Continue to Site

# Capacitors @ very low frequency ripple current; deriving RMS from transient

#### kellogs

##### Member
One of the transients I need to withstand takes the shape of an ~200 us spike every 0.5 s. How should I treat this ripple equivalent in my caps ?

1. Treat it like some sort of 5 kHz ripple and somehow account for the rest of these 0.5 s being devoid of any current

2. Calculate an RMS value the usual way and treat it like a 2 Hz ripple current.

3. ?

Second way is quite hard as the sort of MLCC caps I am looking into have their ESR specced at frequencies upwards of 1 kHz. I did derive an ESR value for 2 Hz and it was not good at all with respect to temperature rise.
Naturally I went for film caps - these being ESR-specced from 100 Hz and upwards. They are generally suitable for the kind of current RMS ripple that I need, but they are soooo huge! And a bit expensive. Also there is something bothering me about this approach...

Need to know the current capability of the spike, i.e. what is the source impedance of the spike?.
Without that you can't calculate the ripple of the spike in a capacitor.

Source impedance is 10 ohm, but i have added 100 ohm externally. Ran a simulation, done some shape approximations and came up with

$sqr(I_{rms}) = 0.000865$

Extrapolated an ESR value from the curve at https://ksim3.kemet.com/capacitor-simulation for the R46KN4100JHP1M cap like so:

$a = ln(y_2/y_1) /{ ln (x_2/x_1) } = ln (100/400) / ln(700 / 107) ~= -0.74$

For 2 Hz:

$ESR = 100 mOhm / [(700 Hz/ 350 Hz)^{-0.74}] ~= 7.63 ohm$

And thus:

$P_{rms} = 6.6 mW$

Which gives a small temperature rise on that polypropylene cap. Does it look about right ?

The MLCC's ESR, extrapolating its curve given by its manufacturer, is just huge at 2Hz - in the kOhm range

Last edited: