An inverter consists of QN connected to VDD = VGS = 2.5 V by RD. Given Vt = 0.5 V, K’n =100E-6 A/V.V. If ID =50 E-6 A, and VOL = 100 mV. Calculate: rDS, W/L of QN, RD, and P dissipation.

Question

An inverter consists of QN connected to VDD = VGS = 2.5 V by RD. Given Vt = 0.5 V,  K’n =100E-6 A/V.V.  If  ID =50 E-6 A, and VOL = 100 mV. Calculate:  rDS, W/L of QN, RD, and P dissipation.

Answer

None of the above

Answer

2 k ohm, 2, 58 K ohm, and 150E-6 W

Answer

2 k ohm, 2.5, 48 K ohm, and 125E-6 W

Answer

2 k ohm, 1.67, 68 K ohm, and 175E-6 W

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An inverter consists of QN connected to VDD = VGS = 2.5 V by RD. Given Vt = 0.5 ...

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