Lecture 1 Intro To Op Amps
Differential amplifiers
A differential (or difference) amplifier is a circuit used for amplifying a voltage difference between two input signals while rejecting signals that are common to both inputs.
IRE = based on approximation IC ≈ IE then IC1 = IC2 = IRE/2 therefore, VC1 = VC2 = VCC – IC1R1
Modes of Signal Operation:
Single-ended input – input signal is applied to either input with the other input connected to ground Differential or double-ended input – two opposite polarity input signals are applied.
Common-mode input – same signal is applied to both inputs.
Single-ended input:
DC Analysis:
Loop 1:
‒VBE – VE = 0
VE = ‒VBE = ‒ 0.7V and IE1 = IE2 since both currents combine in RE,
IE1 = IE2 = IRE/2
Loop 2:
‒ IRERE + VEE + VE = 0
Introduction to Operational Amplifiers (M. LATINA)
p. 1
Differential input:
Example: A certain differential amplifier has a differential voltage gain of 2000 and a commonmode gain of 0.2. Determine the CMRR and express in dB.
Differential Gain:
Common-mode input:
Common-mode signal:
‒ signal that drives both inputs of a differential amplifier equally.
‒ these are interference, static and other kinds of undesirable signals picked-up by the circuit. Common Mode Rejection Ratio(CMRR)
‒ measure of an amplifier’s ability to reject common-mode signals.
vin(d) = vin1 ‒ vin2
From Loop 1: vin1 ‒ ie1re’ ‒ (ie1 + ie2)RE = 0 vin1 = ie1(re’ + RE) + ie2RE (1) vin2 ‒ ie2re’ ‒ (ie1 + ie2)RE = 0 vin2 = ie2(re’ + RE) + ie1RE (2) express in terms of the current: from (2) ie2 = substitute ie2 in equation 1: vin1 = ie1(re’ + RE) +
Introduction to Operational Amplifiers (M. LATINA)
RE
p. 2
which makes ie1 = do the same to compute for ie2 ie2 = at the output side: vout(d) = vc1 ‒ vc2
= RC (ic1 ‒ ic2)
= RC (ie1 ‒ ie2)
= RC
‒
Example: For the circuit shown, calculate:
(a) ICQ and VCEQ
(b) Av(d) and Acm
(c) CMRR
Solution:
Simplify to obtain
Av(d) =
=
*true for balanced output
(vout(d) = vc1 ‒ vc2)
Av(d) =
=
*true for unbalanced output
A differential (or difference) amplifier is a circuit used for amplifying a voltage difference between two input signals while rejecting signals that are common to both inputs.
IRE = based on approximation IC ≈ IE then IC1 = IC2 = IRE/2 therefore, VC1 = VC2 = VCC – IC1R1
Modes of Signal Operation:
Single-ended input – input signal is applied to either input with the other input connected to ground Differential or double-ended input – two opposite polarity input signals are applied.
Common-mode input – same signal is applied to both inputs.
Single-ended input:
DC Analysis:
Loop 1:
‒VBE – VE = 0
VE = ‒VBE = ‒ 0.7V and IE1 = IE2 since both currents combine in RE,
IE1 = IE2 = IRE/2
Loop 2:
‒ IRERE + VEE + VE = 0
Introduction to Operational Amplifiers (M. LATINA)
p. 1
Differential input:
Example: A certain differential amplifier has a differential voltage gain of 2000 and a commonmode gain of 0.2. Determine the CMRR and express in dB.
Differential Gain:
Common-mode input:
Common-mode signal:
‒ signal that drives both inputs of a differential amplifier equally.
‒ these are interference, static and other kinds of undesirable signals picked-up by the circuit. Common Mode Rejection Ratio(CMRR)
‒ measure of an amplifier’s ability to reject common-mode signals.
vin(d) = vin1 ‒ vin2
From Loop 1: vin1 ‒ ie1re’ ‒ (ie1 + ie2)RE = 0 vin1 = ie1(re’ + RE) + ie2RE (1) vin2 ‒ ie2re’ ‒ (ie1 + ie2)RE = 0 vin2 = ie2(re’ + RE) + ie1RE (2) express in terms of the current: from (2) ie2 = substitute ie2 in equation 1: vin1 = ie1(re’ + RE) +
Introduction to Operational Amplifiers (M. LATINA)
RE
p. 2
which makes ie1 = do the same to compute for ie2 ie2 = at the output side: vout(d) = vc1 ‒ vc2
= RC (ic1 ‒ ic2)
= RC (ie1 ‒ ie2)
= RC
‒
Example: For the circuit shown, calculate:
(a) ICQ and VCEQ
(b) Av(d) and Acm
(c) CMRR
Solution:
Simplify to obtain
Av(d) =
=
*true for balanced output
(vout(d) = vc1 ‒ vc2)
Av(d) =
=
*true for unbalanced output