# Capacitor Ripple Voltage

1. Sep 22, 2014

### Xyius

1. The problem statement, all variables and given/known data
This homework problem is for my power electronics course I am currently taking.

The network shown to the right (SEE ATTACHED IMAGE) is used to
study the output voltage ripple of particular
types of dc-to-dc converters. Note that $\tilde{i}(t)$
and $\tilde{v}_C(t)$ represent ripple quantities that
have zero average value – they are not
phasors. Sketch the capacitor ripple voltage,
assuming the capacitor is 10 $\mu$F and is …
a. ideal.
b. has an ESR of 0.2 Ohms

2. Relevant equations

Equation 1:
$i(t)=C\frac{d}{dt}v(t)$

Equation 2:
$ESR=R_{lead}+\frac{1}{R_{leakage}(\omega C)^2}$

3. The attempt at a solution

So for part A, I believe all I do is solve Equation 1 for voltage above by integration of a piecewise function that defines the current i(t). Not so bad.

Part B is what is confusing me. If the capacitor now has an associated resistance with it, it can essentially be modeled as a capacitor and resistor in series. To me this means that an impedance with a real and imaginary part is created. But this is confusing because the problem says they are not phasors, and this is apparent by what the solution will be for the voltage for part A (a piecewise function made up of quadratic expressions). I am not sure how to incorporate this resistance in what the ripple voltage will look like.

#### Attached Files:

• ###### prob1EE413.png
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2. Sep 22, 2014

### rude man

In part (a), Vc = (1/C)integral of i dt.
What is the corresponding expression for part (b) for which ESR = R?

3. Sep 22, 2014

### Xyius

That is what is confusing me. That equation is derived in the following way.

$q(t)=C v(t)$
$\frac{d}{dt}q(t)=\frac{d}{dt}C v(t)$
$i(t)=C \frac{d}{dt}v(t)$

When now there is a resistance, I am not sure which ones of these quantities changes. My only thought is to say that when there is a series resistance, it adds a voltage drop of $i(t)R$ So the total voltage drop becomes the following.

$V=\frac{1}{C}q+IR$
$\frac{d}{dt}V=\frac{1}{C}\frac{d}{dt}q+\frac{d}{dt}IR$
$\frac{d}{dt}V=\frac{1}{C}I+\frac{dI}{dt}R$

Integrating both sides with respect to t gives the following.

$V=\frac{1}{C}\int I dt + IR$

Writing it in the problems notation it would be.

$\tilde{v}(t)=\frac{1}{C}\int \tilde{i}(t) dt + \tilde{i}(t)R$

Would this be correct???

4. Sep 22, 2014

### rude man

Yes.
Since you already have the graph for R=0 it should be very easy to add the IR term, right?

5. Sep 22, 2014

### rude man

6. Sep 22, 2014

### Xyius

This is what I got (See attached). The top is the ideal case. Seems correct. For the second part, I simply added 0.2 multiplied by the piece-wise functions I got earlier for the current. Was this the correct way to do this?

#### Attached Files:

• ###### 2prob1EE413.png
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Views:
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7. Sep 22, 2014

### rude man

This may be OK but the areas over and under the curve for one cycle should be the same. That's because the dc (average) content of i(t) = 0 and so the capacitor voltage must also have zero dc component over 1 cycle. And of course so must Ri(t).

In graphing parts (a) and (b) did you have Vc(t=0) = 0? If so, then V(0) = 0 for part (a) and V(0) = -R for part (b).

8. Sep 23, 2014

### Xyius

When I was solving the integral equation I set Vc(t=0) to zero. Hope that was okay.

Thank you for all your help!

9. Sep 23, 2014