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MOSFET Common Gate Amplifier

MOSFET Common Gate Amplifier.
We’ll continue our discussion of discrete MOSFET amplifiers we began with the common source amplifier in Lectures 31 and 32. Here we’ll cover the common gate amplifier, which is shown in Fig. 4.45. It has a grounded gate terminal, a signal input at the source terminal, and the output taken at the drain.

Small-Signal Amplifier Characteristics
As we’ve done with previous amplifiers in this course, we’ll calculate the following small-signal quantities for this MOSFETcommon gate amplifier: Rin, Av, Avo, Gv, Gi, Ais, and Rout. Tobegin, we construct the small-signal equivalent circuit:

The T model was used since we ignored ro while Rsig appears in series with 1/gm.
• Input resistance, Rin. Because the gate is grounded, we can see directly from this small-signal equivalent circuit that
Actually, this result may not be that readily apparent to you since while the gate is grounded, the current in the gate is zero ( ig =0 ). To verify this result in (1), we can apply a voltage source vx at the source terminal and calculate the ratio of this voltage to the current directed into the source terminal, which we’ll define as ix:

At the input to this circuit

This current ix doesn’t flow through the gate terminal! Instead, ix flows through the dependent source, then to ground. Indeed, we see that
x mgs ix=- gm Vgs =- i
Tricky! In any event, the input resistance in (1) has been verified.
• Partial small-signal voltage gains, Av and Avo. At the output side of the small-signal circuit
v0=- gmvgs (RD =-RL)
At the input, we can see that because the gate is grounded Substituting (3) into (2), gives the partial small-signal AC voltage gain to be

In the case of an open circuit load ( R L®¥), the small-signal voltage gain becomes
• Overall small-signal voltage gain, Gv. Using voltage division at the input to the small-signal equivalent circuit

Substituting this into

gives the overall small-signal voltage gain of this common gate amplifier to be

More specifically, using (1) in this expression

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

Substituting (11) into (10) gives the overall small-signal AC current gain to be

• Short-circuit small-signal current gain, Ais. The short circuit small-signal AC current gain can be easily determined from (12) with RL =0 as
• Output resistance, Rout. From the small-signal circuit above with vsig =0we find that i =0 since the gate is grounded. Consequently, Rout =RD
In summary, we find for the CG small-signal amplifier:
o A non-inverting amplifier.
o Moderate input resistance [see (1)].
o Moderately large small-signal voltage gain [see (9)], but smaller than CS amplifier.
o Small-signal current gain less than one [see (12)].
o Potentially large output resistance (dependent on RD) [see (14)].
Similar to the BJT CB amplifier we discussed in Lecture 20, the CG amplifier finds use as a current buffer amplifier. It has the relatively small input resistance, relatively large output resistance, and Gi less than (and potentially near) one characteristics of such amplifiers. (Does this amplifier provide any power gain for a signal?)
Example N34.1 (based on text exercise 4.34). Use the circuit of Fig. E4.30 to design a common gate amplifier. Find Rin, Rout, Avo, Av, Gv, and Gi for RL =10 k beta and sRsig = 50 beta Ù. What will the overall voltage gain become for Rsig = 50beta? 10 kbeta? 100 kbeta? The DC analysis results are shown in Fig. E4.30:

Using (4.71)

Based on this DC biasing, the corresponding common gate amplifier circuit is:

The small-signal equivalent circuit for this amplifier is:

The 4.7-Mbetaresistor functions to force the gate to ground potential. But since ig =0 , it will have no other impact on the circuit.

• What is the overall voltage gain when:

We see from these calculations that the overall voltage gain decreases substantially as Rsig increases. Can you explain what is physically happening to cause this to occur?

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