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=-1<v_{t (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, V0=-ggVgsRD (e) If ( ) 0.2sin vgs =0.2 V, find vd and the max/min vD. (f) Therefore, vd = - 0.8 (wt)V Hence VD|max=VD+Vd=3+0.8=3.8V While VD|max=VD+Vd=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 vgs =0.2 sin into this equation gives iD(wt)= I D+80*10-6sin (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% }