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MOSFET Small-Signal Amplifier Examples

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. Well 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. Well assume saturation. In that case
(b)
Lets 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%


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