Lecture 6–Active Low Pass and High Pass Filters:

Lecture 6– Active Low Pass and High Pass Filters:

1. Active Low Pass Filter:

· An active filter is a filter circuit that uses and op amp.

· An active filter can produce gain, while a passive filter cannot. So an active filter can make the output bigger than the input (have positive dB on a Bode plot), while a passive filter can only make the output less than or equal to the input (have negative dB).

· Bode Plots: Show the magnitude and phase response on a logarithmic (log – base 10)

scale for frequency like in Figure 1.

So that small gain and big gain can be

shown on one graph

· Decibel: dB = 20 x log (Vout/Vin)

· decade = 10:1 times ten ratio.

· Cutoff frequency (Wc) = -3 dB frequency,

because 20xlog(1/ Ö2)= -3 dB

A. Inverting active low pass filter circuit: Circuit 1

· In the inverting type of active low pass filter,

the output can become smaller than the input.

· The inverting low pass filter has a capacitor in

the feedback, like the integrator.

· Equation for the inverting active low pass:

starting with the idea that the input signals (Vin)

will be sinewaves.

Then using the inverting amplifier equation:

Vout = -(Rf/ Ri) x Vin

Since the input are sinewaves then we can use Zc = 1/ jwC instead of Rf.

So Vout = -Vin / (jwC x Ri) and Vout/ Vin = -1 / (jwC x Ri)

· From this equation, we can see that as the frequency (w=2πF) increases, then Vout will decrease.

Example 1: Given Circuit 1, the op amp with +Vs = 15V and -Vs = -15V, Ri= 10KW, and C=0.1uF. Find Vout when Vin = 5Vsin(200t), Vin = 5Vsin(2000t), Vin = 5Vsin(20000t)?

Draw the Bode plot for Vout/ Vin?

Freq.K rad/s

When Vin = 5Vsin(200t): Vout = -Vin / (jwC x Ri) = j5V / (200 x 0.1u x 10K) = 25 <90 Which will be limited to Vout = 15V

Vout / Vin = 15 < 90 / 5V = 3 so dB = 20 x log (3) = 9.5dB

When Vin = 5Vsin(2000t): Vout = -Vin / (jwC x Ri)

= j5V / (2000 x 0.1u x 10K) = 2.5 <90

Vout / Vin = 2.5 < 90 / 5V = 0.5 so dB = 20 x log (0.5) = -6dB

When Vin = 5Vsin(20000t): Vout = -Vin / (jwC x Ri)

= j5V / (20000 x 0.1u x 10K) = 0.25<90

Vout / Vin = 0.25 < 90 / 5V = 0.05 so dB = 20 x log (0.05) = -26dB Bode plot in Figure 2.

B. Non-inverting active low pass filter circuit: Circuit 2

· In the non-inverting active low pass filter,

the output cannot be smaller than the

input (no negative dB).

· The non-inverting low pass filter has a capacitor

In the feedback.

· Equation for the non-inverting active low pass:

starting with the idea that the input signals (Vin)

will be sinewaves.

Then using the non-inverting amplifier equation:

Vout = (1+ Rf/ Ri) x Vin

Since the input are sinewaves then we can use Zc = 1/ jwC instead of Rf.

So Vout = Vin (1 + 1/ jwC x Ri) and Vout/ Vin = 1 + (1 / jwC x Ri)

· From this equation, we can see that as the frequency (w=2πF) increases, then Vout will decrease, but it will never be smaller than Vin.

Example 2: Given Circuit 2, the op amp with +Vs = 15V and -Vs = -15V, Ri= 2KW, and C=0.47uF. Find Vout when Vin=50mVsin(100t),Vin=50mVsin(1000t),Vin=50mVsin(10000t)?

Draw the Bode plot for Vout/ Vin?

When Vin = 50mVsin(100t): Vout = Vin (1 + 1/ jwC x Ri)=

50mV x (1+ 1/ j100 x 0.47u x 2K) = 50mV (1 – j10.6) =

50mV x 10.6 < -85 = 0.53V < -85

Vout / Vin = 0.53 < -85 / 50mV = 10.6 < -85

so dB = 20 x log (10.6) = 20.5dB

Freq.K rad/s

When Vin = 50mVsin(1000t): Vout = Vin (1 + 1/ jwC x Ri)

= 50mV x (1+ 1/ j1000 x 0.47u x 2K)

= 50mV (1 – j1.06) = 50mV x 1.46 < -47 = 73mV < -47

Vout / Vin = 73mV < -47 / 50mV = 1.46 < -47

so dB = 20 x log (1.46) = 3.2dB

When Vin = 50mVsin(10000t): Vout = Vin (1 + 1/ jwC x Ri) = 50mV x (1+ 1/ j10000 x 0.47u x 2K) = 50mV (1 – j0.106) = 50mV x 1 < -6 = 50mV < -6

Vout / Vin = 50mV < -6 / 50mV = 1 < -6 so dB = 20 x log (1) = 0dB Bode plot in Figure 3.

1. Active High Pass Filter:

C. Inverting active high pass filter circuit: Circuit 3

· In the inverting type of active high pass filter,

the output can become smaller than the input.

· The inverting high pass filter has a capacitor in

Series with the input, like the differentiator.

· Equation for the inverting active high pass:

starting with the idea that the input signals (Vin)

will be sinewaves.

Then using the inverting amplifier equation:

Vout = -(Rf/ Ri) x Vin

Since the input are sinewaves then we can use Zc = 1/ jwC instead of Ri.

So Vout = -Vin x jwC x Rf and Vout/ Vin = - jwC x Rf

· From this equation, we can see that as the frequency (w=2πF) increases, then Vout will increase.

Example 3: Given Circuit 3, the op amp with +Vs = 15V and -Vs = -15V, Rf= 8KW, and C=0.22uF. Find Vout when Vin=20mVsin(500t),Vin=20mVsin(5000t),Vin=20mVsin(50000t)?

Draw the Bode plot for Vout/ Vin?

When Vin = 20mVsin(500t): Vout= -Vin x (jwC x Rf) =

-j20mVx 500x 0.22ux 8K)= -j17.6mV

Vout / Vin = 17.6mV < -90 / 20mV = 0.88 so

dB = 20 x log (0.88) = -1.1dB

When Vin = 20mVsin(5000t): Vout= -Vin x (jwC x Rf) =

-j20mVx 5000x 0.22ux 8K)= -j176mV

Vout / Vin = 176mV < -90 / 20mV = 8.8 so

dB = 20 x log (8.8) = 18.9dB

When Vin = 20mVsin(50000t): Vout= -Vin x (jwC x Rf) = -j20mVx 50000x 0.22ux 8K)= -j1.76V so Vout / Vin = 1.76V < -90 / 20mV = 88 so dB = 20 x log (88) = 38.9dB Figure 4

D. Non-inverting active high pass filter circuit: Circuit 4

· In the non-inverting active high pass filter,

the output cannot be smaller than the

input (no negative dB).

· The non-inverting low pass filter has a capacitor

In series with the feedback resistor to ground.

· Equation for the non-inverting active high pass:

starting with the idea that the input signals (Vin)

will be sinewaves.

Then using the non-inverting amplifier equation:

Vout = (1+ Rf/ Ri) x Vin

Since the input are sinewaves then we can use Zc = 1/ jwC instead of Ri

So Vout = Vin (1 + jwC x Rf) and Vout/ Vin = 1 + jwC x Rf

· From this equation, we can see that as the frequency (w=2πF) increases, then Vout will increase, but it will never be smaller than Vin.

Example 4: Given Circuit 4, the op amp with +Vs = 15V and -Vs = -15V, Rf= 3KW, and C=0.7uF. Find Vout when Vin=0.2Vsin(150t),Vin=0.2Vsin(1500t),Vin=0.2Vsin(15000t)?

Draw the Bode plot for Vout/ Vin?

When Vin = 0.2Vsin(150t): Vout= Vin x (1+ jwC x Rf) =

0.2Vx (1+ j150x 0.7u x 3K)= 0.2 (1 + j0.3) =

0.2 x 1.04 <17 = 0.21 <17

Vout / Vin = 0.21V < 17 / 0.2V = 1 so

dB = 20 x log (1) = 0dB

When Vin = 0.2Vsin(1500t): Vout= Vin x (1+ jwC x Rf) =

0.2Vx (1+ j1500x 0.7u x 3K)= 0.2 (1 + j3) =

0.2 x 3.2 <72 = 0.64 <72

Vout / Vin = 0.64V <72 / 0.2V = 3.2 so

dB = 20 x log (3.2) = 10dB

When Vin = 0.2Vsin(15000t): Vout= Vin x (1+ jwC x Rf) = 0.2Vx (1+ j15000x 0.7u x 3K)= 0.2 (1 + j31.5) = 0.2 x 31.5 <88 = 6.3 <88 so Vout / Vin = 6.3V <88 / 0.2V = 31.5 so

dB = 20 x log (31.5) = 30dB Bode plot in Figure 5

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