written 2.5 years 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