# Exercise repeat the above example if is degenerated

This preview shows pages 670–674. Sign up to view the full content.

Exercise Repeat the above example if is degenerated by a resistor of value . Example 12.29 Analyze the circuit of Fig. 12.56(a), assuming that is not very large. V DD M V 1 M 2 R D2 R D1 b in R F out RF i V DD M V 1 M 2 R D2 R D1 b in out (c) (a) (b) V I X R F R F R F V I 2 1 V I Figure 12.56 Solution As a voltage-current feedback topology, this circuit must be handled according to the rules in Figs. 12.51(b) and 12.52(b). The forward amplifier is formed by , , , and . The feedback network simply consists of . The loop is opened as shown in Fig. 12.56(b), where,

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
BR Wiley/Razavi/ Fundamentals of Microelectronics [Razavi.cls v. 2006] June 30, 2007 at 13:42 646 (1) 646 Chap. 12 Feedback from Fig. 12.51(b), the output port of the sense duplicate is shorted . Since splits between and , we have (12.135) Noting that , we write (12.136) The open-loop input and output impedances are respectively given by (12.137) (12.138) To obtain the feedback factor, we follow the rule in Fig. 12.52(b) and construct the test circuit shown in Fig. 12.56(c), obtaining (12.139) (12.140) Note that both and are negative here, yielding a positive loop gain and hence confirming that the feedback is negative. The closed-loop parameters are thus expressed as (12.141) (12.142) (12.143) where is given by Eq. (12.136). Exercise Repeat the above example if is replaced with an ideal current source. Example 12.30 Analyze the circuit of Fig. 12.57(a), assuming is not small, , and the electronic device has an impedance of .
BR Wiley/Razavi/ Fundamentals of Microelectronics [Razavi.cls v. 2006] June 30, 2007 at 13:42 647 (1) Sec. 12.7 Effect of Nonideal I/O Impedances 647 M I SS M M M in V DD M 1 I out R M 3 4 5 6 M I SS M M M in DD M 1 I out 3 4 5 6 V V R L R M R M X X r O1 M I SS M M M in V DD M 1 3 4 5 6 V R M R M X r O1 i X V X (c) (a) (b) Laser V Figure 12.57 Solution This circuit employs current-voltage feedback and must be opened according to the rules shown in Figs. 12.51(d) and 12.52(d). The forward amplifier is formed by and - , and the feedback network consists of . Depicted in Fig. 12.52(d), the open-loop circuit contains two instances of the feedback network, with the output port of the sense duplicate and the input port of the return duplicate left open. The open-loop gain , and (12.144) To calculate , we note that the current produced by is divided between and : (12.145) where the negative sign arises because flows out of the transistor. The open-loop gain is therefore equal to (12.146) The output impedance is measured by replacing with a test voltage source and measuring the small-signal current [Fig. 12.57(c)]. The top and bottom terminals of respectively see an

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
BR Wiley/Razavi/ Fundamentals of Microelectronics [Razavi.cls v. 2006] June 30, 2007 at 13:42 648 (1) 648 Chap. 12 Feedback impedance of and to ac ground; thus, (12.147) (12.148) The feedback factor is computed according to the rule in Fig. 12.52(d): (12.149) (12.150) Forming , we express the closed-loop parameters as (12.151) (12.152) (12.153) where is given by Eq. (12.146). Exercise Construct the Norton equivalent of the entire circuit that drives the laser.
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### What students are saying

• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

Kiran Temple University Fox School of Business ‘17, Course Hero Intern

• I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

Dana University of Pennsylvania ‘17, Course Hero Intern

• The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

Jill Tulane University ‘16, Course Hero Intern