ECE 342 Solve problem 3 and 4

 

ECE 342

 

 

 

 

 

 

Problem Set #9

 

 

 

Due: 5 P.M. Wednesday, October 28, 2015

 

 

 

Fall Semester 2015

 

 

 

 

 

 

 

Reading Assignment: Sections 6.6.3, 6.6.4, 6.6.6, 5.6.6

 

 

 

  1. Consider the circuit shown below,

 

 

 

 

(a) Find Gv (vout/vsig) for IBIAS values of 0.1, 0.2, 0.5, 1 and 1.25 mA.

 

(b) Why isn’t Gv a linear function of IBIAS?

 

(c) For IBIAS = 1mA, what is the maximum allowed value of Vsig, where vsig = Vsig*sin( t)? The circuit must provide linear amplification.

 

 

 

 

 

 

10V

 

 

 

 

 

vsig ∞ VCC vout ∞ IBIAS RB=480kΩ -VEE RL=100kΩ ∞ Rsig=10kΩ RC=10kΩ ß= 100 VA=25V

 

-10V

 

 

 

  1. Consider the circuit shown below,

 

 

 

 

(a) Choose the value of Re such that Gv is maximized, subject to the constraint that vbe ≤10mV and the transistor stays in the active mode. What is the value of Gv?

 

(b) If βdrops by 10%, what is the new value of Gv? All other design variables are unchanged.

 

 

 

 

vsig=0.05sin(ωt) ∞ +5V vout ∞ 0.2mA 100kΩ 20kΩ ∞ 20kΩ 20kΩ ß= 100 VA=∞ Re

 

-5V

 

 

 

 

  1. Find Rin, Rout, Gv, and the overall current gain io/isig (“Gi”). You are given that β100 and VA=∞ VBE,ON = 0.7V.

 

 

 

 

100kΩ vsig ∞ 5V -5V 3.3kΩ 2kΩ vo io ii

  1. Choose the value of IBIAS such that vout is a sinusoidal signal with amplitude 200 mV or larger.

 

 

 

 

5V 10kΩ vout 75Ω ∞ vsig=0.5sin(ωt) VIN=3V IBIAS ß= 100 VA=∞

 

  1. Consider the circuit shown below,

 

 

 

(a) Find the dc bias point and small-signal model parameters. Assume λ= 0.

 

 

 

 

(b) Find Rin, Rout, Avo and Gv.

 

(c) Repeat part (b) for the case that λ= 0.03. You will have to recalculate ro.

 

(d) Finally, you will consider the body effect; i.e., you will no longer assume that VB = VS but, instead, that VB = -VSS, which is -5V in this circuit. You are given γ= 0.4V1/2 and 2ϕF = 0.6V. You may assume λ= 0 for simplicity. First, you need to recalculate the dc bias point. You may iteratively solve for VS and Vt, using ( ) and (√√). Do not iterate more than 2 or 3 times, as this should be sufficient to obtain the value of Vt with less than 10mV error. Then, redraw the small-signal model, now including the gmb current source shown in Fig. 5.62 of the textbook. The value of gmb can be found using equations (5.110) and (5.111). You are to calculate the value of Gv.

 

 

 

 

 

 

1MΩ vsig ∞ 5V 5kΩ vout 4.7MΩ -5V ∞ 0.5mA Vto=0.75V k=2mA/V2

 

 

 

 

 

Attachments:ECE 342 Problem Set #9 Due: 5 P.M. Wednesday, October 28, 2015 Fall Semester 2015 Prof. E. Rosenbaum Prof. T. Trick Reading Assignment: Sections 6.6.3, 6.6.4, 6.6.6, 5.6.6 1. Consider the circuit shown below, (a) Find Gv (vout/vsig) for IBIAS values of 0.1, 0.2, 0.5, 1 and 1.25 mA. (b) Why isn’t Gv a linear function of IBIAS? (c) For IBIAS = 1mA, what is the maximum allowed value of Vsig, where vsig = Vsig*sin( t)? The circuit must provide linear amplification. vsig ∞ VCC vout ∞ IBIAS RB=480kΩ -VEE RL=100kΩ ∞ Rsig=10kΩ RC=10kΩ ß= 100 VA=25V 2. Consider the circuit shown below, (a) Choose the value of Re such that Gv is maximized, subject to the constraint that vbe ≤ 10mV and the transistor stays in the active mode. What is the value of Gv? (b) If β drops by 10%, what is the new value of Gv? All other design variables are unchanged. vsig=0.05sin(ωt) ∞ +5V vout ∞ 0.2mA 100kΩ 20kΩ ∞ 20kΩ 20kΩ ß= 100 VA=∞ Re -5V 10V -10V 3. Find Rin, Rout, Gv, and the overall current gain io/isig (“Gi”). You are given that β=100 and VA=∞. VBE,ON = 0.7V. 100kΩ vsig ∞ 5V -5V 3.3kΩ 2kΩ vo io ii 4. Choose the value of IBIAS such that vout is a sinusoidal signal with amplitude 200 mV or larger. 5V 10kΩ vout 75Ω ∞ vsig=0.5sin(ωt) VIN=3V IBIAS ß= 100 VA=∞ 5. Consider the circuit shown below, (a) Find the dc bias point and small-signal model parameters. Assume λ = 0. (b) Find Rin, Rout, Avo and Gv. (c) Repeat part (b) for the case that λ = 0.03. You will have to recalculate ro. (d) Finally, you will consider the body effect; i.e., you will no longer assume that VB = VS but, instead, that VB = -VSS, which is -5V in this circuit. You are given γ = 0.4V1/2 and 2ϕF = 0.6V. You may assume λ = 0 for simplicity. First, you need to recalculate the dc bias point. You may iteratively solve for VS and Vt , using ( ) and (√ √ ). Do not iterate more than 2 or 3 times, as this should be sufficient to obtain the value of Vt with less than 10mV error. Then, redraw the small-signal model, now including the gmb current source shown in Fig. 5.62 of the textbook. The value of gmb can be found using equations (5.110) and (5.111). You are to calculate the value of Gv. 1MΩ vsig ∞ 5V 5kΩ vout 4.7MΩ -5V