# Band pass filter question

#### Saviour Muscat

Joined Sep 19, 2014
159
Hello all,
I have a simple task here but I am a little bit confused about amplifier characteristics when saying input impedance @1kHz 1Kohm,
Is it saying simply input of 1KHz signal with signal generator input impedance at 1Kohm?
TY
Saviour

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#### WBahn

Joined Mar 31, 2012
26,398
It's saying that if you put your circuit in a black box with two input terminals sticking out and someone hooks up your circuit as the load to theirs, that at a frequency of 1 kHz it should look to them as if what was in that box was a simple impedance of 1 kΩ.

#### Saviour Muscat

Joined Sep 19, 2014
159
I did the sketch kindly can you correct It?

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#### Saviour Muscat

Joined Sep 19, 2014
159
Please could you provide sketches to be sure that understood well your explanation because my diagram I think is wrong
Thank you again
Saviour

#### Papabravo

Joined Feb 24, 2006
16,473
Please could you provide sketches to be sure that understood well your explanation because my diagram I think is wrong
Thank you again
Saviour
The diagram is correct, but you CANNOT measure it with a DC ohmmeter. It is an AC impeadance with a magnitude of 1KΩ. If you measure the impeadance at another frequency it might be the same or different.

#### Saviour Muscat

Joined Sep 19, 2014
159
If I understood well the total impedance of C1 and R1 should be 1Kohm kindly can you confirm?

#### LvW

Joined Jun 13, 2013
1,279
If I understood well the total impedance of C1 and R1 should be 1Kohm kindly can you confirm?
How would you calculate the "total impedance of C1 and R1..." ?
What does it mean mathematically?

#### Saviour Muscat

Joined Sep 19, 2014
159
Z=root(R1^2+XC1^2)@1kHz am I correct?

#### WBahn

Joined Mar 31, 2012
26,398
Z=root(R1^2+XC1^2)@1kHz am I correct?
That's correct. So that's one of your design constraints. Whatever you pick for R and C have to satisfy that constraint.

#### Saviour Muscat

Joined Sep 19, 2014
159
Many thanks to all
Saviour

#### RBR1317

Joined Nov 13, 2010
649
Z=root(R1^2+XC1^2)@1kHz am I correct?
Your calculation seems based on the assumption that the impedance is calculated through R1 & C1 to ground. What is it about this op-amp circuit that allows that assumption to be true? Note: the same may not be true for other active filter designs, such as Bessel filters.

#### MrAl

Joined Jun 17, 2014
8,359
Hello all,
I have a simple task here but I am a little bit confused about amplifier characteristics when saying input impedance @1kHz 1Kohm,
Is it saying simply input of 1KHz signal with signal generator input impedance at 1Kohm?
TY
Saviour

Hi,

Have you learned how to design filters like this using a Bode plot?
I mean an actual Bode plot as originally suggested by Bode not an actual frequency response curve.
That makes it a lot easier to select the values so i hope you did.
When the frequencies are this spread out that works really well.

#### Saviour Muscat

Joined Sep 19, 2014
159
Hi all,
sorry for the delay because I was at work today
I have drawn the bode plot(with Orcad because LTSpice doesn't have UA741, TL072, and TL082 library) as shown figures attached and I did the calculations to find R1, R2, C1, and C2 which are:1k,100k,7.95uF, and 0.1592nF respectively. Equations used
Z=1K=root(R1^2+XC1^2)@1KHz-----1
100=-R2/R1@40dB------2
wL=1/C1*R1@20Hz------3
wH=1/C2*R2@10KHz------4
I checked my results several times with different opamps and the best(TL072 figure1) results cames out that at Lower cutoff frequency (Figure2) the gain is about 37dB but at Higher cutoff frequency(figure3) at 10KHz the gain is about 36dB, the latter result is normal or cannot be accepted (-3dB)?
TY
Saviour

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#### MrAl

Joined Jun 17, 2014
8,359
That looks good actually. If you went with more accurate values you would see more accurate results that's all. Ultimately this is still a bit of an approximation but it gets very close.