Frequency Response of an Amplifier

1. Frequency Response of an Amplifier
Dr. Varun Kumar
Dr. Varun Kumar (IIIT Surat) 1 / 9

2. Outlines
1 Frequency response of an amplifier
2 Step response of an amplifier
Rise time
3 Bandpass of cascaded stage
Dr. Varun Kumar (IIIT Surat) 2 / 9

3. Frequency Response of an Amplifier
Low frequency response → High pass filter
From above circuit, we find using the complex variable s.
V0(s) =
Vi (s)R1
R1 + 1
sC1
= Vi (s)
s
s + 1
R1C1
(1)
Dr. Varun Kumar (IIIT Surat) 3 / 9

4. Continued–
⇒ Gain or transfer function
V0(jω)
Vi (jω)
= A(jω)

6. s=jω
=
1
1 − j(fL
f )
(2)
where fL = 1
2πR1C1
⇒ The magnitude |AL(jω)| and phase lead can be expressed as
AL(jω) =
1
p
1 + (fL/f )2
and θL = arctan
fL
f
(3)
⇒ If f = fL → AL(jω) = 1
√
2
→ 3 dB loss in power
⇒ If f → ∞, AL(jω) = 1 → Maximum gain or a high pass filter
Dr. Varun Kumar (IIIT Surat) 4 / 9

7. High frequency response
High frequency response → Low pass filter
From above circuit, we find using the complex variable s.
V0(s) =
Vi (s) 1
sC2
R2 + 1
sC2
= Vi (s)
1
1 + sR2C2
(4)
V0(jω)
Vi (jω) = A(jω)

9. s=jω
= 1
1+j( f
fH
)
, where fH = 1
2πR2C2
Dr. Varun Kumar (IIIT Surat) 5 / 9

10. Continued–
⇒ The magnitude |AH(jω)| and phase lead can be expressed as
AH(jω) =
1
p
1 + (f /fH)2
and θH = − arctan
f
fH
(5)
⇒ If f = fH → AH(jω) = 1
√
2
→ 3 dB loss in power
⇒ If f = 0, AL(jω) = 1 → Maximum gain or a low pass filter
Dr. Varun Kumar (IIIT Surat) 6 / 9

11. Step response of an amplifier
Rise time (tr ): The response of the low pass circuit in time domain can
be expressed as
V0(t) = V (1 − e
− t
R2C2 ), where V → step input (6)
Here fH = 1
tp
here tp → pulse width
Dr. Varun Kumar (IIIT Surat) 7 / 9

12. Bandpass of cascaded stages
⇒ Let the high 3 dB frequency for n cascaded stage is f ∗
H and equal the
frequency for which the overall voltage gain drop 3 dB.
⇒ To obtain the overall transfer function of interacting stages, the
transfer gain multiplied together.
⇒ If each stages has dominant poles and if the high 3 dB frequencies are
fHi
∀ i = 1, 2, ..., n.
⇒ f ∗
H can be calculated as
1
p
1 + (f ∗
H/fH1 )2
×
1
p
1 + (f ∗
H/fH2 )2
×....×
1
p
1 + (f ∗
H/fHn )2
=
1
√
2
(7)
⇒ If fH1 = fH2 = ..... = fHn = fH then
h 1
1 + (f ∗
H/fH)2
in/2
=
1
√
2
⇒
f ∗
H
fH
=
p
21/n − 1 (8)
Dr. Varun Kumar (IIIT Surat) 8 / 9

13. Continued–
⇒ On the other side, for lower cut-off, if fL1 = fL2 = ..... = fLn = fL then
f ∗
L can be expressed as
h 1
1 + (fL/fL∗ )2
in/2
=
1
√
2
⇒
f ∗
L
fL
=
1
p
21/n − 1
(9)
⇒ If poles are not widely spaced then equivalent
1
fH
= 1.1
s
1
f 2
1
+
1
f 2
2
+ ... +
1
f 2
n
(10)
⇒ If rise time of isolated individual stages are tr1 , tr2 , ...., trn and if the
input waveform rise time of signal is tr then
tr =
q
t2
r1
+ t2
r2
+ ..... + t2
rn
(11)
Dr. Varun Kumar (IIIT Surat) 9 / 9