Band Pass Filter: Calculate C1=C2 for 17.3 kHz

In summary: You have a sign error in the second term on the right side of the equation).And keep in mind that V1 is really just Vout - V1, so you can express the second equation in terms of Vout and V1. That will help you to solve for Vout faster. (I'd rather that you keep working on this than to just post the answer and deny you the satisfaction of solving it yourself! :) )You should solve for Vout, but you'll see that it's very easy to solve for V1 at the same time! :)I'm not sure where the sign error is. I wrote the equation as :(V1-Vin)/R1 = (
  • #1
sticky430
3
0

Homework Statement



29ntq42.jpg


Given: K=17, R1=R2=6,247 Ohm

Calculate the value of C1=C2=C to set the center frequency of the filter to 17.3 kHz.

Homework Equations



w0 = 1/[C*sqr(Rth*KR2)] from lecture notes
R1=Rth

The Attempt at a Solution



w0 = 1/[C*sqr(Rth*K*R2)]
C= 1/[w0*sqr(Rth*K*R2)]
C= 1/[2*pi*17.3k*sqr(6247*17*6247)]
C= 3.57e-10 F

The equation above is from a similar circuit that doesn't have R2. I wasn't sure if it still would work. Looking at the circuit above. It seems like R2 does not impact the function of the circuit.

Any help would be appreciated.

Thanks.
 
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  • #2
sticky430 said:

Homework Statement



29ntq42.jpg


Given: K=17, R1=R2=6,247 Ohm

Calculate the value of C1=C2=C to set the center frequency of the filter to 17.3 kHz.


Homework Equations



w0 = 1/[C*sqr(Rth*KR2)] from lecture notes
R1=Rth


The Attempt at a Solution



w0 = 1/[C*sqr(Rth*K*R2)]
C= 1/[w0*sqr(Rth*K*R2)]
C= 1/[2*pi*17.3k*sqr(6247*17*6247)]
C= 3.57e-10 F

The equation above is from a similar circuit that doesn't have R2. I wasn't sure if it still would work. Looking at the circuit above. It seems like R2 does not impact the function of the circuit.

Any help would be appreciated.

Thanks.

If the equations are for a circuit without an R2, then why is there an R2 in the equations? Maybe I'm being dense.

Do you understand how to write the KCL equations for an opamp circuit with feedback? That's how you can solve for the transver function of this circuit.

EDIT -- Welcome to the PF, BTW!
 
  • #3
The equation uses K*R2 which is equal to R2 in the similar circuit.

No, I don't know how to write KCL equation for opamp circuit with feedback. It's been a very long time since I used it. Can you help me get started?
 
  • #4
sticky430 said:
The equation uses K*R2 which is equal to R2 in the similar circuit.

No, I don't know how to write KCL equation for opamp circuit with feedback. It's been a very long time since I used it. Can you help me get started?

In this case, the equations are a mix of KCL and something else. The theme is to figure out how many unknowns you are dealing with, and to see that you can write the same number of unique equations to help you to solve for those unknowns.

So in this circuit, Vout is an unknown, and (call it) V1 is an unknown. Call the voltage at the intersection of R1 and C2 "V1". You might initially be inclined to also call the V- input to the opamp a variable, but it's not. Why not?

So write 2 equations for these two unknown voltages. The equation for V1 is just a KCL equation, but the equation for Vout is a slight variation. Still very solvable. Does that help?
 
  • #5
berkeman said:
In this case, the equations are a mix of KCL and something else. The theme is to figure out how many unknowns you are dealing with, and to see that you can write the same number of unique equations to help you to solve for those unknowns.

So in this circuit, Vout is an unknown, and (call it) V1 is an unknown. Call the voltage at the intersection of R1 and C2 "V1". You might initially be inclined to also call the V- input to the opamp a variable, but it's not. Why not?

So write 2 equations for these two unknown voltages. The equation for V1 is just a KCL equation, but the equation for Vout is a slight variation. Still very solvable. Does that help?

V- does not need to be called a variable because the V+ is connected to ground thus V- is a virtual ground too. Right?

The KCL equation I got for node V1 is:
KCL1: (V1-Vin)/R1 = (Vout - V1)/(1/sC) + V1/(R2 + (1/sC)).

For Vout node i got:
KCL2: Vout/KR2 + (Vout - V1)/(1/sC) = 0

Are those equation correct? So, now I use KCL2 and solve for Vout? Then substitute it into KCL1 and solve for what?
 
  • #6
sticky430 said:
V- does not need to be called a variable because the V+ is connected to ground thus V- is a virtual ground too. Right?

The KCL equation I got for node V1 is:
KCL1: (V1-Vin)/R1 = (Vout - V1)/(1/sC) + V1/(R2 + (1/sC)).

For Vout node i got:
KCL2: Vout/KR2 + (Vout - V1)/(1/sC) = 0

Are those equation correct? So, now I use KCL2 and solve for Vout? Then substitute it into KCL1 and solve for what?

Correct about V- = 0V.

You have a sign error in your first equation. Write KCL equations as the sum of all currents out of a node = 0.
 

1. What is a band pass filter?

A band pass filter is an electronic circuit that allows a specific range of frequencies to pass through while filtering out all other frequencies. It is commonly used in audio and radio frequency applications to isolate a specific frequency range.

2. How does a band pass filter work?

A band pass filter works by using a combination of capacitors and inductors to create a resonant circuit. This circuit allows a specific range of frequencies to resonate, while all other frequencies are attenuated.

3. Why is it important to calculate C1=C2 for a 17.3 kHz band pass filter?

C1 and C2 are the two capacitors in a band pass filter that determine the cutoff frequency. In order for the filter to work properly and pass the desired frequency range, C1 and C2 must be properly calculated and matched.

4. What factors affect the calculation of C1=C2 for a 17.3 kHz band pass filter?

The calculation of C1=C2 is affected by the desired cutoff frequency, the values of the inductors used in the circuit, and the overall design of the filter. It is important to use precise components and follow the proper calculations to ensure the desired frequency range is achieved.

5. Can I use any value for C1 and C2 in a 17.3 kHz band pass filter?

No, the values of C1 and C2 must be carefully calculated and matched in order for the filter to work properly. Using incorrect values can result in a filter that does not effectively isolate the desired frequency range.

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