DC Ripple and the 80000 number?

[georgeburns]

I noticed on several threads that people are using the formulas:

(80,000 * I) / V = C [for 60Hz]

and

(100,000 * I) / V = C [for 50Hz]

to determine capacitor requirements.

My question is... where did the magic number "80,000" come from? Is this some electronics standard? or is there some calculation that produces the 80,000? I would think so, as there is a different number for 50 Hz systems.

So how was this number arrived at?

Thanks for your time.

-guy that gets shocked a lot.

[georgeburns]

Well... through a couple of hours or googling... I think I found something closer to the truth.

Capacitor minimum microfarads = (amps required * (1*(2/Hertz from your local system))/(Your ripple maximum * voltage)) * 1,000,000 [to bring it to microfarads]

Allowable ripple for most stepper power supplies is 10% or .1 from what I have read.

So the actual number for 60 Hz should be 83,333 as opposed to 80,000?
1*(2/60hz) = 8333.333333 / max ripple of .1 = 83,333

The number "100,000" for a 50 Hz system works out perfect.

Does this sound right?

Can anyone good at algebra show me how to switch the formula around to get the 80,000? In return I will post a picture of my linesman pliers that I cut a live 20 amp 110V circuit with.

[bunalmis]

I = C * dv/dt

C = I * dt/ dv

DeltaV is a ripple voltage.

T 1/2 of period.

If you want to 1v ripple for 60Hz and 1A;

Delta T=1/(2*60)=8.3 msec.

Delta V=1v

C= 1* 8.3E-3 / 1 = 8.3E-3 Farad = 8300uF

If you want 0.1V ripple you need 83000uF.

This is approximated formula !

0.1V too small ripple for motor drivers. But driver character importand.

For 50V motor driver supply 1V ripple very good and you need only 8300uF.

[Mariss Freimanis]

This is an equation I generated. Doing violence to the differential equation I = C dv/dt (treating it as it it were an algebraic expression) gives:

C = I * dt/dv
I = The maximum power supply current in Amperes
dt = 8.333 milliseconds at 60Hz, 10 milliseconds at 50Hz
dv = 10% of power supply's open circuit DC voltage

Putting it altogether gives:

C = I * (.008 seconds) / 0.1 * Volts

Since C is in Farads, multiply by 1 million to get "uF = 8,000 * I / 0.1 Volts" and reduce the denominator to get:

uF = 80,000 * I / V for 60Hz and uF = 100,000 * I / V for 50Hz

Mariss

[Mariss Freimanis]

I didn't see the earlier 2 posts; I've had this page open for 1 day.:-)

83,333 is the correct number if the cap charges instantly. It doesn't because the source is a rectified sine function. Making the number 80,000:

1) Accounts for the 120Hz period minus the charge time.
2) It is a much easier number to remember.

Mariss

[georgeburns]

Thanks Mariss... loving the 203Vs by the way.

You wrote,

uF = 80,000 * I / V for 60Hz and uF = 100,000 * I / V for 50Hz

Then you wrote...

83,333 is the correct number if the cap charges instantly. It doesn't because the source is a rectified sine function. Making the number 80,000: 1) Accounts for the 120Hz period minus the charge time. 2) It is a much easier number to remember.

So...

If the 80,000 compensates for the 120Hz period minus the charge time, then why is 100,000 left alone?

What is the math for the 120Hz period minus the charge time?

Hope I'm not asking for a lesson in string theory. I really just hate doing/seeing stuff (the power supply is moving the steppers to move the motors) and not understanding how it works. I think it is a mental disorder of some type.

[Mariss Freimanis]

OK; because they are easy numbers. Consider the tolerances for the answer. A 10% ripple voltage isn't written in stone anywhere as being the perfect, exact number. 9% is just as good as 11% in the greater scheme of things. The equation generates the minimum acceptable capacitor value, 3 times that value is the maximum practical size. That's quite a spread.

Mariss

[georgeburns]

Originally Posted by Mariss Freimanis OK; because they are easy numbers. Consider the tolerances for the answer. A 10% ripple voltage isn't written in stone anywhere as being the perfect, exact number. 9% is just as good as 11% in the greater scheme of things. The equation generates the minimum acceptable capacitor value, 3 times that value is the maximum practical size. That's quite a spread. Mariss

Thanks for putting into perspective.

Just a side question... what is the ripple tolerances of your gecko drives and what would be the ill effects of going in either direction of too much or too little ripple if any?

[Mariss Freimanis]

There are no ill effects as far as the drives are concerned. If the ripple voltage is very large, say 20V peak-to-peak on a 50VDC supply (no-load voltage), the motor performance will be limited by the trough voltage of 30VDC (50V - 20V). You may as well have used a 30VDC power supply.

If the capacitor is much bigger, then your wallet is emptier than it should be. Secondly, the power factor for the supply becomes unnecessarily large.

Mariss