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HW3-solution

# HW3-solution - EE456 HW3 SOLUTIONS Figure 1 shows an NMOS...

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EE456 HW3 SOLUTIONS Figure 1 shows an NMOS inverter with resistive load. Use 0.25um TSMC 2.5V CMOS model parameters, i.e., for NMOS, VT0=0.456V, UO=0.0339m 2 /Vs, and TOX=5.8e-9m. Figure 1. Resistive-load inverter a. ) Qualitatively discuss why this circuit behaves as an inverter. SOLUTIONS For Vin < V T , M1 is in cutoff regime, thus I=0 and Vout=2.5V. For Vin > V T , M1 is conducting and Vout=2.5V - (I*R). This in turn gives a low Vout and the input signal is inverted. b. Calculate V OH , V OL , V IH , V IL , V M , NM H , and NM L Assuming negligible leakage, when \Vin < V T , transistor M1 is off and V OH =2.5V. For Vin=2.5V, assume M1 is in the linear region, and because V DS is negligible in the linear region, channel-length modulation can be ignored.

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Therefore V OL =26.89mV. Checking the assumption: V GS -V T (=2.044V), and V DS =26.89mV, thus, M1 was correctly assumed to be in the linear region. To find V M , set the resistor current equal to the NMOS current, with an input and output voltage of V M . M1 is in the saturation region Therefore V M =0.735V.
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