g NMOS PMOS p x is 123 for the each MOSFET DDA A DD bias un

g NMOS & PMOS p x is 1,2,3 for the each MOSFET. DDA A DD bias unC, 50 HAJV VTN 0 V, VTp 0.7 VANE 0.05 V 0.07 V LN LP 0.25 200 HA, and VDD 4 V. A gdi utput

Solution

Solution:- The given circuit is the basic current mirror circuit using Mosfets.

Since Width of Mosfet is not given, therefore in this solution Width of Mosfet is assumed to be 0.25um for simplicity. If different width is required simply change the slight calculations.

And all Mosfets are assumed to be in Working in Saturation Region.

So,

Id₂= Id₁(W/L)₂/(W/L)₁

Since Length of each transistor is 0.25 um. Therefore

(W/L)₁=(W/L)₂= 1

Therefore Id₂ =Id₁

Drain Current across Mosfet in Saturation Regioin is given by

Id= µ_nC_ox(W/L)(Vgs-Vt)²

In this question

Id₂ =Id₁= I_bias= 200uA

For NMOS Transistor Q3

Source is at ground (Vgs=Vg)

Vgs= Vg

200= 100(Vgs- 1)²

V_gs(Q1) =2.41 V

For PMOS Transistor Q2

Here Vs= Vdd = 4V

I_d= µ_pC_ox(W/L)(Vgs-Vt)²

Also Same current will flow across transistor Q2 and Q3

Therefore, Id(Q2) = Id(Q3)= 200 uA

200= 50(Vg-4-(-0.7)2

Therefore Vg2= 2+4.7= 6.7

So, Vgs= 4.7-4= 0.7V

For Transistor Q1

Since same Current is flowing across both PMOS and Vdd is also same

Therefore

Vgs(Q1)= Vgs(Q2)= 0.7V