply, would this be correct?

(5 X 0.020A) / (12V x 1000Hz) = 0.1 / 12000 = 0.000008F so a 8uF Cap is correct, no???

If all is well and good with the math, the only thing i am confused with is the first number 5! is that for the amount of time constants?

This is just an example question for learning purposes , and i don't really plan to smooth a 1000Hz signal.

Thanks!!!

Hi,

It's not typical to see the DC supply voltage in a formula like this, and that's because the DC level has nothing to do with the smoothing of the ripple. I have to question this formula because of that reason.

It might be better to start with the formula for the change in voltage across a capacitor given a change in current:
dv=I*dt/C

This simply states that the change in voltage 'dv' is equal to the current 'I' times the time period 'dt' divided by the capacitance. This means that in a given amount of time (dt) with a current level of I amps and a capacitance of C Farads we'll see a change in voltage of dv volts.

What this is telling us is that we need a certain amount of capacitance in order to keep the change in voltage below a certain level, and that level is usually what is considered to be an acceptable amount of ripple voltage.

Since the capacitor will be charging and discharging, dt will be equal to one half the period, and dv will be peak to peak volts. Thus if we decide on an acceptable level of ripple and we know the current and the time period we can calculate the needed capacitance by rearranging the formula a little:
C=I*dt/dv

Here we have solved for C. Note the similarity to your original formula, except as expected this new formula does not have the DC voltage level in it but only the 'change' in voltage level, which we are calling the peak to peak ripple voltage.

Now lets look at an example...

Say we deem 1v peak to peak ripple acceptable, with a current of 20ma and frequency of 1kHz. This means the period dt=1/(1000*2)=1/2000 and dv=1. Plugging that in, we get:
C=I*(1/2000)/1=I/2000

Now knowing that I is 0.020 we plug that in and we get:
C=0.020/2000=0.00001 Farads, which is 10uf.

In this case we came up with something close to the original formula, but that was just a coincidence.

Also keep in mind that these kinds of formulas are always approximations as the true quantities and results also depend on the characteristics of the other parts in the system such as the diodes and transformer.

 

[arakaru]

i just watched formulas in website u mentioned in old post and check why 5 is there. thank you for the explanation.i want to know whether i can smooth a voltage in to 24V with 10 A current using 50 Hz AC. so what will be the capacitance i need.

 

[ronv]

It depends on a few things, but mostly on how smooth is smooth.

For a full-wave rectifier:

Vpp= I/2FC


For a half-wave rectification:

Vpp= I/FC


where

Vpp is the peak-to-peak ripple voltage
I is the current in the circuit
f is the frequency of the ac power
C is the capacitance

 

[arakaru]

will the value given by that OK with a big current of 10A

 

[MrAl]

Hi again,

You cant go wrong with the simple equation
dv=i*dt/C

because it gives you an idea what is going on in the circuit and tells you approximately what kind of ripple you will get.

None of these approximations are exact of course so a while back i developed a program to calculate the ripple based on all the more important aspects of the circuit and included spice models for a 1 amp diode and 3 amp diode but allowed the user to enter in their own spice model for the diodes. If anyone is interested i'll try to dig up this old program so you can do some quick and fairly accurate calculations to determine a reasonable size capacitor. This is not an approximation formula but an actual very very fast spice analysis that provides results in much less than 1 second.