Building a low pass active filter

[jb885]

I want to build a first order low pass filter using an inverting opamp with a gain of 10 below the bandwidth, which should be about 1.7kHz. How do I calculate suitable values of the two resistors and 1 capacitor?(Only these three components will be in the filter)?

 

[lebevti]

your transfer function should be something like H(s) = A/(s+ 1/w)

i think u have a cap and res. in parallel in the feedback branch, and another resistor into the inverting input

calculate the transfer function of that configuration and see if the form matches what i said above (i.e., 1 pole)
I'm too lazy to figure it out now

 

[jb885]

I know the transfer function but could you tell me what information I get from it? How do I proceed in choosing the values?

 

[Winterstone]

your transfer function should be something like H(s) = A/(s+ 1/w)
i think u have a cap and res. in parallel in the feedback branch, and another resistor into the inverting input
calculate the transfer function of that configuration and see if the form matches what i said above (i.e., 1 pole)
I'm too lazy to figure it out now

There is a slight error in the general form for the transfer function (typo?):
It is H(s)=Ao/(1+s/wo) with wo=1/tau .

 

[jb885]

I found the transfer function to be R2/(R1+R1jwCR2). How would I proceed?

 

[Winterstone]

I found the transfer function to be R2/(R1+R1jwCR2). How would I proceed?

Divide numerator and denominator by R2 - and you can compare the general form with your specific form.
That means: You can identify the time constant T=1/wo (with wo= 3dB cut-off).

 

[MrAl]

I want to build a first order low pass filter using an inverting opamp with a gain of 10 below the bandwidth, which should be about 1.7kHz. How do I calculate suitable values of the two resistors and 1 capacitor?(Only these three components will be in the filter)?

Hello there,


With R1 being the input resistor and R2 being the feedback resistor and R2 in parallel with the cap, the 3db down point is:
w=1/(R2*C)

and the midband gain is:
A=R2/R1

So for example with a R2=10k feedback and R1=1k input resistor, and a cap of 1uf, we would see the output 3db down at the frequency of about 15.9Hz and somewhat lower than that we'd see a gain of 10. So the gain would be 10 until we get close to 15.9Hz and then it would start falling.

The midband gain is R2/R1 so that's 10k/1k=10.
The cutoff frequency is 1/(2*pi*R2*C)=15.9Hz.

Note this is only a first order LP filter. Usually op amp filters are made to be 2nd order, but that's not mandatory.

 

[jb885]

0000000

 

[jb885]

but why setting the bandwidth at 15.9kHz instead of 15.9Hz won't yield a constant gain of 10?

 

[Winterstone]

but why setting the bandwidth at 15.9kHz instead of 15.9Hz won't yield a constant gain of 10?

Who has made this claim? Just modify the capacitor value.

 

[jb885]

I mean it maintains a gain around 10 (7.something, 8.something) but not 10 with a 10nF capacitor unless it's not the correct one

 

[Winterstone]

I mean it maintains a gain around 10 (7.something, 8.something) but not 10 with a 10nF capacitor unless it's not the correct one

I don't understand (probably, it is my fault) your reply - nevertheless, the gain is defined by the resistor ratio only. And the cut-off frequency is defined by the time constant, which can be modified via C without touchuing the gain value.

 

[jb885]

OK. I understand what you mean. What about measuring the gain at a particular frequency? I derived the voltage transfer function but how do I use it to derive the gain in dB at a particular frequency?

 

[Winterstone]

OK. I understand what you mean. What about measuring the gain at a particular frequency? I derived the voltage transfer function but how do I use it to derive the gain in dB at a particular frequency?

I like to inform you that the transfer function gives you the output-input-ratio (gain) with the angular frequency as a running parameter.
Since this function is complex you have to discriminate between magnitude and phase. Do you need more infos? (I don't think so).

 

[jb885]

No that would be enough! I don't think I would enjoy getting deeper into this! Thanks for everything

 

[MrAl]

Hello again jb885,


If you use an input resistor of 936 ohms and a feedback resistor of 9362 ohms and a capacitor of 0.01uf you'll get a gain of 10 in the passband and the cutoff frequency will be 1.7kHz.

The input impedance to this amp will only be 936 ohms however, so just in case you need a higher input impedance you can use a 9362 ohm input resistor and a 93.6k feedback resistor, and a cap of 0.001uf instead. So to get higher input impedance you multiply the two resistor values by a factor X and divide the cap value by that same value.