E40M Review Part 1 M. Horowitz, J. Plummer, R. Howe 1 Current - PowerPoint PPT Presentation

E40M Review – Part 1 M. Horowitz, J. Plummer, R. Howe 1

Current • Current – It is the flow of charge – All devices, and wires (nodes) are charge neutral i 1 • Current into a device = current flowing out of device i 3 i 2 • All devices must have at least two terminals – The arrows are only assumed current directions. Calculated or measured currents may be + or - . • KCL – Sum of current flowing into node or device is i 1 = i 2 + i 3 zero • Series devices must have the same current M. Horowitz, J. Plummer, R. Howe 2

Example: Kirchhoff’s Current Law (KCL) i 3 i 6 i 1 i 2 i 8 i 4 i 7 i 5 i 10 i 9 i 1 = i 2 = i 3 = and i 4 = i 8 = M. Horowitz, J. Plummer, R. Howe 3

Voltage • Voltage – It is the potential energy for electricity V 1 – It is always relative (only defined between - two points) + – Must be consistent - V 2 V 3 • Voltage around any loop must sum to zero + – + and – signs are only assumed - + polarities, actual calculated or measured voltages may be + or - . • KVL V 1 + V 2 – V 3 = 0 – Sum of device voltages around any loop is zero • Parallel devices must have the same voltage M. Horowitz, J. Plummer, R. Howe 4

Example: Kirchhoff’s Voltage Law (KVL) Loop #1: Loop #2: M. Horowitz, J. Plummer, R. Howe 5

Using KCL and KVL • Find the current, and voltages for the circuit below V 1 V 3 - - + + + + + 8V 100mA 3V i 1 5V i 3 i 2 - - - 50mA M. Horowitz, J. Plummer, R. Howe 6

Power • Power = iV Measured in Watts (= Volt *Amp) Absorb or Provide – It is the flow of energy Power? • Energy is measured in Joules • Watts = Joules/sec -1 A 1 A • For devices that absorb energy power flows into device - + – Current flows from higher to lower voltage -5V 2V • For devices that supply energy, power flows - + out of device – Current flow from lower to higher voltage • Remember the above description is about the higher voltage – And not the pin which has a + label • And a negative voltage across it M. Horowitz, J. Plummer, R. Howe 7

Electrical Devices • We learned about many different Current: i electrical devices. + Inductors Capacitors Device Resistors Diodes 5 V Voltage Source – Transistors Light Emitting Diodes Motors M. Horowitz, J. Plummer, R. Howe 8

Electrical Devices – Some Properties • Charge neutral; i.e ., charge entering = Current: i charge leaving – Batteries or power supplies separate i IN charge but the overall device is still charge neutral + • The net current into any device is Device 5 V Voltage always zero , so i IN = i OUT Source – – Current that flows into one end of a wire must flow out the other – Often called KCL (Kirchhoff’s Current i OUT Law) • Dissipate power (P = i · V) M. Horowitz, J. Plummer, R. Howe 9

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 i-V curves for these devices. i i + + i + , v - i + , v + V i 2 1 – v – 3 4 i - , v + i - , v - i – + V M. Horowitz, J. Plummer, R. Howe 10

Resistor Circuits Find the resistance between node a and node b M. Horowitz, J. Plummer, R. Howe 11

The Power of Redrawing a Circuit 1 k Ω + 1 k Ω i 2k Ω - 1 k Ω M. Horowitz, J. Plummer, R. Howe 12

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 13

Example: Nodal Analysis 2V Compute the node voltages 1 k Ω and branch currents. - + + 5V 4k Ω 2k Ω - 1mA M. Horowitz, J. Plummer, R. Howe 14

Superposition For Linear Circuits • Reason: – Resistors, voltage, and current sources are linear – Resulting equations are linear • What’s the benefit? – Superposition enables the analysis of several simpler circuits in place of one complicated circuit M. Horowitz, J. Plummer, R. Howe 15

Example: Superposition 2V 1 k Ω Compute V A using superposition. A - + + 5V 4k Ω 2k Ω - 1mA M. Horowitz, J. Plummer, R. Howe 16

Electrical Devices – Diodes • Diode is a one-way 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 1V 2V 10mA R1 50 Ω 50 Ω M. Horowitz, J. Plummer, R. Howe 20

Example 2: Diode Circuit • Don’t really want to randomly choose diode state in this case 1k + Vx - 1mA 3k +Vy - M. Horowitz, J. Plummer, R. Howe 21

Electrical Devices - Capacitors • What is a capacitor? – It is a new type of two terminal device – It is linear • Double V, you will double I – We will see it doesn’t dissipate energy • Stores energy • Rather than relating i and V – Relates Q, the charge stored on each plate, to Voltage – Q = CV – Q in Coulombs, V in Volts, and C in Farads • Like all devices, it is always charge neutral – Stores +Q on one lead, -Q on the other lead M. Horowitz, J. Plummer, R. Howe 22

Capacitors Only Affect Time Response not Final Values • Capacitors relate i to dV/dt This means if the circuit “settles down” and isn ’ t changing with • time, a capacitor has no effect (looks like an open circuit). @t = 0 @t = ∞ M. Horowitz, J. Plummer, R. Howe 23

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 = 2 pF. Will the inverter produce “1” and “0” values at its output, if “1” means > 4V and “0” means < 1V? 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 Δ t inductors look like current sources – They can supply very large voltages (+ or -) – And not change their current • But for large Δ t – Inductors look like short circuits (they are a wire) M. Horowitz, J. Plummer, R. Howe 25

Example: RL Time Domain Analysis 5V If the switch opens at t = 0 after being closed for a long time, what is the voltage at node A at t = 0 + ? 1 mH 10k Ω A 1 k Ω M. Horowitz, J. Plummer, R. Howe 26

Example: RL Time Domain Analysis 5V After the switch opens at t = 0, how long does it take the voltage at node A to decrease to 1V? You can assume that 1 mH 10k Ω the time dependence of the current decay is exponential as was the case for RC circuits discussed in class. A 1 k Ω M. Horowitz, J. Plummer, R. Howe 27