Equation for Breakpoint Frequency in High and Low Active Filters

[Midas_Touch]

How do I derive the equation of the breakpoint frequency f = 1/2*pi*R*C for a high active filter and a low active filter? Do I use Vin/Vout?

R= resistance
C = Capacitance
Vin = Voltage input
Vout = Voltage output

 

[SGT]

How do I derive the equation of the breakpoint frequency f = 1/2*pi*R*C for a high active filter and a low active filter? Do I use Vin/Vout?
R= resistance
C = Capacitance
Vin = Voltage input
Vout = Voltage output

The breakpoint is the frequency where the reactive impedance [tex]j2\pi fC[/tex]is equal in absolute value to the real impedance R.

 

[ravenprp]

The equation for the breakpoint frequency, also known as the cutoff frequency, in high and low active filters can be derived by using basic principles of circuit analysis and the transfer function of the filter. Let's first define what high and low active filters are.

High active filters are those in which the op-amp or active element is placed in the feedback loop, while in low active filters, the active element is placed in the input path. This placement of the active element affects the transfer function of the filter, and hence the breakpoint frequency.

To derive the equation for the breakpoint frequency in a high or low active filter, we need to use the transfer function of the filter. The transfer function is the ratio of the output voltage to the input voltage, and it is expressed as Vout/Vin.

For a high active filter, the transfer function is given by:

Vout/Vin = -R2/R1

Where R2 and R1 are the resistances in the feedback and input paths, respectively.

Similarly, for a low active filter, the transfer function is given by:

Vout/Vin = -R2/R1

Where R2 and R1 are the resistances in the input and feedback paths, respectively.

Now, we know that at the breakpoint frequency, the output voltage is 70.7% of the input voltage, i.e., Vout = 0.707*Vin. Substituting this in the transfer function equations, we get:

For high active filter: 0.707*Vin/Vin = -R2/R1

For low active filter: 0.707*Vin/Vin = -R2/R1

Simplifying these equations, we get:

For high active filter: 0.707 = -R2/R1

For low active filter: 0.707 = -R2/R1

Now, we know that the cutoff frequency occurs when the magnitude of the transfer function is equal to 1/sqrt(2). So, we can set the magnitude of the transfer function in the above equations to 1/sqrt(2) and solve for the unknown variable, which in this case is the product of R and C.

For high active filter: 1/sqrt(2) = R2/R1

For low active filter: 1/sqrt(2) = R2/R1

Solving for R2/R1, we get:

For high active filter: R