LC filter in a design, why not an RC filter?

Hello everyone,

I am an electronics/ICT student trying to learn more about analog design. At my university lots of digitl and DSP is taught, just only basic analog design.

I spotted a schematic online (halfway the page) wondering what the purpose of L1&C3 could be? I suppose it is a low pass filter? But why an inductor instead of a resistor? Something about current?

I was thinking that maybe the spinning of a propellor can induce a current in the measuring circuit? And that is why you have to filter out the higher harmonics?

The fact is when I would design such a thing (or trying to ) I would never come up with this? Unless it does not function at all. And if I would determine influences of harmonics I would build an RC (because we did not learn anything about LC-circuits).

Hope someone can extend my knowledge and broaden my view

Thank you for any input.

An LC filter is far better than an RC one - check the 'slope (Db's per octave) to see the advantages.

As previously stated, the inductor is better, it has a higher reactance at higher frequencies and so the LPF has higher attenuation.
Also the inductor has a very low DC resistance.

Now consider what is happening in that circuit.
It looks as though two digital outputs are being used to provide the power for the ANALOG parts of the circuit.
So, a low DC resistance is important to minimise voltage drop when current flows through the inductor.

I guess that the LPF is there to prevent any high frequency noise for the micro controller getting into the analogue circuits.

JimB

It looks as though he is using two output pins paralleled to provide the Analog power supply to the OPA circuits, I dont see any other Analog supply source.

The L/C is filtering the pins output.

Hello there,

An ideal low pass filter would pass all frequencies up to the frequency of interest, then block all frequencies above that. The single RC section low pass filter is a first order filter while the LC filter is a second order filter. The second order filter has a sharper response so the 2nd order low pass filter gets closer to the ideal filter as the order increases.

However, using two RC sections where the R and C of the second section are R2=R1*a and C2=C1/a where a is a large constant, we also see a second order response. So in some cases a double section RC low pass filter can act exactly like an LC filter (actually an RLC filter because there is always some R in series with L).. But the catch here is that with the proper values the RLC filter can have a sharper cutoff with the tradeoff being that the gain will not always be equal to 1 in the passband.

So in short, the LC filter can be made sharper than the dual RC filter, and it can also mimic a bandpass filter to some extent where we can get a very narrow bandwidth if we really wanted that. It would be exactly the same as a bandpass filter, but it can be similar with the right values. On the other hand, a dual RC filter makes a rather poor bandpass filter as compared to the LC.

So when you see an LC filter like that you have to ask if it is being used as a low pass filter or a bandpass filter. The way to tell is from what the application appears to be doing. In this case if the output is going to be a wide range of frequencies then it is probably being used as a low pass filter, but if the output is supposed to be a single frequency then it may be being used as a bandpass filter.

But used as a low pass filter the cutoff can be made much sharper than with a dual RC low pass filter so that makes it better at attenuating unwanted frequencies.

To transform an RLC filter into a filter with the same characteristics as that of a dual RC filter, make the two values as:
C=2*c*r/R
L=c^2*r^2/C

where L, C, and R are values in the RLC filter, and r and c are the values used in the first section of the RC low pass filter. The second section of the RC filter is assumed to have values:
r2=r*a
c2=c/a
where a is a number greater than 10.

Note that we can get the RLC filter to act just like the dual RC filter, but we can not go the other way, we can not always get the dual RC filter to act like the RLC filter.

ericgibbs said:

It looks as though he is using two output pins paralleled to provide the Analog power supply to the OPA circuits, I dont see any other Analog supply source.

The L/C is filtering the pins output.

Click to expand...

Yeah that's the only way that circuit makes sense. It's powered from a 3v cell with no regulator, and the analogue parts are normally powered down. The 2 joined micro pins will supply power to the analogue circuit.

As far as I can see the inductor is only to eliminate the high current surge when the micro turns on the analogue power. The two micro pins would either be on (to take a reading) or off, when not needing to take a reading. At 3v Vdd those pins would not be modulated, so it's not really a "low pass filter" it's a "turn on surge reducer".

Mr RB said:

Yeah that's the only way that circuit makes sense. It's powered from a 3v cell with no regulator, and the analogue parts are normally powered down. The 2 joined micro pins will supply power to the analogue circuit.

As far as I can see the inductor is only to eliminate the high current surge when the micro turns on the analogue power. The two micro pins would either be on (to take a reading) or off, when not needing to take a reading. At 3v Vdd those pins would not be modulated, so it's not really a "low pass filter" it's a "turn on surge reducer".

Click to expand...

There's not really any 'surge' to reduce, more likely it to prevent high frequency noise from the processor getting to the opamps?.

I would suggest the reasoning behind using a choke isn't so much because it's a better filter, but because it exhibits a lower voltage drop across it, and you haven't got any to waste.

Nigel Goodwin said:

There's not really any 'surge' to reduce, more likely it to prevent high frequency noise from the processor getting to the opamps?.

Click to expand...

Or perhaps these pins along with the choke and capacitor are a switching buck power supply for the analogs. Looks like it has a voltage sampling divider right next to it. Why would you have it if you don't switch. Can't understand why it is connected to the right side of the choke, not to ANALOG though.

I agree that it looks like it's part of a switching power supply to generate the ANALOG power for the analog circuits. The microprocessor apparently generates a PWM digital output signal which is converted to DC by L1 and C3. The analog current is likely low enough that the processor digital output can directly deliver enough current for the analog circuits.

samovar said:

Or perhaps these pins along with the choke and capacitor are a switching buck power supply for the analogs. Looks like it has a voltage sampling divider right next to it. Why would you have it if you don't switch. Can't understand why it is connected to the right side of the choke, not to ANALOG though.

Click to expand...

Wouldn't there be a rectifier diode in that case? - similar to how a PICKit programmer works.

thunderbird said:

Hello everyone,

I am an electronics/ICT student trying to learn more about analog design. At my university lots of digitl and DSP is taught, just only basic analog design.

I spotted a schematic online (halfway the page) wondering what the purpose of L1&C3 could be? I suppose it is a low pass filter? But why an inductor instead of a resistor? Something about current?

I was thinking that maybe the spinning of a propellor can induce a current in the measuring circuit? And that is why you have to filter out the higher harmonics?

The fact is when I would design such a thing (or trying to ) I would never come up with this? Unless it does not function at all. And if I would determine influences of harmonics I would build an RC (because we did not learn anything about LC-circuits).

Hope someone can extend my knowledge and broaden my view

Thank you for any input.

Click to expand...

Keeping it simple, A capacitor on it's own smooths out the wobble in the supply working against the resistor. An inductor instead of the resistor prevents a certain about of wobble getting through in the first place. so the cap has less to do.
The two interact to reduce the total wobble in the circuit.The capacitor will reduce the wobble on its own with the resistor by a certain amount. The inductor will also do the same. Combined and the wobble is reduced further.

Nigel Goodwin said:

There's not really any 'surge' to reduce, more likely it to prevent high frequency noise from the processor getting to the opamps?.

I would suggest the reasoning behind using a choke isn't so much because it's a better filter, but because it exhibits a lower voltage drop across it, and you haven't got any to waste.

Click to expand...

There's no high frequency noise as the outputs would not be PWM'ed they are either on or off. PWMing the outputs would make the analogue voltage less than the unregulated 3v Vdd. The analogue circuit uses a LM358 which is barely ok running at 3v, so there is no gain from PWMing and reducing the voltage.

The "surge" would be whatever current is needed for the analogue voltage rail filter caps, or anything that would draw startup current. I've done similar circuit myself with the PIC pins turning something on by supplying the power, instead of using a resistor to reduce the surge (which would decrease the voltage) I used a small inductor.

Mr RB said:

The "surge" would be whatever current is needed for the analogue voltage rail filter caps, or anything that would draw startup current.

Click to expand...

Which is minimal, and presumably no problem at all? - there's only a 0.1uF on the rail, and the internal impedance of the chip would limit any tiny surge.

The switch-mode voltage increase sounds a nicer idea, apart from the lack of a rectifier.

Hi,

I assumed it was a low pass filter too and was used to smooth out the PWM used to drive the 'meter' or other device that will use the output as some sort of signal.

I agree with Nigel in that the output can not be a constant DC value that is higher than the supply output on the pin without a diode. To get a DC boost circuit you have to have a diode.

To get an AC boost circuit you dont need a diode, but rather a constant frequency. You can get a pretty high AC boost too. Thing is though, why would the output 'analog' need an AC signal? I suppose it is possible but i dont see it, at least not until someone points out what that 'analog' output is actually used for. Unless maybe they assume the 'analog' output goes to a fast rectifier diode so the load performs it's own rectification.

hi Al,.

As the project is 'button' powered, the MCU outputs that are used as an 'analog' power source can be switched off when not in use, thus conserving batter power.

Its DC pin powered, not switched/pulsed AC in this application.

Eric

ericgibbs said:

hi Al,.

As the project is 'button' powered, the MCU outputs that are used as an 'analog' power source can be switched off when not in use, thus conserving batter power.

Its DC pin powered, not switched/pulsed AC in this application.

Eric

Click to expand...

That's what we all assumed - but the mention of the potential divider fed from the output was an interesting point.

But it's probably just a battery test facility, with the potential divider switched out to conserve power.

Nigel Goodwin said:

That's what we all assumed - but the mention of the potential divider fed from the output was an interesting point.

But it's probably just a battery test facility, with the potential divider switched out to conserve power.

Click to expand...

Perhaps charging the capacitor takes some time, so they use the divider to figure out when it's charged and and it's Ok to start using analogs. Still would be better on the other side of the choke.

Nigel Goodwin said:

Which is minimal, and presumably no problem at all? - there's only a 0.1uF on the rail, and the internal impedance of the chip would limit any tiny surge.
...

Click to expand...

Yeah, agreed it is minimal, but the datasheet has a spec for max capacitive loading of a pin, on PIC's it is about 50pF.

If the digital pin is directly switching a cap larger than 50pF then technically it should have a resistor (or inductor) to limit surge current and remain within spec.

I've used a similar inductor trick, although I generally put a lot more than 0.1uF on the switched voltage rail (so there is a greater need for the inductor than in this case).

Would anyone who plays with simulators care to do a simulation?.

Be interesting to see.

Nigel Goodwin said:

Would anyone who plays with simulators care to do a simulation?.

Be interesting to see.

Click to expand...

hi Nigel,

Sounds fun, any particular configuration or parameters would you like to see.

Eric