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
Vt=1.5V,KaW|L=0.25 MA /V2, and VA=50 V
The first step is to determine the DC operating point. The DC equivalent circuit is:
Since Vg VD =0 < the MOSFET is operating in the saturation mode if ID =0 So, neglecting ro and assuming operation in the saturation mode
For this DC circuit
Notice in the circuit that VDS= VGS so that this last equation
Becomes
ID=0.125(VDS-1.5)2
Also, by KVL
VDS=15-RDID=15-10,000 ID
Substituting (2) into (1)
ID=1.25*10-4(15-10000ID-1.5)2
Solving this equation gives
ID=1.06 mA Þ VDS=4.4 V (=Vgs
ID=1.72 mA Þ VDS=2.2 V (=Vgs
This latter result is not consistent with the assumption of
operation in the saturation mode since Vt < VGS = 1.5V. So the proper solution for ID is the first ( ID =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 RG >> rt0|| RD|| RL )
V0-gmVgs(r0||RD||RL)
At the input notice that vgs = Vi. Therefore
Notice that the assumption RG|| r0|| RD|| RL is met and hugely exceeded since 10 MO >> 4,521 O.
For the input resistance Rin calculation, we cannot set 0 gs vgs =0 and subsequently open circuit the dependent current source – since this would artificially force Rin =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
Rout =||R0 ||RD|| 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 Vt =1 V, kn ' =20 ?A/V2 2, W/L= 20, and . l = 0.
(a) Determine ID and VD. We see from the circuit that Vt >Vt . 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:
VGD=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,
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%
