**1 Answer**

written 22 months ago by |

Kn1 = Kn2 = 0.1 mA/$V^2$

Kn3 = Kn4 = 0.3 mA / $V^2$

VTN = 1V

$\lambda$ = 0 for M1 , M2 , M3 and

$\lambda$ = 0.01/v for M4

calculate I1 and VGS

The current flowing through 30k $\Omega$ is I1 applying KVL to the loop,

10 – 30 k I1 – VGS4 + 10 = 0

I1 = $\frac{20-VGS4}{30}$ -------- (1)

Currrent I1 is the drain current flowing through M4, hence

I1 = Kn4 $(VGS4 – Vr )^2$

= 0.3 m $(VGS4 – 1)^2$ → (2)

Equating equation (1) and (2), we get

$\frac{20-VGS4}{30K}$

= 0.3m $(VGS4 – 1)^2$

20-VGS4 = 9 $VG^2$S4 – 18 VGS4 + 9

$\therefore$ 9 $VG^2$S4 – 17 VGS4 – 11 = 0 ;

We get VGS4(1) = 2.4 V or

VGS4 (2) = - 0.5V

We consider VGS4 = 2.4 V

I1 = 0.3 m $(2.4 – 1)^2$

I1 = 0.588 mA

IQ is actually the drain current of $M_4$. Since $M_4$ and $M_3$ are identical in all aspects, the drain current will also be same.

$\therefore$ IQ = I1 = 0.588 mA