A resistive load inverter in two logic states is given as follows: Let

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A resistive load inverter in two logic states is given as follows: Let VTN = 0.6 V. For the circuit shown in the figure, VL = 0.6 V, VH = 2.4 V, VDD = 2.4 V, Kn′ = 100 μA/V2, VDS = 0.5, VTN = 0.6 V and power P = 0.0003 W. Find the new values of R and (W/L)S assuming that the power dissipation remains the same? R = Ω (W/L)S = 1/

A resistive load inverter in two logic states is given as follows: Let VTN = 0.6 V. For the circuit shown in the figure, VL = 0.6 V, VH = 2.4 V, VDD = 2.4 V, Kn′ = 100 μA/V2, VDS = 0.5, VTN = 0.6 V and power P = 0.0003 W. Find the new values of R and (W/L)S assuming that the power dissipation remains the same? R = Ω (W/L)S = 1/

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Required information A resistive load inverter in two logic states is given as follows:

Let

V_{T N} = 0.6 V

.

For the circuit shown in the figure,

V_{L} = 0.6 V, V_{H} = 2.4 V, V_{D D} = 2.4 V, K_{n^{'}} = 100 μ A / V^{2}, V_{DS} = 0.5, V_{T N} = 0.6 V

and power

P = 0.0003

W.

Find the new values of

R

and (W/L)S assuming that the power dissipation remains the same?

R = ◻ Ω

{(\frac{W}{L})}_{S} = 1 /

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