Question

Consider a long-channel NMOS transistor with source and body connected together. Assume that the electron mobility is independent of V_GS and V_DS. Given, g_m = 0.5 μA/V for V_DS = 50 mV and V_GS = 2 V,

g_d = 8 μA/V for V_GS = 2 V and V_DS = 0 V,

Where \({{\rm{g}}_{\rm{m}}} = \frac{{\partial {{\rm{I}}_{\rm{D}}}}}{{\partial {{\rm{V}}_{{\rm{GS}}}}}}\)  and \({{\rm{g}}_{\rm{d}}} = \frac{{\partial {{\rm{I}}_{\rm{D}}}}}{{\partial {{\rm{V}}_{{\rm{DS}}}}}}\) 

The threshold voltage (in volts) of the transistor is ________.

Answer 1

Concept:

Condition of Operation of NMOS in linear region:

V_DS < [V_DS - V_T]

‘OR’

V_DS < V_GS

And the current equation is:

 I_D = K_n’ [V_GS - V_T] V_DS

Calculation:

Given:

g_m = 0.5 μA/V for V_ds = 50 mV, V_GS = 2V,

g_d = 8 μA/V for V_GS = 2V, V_DS = 0 V

Since,

V_GS > V_DS, MOSFET is in linear operation,

So, I_D = K_n’ (V_GS - V_T) V_DS

\(\frac{{\partial {I_D}}}{{\partial {V_{GS}}}} = K_n'\;{V_{DS}}\) 

\({g_m} = K_n'\;{V_{DS}}\)                 

\(\therefore {g_m} = \frac{{\partial {I_D}}}{{\partial {V_{GS}}}}\) (Given)

So, .5 × 10^-6 = K_n’ [50 × 10^-3]

⇒ K_n’ = 10^-5

Now,

\(\frac{{\partial {I_D}}}{{\partial {V_{DS}}}} = K_n'\left[ {{V_{GS}} - {V_T}} \right]\) 

\({g_d} = K_n'\;\left[ {{V_{GS}} - {V_T}} \right]\) 

\(\therefore {g_d} = \frac{{\partial {I_D}}}{{\partial {V_{DS}}}}\)  (Given)

So, 8 × 10^-6 = 10^-5 [2 - V_T]

V_T = 1.2 V