• University of California, Irvine

• EECS 170c

Eq. 1.b. 𝑔 𝑚 = 2 * 𝐼 𝐷𝑆 / (𝑉 𝐺𝑆 − 𝑉 𝑡 ) = (2 * 3. 15𝐸 − 4) / (1. 63 − 1. 40) = 2. 73𝐸 − 3 Result: The given 𝑔 in LTSpice is 2.77E-3. The percent difference is about 1.45%, which is 𝑚 very low and shows that the calculation and the model both provide similarly accurate small signal parameter 𝑔 . 𝑚 Eq. 1.c. 𝑟 0 = 1 / λ * 𝐼 𝐷 = 1 / (. 005 * 3. 15𝐸 − 4) = 634920. 63 Result: There was no 𝑟 present on the output listing, therefore there is nothing to compare the 0 𝑟 small signal parameter to. 0 d. Figure 1.c. Transient Analysis of Common-Drain Amplifier with Vin Sine Wave and DC Component Equal to Vin (eq1), or 3.2 V. Observations: The main thing observed is that both curves are almost identical. The only difference is that the amplitude of Vin is greater than the amplitude of Vout by less than .001 V. e. Figure 1.d. Transient Analysis of Common-Drain Amplifier with Vin Sine Wave and DC Component Equal to Vin (eq1) Increased by 10 mV, so 3.21 V