1. Synergy of Design and Analysis Product Specifications Circuit Design Circuit Structure (topology)- creative part Determination of Parameter Values (R, L, C, transistor sizes, etc.) – mathematical optimization Circuit Analysis Check against specifications If specs are met, done Otherwise, improve/modify the design until specs are met EE101 Fall 2012 Lect 2- Kang1

2. Electronic Circuits consisted of R, L, C, Independent Sources, Dependent Sources, Transistors, OP Amps, etc.. EE101 Fall 2012 Lect 2- Kang2 Circuit Design involves (1) Determination of a Structure/Topology for Right Components, and then (2) Determining Component Values

3. Design an Ohmmeter that measures the Resistance (R x ) of Resistors EE101 Fall 2012 Lect 2- Kang3 I fs = I m when R x = 0 Measure using (E/I- Ri), where Ri is the internal resistance in series. Using an Ammeter

4. EE101Lect2- Basic Concepts for Electronic and Electric Circuits EE101 Fall 2012 Lect 2- Kang4

5. Current Through R3? EE101 Fall 2012 Lect 2- Kang5

6. Simple Example of Resistive Circuits Problem: [Q] Find is the current through R 3 [Q] Find the power dissipated in R 3 SOLUTION: Let i3 (downward direction) be the current through R3. Let i2 (downward direction) be the current through R2. Let i1 (left to right direction) be the current through R1. Then i1=i2+i3 (Eq.1) Vs (=90V) = VR1+ VR2(=VR3)=10 (i2+i3)+60 X i3 (Eq.2) We are mainly interested in iR3. So how do we express i2 in terms of i3? The voltages across both R2 and R3 are same. VR2=VR3 and thus i2=60 X i3/30=2 X i3 (Eq.3) From Eqs. 2 and 3, 90=10(3 X i3)+60 X i3=90 X i3, thus i3=1 Amp. (Ans) The power dissipated in R3 is P3= V3x i3= (60x1) x 1= 60 Watt. (ans) This is like a 60W light bulb case, although real-life light bulbs are for AC (alternating current) of 110- 120V, whose average value is close to 90V DC. EE101 Fall 2012 Lect 2- Kang6

7. Some Observations (KVL, KCL) EE101 Fall 2012 Lect 2- Kang7

8. Some Basic Mathematics EE101 Fall 2012 Lect 2- Kang8

9. Examples of Dependent Sources A device model for Bipolar Junction Transistor (BJT) shown in Fig. 3.40 of our textbook. B C E BC E V BE βiBβiB iBiB + - Diamond symbol stands for dependent sources. In this example, it is current- controlled current source (CCCS) whose magnitude depends on other current, iB. In contract independence sources are marked in circle or by the battery (DC) symbol. EE101 Fall 2012 Lect 2- Kang9

10. Examples of Dependent Sources (continued) Voltage-controlled voltage source (VCVS) Let us study the case of an operational amplifier (OP Amp) shown in Fig. 5.4 of our textbook. + RiRi RoRo A v d vdvd _ + + _ _ a b c a b c Ri = input resistance, usually very large = almost infinite (∞). Ro = output resistance, usually very small. A = amplification factor, usually very large in thousands or even more. EE101 Fall 2012 Lect 2- Kang10

11. How doe we handle dependent sources? EE101 Fall 2012 Lect 2- Kang11

12. A Resistive Circuit with both Independent and Dependent Sources We can apply KCL to find v1, v2, v3: Node 1: is+2ix=(v1-v2)/R1= is+2 (v3-v2)/R3 (1) Node 2: (v1-v2)/R1 + (v3-v2)/R3=v2/R2(2) Node 3: 3 (v3-v2)/R3= -v3/R4(3) Here the values of is, R1, R2, R3, R4 are known. So, we have 3 (independent) equations with 3 unknowns of v1, v2, v3 and thus can find node voltages. (matrix determinant method can be used.) After finding these voltages, we can also find currents (ik) through all branch Elements and power dissipation in each element (Pk=vk x ik). EE101 Fall 2012 Lect 2- Kang12

13. . EE101 Fall 2012 Lect 2- Kang13

14. EE101 Fall 2012 Lect 2- Kang14