power supply and smoothing capacitor calculations

[Simon]

Yes it's assignment time again and that means I am reading my module material a second time 

So cut a long story short attached is the lesson. I cannot see any calculation given that allows me to calculate the average load voltage without knowing the load current and to know the load current I need to know the load voltage. So yes I have become a dog chasing my own tail.

I hoped the example on page 20 (Q2) solved on page 23 would shed some light on the matter but it seems they too are dogs chasing their own tails as "x"is solved by a formula that has "x"in it and my result of that using MS mathematica is that x=0.....

[Simon]

Correction Vrpp is 7.46 I had entered the calculation wrongly. But is there a simpler way ?

[Jwillis]

Although some would say that this is over simplified ,it works.Because the calculated values will never match the standard values ,a  very high degree of accuracy is irreverent. I use the 70% rule as that the capacitor supplies current to the load approximately 70% of the cycle .

C = capacitance in farads, I = current in amps, ?V = peak-to-peak ripple voltage, F = ripple freq in hZ

So putting it together we get C = 0.7 * I /(?V * F)  for example  for  4 amp at 40volt  C=0.7*4/40*120=0.000583 Farads =583uF

Of course this is for zero load.
Unless you know what your load in ohms is you can't really calculate the ripple under load.

[Simon]

Well I did my calculations and came up with sensible numbers which are not expected to match the simulation results as I understand that the math presented in the module uses approximations.

[fourtytwo42]

There seems to be an awfull lot of maths in that paper! For a lifetime I have used C = (IT)/V or any juxtopision to get at the unknown where V is the p-p ripple voltage, I the load current and T the time between charge pulses.
Maybe if I think about it your problem is related to knowing I for a specific load resistance as I depends upon V that is in turn dependant upon I Generally I resolve that issue by guestimating an initial I if needed however most times I is fixed in a specification somewhere rather than the load being a specified R.

[Simon]

It's a classical case of being asked to solve something no one bothers to solve. as pointed out it can be approximated to an RC circuit and and a calculation made for C in terms of discharge voltage over one cycle. You always end up over calculating C and then rounding it to the nearest E6 value so the ripple can never be more than estimated. Oh well I suppose they need some excuse to pass the dumb students that simply need to change the numbers in the example to solve the assignment.