• The d.c. analysis means to obtain the operating point values i.e. ICQ and VCEQ for the transistors used. The supply voltages are d.c. while the input signals are a.c., so d.c. equivalent circuit can be obtained simply by reducing the input a.c. signals to zero.
D.C. Analysis of Differential Amplifier
•
The d.c. analysis means to obtain the operating point values i.e. ICQ
and VCEQ for the transistors used. The supply voltages are d.c.
while the input signals are a.c., so d.c. equivalent circuit can be obtained
simply by reducing the input a.c. signals to zero. The d.c. equivalent circuit
thus obtained is shown in the Fig. 8.7.1. Assuming RS1 = RS2,
the source resistance is simply denoted by RS.
•
The transistors Q1 and Q2 are matched transistors and hence for such a matched
pair we can assume :
i)
Both the transistors have the same characteristics.
ii)
REI = RE2 hence RE = RE1 || RE2
.
iii)
RC1 = RC2 hence denoted as RC.
iv)
|VCC| = |VEE| and both are measured with respect to
ground.
•
As the two transistors are matched and circuit is symmetrical, it is enough to
find out operating point ICQ and VCEQ, for any one of the
two transistors. The same is applicable for the other transistor.
•
Applying KVL to base-emitter loop of the transistor Q1
From
the equation (8.7.6), we can observe that
i)
RE determines the emitter current of Q1 and Q2
for the known value of VEE
ii)
The emitter current through Q1 and Q2 is independent of
collector resistance RC.
•
Now let us determine VCE. As IE is known and IE
≅ IC, we can
determine the collector voltage of Q1 as
VC
= VCC – IC – IC RC …. (8-7.7)
•
Neglecting the drop across RS, we can say that the voltage at the emitter of Qi
is approximately equal to -VBE. Hence the collector to emitter voltage is
•
Hence IE = IC = ICQ while VCE = VCEQ
for given values of VCC andVEE.
Key
Point : In the equation (8.7.6), the sign of
VEE is already considered to be negative, while deriving it. Hence while using
this equation to solve the problem, only the magnitude of VEE should be used
and negative sign of VEE should not be used again.
•
Thus for both the transistors, we can determine operating point values, using
equations (8.7.6) and (8.7.8). With the same biasing arrangement, the d.c.
analysis remains same for all the four possible configurations of differential
amplifier.
Key
Point : The d.c. analysis and expressions for ICQ and
VCEQ remain same for all the differential amplifier circuit configurations.
Ex.
8.7.1 Calculate the operating point values for the circuit shown in the Fig.
8.7.2.
•
The sign of VEE is already considered while deriving the equation hence VEE
must be replaced by 12 V. And assume VBE = 0.7 V.
Electron Devices and Circuits: Unit IV: Multistage and Differential Amplifiers : Tag: : - D.C. Analysis of Differential Amplifier
Electron Devices and Circuits
EC3301 3rd Semester EEE Dept | 2021 Regulation | 3rd Semester EEE Dept 2021 Regulation