Question on LC filtering

Thread Starter

cmartinez

Joined Jan 17, 2007
6,461
I've successfully been using caps to filter transients going into digital circuitry in a configuration such as this:

Image00001.jpg
The way I understand it, the caps are there to provide current whenever the supply falters. The smaller cap is supposed to have a faster response than the bigger one, and it's there to tend to very short transients. The bigger cap takes care of the longer lasting ones. The diode is there to prevent the current supplied by the caps from going back into the power supply.

But then @ScottWang posted this diagram:

Untitled.png

My question is (and please pardon my ignorance, some of the basics are still not entirely clear to me) what are the inductors there for? Is it because the caps are there to provide current when the voltage falters, whilst the inductors provide voltage when current falters? Or am I saying nonsense because voltage and current are always interdependent?

I've been thinking about simulating this circuit and play with it just to see what happens. But then again the simulation won't be able to explain to me the why of things in plain english.
 

Papabravo

Joined Feb 24, 2006
12,389
The series L's provide a low pass filtering effect in conjunction with the C's. The Impedance of a series L increases with frequency, so fast transients will be attenuated before they even get to the C's which will shunt them to GND or supply. The next level would be a pi-filter consisting of a shunt C, a series L, followed by a shunt C. This type of pi-filter is very common on high output current supplies.

https://electronicscoach.com/pi-filter.html The English here is a bit confusing
https://www.electronics-notes.com/articles/radio/rf-filters/constant-k-simple-low-pass-lc-rf-filter-design-calculations.php
 
Last edited:

rsjsouza

Joined Apr 21, 2014
201
You instinctively know that capacitors will filter high voltages - in other words, they resist to voltage variations from the surge.

Now, apply the same intuition to inductors, but instead of voltage you use current.

When a surge comes, an increase in voltage (where the capacitors will resist) will also come with an increase in current (where the inductors will also resist). Therefore, they will be working together to avoid damage.
 

rsjsouza

Joined Apr 21, 2014
201
... and by impedance, you're saying that they act as variable resistors that become active only when fast changes in voltage over time happens?
They don't "become active only when..." They are always active, but their impedance ( "resistance" ) is too small at lower frequencies to influence the circuit. When a higher frequency variation (surge) happens, their impedance becomes very high across its terminals.
 

Papabravo

Joined Feb 24, 2006
12,389
... and by impedance, you're saying that they act as variable resistors that become active only when fast changes in voltage over time happens?
Saying variable resistors is perhaps a bit simplistic. The big difference is that impedances don't consume power the way a resistance does.

\[Z_L\;=\;j\omega L\]
where
\[\omega\;=\; 2\pi f\]
 

Thread Starter

cmartinez

Joined Jan 17, 2007
6,461
Saying variable resistors is perhaps a bit simplistic. The big difference is that impedances don't consume power the way a resistance does.

\[Z_L\;=\;j\omega L\]
where
\[\omega\;=\; 2\pi f\]
But heat (as is the case with resistors) will also be generated whenever an inductor suppresses transients, right?
 

OBW0549

Joined Mar 2, 2015
2,982
My question is (and please pardon my ignorance, some of the basics are still not entirely clear to me) what are the inductors there for? Is it because the caps are there to provide current when the voltage falters, whilst the inductors provide voltage when current falters?
Serving as a charge reservoir to maintain voltage during transient dropouts is only one thing the capacitors are there for; they also serve to filter out high-frequency noise on the power by shunting it to ground through their low impedance at high frequency. The series inductors shown in the second diagram aid this noise reduction because of their high impedance at high frequencies.
 

Thread Starter

cmartinez

Joined Jan 17, 2007
6,461
Another question. Why is there a second inductor in Scott's diagram splitting the ground line? Wouldn't it be best in that case to have a single ground connection (the one on the left of the circuit) and do away with the second ground connection on the right? Something like this:

Untitled.png
 

Papabravo

Joined Feb 24, 2006
12,389
But heat (as is the case with resistors) will also be generated whenever an inductor suppresses transients, right?
For an ideal inductor the answer is no. A real inductor has a small series resistance that will dissipate some heat according to:

\[P_D \;=\; i^{2}R_{ser}\]

For a series of low duty cycle fast transients, such as from an SMPS, this would be a fairly minor consideration.
 

Papabravo

Joined Feb 24, 2006
12,389
Another question. Why is there a second inductor in Scott's diagram splitting the ground line? Wouldn't it be best in that case to have a single ground connection (the one on the left of the circuit) and do away with the second ground connection on the right? Something like this:

It is a defensive strategy if you don't know the source of the disturbance. Something yanking your ground around can be just as upsetting as something doing the same thing to Vcc. You should use two separate ground symbols in this case: call them power GND and Signal GND. Power GND enters your board at ONE PLACE ONLY. Everything else on the board is signal GND.
 

OBW0549

Joined Mar 2, 2015
2,982
Another question. Why is there a second inductor in Scott's diagram splitting the ground line?
Presumably, it's there because the external system ground has noise on it that needs to be rejected. Sometimes you see both Vdd and GND with series inductors, most of the time not. There is no universal solution, and no single "right" way.
 

crutschow

Joined Mar 14, 2008
23,314
To summarize, the impedance (not resistance) of an inductor goes up with frequency, and the impedance of a capacitor goes down with frequency.
So at DC the impedance of an ideal inductor is zero, and at an infinite high frequency the impedance of an ideal capacitor is zero.
 

Papabravo

Joined Feb 24, 2006
12,389
To summarize, the impedance (not resistance) of an inductor goes up with frequency, and the impedance of a capacitor goes down with frequency.
So at DC the impedance of an ideal inductor is zero, and at an infinite high frequency the impedance of an ideal capacitor is zero.
Yup, that is the way I see it.
 

Thread Starter

cmartinez

Joined Jan 17, 2007
6,461
To summarize, the impedance (not resistance) of an inductor goes up with frequency, and the impedance of a capacitor goes down with frequency.
So at DC the impedance of an ideal inductor is zero, and at an infinite high frequency the impedance of an ideal capacitor is zero.
man ... too bad there's not a "double like" button in this place :) ... many thanks!
 
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