A differential amplifier is the basic unit with which the Operational Amplifier^{1} is built. This amplifier is very useful when there is a need to amplify low amplitude small signals. If the signal to be amplified is very small, then there exists the possibility that noise will impair the output signal because the noise will be amplified in the same ratio as the input signal. The differential amplifier's ability to reject commonmode signals (identical signals on both inputs) – like thermal noise – makes it an extremely useful amplifier for the purpose of rejecting the noise.
Figure 21.1: Differential Amplifier
A typical differential amplifier circuit is shown in Figure 21.1 (a) and the schematics in Figure 21.1(b). The circuit is a symmetric circuit built with two identical transistors, two identical collector resistors (R_{c}), and one emitter resistor (R_{E}). In general two power supplies are needed to bias the amplifier, however, the circuit can also be operated with only one supply. It is not necessary to have capacitors at the input terminals, or resistors from base to ground^{2}. The amplifier, as you can see, has two inputs and two outputs. Because of this, there are three different ways (modes) that the amplifier can be operated. We will discuss these modes of operation in Section 21.3.2.
^{1} We will introduce OpAmps in the next experiment.
^{2} However, in some instances it may be convenient to use capacitors and base resistors for specific applications.
DC Operation
To determine the DC bias operating point of the amplifier we ground the two AC inputs. This is shown in Figure 21.2. The DC voltages at the bases of the two transistors become zero (V_{B} = 0_{V}).
Figure 21.2: DC Bias of Differential Amplifier Circuit
Writing a mesh equation around the baseemitter loop we get
Solving for I_{E}, we get
where we made use of the fact that . Now, assuming that the two collector resistors are equal, and the transistors are identical, we can write
  (21.3)

Knowing the values of these currents, we can determine any other DC parameter that we may need.
AC Operation
To operate the differential amplifier in AC mode we apply separate AC signals at each input (v_{i}_{1} and v_{i}_{2}). This will result in separate output signals at v_{o}_{1} and v_{o}_{2}. Because there are two inputs we can operate the amplifier in three different modes:singleended, doubleended and commonmode.
◊ Singleended mode. In this mode only one input signal is applied at either input terminal while the other input is grounded. Eventhough there is only one signal at the input, due to the fact that both transistors are connected to the same emitter resistor, there will be output signals at both collectors (v_{o1} and v_{o2}). These output signals will have the same amplitude but different polarity. If the input signal is applied at input v_{i}_{1}, the output signal at v_{o1} will be inverted respect with to the input signal and the output signal at v_{o2} will be in phase with the input signal. If the input signal is applied at v_{i}_{2} keeping v_{i}_{1} grounded, the reverse operation will occur. Both signals, however, will be amplified by the same amount. The voltage gain in this mode is given by
  (21.4)

where v_{o} is the output voltage at any of the two outputs (v_{o}_{1} or v_{o}_{2}, v_{i}) is the input voltage, and r_{e} is the transistor emitter resistance given by
◊ Doubleended mode. This mode of operation occurs when two different input signals are applied. We define the differential voltage v_{d} by
  (21.5)

At each output terminal an output voltage will result. Both output voltages are equal. The voltage gain of the differential amplifier working in this mode is given by
  (21.6)

where v_{o} is the output voltage at any of the two outputs (v_{o}_{1} or v_{o}_{2}), vd is the differential input voltage given by Eq. 21.5, and r_{i} is the transistor input resistance given by
Notice that Equations 21.6 and 21.4 represent the same quantity.
◊ Commonmode. If the two input signals are equal, according to Eq. 21.6, the voltage gain should be zero resulting in an output voltage equal to zero as well. However, due to the impossibility of matching the two transistors exactly, the output voltage at any output terminal will not be equal to zero. It can be shown^{3} that the voltage gain of the differential amplifier when operating in commonmode is given by
  (21.7)

where v_{o} is the output voltage at any one of the output terminals (v_{o}_{1} or v_{o}_{2}) and v_{c} is the common input voltage which can be calculated using the following equation:
  (21.8)

Normally r_{i} is much smaller than , so we can approximate the denominator of Eq. 21.7 by its second term. Also, we know that for any transistor. Thus, the commonmode gain can be approximated by
  (21.9)

^{3} This will be left as an exercise to the reader.
Constant Current Source
A differential amplifier should have a large value for A_{d}, the differential voltage gain, and a small value for A_{c}, the commonmode gain. The reason for this is because we want to have small amplification for voltages that are equal at both inputs (like noise, for instance). The ability to reject the common inputs is what makes the differential amplifier such a useful amplifier.
Looking at Eq. 21.7 or Eq. 21.9 we see that in order to make A_{c} as small as possible we should have R_{E} as large as possible. We cannot, however, choose a value for R_{E} as big as we want because by doing so the DC operating current I_{E} will become very small, driving the Qpoint of the transistors close to cutoff. This, in turn, will defeat the purpose of the amplifier.
Figure 21.3: Differential Amplifier with Constant Current Source
In order to achieve a small value for A_{c} we must have a reasonable collector current and, at the same time, a high R_{E} (or an equivalent). These two conditions can be achieved if, instead of R_{E} as the common resistance, we use a constant current source. As you should remember a current source has a very high resistance (impedance) and the current produced by the source is constant and its value can be adjusted to any desired value. A practical constant current source can be constructed with some resistors and a transistor as is shown in Figure 21.3. In this circuit we have replaced the emitter resistor with a constant current source built with three resistors (R_{1}, R_{2} and R_{3}) and one transistor (Q_{3}).
The current I_{c} indicated in the diagram is what in Fig. 21.2 we call I_{E}. This current is approximately equal to I_{E} in Fig. 21.3. By a simple analysis it can be shown that for the constant current source circuit in Fig. 21.3 the emitter (and collector) current is given by
  (21.10)

where
Notice that by properly choosing the values of R_{1}, R_{2}, R_{3}and V_{EE} we can have practically any desired current value.
The internal resistance of this type of source is approximately equal to the output resistance of the transistor, r_{o}, normally given by the manufacturer. As you should know^{4} the output resistance of transistors are normally very large. With the emitter resistor replaced with the constant source, the equation for the commonmode gain is given by
  (21.11)

  (21.12)

4 If you do not know or if you don’t remember, for God’s sake go back and review the topics about the equivalent circuit of transistors!
CommonMode Rejection Ratio
A measure that indicates the ability of the differential amplifier^{5} to reject identical input voltages at each input is the Common–Mode Rejection Ratio (CMRR) which we define as the ratio of the differentialmode gain to the commonmode gain, or in equation form
  (21.13)

The CMRR can be expressed in decibels as
  (21.14)

A good amplifier will have A_{d} very large and A_{c} very small. Therefore, according to Eq. 21.13 the CMRR of such an amplifier will be very large. This means that different signals applied at the inputs will be greatly amplified whereas equal signals applied at the inputs will not be amplified. Ideally this ratio is infinity.
^{5}We use the same concept with OpAmps, as well.
© 2024 GlobalSpec, Inc.