please explain why u do what u do; Consider the circuit shown in Fig. 1, with NM

Question

please explain why u do what u do;

Consider the circuit shown in Fig. 1, with NMOS body terminal connected to the source terminal, and Vt0=O.4V,lambda=0, kn =4mA/V2. Analyze the circuit to determine the voltages at all the nodes and currents through all the branches. At what value of Vcc will the transistor be at the edge of the triode (linear) region? At what value of RD will the transistor be at the edge of the triode region? At what value of Rs will the transistor be at the edge of the triode region?

Explanation / Answer

as there is no current into the gate of the NMOS,

voltage at gate=(1.2/(50k+100k) )*100k=0.8 volt

assuming the NMOS to be in saturation region,

voltage at source let be v.

let current be i mA.

then v=10*i

Vds=1.8-15*i

Vgs-Vt=0.8-10*i-0.4=0.4-10*i

i=4*(0.4-10*i)^2

i=4*(100*i^2-8*i+0.16)

i=0.03 mA

so voltage at source=10*i=0.3 volt

Vds=1.35 volt

b)let value of Vcc be V for which it is in the edge of triode region.

then Vds=Vds(sat)=Vgs-Vt

i.e

V-15*i=0.4-10*i

V=0.4+5*i

now i=0.03 mA

ao V=0.4+5*0.03=0.55 volts

c)let value of Rd be R kiloohms

then Vds=1.8-(R+10)*i

at edge of triode region,

Vds=Vgs-Vt

1.8-(R+10)*i=0.4-10*i

1.4=R*i

R=1.4/i=46.67

so value of Rd=46.67 kiloohms

d)

let value of Rs be R kiloohms

then Vds=1.8-(R+5)*i

at edge of triode region,

Vds=Vgs-Vt

1.8-(R+5)*i=0.4-10*i

1.4=(R-5)*i

R=51.67

so value of Rd=51.67 kiloohms

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