still losing capacitor in power circuit

why did i still lose capacitor C5 in this circuit , it is 100uF 25 v and it is working on 50 watt power source connected to the drain of the mosfet , i replaced it with 470uF and it is also heated till boiling . without this capacitor the circuit oscillates and give wrong readings

Have you calculated the ripple current in that cap?

There is not really enough information on your diagram to be able to do it (Switching frequency and duty cycle, operating voltage of the cell).

It would not surprise me if the ripple current was half the average cell current, which could be several amps, and I bet that cap is not rated for that current level.
What is the ESR rating of that cap? Ripple current rating of the cap?

Also you should have several low esr ceramics across that cap (1uF X7R would probably be ok), connected directly between the fet ang ground (very short traces).

Usually when building a switcher at any kind of power level you need several parallel bus caps even if using low esr types.

73 Dan

switching frequency is 100Khz , duty cycle is not constant as it is clear my source is solar cell it has variable output voltage so does the duty cycle. operating voltage from 14 to 20 v
current is ranging from 0 to 3 amps and the ripple is very small may be 0.3 amp.it is electrolytic capacitor and i don't know its ESR

Sketch a waveform of the converter input current and estimate the AC current RMS value. I would expect something around 1 or 1.5 A. Only a low-ESR electrolytic capacitor or multiple capacitors in parallel will be able to stand it.

C5 is the input capacitor.
In a buck converter, the input current is discontinuous. The current's peak to average ratio could easily be be 3:1 (or maybe higher).

You REQUIRE several low-ESR caps in parallel, purchased from reputable capacitor vendors (Nichicon, Elna, Teapo).

what about 3 capacitors in parallel of 100uf ? will it do the job as this is the available here .

If you are talking about standard ("high" ESR) capacitors with an AC current rating of a few 100 mA, you'll better consider 5 or10 capacitors in parallel.

What are the ripple current ratings of those caps? Also what is the ESR@100KHz (Both numbers will be in the data sheet for the capacitor)?

Regards, Dan.

something like this will be ok https://www.altronics.com.au/index.asp?area=item&id=R6124

mshh:

This is one of those instances, where careful reading of the capacitor's datasheet and either measurement or calculation of your worst case ripple current is a must.

You could go the brute-force way, which is to parallel 5 to 10 caps as FvM suggested.

I still didn't know the reason for heat , is it ripple ? no the ripple is normal .is it because of high ESR?

Look at it in most most simple way: Consider the capacitor is perfect and lossless and therefore has no power loss and always runs cold. Now consider the reality that a capacitor also has resistance (ESR) which you can think of as a real resistor in series with the perfect capacitor. The current flowing into the capacitor as it charges and flowing out of it as it discharges is all passing through that resistor. When current flows through a resistance, a voltage is dropped and power is lost. It is that power that causes the heating you observe.

The cure is to use a capacitor constructed with a low ESR, they are available. Lower resistance means less losses in it and lower heat production -OR- you can use several standard ESR capacitors in parallel. Each will dissipate a portion of the heat so individually they keep within a safe temperature range.

The link you posted is rated at 325mA although the web page doesn't state at what frequency. The ESR increases with frequency so the figure quoted may be higher than you would achieve in practice. In any case, always allow a large safety margin as the ESR increases as the capacitor gets older. If FvM's assumption about the current being 1.5A is correct (nobody knows for sure what it is) you need at least 1500/325 = 5 capacitors in parallel to reach safety. Personally, in this situation I would use maybe 8 or more.

Brian.

I still didn't know the reason for heat

Click to expand...

Do you know how to calculate resistive losses? P = I²R

R is in this case the capacitor's ESR and I the RMS value of the current flowing through the capacitor. Assume an converter output current of 3 A, worst case duty cycle of 50% and all input ripple current consumed by the capacitor, the you get 1.5 A rms. Means 5 capacitors if each is rated 0.3 A.