EXPERIMENT 2
THE BJT DIFFERENTIAL AMPLIFIER
I.
OBJECTIVES
• To determine the gain and CMRR for the BJT differential amplifier.
II. INTRODUCTION AND THEORY
The differential amplifier, or differential pair, is an essential building block in all integrated amplifiers.
In general, the input stage of any analog integrated circuit with more than one input consists of a
differential pair or differential amplifier. The basic differential pair circuit consists of two-matched
transistors Q1 and Q 2 , whose emitters are joined together and biased a constant current source I as
shown in Figure 1. The operation mode of the differential amplifier is defined according to the type of
the input signal, for example large or small input signal, polarity of the input signals.
Figure 1
Three important characteristics of the differential input stage are: the common-mode rejection ratio
CMRR, the input differential resistance Rid , and the differential-mode gain Ad .
THE DIFFERENTIAL –MODE GAIN
Let vB1 − vB 2 = vid , then v01 = − Ad vid and v02 = Ad vid . For perfectly matched transistors pair, the
differential gain is given by
1
v
Ad = O 2 = g m RC
vid 2
Note that the above equation is obtained using half-circuit concept that is one half of the circuit was
IC
used to conduct the small-signal analysis. The trans-conductance of either transistor is g m =
where
VT
is the thermal voltage VT ≅ 25mV .
1
THE COMMON –MODE REJECTION RATIO
Let vB1 = vB 2 = vcm in Figure 1 where the voltage vcm is called common-mode voltage. Assume that the
two transistors Q1 ,Q2 are perfectly matched. It follows that the current I is divided equally between
the two transistors and remain so as long as the transistors are in active region. The voltage at each
collector will be VCC − 0.5αIRC , and the difference VO1 − VO 2 = 0 . Any change in vCM will not affect the
balance of the emitter current in both transistors and the collector voltages remain the same this means
that the differential pair rejects common-mode input signal. The common-mode gain is given by
ACM =
vO 2
R
≈− C
vcm
2 RE
Basically the differential amplifier is designed to amplify differential signal, this requires Ad >> Acm .
The ideal differential amplifier has Ad ≈ ∞, and Acm ≈ 0 . The Common-Mode Rejection Ratio CMRR
is used as a measure of the differential amplifier performance. It is defined as
CMRR =
Ad
Acm
Substitute the values of the differential and common gains in the above equation
CMRR = − g m RE
As we can see from the above equation increasing the value of R E , the CMRR will increase, in other
words the performance of the differential amplifier can be improved by simply increasing the emitter
resistance. A common practice is to use a current source to replace RE , the results will be high
CMRR. To avoid dealing with large numbers the CMRR is expressed in dB as given below
(CMRR )dB = 20 log
Ad
Acm
PRACTICAL INPUT- OUTPUT SIGNALS
So far we have assumed that the input signal is present in either common-mode or differential-mode.
In practice the input signals can be decomposed into common-mode and differential-mode
components.
The output signal will be given in general by
vO 2 = Ad vid + Acm vcm = Ad (vB1 − vB 2 ) + Acm (
2
vB1 + vB 2
)
2
DIFFERENTIAL-MODE INPUT- OUTPUT RESISTANCE
Looking into the collector of either transistor, assume that the transistor load is a passive load as shown
in Figure 1. The differential-mode output resistance is simply the output resistance of common-emitter
stage and equals RC . If RC is replaced by an active load (current mirror), the output resistance
will be rO as explained in Experiment 1.
The differential–mode input resistance Rid is the resistance seen by the differential signal vid (i.e.
V
looking into the base of the BJT) and is given by
Rid = 2rπ , where rπ = T .
IB
VO2
Vs
VB1
Figure 2 – The Differential Amplifier
III PROCEDURE:
Experiment Equipments:
•
•
•
•
•
KL2101 Linear Circuit Lab
Resistors: 100kΩ, 3x47kΩ
Signal Generator
Oscilloscope
CA3046 General Purpose NPN Transistor Array
3
IV LAB PROCEDURE:
1. Build the differential amplifier shown in Figure 2.
2. Before applying Function Generator input, make sure that you have the minimum voltage
under -40 dB attenuation with 1 kHz frequency!! (First press -20 dB ATT button, and also pull
-20 dB knob located on the right hand side.)
3. By connecting Ch1 to the output of the amplifier VO 2 (Collector Voltage), increase the input
voltage until you have the maximum possible undistorted waveform.
4. Measure the peak-to-peak values of the followings.
(VS ) pp
(VB1 ) pp
(VO 2 ) pp
5. Calculate the differential gain Ad and input resistance Rid by using the results in (4)
Ad = ________
Rid = ________
6. Modify the circuit connections as follows: disconnect the ground connection at point VB2 and
connect this point to VB1. This circuit configuration is known as differential amplifier with a
common-mode input signal.
7. Measure and record the peak-to-peak values of the following.
(VS ) pp
(VB1 ) pp = (Vcm ) pp
(VO1 ) pp
(V O 2 ) pp
8. Calculate the common-mode gain Acm and the common-mode input resistance Ricm as in (5).
Calculate the CMRR (dB) as above.
Acm = ______
Ricm = ______
( CMRR )dB = _______
9. Compare the input resistance and gains for eachcase.
4