Question 9.1: Parameters for a depletion-mode MOSFET Assume that a deplet...
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 \times 10^{15} cm^{-3} , and the body-effect parameter \gamma = 0.5 V^{1/2} . The MOSFET is connected in the circuit shown, and a current equal to 30 μA is measured with the supply voltage V_{ss} = 0 . When V_{ss} is increased to 1 V, the current is reduced to 23.1 μA.
I_D=\mu _nC_{ox}\frac{W}{L} \left[\left(V_G-V_T-\frac{1}{2}V_D \right)V_D \right] (9.1.5)
I_{Dsat}=\mu _nC_{ox}\frac{W}{2L} (V_G-V_T)^2 (9.1.6)
Calculate the threshold voltage V_T(0) when the source voltage V_s = 0 V, and also find the prefactor \mu _nC_{ox} \ W/L in Equations 9.1.5 and 9.1.6.
Because V_{DS} ≥ 4 V for both measurements and V_{GS} = 0 V in the circuit, we assume that V_{DS} > (V_{GS} – V_T) so that Equation 9.1.6 can be applied to both measurements. We will check the calculated value of V_T 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 \left|\phi _p\right| . From the given dopant concentration in the substrate, we calculate \left|\phi _p\right| = 0.3 V (Equation 4.2.9b).
\Delta V_T=\frac{\sqrt{2\epsilon _sqN_a} }{C_{ox}} \left(\sqrt{2\left|\phi _p\right|+\left|V_{SB}\right| }-\sqrt{2\left|\phi _p\right| } \right) \\ =\gamma \left(\sqrt{2\left|\phi _p\right|+\left|V_{SB}\right| }-\sqrt{2\left|\phi _p\right| } \right) (9.1.11)
\phi _p=\frac{-kT}{q} \ln\frac{N_a}{n_i} (4.2.9b)
At V_{ss} = 0 V, the source is at 0 V and
30=\mu _nC_{ox}\frac{W}{2L} \left[0-V_T(0)\right]^2At V_{ss} = 1 V, the source is at 1 V and, from Equation 9.1.11,
\Delta V_T=0.5\left[\sqrt{0.6+1}-\sqrt{0.6} \right] =0.245 \ VBecause the body effect makes the threshold voltage less negative for this n-channel MOSFET, V_T(1) = V_T(0) + 0.245 . Using the measured current at V_{ss} = 1 V, we have
23.1=\mu _nC_{ox}\frac{W}{2L} \left[0-(V_T(0)+0.245)\right] ^2Hence,
\left[\frac{I_D(0)}{I_D(1)} \right] ^{1/2}=1.14=\frac{-V_T(0)}{-V_T(0)-0.245}From this equation, we calculate V_T(0) = -2 V. Using either measured current value, we have \mu _nC_{ox}W/L=15 \mu A \ V^{-2} . The calculated value of the threshold voltge is consistent with our assumption that the MOSFET is in saturation because V_{DS} > (0 – V_T) for either of the measured conditions.