Parameters for a depletion-mode MOSFET Assume that a depletion-mode n-channel MOSFET can be described by Equations 9.1.5 and 9.1.6 if a negative value is used for the threshold voltage. The substrate doping is 1.63 × 10^15 cm^-3, and the body-effect parameter γ = 0.5 V^½. The MOSFET is connected

Question

Parameters for a depletion-mode MOSFET

Assume that a depletion-mode n-channel MOSFET can be described by Equations 9.1.5 and 9.1.6 if a negative value is used for the threshold voltage. The substrate doping is [latex] 1.63 imes 10^{15} cm^{-3} [/latex], and the body-effect parameter [latex]\gamma = 0.5 V^{1/2} [/latex]. The MOSFET is connected in the circuit shown, and a current equal to 30 μA is measured with the supply voltage [latex] V_{ss} = 0 [/latex]. When [latex] V_{ss} [/latex] is increased to 1 V, the current is reduced to 23.1 μA.

[latex] I_D=\mu _nC_{ox}\frac{W}{L} \left[\left(V_G-V_T-\frac{1}{2}V_D ight)V_D ight] [/latex] (9.1.5)

[latex] I_{Dsat}=\mu _nC_{ox}\frac{W}{2L} (V_G-V_T)^2 [/latex] (9.1.6)

Calculate the threshold voltage [latex] V_T(0) [/latex] when the source voltage [latex] V_s = 0 [/latex] V, and also find the prefactor [latex] \mu _nC_{ox} \ W/L [/latex] in Equations 9.1.5 and 9.1.6.

Accepted Answer

Because [latex] V_{DS} ≥ 4 [/latex] V for both measurements and [latex] V_{GS} = 0 [/latex] V in the circuit, we assume that [latex] V_{DS} > (V_{GS} - V_T) [/latex] so that Equation 9.1.6 can be applied to both measurements. We will check the calculated value of [latex]V_T [/latex] later to see that this assumption is true. Because the source voltage changes, we must consider the body effect using Equation 9.1.11, and we therefore need a value for [latex] \left|\phi _p ight| [/latex]. From the given dopant concentration in the substrate, we calculate [latex] \left|\phi _p ight| [/latex] = 0.3 V (Equation 4.2.9b).

[latex] \Delta V_T=\frac{\sqrt{2\epsilon _sqN_a} }{C_{ox}} \left(\sqrt{2\left|\phi _p ight|+\left|V_{SB} ight| }-\sqrt{2\left|\phi _p ight| } ight) \ =\gamma \left(\sqrt{2\left|\phi _p ight|+\left|V_{SB} ight| }-\sqrt{2\left|\phi _p ight| } ight) [/latex] (9.1.11)

[latex] \phi _p=\frac{-kT}{q} \ln\frac{N_a}{n_i} [/latex] (4.2.9b)

At [latex] V_{ss} = 0 [/latex] V, the source is at 0 V and

[latex] 30=\mu _nC_{ox}\frac{W}{2L} \left[0-V_T(0) ight]^2 [/latex]

At [latex] V_{ss} = 1 [/latex] V, the source is at 1 V and, from Equation 9.1.11,

[latex] \Delta V_T=0.5\left[\sqrt{0.6+1}-\sqrt{0.6} ight] =0.245 \ V [/latex]

Because the body effect makes the threshold voltage less negative for this n-channel MOSFET, [latex] V_T(1) = V_T(0) + 0.245 [/latex]. Using the measured current at [latex] V_{ss} = 1 [/latex] V, we have

[latex] 23.1=\mu _nC_{ox}\frac{W}{2L} \left[0-(V_T(0)+0.245) ight] ^2 [/latex]

Hence,

[latex] \left[\frac{I_D(0)}{I_D(1)} ight] ^{1/2}=1.14=\frac{-V_T(0)}{-V_T(0)-0.245} [/latex]

From this equation, we calculate [latex] V_T(0) = -2 [/latex] V. Using either measured current value, we have [latex] \mu _nC_{ox}W/L=15 \mu A \ V^{-2} [/latex]. The calculated value of the threshold voltge is consistent with our assumption that the MOSFET is in saturation because [latex] V_{DS} > (0 - V_T) [/latex] for either of the measured conditions.

Question 9.1: Parameters for a depletion-mode MOSFET Assume that a deplet...

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