# Electronic – making the DC/DC converter’s capacitor to blow up

boostcapacitordc/dc converteresrpower-dissipation

I'm having some capacitors blown up and I am not sure what's the cause of this. It is definitely NOT OVERVOLTAGE and NOT in WRONG POLARIZATION. Let me introduce the scenario:

I have designed a double cascaded Boost converter using this scheme:

Vout can be obtained from: \\$\ Vout=Vin/(1-D_\max)^2\\$ where \\$D_ \max\\$ is the maximum duty cycle.

I want to step-up an input voltage of 12V into a 100V output voltage. My load is 100Ω, hence it would be dissipating 100W. If I consider no losses (I know I'm being TOO idealist, calm down), the input voltage source will deliver 8.33A

We can split the circuit into two stages, the first stage's ouput is the second stage's input. Here comes my problem:

C1 is blowing up when the voltage accross it reaches aproximately 30V. C1 is rated for 350V and it's a 22uF electrolytic capacitor (radial) 10×12.5mm. I am totally sure the polarization is right.

The second stage's input current should (ideally) be around 3.33A (in order to keep the 100W with 30V for this stage). I know the current might be higher, but it's a good aproximation for this purpose. The switching frequency is 100Khz.

For some reason the cap blows up and I don't really know why. Of course that when this happens the cap (dead) is hot.

May it be an effect of the ESR? This cap has a 0.15 Dissipation Factor at 1kHz. \\$|X_c|= 1/(2*pi*100Khz*22uF) =0.07234Ω \\$
So \\$ESR=0.15*0.07234= 0.01Ω\\$ (DF would also increase for a higher frequency) for C1.

Since L2 is pretty large, I would expect C1 to deliver a pretty constant current equal to the second stange's input current (3.33A) so the power dissipated in ESR is supposed to be around: \\$3.33A^2 * 0.01Ω = 0.11W\\$

Can this make it too hot and explode? I doubt it….

• D1 is a schottky 45V diode
• I tried two different capacitors: 160V 22uF that blown up, and then I tried the 350V 22uF which also blown up.
• Measuring the current in the cap would be difficult due to PCB layout
• Both the first and second MOSFET has a small snubber RC network. I don't think it could cause any problem in C1.

## I am waiting for your ideas!

EDIT n°1=
L1 is pretty large, ripple is only 1% of the rated input current (let's say 100W/12V = 8.33A) so que can assume it's almost like a constant current at the input of stage 1. For stage 2 inductor current ripple is less than 5%, we can also think it's a constant current). When MOSFET 1 is turned ON, around 8.33A goes through it, but when it's turned off, that current (we said "practically constant") would go through D1. We can say current in the capacitor would be \\$ I_{D1} – I_{L2} \\$ . Then we finally find that the peak current in C1 must be in the order of \\$ 8.33A – 3.33A = 5A\\$. Pretty much current! and it would dissipate \\$5A^2 *0.01Ω = 0.25W \\$ … but looks not so much power dissipated in the ESR.

As someone said, I might also consider the internal inductance of the cap, but i think this wouldn't be a cause of power dissipation (we know inductors store energy but don't make it into heat) Anyways, despite of the calculation above was very simplified and it might be a little higher power dissipated, I still wonder if it's enough to make it boil and explode!