Question

Find the Q-point of the transistor in the figure below, for R1=100KOhms, R2=220Kohms, R3=24KOhms, R4=12KOhms and VDD=12V. Assume that
ΤΟ
=1 V, K'=$0.1mA/V^{}2$, $\gamma$ =0,$\lambda$=0 and W/L=5/1

Answer 1

Consider the following circuit diagram. 220 k92 R 12 k22 R + V DD 12 V 100 ke2 R, 24 k92 R; Figure 1

To determine the Q-point for the transistor of given gate bias circuit. Simplify the circuit the gate bias circuit and replace by its Thevenin equivalent circuit, the expression of Thevenin voltage Vo is shown below. VEQ = VDD R R, + R₂ Here, Circuit resistances R, is 100 k 2 Circuit resistances R, is 220 k12, Supply voltage is Vpis 12 V. Substitute 100 kN for R, 220 ks for Ry, and 12 V for Vpp in expression of Veg VEp = (12 V) 100 k22 100 kQ2 +220 k 2 = 3.75 V

REO Thevenin equivalent resistance (Ro) is determined by, RR R,+R Substitute 100 k22 for R, 220 k 2 for R, in this expression. (100k 2) (220 k.) 100 k22 +220 k12 = 68.75 k 2 REQ

Thevenin's equivalent circuit of a four resistor bias network is shown in Figure 2: R) = 12 k12 REO V DD 12 V + V EQ R$ 24 k2

For the above given circuit the expression of gate source voltage (VG) is given by Vos = V0 -1,R Here, the drain current is 1, and source resistance (Rs) is 24 k12, so the above expression is transformed to the following expression: Vos = 3.75 V-1, (24 kN) (1) = 3.75 - 240001, Also, for the above given circuit the expression of source bulk voltage (V) is given by Vse = 1, RS Here, the drain current is 1, and the source terminal resistance (Rs) is 24 kN2, so the above expression is transformed to the following expression: VS8 = 1, (24000 22) (2) = 240001

The expression of threshold voltage (V) is given by, Vyv = Vro+Y(JV sp +261 – 20:) Here, Body effect parameter (y) is 0.6 VV, Surface potential parameter (20.) is 0.6 V, Zero-substrate bias value (vro) is 1V. Substitute the corresponding values in expression of threshold voltage (V) Vrv =1+0.6(VV3+0.6 - V0.6) (3)

The value of trans conductance parameter (K,) can be determined in the following way: K=K L Here, Trans conductance parameter (K) is 1004A/V?, 5 Channel width to channel length ratio is mia L So, the value of trans conductance parameter (K) is calculated as below: K. = (100 A/V") () = 500 x 10 Now, the expression of drain current (1) is given by K 1) (Vos-VN) 2 Substitute 500x10 for K, in this expression: 500x106 1 (Vas - VN) = 250x10° (Vas - VN)

Also, from the circuit in Figure 2 the expression of drain source voltage (Vps) is given as follows: Vps = Vpp -1, (R, +R) Here, the source terminal resistance (Rs) is 24 kN and the drain terminal resistance (R) is 12 k22 Therefore, the expression of drain source voltage will be transformed to the following below expression: Vps = 12-1, (12x10° + 24x10') (5) = 12–(36x103)1,

Now, solve equations (1), (2), (3), (4) and (5) iteratively for the different values of drain current (1) Take 1. = 70uA and calculate the required parameters as follows: From equation (1) the value of gate source voltage can be calculated as follows: Vos = 3.75 - 240001 = 3.75 - 24000(70×10) = 2.07 V From equation (2) the value of V. can be calculated as follows: Vs8 = 240001 = 24000(70x10“) = 1.68 V

From equation (3) the value of Vr can be calculated as follows: Vyv =1+0.6(JV5 +0.6 - V0.6) Substitute 1.68 V for Vin this expression. VIN =1+0.6(71.68 +0.6 - 10.6 = 1+0.6(0.7355) = 1.4413 V SB

From equation (4) the value of 1, can be calculated as follows: 15 = 250x10*(V6 - VN) Substitute 1.4413 V for Vzx and 2.07 V for Vos in this expression 15 = 250x10°(2.07 -1.4413) =98.81uA

From equation (5) the value of Vps can be calculated as follows: Pos =12 – 360p Substitute 98.81x10® A for 1, in this expression. Vps = 12–(36x10°)(98.81x10“) = 8.44 V The calculated value of 1, (98.81 ua) is different from the assumed value of 1, (70 uA). thus, the assumed value of I, is incorrect.

Now, for different value of 1), the iteration in equations (1), (2), (3), (4), and (5) are shown in tabular form below: 1. VSB VGS VT 1 DS 71x10 1.704 2.046 1.4460 90x10% 8.76 73.1x10 1.754 1.996 1.4558 72.95x109.37 74x10 1.776 1.974 1.4601 66.02x10 9.62

The calculated value of 1, (72.95 ua) is approximately equal to the assumed value of 1 (73.1uA), therefore the assumed value of 1, is correct. Hence, the Q-point for the given transistor is (72.95 uA, 9.37 V)

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