E40M Review - Part 1 - PDF Free Download

E40M Review Part 1 Topics in Part 1 (Today): KCL, KVL, Power Devices: V and I sources, R Nodal Analysis. Superposition Devices: Diodes, C, L Time Domain Diode, C, L Circuits Topics in Part 2 (Wed): MOSFETs, CMOS Circuits, Logic Gates Binary Numbers Time Division Multiplexing Frequency Domain Circuits, Impedance Filters, Bode Plots Op Amps M. Horowitz, J. Plummer, R. Howe 1

KCL and KVL KCL Sum of current flowing into node or device is zero KVL Sum of device voltages around any loop is zero V 1 i 1 i 2 i 3 i 1 = i 2 i 3 V 3 V 2 V 1 V 2 V 3 = 0 M. Horowitz, J. Plummer, R. Howe 2

Using KCL and KVL Find the current, and voltages for the circuit below V 1 V 3 100mA 3V i 1 i 2 5V i 3 50mA 8V M. Horowitz, J. Plummer, R. Howe 3

Power Power = iv Measured in Watts (= Volt *Amp) It is the flow of energy Energy is measured in Joules Watts = Joules/sec Absorb or Provide Power? 1 A 1 A For devices that absorb energy power flows into device Current flows from higher to lower voltage 2V 5V For devices that supply energy, power flows out of device Current flow from lower to higher voltage M. Horowitz, J. Plummer, R. Howe 4

Electrical Devices Current: i We learned about many different electrical devices. 5 V Voltage Source Device Resistors Diodes Inductors Capacitors Light Emitting Diodes Motors Transistors M. Horowitz, J. Plummer, R. Howe 5

Electrical Devices Some Properties Current: i i IN Charge neutral; i.e., charge entering = charge leaving 5 V Voltage Source Device Batteries or power supplies separate charge but the overall device is still charge neutral The net current into any device is always zero, so i IN = i OUT i OUT Dissipate power (P = i V) M. Horowitz, J. Plummer, R. Howe 6

Electrical Devices Voltage Source, Current Source, Resistor Note that the energy is dissipated by the device in quadrants 1 and 3, and power is generated by the device in quadrants 2 and 4. Sketch the iv curves for these devices. i i V i i, v 2 1 i, v v 3 4 i, v i V i, v M. Horowitz, J. Plummer, R. Howe 7

Resistor Circuits 10KΩ 10KΩ A What is the resistance between nodes A and B? 20KΩ 10KΩ 20KΩ 10KΩ 10KΩ B M. Horowitz, J. Plummer, R. Howe 8

The Power of Redrawing a Circuit 1kΩ 2kΩ i 1kΩ 1kΩ M. Horowitz, J. Plummer, R. Howe 9

Nodal Analysis: The General Solution Method 1. Label all the nodes (V A, V B, or V 1, V 2, etc.), after selecting the node you choose to be Gnd. 2. Label all the branch currents (i 1, i 2, etc.) and choose directions for each of them 3. Write the KCL equations for every node except the reference (Gnd) Sum of the device currents at each node must be zero 4. Substitute the equations for each device s current as a function of the node voltages, when possible 5. Solve the resulting set of equations M. Horowitz, J. Plummer, R. Howe 10

Example: Nodal Analysis 1kΩ 2V Compute the node voltages and branch currents. 5V 4kΩ 2kΩ 1mA M. Horowitz, J. Plummer, R. Howe 11

Example: Nodal Analysis 1kΩ A 2V B Compute the node voltages and branch currents. 5V i 1 i 2 4kΩ i 3 i 4 2kΩ 1mA At node A: i 1 = i 2 i 3 1. 5 V A 1kΩ = V A 4kΩ i 3 At node B: i 3 = i 4 1mA Substituting 2. into 1. 5 V A 1kΩ 5 V A 1kΩ = V A 4kΩ V A 2V 2kΩ 1mA V A = V A 4kΩ V B 2kΩ 1mA 7 4kΩ = 7mA V A = 4V 2. i 3 = V B 2kΩ 1mA 3. V A = V B 2V V A = 4V, V B = 2V, i 1 = 1mA, i 2 = 1mA, i 3 = 0, i 4 = 1mA M. Horowitz, J. Plummer, R. Howe 12

Superposition For Linear Circuits Calculate the response of the circuit for each independent source at a time, with the other s turned off What happens when we turn off a source? Voltage sources: have 0 V (are shorted replace by a wire) Current sources: have 0 current (are opened replace by a broken wire) I V X = I V I X V shortcircuited so V = 0 opencircuited so I = 0 M. Horowitz, J. Plummer, R. Howe 13

Example: Superposition 1kΩ A 2V Compute V A using superposition. 5V 4kΩ 2kΩ 1mA M. Horowitz, J. Plummer, R. Howe 14

Example: Superposition 1kΩ A Compute V A using superposition. 5V 4kΩ 2kΩ 4KΩ 2kΩ = 4 3 kω V A1 = 5V 4 3 kω 4 kω 1kΩ 3 = 20 7 V 1kΩ A 2V 4kΩ i 2kΩ 4KΩ 1kΩ = 4 5 kω i = 2V 2kΩ 4 = 10 5 kω 14 ma = 5 7 ma V A2 = 5 7 ma 4 5 kω = 4 7 V M. Horowitz, J. Plummer, R. Howe 15

Example: Superposition 1kΩ A 4KΩ 2kΩ 1kΩ = 4 7 kω 4kΩ 2kΩ 1mA V A3 = 1mA 4 7 kω = 4 7 V V A = 4 7 V 4 7 V 20 7 V = 4V M. Horowitz, J. Plummer, R. Howe 16

Electrical Devices Diodes Diode is a oneway street for current Current can flow in only one direction An idealized diode model Is a voltage source for positive current Voltage drop is always equal to Vf for any current Is an open circuit for negative current Current is always zero for any voltage M. Horowitz, J. Plummer, R. Howe 17

Electrical Devices Solar Cells Incoming photons create current. If no external current path (i = 0), current flows through diode. i i Open Circuit Voltage v Short Circuit Current Maximum power provided 0.5 0.3 0.1 0.1 0.3 0.5 0.7 0.9 M. Horowitz, J. Plummer, R. Howe 18

Solving Diode Circuits Look at the circuit, guess the voltages and/or currents From this, guess whether the diode will be on or off If you can t estimate anything, just guess the diode state(s) Assume your guess was right Solve for the voltages in the circuit Then check your answer If you guessed the diode was off, Look at the resulting diode voltage Check to make sure it is less than V f If you guessed that the diode was on You fixed the voltage to be V f So check to make sure the current is positive If your guess was wrong, change the guess and resolve M. Horowitz, J. Plummer, R. Howe 19

Example: Diode Circuit 10mA 1V R1 50Ω 2V 50Ω M. Horowitz, J. Plummer, R. Howe 20

Example: Diode Circuit 10mA 1V R1 50Ω A i 1 i 2 2V 50Ω The current source is in the right direction to forward bias the diodes, so we might guess that both diodes are ON. 10mA = i 1 i 2 V A = 1V i ( 1 50Ω) = 2V i ( 2 50Ω) 10mA = 1V i 1 = 1V i ( 2 50Ω ) 50Ω = 1V 50Ω i 2 50Ω 2i 2 = 20mA 2i 2 so that i 2 = 5mA Clearly this is incorrect, so i 2 must = 0, i = 10mA and V = 1V ( 10mA) ( 50Ω) = 1.5V. 1 A M. Horowitz, J. Plummer, R. Howe 21

Example 2: Diode Circuit Don t really want to randomly choose diode state in this case 1k 1mA Vx 3k Vy M. Horowitz, J. Plummer, R. Howe 22

Example 2: Diode Circuit B E A C D Don t really want to randomly choose diode state in this case. The red line shows a reasonable guess as to the current path since this is through 3 forward biased diodes. If the forward voltage on each diode is 0.6V, then the node voltages are V A = 5.8V, V B = 5.2V, V C = 4.2V, V D = 3.6V, V E = 0.6V. The other three diodes are all reverse biased so this is a selfconsistent solution. M. Horowitz, J. Plummer, R. Howe 23

Electrical Devices Capacitors What is a capacitor? It is a new type of two terminal device It is linear It doesn t dissipate energy Rather than relating i and V Relates Q, the charge stored on each plate, to Voltage Q = CV So if Q = CV, and i=dq/dt i= C dv/dt Z C = 1/ j 2πFC so at low F it is an open circuit, at high F it is a wire. M. Horowitz, J. Plummer, R. Howe 24

Electrical Devices Inductors An inductor is a new type of two terminal device It is linear double V and you will double i Like a capacitor, it stores energy Ideal inductors don t dissipate energy Defining equation: V = L di/dt L is inductance (in Henrys) For very small Δt inductors look like current sources They can supply very large voltages ( or ) And not change their current so at low F, it looks like a wire, at high F an open circuit. Z L = j 2πFL M. Horowitz, J. Plummer, R. Howe 25

Example: Asymptotic Circuit Analysis i(t) i(t) = (2 A) cos(2πft) Find the power provided by the 2A current source when F = 0 Hz and when F à infinity M. Horowitz, J. Plummer, R. Howe 26

Example: Asymptotic Circuit Analysis: F = 0 Hz F = 0 Hz à current source is 2 A, DC v i M. Horowitz, J. Plummer, R. Howe 27

Example: Asymptotic Circuit Analysis: F à Hz i(t) v i M. Horowitz, J. Plummer, R. Howe 28

Example: RC Time Domain Analysis 5V A CMOS inverter is driven with a 1 GHz square wave input. Assume the transistor R on = 250 Ω and C = 5 pf. Will the inverter produce 1 and 0 values at its output, if 1 means > 4V and 0 means < 1V? i RES = V out /R 1 i CAP = CdV out /dt The output waveform will be symmetric. We need to see if it can reach 4V in ½ of the input period. F = 1 GHz, so ½ period = 0.5 nsec V = 5V 1 e t/r C 1 out So the answer is NO. dv out dt ( ) = 5V 1 e = V out R 1 C 0.5x10 9 ( ) ( 250) 5x10 12 = 1.64V M. Horowitz, J. Plummer, R. Howe 30

Example: RL Time Domain Analysis 10kΩ 5V 1mH If the switch opens at t = 0 after being closed for a long time, what is the voltage at node A at t = 0? A 1kΩ M. Horowitz, J. Plummer, R. Howe 31

Example: RL Time Domain Analysis If the switch opens at t = 0 after being closed for a long time, what is the voltage at node A at t = 0? When the switch is closed, the current flows through the 1kΩ resistor. i = 5V 1kΩ = 5mA When the switch opens this current has no where to go except through the 10kΩ resistor, so the voltage across the 10kΩ resistor will need to be V 10kΩ (t = 0 ) = ( 5mA) ( 10kΩ) = 50V So the voltage at node A will need to be 55 V. M. Horowitz, J. Plummer, R. Howe 32

Example: RL Time Domain Analysis 10kΩ 5V 1mH After the switch opens at t = 0, how long does it take the voltage at node A to decrease to within 1V of its final value? A 1kΩ M. Horowitz, J. Plummer, R. Howe 33

Example: RL Time Domain Analysis After the switch opens at t = 0, how long does it take the voltage at node A to decrease to within 1V of its final value? V R V L = 0 Ri t ( ) L di ( t ) dt = 0 L R di dt i t ( ) = 0 i( t) = I 0 e R L t = 5e R L t ma V 10kΩ = 1V when i = 0.1mA, so 0.1= 5e 10 4 10 t 3 or ln 0.1 5 = 104 10 t 3 = 1.609 = 10 7 t or t = 1.6x10 7 sec = 0.16µsec M. Horowitz, J. Plummer, R. Howe 34