# Calculate Ripple Voltage

Discussion in 'Homework Help' started by RoveyDoveyGrovey, Feb 12, 2011.

1. ### RoveyDoveyGroveyNew Member

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Basically i've been asked to prove that by changing the capacitor value in the circuit (see attached) this equation is true (Simulated in PSpice)

Vripple = (i/c) * t

or

Vripple = i/(cf)

where,
i = current
c = capacitance
t = time (1/freq)

As this is a half wave rectifier with an input voltage of 10vpkpk at 50Hz, with c= 10uF the output goes up to about 9v and then down to 0v in the simulation, so the ripple is about 9v.

I measured the current as about 187mA at its peak, c = 10uF and t = 1/50Hz

stick that in the equation and Vripple = 374v I think not!!

The only thing I can think of is that im measuring the current in the wrong place, but i've tried everywhere and I still get a stupid answer.

Obviously if you rearrange the equation,

Vripple*cf = i

9v*10uF*50Hz = 4.5mA

I cant find 4.5mA anywhere in the circuit.

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2. ### RCinFLAWell-Known Member

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Not sure how far you are into calculus but couple of equations that apply here.

First is decay of voltage for the R-C output which will follows Vout = Vpk*(exp^-(t/RC)).
Second equation is current through cap which follows I = C * delta V / delta t

Vpk will be the peak of the input sinewave (minus diode forward conduction voltage drop).
The delta time will be time between voltage peaks that recharge the capacitor.

For small amount of ripple voltage you can assume capacitor current is constant, therefore use I = CdV/dt, rearranging, dV= dt*I/C.
where I is the current consumed by load resistor.
For half wave rectifier the delta t will the 1/f so you can write equation as dV = (1/f) * I /C, or ripple V = I/(f*C)

With small amount of ripple voltage the current I through load resistor will be Vpk/Rload.

The above equation loses accuracy at large ripple because the RC discharge will follow an exponential dropoff in voltage, not the linear approximation for small amounts of ripple voltage.

If you want to get real picky, at higher ripple the delta time of decay will be shortened as the decay ceases when the rising sinewave voltage cause diode to start conduction before the time where the sinewave peaks.

Last edited: Feb 12, 2011
3. ### MikeMlWell-Known MemberMost Helpful Member

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The peak current through the diode is limited only by the source resistance (zero in your ideal case), the di/dt slope of the diode, and the effective series resistance of the filter capacitor (zero in you case). Hint: it will be high, for a short time.

The capacitor will charge to the peak of the voltage source minus the forward drop of the diode.

The capacitor will discharge into the resistor(with a classic exponential discharge curve) until the next positive peak, about 20ms later.

Here is what LTSpice shows as CF varies from 10uF to 1000uF:

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5. ### MrAlWell-Known MemberMost Helpful Member

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Hello there,

Whenever we use an equation in electronics we always have to remember that the mathematics is an abstract concept which we intend to apply to a circuit. If we dont apply it correctly it wont work at all, and part of that means understanding the limits that define the validity of the equation. There are always limits, and sometimes we wont be told beforehand so we have to investigate a little and see what we find.
In this application you want to use an equation that approximates the ripple voltage that will appear across the cap, and you have calculated a value that is greater than the supply voltage in fact. What this means is that we certainly hit some sort of limit where the equation doesnt work anymore.
If you try to visualize what is happening in the circuit you'll notice that the RC time constant is very small. So small that the cap does not even work as a filter very well and instead it discharges faster than the input sine wave falls. That means the sine wave falls and the cap voltage simply follows it, so the ripple isnt even worth mentioning yet because the circuit really isnt working properly yet. Once we get within the limits of the equation (and the way the circuit should work) we'll start to see values that work out a little better in real life, although they wont be perfect unless even tighter limits are imposed.

With the above in mind, raise the value of the cap to the point where the RC time constant is say at least 10 times the input period, then try the equation again.

Last edited: Feb 12, 2011
6. ### RoveyDoveyGroveyNew Member

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Ok thanks for all the replies, i've done a few more calculations with higher value caps and the equation just about works but it's pretty inaccurate, hopefully this is what he wanted us to find.

7. ### MrAlWell-Known MemberMost Helpful Member

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Hello again,

Yes it is just an estimate, and not very good sometimes. It's quite difficult to calculate this kind of thing with high accuracy because even though it seems simple there is a lot involved, like when the diode conducts and when it doesnt, and how much resistance it offers to the circuit when it is on.
The best way these days is to run a simulation. I had this same problem a while back with a full wave rectifier, but the simulation ran kind of slowish and i wanted it to be faster so i wrote a program dedicated to solving the full wave bridge circuit with diodes, series resistance, and capacitors with esr. It's quite fast, but unfortunately it was not written to do half wave because that is fairly rare these days. Maybe a few modifications would do it.