Electronic – Ripple reduction with LC filter


This seems like it must be a common problem but I don't find a clear answer anywhere, maybe my approach is wrong.

I have a DC-DC regulator in my circuit, which has a ripple of 50mV at 200kHz, the current draw is 500mA, voltage output 5V. I want to add a lowpass LC filter to reduce this.

However I struggle at determining which values I should selected, is there a standard approach in determining these values based on the ripple frequency or otherwise, or how should I go about determining these?


Best Answer

I struggle at determining which values I should selected

It's a low pass filter like this: -

enter image description here

And it could produce several spectral responses as shown below: -

enter image description here

What you don't want is a high resonance peak (low values of zeta, \$\zeta\$) because you might generate an output voltage that could exceed the absolute limits of what you are feeding. Consider the scenario where your target load is a 5 V logic circuit that is easily damaged with 6 V and, if the supply to the input of the filter came-up (activated) quickly, it could cause ringing on the output and the first peak could be disastrous: -

enter image description here

So you need to control zeta. Here's what you do. You choose where the cut-off frequency of the filter is best placed - slightly above DC would give great attenuation of ripple but you will find it really expensive to have a 100 henry inductor. So you set your sights more pragmatically and maybe opt for 20 kHz and from this you can determine ripple reduction because the frequency will reduce in amplitude at 40 dB per decade (standard for 2nd order filters).

So, between 20 kHz and 200 kHz you have a 40 dB reduction or 100:1. This takes your ripple from 50 mV to 500 uV. Maybe you might want a little more reduction so choose a lower frequency.

But you still have to control zeta to prevent overshoot and this is done with resistors either in series with the inductor or resistors in parallel with the output capacitor. Clearly resistors in parallel with C are equivalent to the load so you can trade things off if you know the minimum load current.

The bottom line is there are two formulas that can be used to calculate L, C and R and these are: -

enter image description here

This is for R in series with L (there are other formulas for R||C of course).

F = \$\dfrac{1}{2\pi\sqrt{LC}}\$

In the first formula Q = 1/2*zeta and ideally you want to be aiming for a Q of about 0.5 (called critically damped) to prevent overshoot. So how much R can you tolerate in series? Maybe 0.1 ohms? This now gives you an \$\omega_0L\$ of 0.05 ohms. Operating frequency is 20 kHz therefore L = 4 uH.

From this you can calculate C at 15.8 uF.