MOSFET Small-Signal
Amplifier Examples.
We will illustrate the analysis of small-signal MOSFET
amplifiers through two examples in this lecture.
Example N29.1 (text example 4.10). Determine Av (neglecting the effects of RG), Rin, and Rout for the circuit below given that
V_{t}=1.5V,K_{a}W|L=0.25 MA /V^{2}, and V_{A}=50 V

The first step is to determine the DC operating point. The DC equivalent circuit is:

Since V_{g} V_{D} =0 < the MOSFET is operating in the saturation mode if I_{D} =0 So, neglecting ro and assuming operation in the saturation mode

For this DC circuit

Notice in the circuit that V_{DS}= V_{GS} so that this last equation
Becomes
I_{D}=0.125(V_{DS}-1.5)^{2}
Also, by KVL
V_{DS}=15-R_{D}I_{D}=15-10,000 I_{D}
Substituting (2) into (1)
I_{D}=1.25*10^{-4}(15-10000I_{D}-1.5)^{2}
Solving this equation gives
I_{D}=1.06 mA Þ V_{DS=}4.4 V (=V_{gs}
I_{D}=1.72 mA Þ V_{DS=}2.2 V (=V_{gs}
This latter result is not consistent with the assumption of
operation in the saturation mode since V_{t} < V_{GS} = 1.5V. So the proper solution for ID is the first ( I_{D} =1.6 mA).
Next, we construct the small-signal equivalent circuit. We’ll use the p small-signal model of the MOSFET with ro included:

To compute the small-signal voltage gain, we start at the output (assuming RG is extremely large R_{G} >> r_{t}0|| R_{D}|| R_{L} )
V_{0}-g_{m}V_{gs}(r_{0}||R_{D||}R_{L})
At the input notice that v_{gs} = V_{i}. Therefore

Notice that the assumption R_{G}|| r_{0}|| R_{D}|| R_{L} is met and hugely exceeded since 10 MO >> 4,521 O.
For the input resistance Rin calculation, we cannot set 0 gs v_{gs} =0 and subsequently open circuit the dependent current source – since this would artificially force R_{in} =0 . Rather, we need to determine ii as a function of vi and use this in the definition:

The dependent current source will remain in these calculations. Proceeding, at the input of the small-signal equivalent circuit shown above

Therefore,

Consequently, using this expression we find that

Lastly, to determine the output resistance, we can set 0 gs v = in the small-signal equivalent circuit above, which will open circuit the dependent current source leading to the equivalent circuit:

from which we see that
R_{out} =||R_{0} ||R_{D}|| r =4 =5.2 kO
Example N29.2 (text exercise 4.23). Determine the following quantities for the conceptual MOSFET small-signal amplifier of Fig. 4.34 given that VDD = 5 V, RD = 10 kO, and VGS = 2 V.

The MOSFET characteristics are V_{t} =1 V, k_{n} ' =20 ?A/V^{2} 2, W/L= 20, and . l = 0.
(a) Determine ID and VD. We see from the circuit that V_{t} >V_{t} . Therefore, the MOSFET is operating in the saturation or triode mode. We’ll assume saturation. In that case
(b)
Let’s check if the MOSFET is operating in the saturation
mode:
V_{GD}=2-3=-1t
(c) Determine gm. Using (4.61)
(d)
(c) Determine the voltage gain Av. We begin by first
constructing the small-signal equivalent circuit

Directly from this circuit,
V_{0}=-g_{g}V_{gs}R_{D}

(e) If ( ) 0.2sin v_{gs} =0.2 V, find vd and the max/min vD.
(f)
Therefore, v_{d} = - 0.8 (wt)V
Hence V_{D}|_{max}=V_{D}+V_{d}=3+0.8=3.8V
While V_{D}|_{max}=V_{D}+V_{d}=3-0.8=2.2V
e) Determine the second harmonic distortion. From (4.57) or
(6) in the previous lecture notes, the drain current is given as

or

Substituting sin v_{gs} =0.2 sin into this equation gives
i_{D}(wt)= I _{D}+80*10^{-6}sin (wt) t+8*10^{-6 sin 2(wt)
Using the trigonometry identity
sin2 =1 /2- (wt) 1 /2cos (wt).
this last expression becomes
iD=200+80 sin (wt)+4-4 cos (2wt) uA
iD=204+80 sin (wt)-4 cos (2wt) uA
The first term in iD is ID, the DC current. We see that there is a slight shift upward in value by 4 mA.
The third term in iD is the second harmonic term because it
varies with time at twice the frequency of the input signal.
The second harmonic distortion is
4 \80*100%=5%
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