# How can we control Low pass filter.

#### apollo321j

Joined Dec 30, 2020
23
Hi ,I am studying about Filter Circuits and I cannot find " How to control (adjust) the gain in active low pass filter circuits? " could you explain me ? thank you .

#### ericgibbs

Joined Jan 29, 2010
12,902
hi 321,
As you mention Gain, I assume that a OPA is going to be used as part of the LP filter.??

Do you have a staring point circuit diagram and a LP filter specification.?

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#### apollo321j

Joined Dec 30, 2020
23
Yes thank you for replying ericgibbs there is a sample circuit here and true it has Op-amp .

#### bertus

Joined Apr 5, 2008
21,364
Hello,

Have a look at chapter 3 of the attached PDF.

Bertus

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#### ericgibbs

Joined Jan 29, 2010
12,902
hi 321,
That looks OK for a test run.
What frequency ranges did you have in mind, we should have a design spec.
eg: the frequency point at -3dB and Gain figure.
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#### apollo321j

Joined Dec 30, 2020
23

Hi ericgibbs it's my circuit and bode plot ,I don't understand you clearly .is that what you want ,thank you .

#### ericgibbs

Joined Jan 29, 2010
12,902
hi 321,
Will look over your posted circuit shortly.
As a LP filter, this demo plot is what I would expect to see.
It has a Low Pass bandwidth from 1 thru ~20kHz the -3dB point on the plot
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#### apollo321j

Joined Dec 30, 2020
23
Then eric is my bode plot wrong ? and can you explain more thank you .

#### ericgibbs

Joined Jan 29, 2010
12,902
hi 321,
This is your post #3 circuit, with the damping resistors at [R1/R2]~0.6 giving ~+4db 1 Hz thru #20kHz

Do you have values for the circuit in post #6.?

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#### apollo321j

Joined Dec 30, 2020
23
oh . I am so sorry I write them but pc gived me error then delete them Voltage = 10 Volt, R0= R1 = 1K OHM , Capacitors = 1pikoFarad,R3=5.85 KOHM,R2=10K Ohm .Then , I think I find wrong bode plot my transfer function absurd but I don't know (My transfer function = 1.585/(s^2 + 0.002415 s +1).my question " Design second order active low pass filter with 4 dB gain. Determine cut-off frequency (you are free) and use appropriate circuit elements. Compare the results with the desired paramaters. (cut-off) "

#### ericgibbs

Joined Jan 29, 2010
12,902
hi,
This is your circuit in post #6.
What -3dB point design point did you calculate.??

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Update:

#### apollo321j

Joined Dec 30, 2020
23
Okey I will check it then my 3 db and my -3 dB is approximately View attachment 239465like that.Therefore there is another capacitor in my design .

#### ericgibbs

Joined Jan 29, 2010
12,902
hi,
I guess you know that you should expect a gain of ~+4dB from that type of circuit.
The 5.85k/10k are important.
There are ready calculated Tables for these values.
BTW: Note C2....
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#### apollo321j

Joined Dec 30, 2020
23
No ,I mailed to my teacher he said okey he accept my submission thank you for all helping me ericgibbs.Have a nice day I solve it .

#### ericgibbs

Joined Jan 29, 2010
12,902
hi 321,
When you have time check the maths.
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#### anniel747

Joined Oct 18, 2020
1,041
I never saw polarized resistors before. Is this common now in education?

#### ericgibbs

Joined Jan 29, 2010
12,902
hi 747,
In simulators the current direction in components is specified due to the Net list, so some indication of polarity can avoid mistakes in measuring the current direction in a component.

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@anniel747
R2 has been rotated 180 deg
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R1 N001 0 1k
R2 0 N001 1k
V1 N001 0 10v
.tran 1
.backanno
.end

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#### anniel747

Joined Oct 18, 2020
1,041
hi 747,
In simulators the current direction in components is specified due to the Net list, so some indication of polarity can avoid mistakes in measuring the current direction in a component.

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@anniel747
R2 has been rotated 180 deg
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Ha, ok. It's only a simulator thing.

#### apollo321j

Joined Dec 30, 2020
23
Yes just simulator thing ,normally I use proteus and there is no polarized resistor in there ,I don't use .

#### crutschow

Joined Mar 14, 2008
27,190
I use proteus and there is no polarized resistor in there
All simulated resistors have a current direction polarity, which gives a plus or minus sign when viewing the current in the resistor.
This current polarity direction normally doesn't show on the simulation schematic.