Dear All

Please calculate the cutoff frequency of this circuit. I think the R2 will not depend on the cutoff frequency, but it depends on the other parameter such as pass band gain and quality factor.

 

Can you tell more about why you struggle. You have drawn your figure in some sort of ecad system. Do this tool have a simulator you can use for simulation?

 

Yes I have drown this circuit in LT SPICE. I have the option for simulation, I can get the cutoff frequency by simulating the circuit, but I need the mathematical steps to solve this. I know the Simple LPF cutoff frequency ( 1/(2*pi*R*C)). How R2 is depends on this circuit characteristics.

 

You have to find the transfer function. This is equal to Vout/Vin. In this case I think nodal analysis is the essayist way to solve this problem. Are you familiar with this form of circuit analysis?

 

A clue: The value of R2 will drastically affect the cutoff frequency, it is as important as R1.

 

Are you planning on using the classic 3db level? This is typically ½ the input signal voltage. This question is more important than it looks, because you have a voltage divider whose voltage out is entirely dependent on frequency, and you need to set a point where this magic level (3db) happens.

 

Combined impedance of C1 and R2 is product/sum, thus:

(R2/jwC)/(R2+1/jwC) = R2/(1+jwCR2).

Then use resistive divider equation to find transfer function:

[R2/(1+jwCR2)] / ([R2/(1+jwCR2)] + R1) = R2/(R2+R1(1+jwCR2))

cut off frequency is when denominator real = denominator imaginary

ie when (R2 + R1) = jwCR1R2 (expanding out the denominator)

If R1 = R2 = 1k and C = 1uF

2k = 2pi x f x 1uF x 1k x 1k (w = 2pi x f)

318 Hz

As a rough check, your 2 1k resistors are in parallel with a voltage source, so the cut off freq is the same as a single RC filter with a 500 Ohm resistor and a 1uF cap - 318 Hz according to my Casio..

is this what Linear tech say...?

Interested to know..

Bill Naylor

<snip>
Electronic Kits for Education and Fun

 

Are you planning on using the classic 3db level? This is typically ½ the input signal voltage.

Is not the 3 Db level equal to 1/square root(2) or approximately 0.707

 

-3dB is half the output power or 0.707 times the voltage.

 

Sheesh. I stepped in that one.

 

318 Hz

As a rough check, your 2 1k resistors are in parallel with a voltage source, so the cut off freq is the same as a single RC filter with a 500 Ohm resistor and a 1uF cap - 318 Hz according to my Casio..

is this what Linear tech say...?

Interested to know..

Bill Naylor

<snip>
Electronic Kits for Education and Fun

An LT Spice IV simulation of this circuit agrees precisely with your estimate.

 

Yes I need 3dB bandwidth. The 3dB bandwidth cutoff frequency is my requirement.

 

If the circuit is in cutoff frequency how we can write (R1+R2) = 2Pi*R1R2*f ?

Can you please explain me , Bill Naylor

 

The LT-SPICE simulation tool is showing the Cutoff frequency is = 314.534Hz. this value is based on the 3dB reference point. I have attached the file for the reference. It shows different value than 318hz.

 

The cut-off frequency is equal
Vo/Vin=1/√2=0.7071067812=-3.010299957dB
And for this circuit is equal
Fc=1/ ( 2*PI*(R1||R2*C) ) = 318.3098862Hz
And simulation show:
Fc=318.313Hz

 

The LT-SPICE simulation tool is showing the Cutoff frequency is = 314.534Hz. this value is based on the 3dB reference point. I have attached the file for the reference. It shows different value than 318hz.

As Jony130 pointed out, the cutoff frequency occurs at -3.010299957dB. Another, easier to remember point, is the phase shift. It will be exactly -45.0000 degrees.