MOSFET as an Amplifier. Small-Signal Equivalent Circuit Models

As with the BJT, we can use MOSFETs as AC small-signal
amplifiers. An example is the so-called conceptual MOSFET amplifier shown in

This is only a “conceptual” amplifier for two primary reasons:
1. The bias with VGS is impractical. (Will consider others
later.)
2. In ICs, resistors take up too much room. (Would use
another triode-region biased MOSFET in lieu of RD.)
To operate as a small-signal amplifier, we bias the MOSFET in the saturation region. For the analysis of the DC operating point,
we set v_{gs} =0 so that from (4.22) with l=0

From the circuit
V_{DS}=V_{DD}-I_{D}R_{D}
For operation in the saturation region
V_{GD} £ V_{t}_{Þ}V_{GS}-V_{DS} £ V_{t}
Or
V _{DS} ³ V_{GS}-V_{t}
where the total drain-to-source voltage is

Similar to what we saw with BJT amplifiers, we need make sure that (3) is satisfied for the entire signal swing of vds.
With an AC signal applied at the gate
V_{GS}=V_{GS}+V_{g}
Substituting (4) into (4.20)

The last term in (6) is nonlinear in vgs, which is undesirable for a linear amplifier. Consequently, for linear operation we will require that this term be small:

If this small-signal condition (7) is satisfied, then the total drain current is approximately the linear summation

where

From this expression (9) we see that the AC drain current id is related to vgs by the so-called transistor transconductance, gm:

which is sometimes expressed in terms of the overdrive voltage

Because of the VGS term in (10) and (11), this gm depends on the bias, which is just like a BJT. Physically, this transconductance gm equals the slope of the iDvGS
characteristic curve at the Q point:

In the case that the MOSFET has a non-zero channel-length
modulation coefficient (i.e., 0 .. ), then the drain current is
given from (4.22) to be

Using (13) in (12) then

Lastly, it can be easily show that for this conceptual amplifier in

Consequently, A_{m} g_{v} , which is the same result we found for a similar BJT conceptual amplifier [see (5.103)]. MOSFET Small-Signal Equivalent Models
For circuit analysis, it is convenient to use equivalent smallsignal models for MOSFETs – as it was with BJTs.
The MOSFET acts as a voltage controlled current source. The control voltage is vgs and the output current is iD, which gives rise to this small-signal p model:

Things to note from this small-signal model include:
1. i_{g} =0 and v_{gs} ¹0Þ. infinite input impedance.
2. ro models the finite output resistance. Practically speaking, it will range from 10 k Mbeta
3. From (10) we found (l .=0 )

Alternatively, it can be shown that

which is similar to g _{m}V = 1|V_{t}for BJTs.
One big difference from BJTs is 25 V_{t} mV while V_{ eff } =0.1 V or greater. Hence, for the same bias current gm is much larger for BJTs than for MOSFETs. A small-signal T model for the MOSFET is shown in

Notice the direct connection between the gate and both the
dependent current source and 1/gm. While this model is correct, we’ve added the explicit boundary condition that ig = 0 to this small-signal model. It isn’t necessary to do this because the currents in the two vertical branches are both equal to g_{gs} v_{m} , which means i_{g} =0 . But adding this condition i_{g} =0 to the small-signal model in Fig. 4.40a makes this explicit in the circuit calculations. (The T model usually shows this direct connection while the p model
usually doesn’t.)
MOSFETs have many advantages over BJTs including:
1. High input resistance
2. Small physical size
3. Low power dissipation
4. Relative ease of fabrication.
One can combine advantages of both technologies (BJT and MOSFET) into what are called BiCMOS amplifiers:

Such a combination provides a very large input resistance from the MOSFET and a large output impedance from the BJT.