# 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. https://upload.wikimedia.org/wikipedia/commons/e/eb/Lowpass_Filter_LC.svg

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?

Cheers Sam

• TI AN-2162 immediately jumps to mind. Mar 21 '16 at 18:16

I struggle at determining which values I should selected

It's a low pass filter like this: - And it could produce several spectral responses as shown below: - 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: - 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: - 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.

• Andy, thanks for the well written explanation and introduction to quality and damping factors.
– sam
Mar 21 '16 at 22:19
• This leaves me wondering something else, as my load is an active module with current consumption between 125mA and 500mA. I don't have a figure on the load capacitance, is it safe enough to approximate the load as a resistor for the filters purpose? I guess I could estimate the capacitance at around 50uF, any extra capacitance in parallel of the filter should bring the cut-off which seems good, but how may this load capacitance (or varying load resistance) effect the zeta?
– sam
Mar 21 '16 at 22:38
• Extra capacitance just lowers the cut-off frequency and, if you do the math, lowers the peak of the response so there's no great problem with more capacitance. I probably over-egged understanding capacitance in my explanation in order to cover all bases but it aint a big deal. Assume 100mA current consumption, work out the equivalent resistance in series and see if it makes much difference to the calculations. Mar 21 '16 at 23:26
• @Andyaka, What is the solution if load current is not to exceed a certain maximum? If that maximum current corresponds to a load resistor of a value greater than sqrt(L/C), the response would be underdamped. There will be two complex LC poles due to which phase drops suddenly and loop compensation becomes tricky. Mar 22 '16 at 4:03
• @AdityaPatil the Q formula is in my answer and Q is 0.5 for critical dampening. If Q gets smaller due to extra load it's not a big deal. Mar 22 '16 at 12:10