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

• M. Horowitz, J. Plummer, R. Howe

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E40M Review – Part 1

• M. Horowitz, J. Plummer, R. Howe

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Current

• Current

– It is the flow of charge – All devices, and wires (nodes) are charge neutral

• Current into a device = current flowing out of

device

• 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 zero

• Series devices must have the same current

i1 i2 i3 i1 = i2 + i3

• M. Horowitz, J. Plummer, R. Howe

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Example: Kirchhoff’s Current Law (KCL)

i1 i2 i3 i4 i5 i6 i8 i7 i9 i10 i1 = i2 = i3 = and i4 = i8 =

• M. Horowitz, J. Plummer, R. Howe

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Voltage

• Voltage

– It is the potential energy for electricity – It is always relative (only defined between two points) – Must be consistent

• Voltage around any loop must sum to zero

– + and – signs are only assumed polarities, actual calculated or measured voltages may be + or - .

• KVL

– Sum of device voltages around any loop is zero

• Parallel devices must have the same

voltage

+ + +

• V1

V2 V3 V1 + V2 – V3 = 0

• M. Horowitz, J. Plummer, R. Howe

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Loop #1: Loop #2:

Example: Kirchhoff’s Voltage Law (KVL)

• M. Horowitz, J. Plummer, R. Howe

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Using KCL and KVL

• Find the current, and voltages for the circuit below

3V 100mA

+ + + + +

• V1

5V V3 8V i1 i2 i3 50mA

• M. Horowitz, J. Plummer, R. Howe

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Power

• Power = iV Measured in Watts (= Volt *Amp)

– It is the flow of energy

• Energy is measured in Joules
• Watts = Joules/sec
• For devices that absorb energy power flows

into device – Current flows from higher to lower voltage

• For devices that supply energy, power flows
• ut 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

Absorb or Provide Power?

+

• 2V
• 1 A

+

• 5V

1 A

• M. Horowitz, J. Plummer, R. Howe

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Electrical Devices

• We learned about many different

electrical devices.

Resistors Diodes Inductors Capacitors Light Emitting Diodes Motors Transistors

+ –

5 V

Device Current: i Voltage Source

• M. Horowitz, J. Plummer, R. Howe

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Electrical Devices – Some Properties

• Charge neutral; i.e., charge entering =

charge leaving – Batteries or power supplies separate charge but the overall device is still charge neutral

• The net current into any device is

always zero, so iIN = iOUT – Current that flows into one end of a wire must flow out the other – Often called KCL (Kirchhoff’s Current Law)

• Dissipate power (P = i·V)

+ –

5 V

Device Current: i Voltage Source iIN iOUT

• M. Horowitz, J. Plummer, R. Howe

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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 v i + , v + i - , v + i - , v - i + , v -

V i

1 2 3 4

+ –

V

i

+ –

i

+ –

• M. Horowitz, J. Plummer, R. Howe

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Resistor Circuits

Find the resistance between node a and node b

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The Power of Redrawing a Circuit +

i

• 1

kΩ 1 kΩ 1 kΩ 2kΩ

• M. Horowitz, J. Plummer, R. Howe

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Nodal Analysis: The General Solution Method

1. Label all the nodes (VA, VB, or V1, V2, etc.), after selecting the node you choose to be Gnd. 2. Label all the branch currents (i1, i2, 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

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Example: Nodal Analysis

5V

+

• 2V

+

• 1

kΩ 4kΩ

1mA

2kΩ

Compute the node voltages and branch currents.

• M. Horowitz, J. Plummer, R. Howe

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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

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Example: Superposition

5V

+

• 2V

+

• 1

kΩ 4kΩ

1mA

2kΩ

A

Compute VA using superposition.

• M. Horowitz, J. Plummer, R. Howe

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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

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Electrical Devices - Solar Cells

• 0.5
• 0.3
• 0.1

0.1 0.3 0.5 0.7 0.9

Short Circuit Current Open Circuit Voltage i v i

• Incoming photons create current.
• If no external current path (i = 0), current flows through diode.

Maximum power provided

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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 Vf
• If you guessed that the diode was on
• You fixed the voltage to be Vf
• 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

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Example: Diode Circuit

50Ω 50Ω 1V 2V R1 10mA

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Example 2: Diode Circuit

• Don’t really want to randomly choose diode state in this case

1k 3k 1mA +Vy - + Vx -

• M. Horowitz, J. Plummer, R. Howe

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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

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• 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).

Capacitors Only Affect Time Response not Final Values

@t = ∞

@t = 0

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Example: RC Time Domain Analysis

5V

A CMOS inverter is driven with a 1 GHz square wave input. Assume the transistor Ron = 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

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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

• But for large Δt

– Inductors look like short circuits (they are a wire)

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Example: RL Time Domain Analysis

1 kΩ

10kΩ

1 mH

5V

A If the switch opens at t = 0 after being closed for a long time, what is the voltage at node A at t = 0+?

• M. Horowitz, J. Plummer, R. Howe

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Example: RL Time Domain Analysis

1 kΩ

10kΩ

1 mH

5V

A 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 the time dependence of the current decay is exponential as was the case for RC circuits discussed in class.