Question

Notify Me Bookmark

The circuit shown below is an example of a common-source amplifier biased with a PMOS current mirror. The CMOS process used to fabricate the circuit results in NMOS transistors with unCox = 100 uA/V^2 , lambda_n = 0.1 V^-1 , and VThn = 0.4 V; and PMOS transistors with upCox = 50 uA/V^2, lambda_p = 0.2 V^-1 , and VThp =

The circuit shown below is an example of a common-source amplifier biased with a PMOS current mirror. The CMOS process used to fabricate the circuit results in NMOS transistors with unCox = 100 uA/V^2 , lambda_n = 0.1 V^-1 , and VThn = 0.4 V; and PMOS transistors with upCox = 50 uA/V^2, lambda_p = 0.2 V^-1 , and VThp =

Image text
The circuit shown below is an example of a common-source amplifier biased with a PMOS current mirror. The CMOS process used to fabricate the circuit results in NMOS transistors with unCox = 100 uA/V^2 , lambda_n = 0.1 V^-1 , and VThn = 0.4 V; and PMOS transistors with upCox = 50 uA/V^2, lambda_p = 0.2 V^-1 , and VThp =

Detailed Answer

0 0

Please enter your email here. You will get email when this question is answered by our tutor.

Answer
Doubtrix Logo

This Problem has been solved!

You'll get a detailed, step-by-step and expert verified solution.

Join Doubtrix with 7-days free trial

  • Icon Correct & Verified Answers
  • Icon No AI-Generated Answers
  • Icon Video Solution for Difficult Questions
  • Icon Work With Experts to Reach at Correct Answers
Login Sign Up

Similar/Related Questions

The following figure shows a pseudo-NMOS inverter gate, in which gate-grounded PMOS acts like resistive load. Assume VDD = 2.0V, unCox = 100 uA/V2, upCox = 50 uA/V2, VTHN = 0.4V, VTHP = -0.4V, lambda_n = 0, and lambda_p = 0. When (W/L)n =10 and Vin = VDD, Vout = 0.2V. Determine (W/L)p. Also assume that both NMOS and PMOS operate in linear region although PMOS actually operates in saturation region.
Solution
1 0

unCox = 4*upCox = 400 uA/V^2

The differential amplifier in the figure below is fabricated in a 0.18-um CMOS technology for with unCox = 400 uA/V^2, |Vt| = 0.5 V, and |VA'| = 10 V/um. If the bias current I = 200 uA and all transistors have a channel length twice the minimum and are operating at |vov| = 0.2 V, find W/L for each of Q1, Q2, Q3, and Q4, and determine the differential voltage gain Ad. The differential amplifier of Fig. 9.12(a) is fabricated in a 0.18-um CMOS technology for which un Cox = 4upCox = 400 uA/V^2, |Vt| = 0.5 V, and |VA'| = 10 V/um.
Solution
1 0

Part 2: Analysis of a CMOS Inverter's Dynamic Behavior Objective: Perform hand calculations of switching delays through a CMOS inverter. Consider a CMOS inverter such as the one shown in Figure 2. The delay times, trise and tsunv , will be determined by the current-driving capacities of the PMOS and NMOS transistors, respectively, as well as the magnitude of the load capacitance CL. Figure 2. CMOS Inverter The rise time trice for a CMOS inverter can be calculated as: tnse = CLΔV IDAvg , where IDP wg is the average of the PMOS currents at the 10% point of the low-to-high transition (Vout = Vtow ) and at the 90% point (Voor = Vsow ). Likewise, the fall time tfall for a CMOS inverter can be calculated as: tsanl = CLΔV IDavg , where ID timg is the average of the NMOS currents at the 90% of the high-to-low transition (Vout = VSop ) and at the 10% point (Vous = V10%). ⇒ Calculate the expected rise and fall times, trise and tfall . Use LCM equations with a load capacitance value of 25 pF. Use VT0 = 0.4 V, kn′ = 432 μA/V2, |2ϕϝ| = 0.88 V, γ = 0.2 V1 /2, λ = 0 V−1, WN = 2 μm, and LN = 100 nm for the NMOS device. Use Vpv = −0.4 V, kp′ = 108 μA/V2, |2ϕF| = 0.88 V, γ = 0.2 V1 /2, λ = 0 V−1, Wp = 8 μm, and Lp = 100 nm for the PMOS device.

Solution
0 0