Common Source Amplifier

Common Source Amplifier

Common Source Amplifier.
We’ve studied MOSFET small-signal equivalent models and the biasing of MOSFET amplifiers in the previous three lectures. We’ll now apply those skills by looking closely at three basic MOSFET amplifier types:
1. Common source amplifiers, including configurations with a source resistor (called source degeneration).
2. Common gate amplifiers.
3. Common drain (or source follower) amplifiers.
All of these amplifier types are appropriate for discrete component designs. In the case of IC amplifiers, we’ll also show corresponding designs implemented in CMOS, beginning in
Common Source Small-Signal Amplifier
This type of amplifier is shown in Fig. 4.43a, as biased by a combination of voltage and current sources:

Assuming sufficiently large values for the coupling capacitors (CC1 and CC2) and the bypass capacitor (CS) – so that their reactances are very small at the frequency of operation – the equivalent small-signal circuit for this amplifier is shown in

The text mentions performing the small-signal analysis directly on the amplifier circuit, as illustrated in Fig. 4.43(c). We do not recommend this approach. It is better to take the time and construct the small-signal equivalent circuit, as we’re doing here.
Small-Signal Amplifier Characteristics
As we did when studying BJT amplifiers, we’ll calculate the following quantities for this MOSFET common source amplifier: Rin, Av, Gv, Gi, and Rout.
• Input resistance, Rin. From the small-signal circuit above, and noting that ig =0 , then
Rin = R G
• Partial small-signal voltage gain, Av. From the output side of the small-signal circuit
V 0 = - GmV gs (r0||RD||RL)
while at the input,
V i=Vgs
Substituting (3) into (2), we find that the partial voltage gain as

• Overall small-signal voltage gain, Gv. As we did with BJT amplifiers, we can derive an expression for Gv in terms of Av. By definition,

Applying voltage division at the input of the small-signal equivalent circuit,

Substituting (6) into (5) and using (4) we find

• Overall small-signal current gain, Gi. Using current division at the output in the small-signal model above

while at the input,
Vgs= iiRG Substituting (9) into (8) we find that the overall small-signal current gain is

Notice that as RG ®¥ in this last expression, i G i® ¥. Is thisa problem? (To help answer this question, notice that as RG ®. ¥, ii® 0.) Also, what does Gi as RG ¥ mean for the overall small-signal output power of this amplifier? 

 

• Output resistance, Rout. To calculate the output resistance, we first set vsig =0 , which also means that gm vgs =0 . The input impedance of the dependent current source is infinite. Consequently,
Rout= r0|| RD