Question

Following figure shows a typical circuit for a low-pass filter. An AC input V_i = 10 mV is applied at the left end and the output V₀ is received at the right end. Find the output voltage for ν = 10 k Hz, 1.0 MHz and 10.0 MHz. Note that as the frequency is increased the output decreases and, hence, the name low-pass filter.

Answer 1

Here,
Input voltage to the filter, Vi = 10 × 10⁻³ V
Resistance of the circuit, R = 1 × 10³ Ω
Capacitance of the circuit, C = 10 × 10⁻⁹ F
(a) When frequency, f = 10 kHz
  A low pass filter consists of resistance and capacitance. Voltage across the capacitor is taken as the output.
     Capacitive reactance `(X_C)`is given by,
`X_C = 1/(omegaC) = 1/(2pifC)`

`rArr X_C = 1/(2pixx10xx10^3xx10xx10^-9)`
`rArr X_C = 1/(2pixx10^-4)`

`rArr X_C = 10^4/(2pi) = 5000/pi Omega`
  Net impedence of the resistance-capacitance circuit (Z) is given by,

`Z = sqrt(R^2 + X_C^2`

`rArr Z = sqrt((1 + 10^3) + (5000//pi)^2`

`Z = sqrt (10^6 + (5000// pi)^2`

Current (I₀) is given by,
`I_0 = V_1/Z`
`I_0 = (10xx10^-3)/(10^6 + (5000//pi))`

Output across the capacitor `(V_0)` is given by,
`V_0 = 10^2/sqrt(10^5 + (50//pi)^2 )xx 50/pi`
`rArr V_0 = 1.6124  V = 1.6  mV`

(b)When frequency, f = 1 MHz = 1 × 10⁶ Hz
Capacitive reactance `(X_C)` is given by,
`X_C = 1/(omegaC)`
`rArr X_C = 1/(2pifc)`
`rArr X_C = 1/(2pi xx10^6 xx 10 ^-9 xx 10)`

`rArr X_C = 1/(2pi xx 10^-2)`


`rArr X_C = 100/(2pi)`

`rArr X_C = 500/pi Omega`
 Total impedence (Z) = `sqrt(R^2 + X_C^2`

`Z = sqrt((10^3)^2 + (50//pi)^2`
current `(I_0) = V_1/Z`

`rArr I_0 = (10xx10)^-3 /sqrt(10^6 + (50//pi)^2`

Output voltage `(V_0) = I_0X_c`

`rArr V_0 = 10^-2/sqrt(10^5 +(5//pi)^2 ) xx 50/pi`

`rArr V_0 = 0.16 mV`
(c) When frequency, f = 10 MHz = 10 × 10⁶ Hz = 10⁷ Hz
       Capacitive reactance `(X_C)` is given by,
`X_C = 1/(omegaC)`
`rArr X_C = 1/(2pifC)`
` rArr X_C = 1/(2pixx10^7xx10xx10^-9)`

`rArr X_C = 5/pi Omega`
`Impendence ( Z) = sqrt(R^2 + X_c^2`
`z = sqrt(10^3)^2 + (5//pi)^2`

current `(I_0)= V_1/z`
`I_0 = 10xx10^-3/sqrt(10^6 + (5//pi)^2`
`v_0 = i_0X_C`
`rArr V_0 = 10xx10^-3 /sqrt(10^6 + (5//pi))^2 xx 5/pi`
`rArr V_0 = 16muV`