EE 461G
Homework # 12 Due December 9, 1999
Read Sedra and Smith read sections 6.1, 6.2, 6.3, 6.4, and 6.6 in chapter 6
Do D6.40 and 6.51 in Sedra and Smith. Hint for 6.40, do not let the
a in the equation confuse you. It is a3 and is one since the problem says b is infinite. Hint for 6.51, start by finding VR1 and thus the current in R1.Previous amplifier circuits that have been studied are most suited to a discrete circuit implementation. However, most modern circuits are built as integrated circuits on a single silicon chip. Integrated circuits provide lower cost, higher speed, and lower power dissipation. They also impose constraints on the types and size of components that can be used. For instance, capacitors above ~50 pF would occupy too much area on a chip; similarly, large valued resistors require prohibitively large amounts of space. Since transistors occupy little chip area, they are preferred in integrated circuits. Transistors on the same chip are well matched with almost identical values of
b and other fabrication parameters. This fact makes it possible to implement particular multi-transistor circuits, one of which is the differential amplifier. Two versions of the differential amplifier are shown below.
(a) (b)
In order to maintain a stable bias for changing values of b, a single transistor amplifier requires significant negative feed back. This was accomplished in the basic common emitter amplifier by using a relatively large emitter resistor. This same negative feedback reduces the amplifiers gain, so a bypass capacitor was required to short the emitter resistor to ground for the AC signals to be amplified. The use of a bypass capacitor prohibits the use of the amplifier at low frequencies. In addition, such large resistor and capacitance values can not be built as part of an integrated circuit. A solution to this problem is to add another active device to provide the required low impedance bypass to ground. One example of such a configuration is shown above in circuit (a).
A properly designed differential amplifier will amplify the difference between the two input signals V1 and V2 (V1 - V2) by a large gain factor and suppress the average or common component of the input signals (V1 + V2)/2. The current source Io serves to bias (provide the DC component) both of the transistors. In reality, a current source must be synthesized from various components. The crudest, yet easiest approximation to a current source is a resistor as shown in circuit (b) above.
The two input voltages can be expressed as linear combinations of the differential input Vidm = V1 - V2, and common mode input Vicm = (V1 + V2)/2. These equations constitute a linear transformation from (V1 V2) to (Vidm Vicm). This transformation is invertible so that the above equations can be solved for the input voltages in terms of Vidm, and Vicm. Using superposition, the output voltage for any arbitrary combination of V1 and V2 can be found once the voltage gain expressions for a pure differential input (V1 = -V2, and Vicm = 0) and a pure common mode input (V1 = V2, and Vidm = 0) are known.
The differential mode gain with respect to the output Vout1 is defined as Vout1 / (V1 - V2) and is called Adm-se1; likewise, the differential mode gain Adm-se2 with respect to the output Vout2 is defined as Vout2 / (V1 - V2).
The common mode gain with respect to the output Vout1 is defined as Vout1 / ((V1+V2) / 2) and is called Acm-se1; likewise, the common mode gain Acm-se2 with respect to the output Vout2 is defined as Vout2 / ((V1+V2) / 2). Either output may be found by combining these two gains (using superposition) as follows:
The expressions for the differential and common mode gains can be found using a small signal analysis on the circuit as usual (see Sedra and Smith for sample derivations).
The overall differential mode gain to find the output Vout1-Vout2 is 
and the overall common mode gain is 
. Here "dm" denotes differential mode, "cm" denotes common mode, "se" indicates single ended (Vout to ground). The purpose of the differential amplifier is to amplify the differential mode component of the input and to suppress the common mode component of the input (make the differential mode gain as high as possible and the common mode gain as small as possible). An important characterization of the differential amplifier is its common mode rejection ratio defined as CMRR = 20log |Adm-diff / Acm-diff|.
