Basic Circuit Analysis

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Prof. N. Sujatha Phone: 6513 8079 Email: [email protected] Office: N1.3-B4-07

BG 2104: W1-W7

Basic concepts and basic circuit analysis Principles of semiconductor devices Diodes and applications (self-learning) Bipolar Junction Transistor (BJT)

Contents CA (15%)

Assignment on electric circuits

mailto:[email protected]

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Electrical circuits and analysis

An electrical system

3

Basic electrical quantities

• Electric charge (Q) – units of Coulombs (C)

– quantized in units of charge on electron e = -1.610-19 C - How many electrons would build up 1C charge?

+ - -

4

• Current (I)

– rate of flow of electric charge (water pipe analogy) – I = dQ/dt – units of Ampere = 1C/sec (A)


What motivates the flow of charges / electrons (here)?

5

Electric potential energy: is the energy stored in a system due to accumulation of charges, due the interaction between charges


charge

Electric potential Electric potential energy / unit charge

6

•Voltage (V) difference in the electric potential between 2 points (nodes) units of Volts (V)


7 Voltaic cell- diagrammatic representation

http://en.wikipedia.org/wiki/File:Galvanic_cell_with_no_cation_flow.png

Electron flow Conventional flow (used in this course)


–conventional flow: define the current direction as the direction of flow of positive charges (high potential low potential, opposite to the direction of electron flow. 8

Voltage sources

An ideal voltage source delivers a constant voltage regardless of the current it produces. Its internal resistance is zero.

Current sources

An ideal current source delivers a constant current regardless of the output voltage. Its internal resistance is infinity.

+ –

9

Symbol Convention

ID (VD) means DC current (voltage) only.

id (vd) means AC current (voltage) only.

iD (vD) means DC and AC current (voltage) in superposition.

Voltage source

Current source

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When a current is constant with time, we say that we have direct current, abbreviated as dc. On the other hand, a current that varies with time, reversing direction periodically, is called alternating current, abbreviated as ac.

Similarly, we have dc and ac voltage.

Direct and alternating signal

11

.

Examples of DC and AC currents versus time

I

12

AC currents (or voltages) can have various waveforms

13

0 V 0 V

+ 5 V

V

+ 5 V

- 5 V

AC signal can be superimposed upon a DC signal (value)

14

The basic electrical system: What else?

+

-

Charge flow rate

Voltage Load

15

Types of loads

1. Resistive 2. Capacitive 3. Inductive 4. Combination (practical)

Passive

Active Amplifiers

16

Resistance • Resistance (R)

-units of Ohms (Ω) -conductance G = 1/R in units of Siemens (S)

• Resistors: - is a two-terminal electronic component that resists the flow of current, producing a voltage drop between its terminals in accordance with Ohm's law.

- typical range 1Ω to100MΩ - types: Carbon (cheapest) Metal film (more stable, better accuracy) Wire wound (most accurate and most expensive)

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Ohm’s Law

For a resistor R, as shown below, the voltage drop from point a to b, V=Vab=Va-Vb is given by:

The power dissipated by the resistor is:

RVRIVIP /22 ===

IRV =

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Resistor- Value?

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Equivalent Resistance: in series

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122313

RRI

VVI

VVI

VVRR toteq

+=

−+

−=

−==

R1

R3 R2 Req

Req = R1 + R2 + R3

A series circuit is a single path for electric current through all of its components.

What if multiple resistances in the circuit?

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Equivalent Resistance: in parallel

At point A, I=I1+I2

21

21

212112

1212

/)(/)( RRRR

RVVRVVVV

IVVRR toteq +

=−+−

−=

−==

Req R3 R2 R1

321

1111RRRReq

++=

A parallel circuit has multiple paths for current components to pass separately. A parallel circuit provides the same voltage across all its branchs.

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Equivalent Resistance: combinations of series and parallel

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http://en.wikibooks.org/wiki/Image:Resistorscombo.png

Voltage divider- A series resistor circuit connected to a voltage source

+

==21

22 RR

RVVV inout

In general, ∑

==i

xinRout R

RVVVx

V1

V2

Vin = V1+ V2

+ -

Vin

23

Current divider

Before you move on to the next slide, can you try to solve for i1 and i2 using Ohm’s law and equivalent resistance? 24

Current divider

21

212211 RR

RRiRiRiRiv seqs +====

2121

12

21

21 iii

RRRii

RRRii sss +=

+=

+=

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Current divider

IS R2 V

+

–

R1 I1 I2

R3 I3

j

parSR R

RII

j=∑=+++=

iMpar RRRRR11111

21

In general,

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Circuit diagrams • A circuit diagram represents the interconnections between electrical units in a circuit.

• A circuit diagram is just one representation of the actual circuit, among many possible diagams that can equally represent the circuit.

• A circuit diagram do not reflect the actual physical lengths, sizes, shapes, placement of the electrical units.

• A line (or wire) in a circuit diagram does not necessarily indicates the physical existence of a connecting wire. It simply indicates that the points connected by this line in the diagram is electrically “identical”, i.e., they assumes the same voltage. • A node in diagram is where three or more circuit components are connected together.

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What are the differences between these two circuits? Please think carefully before move to the next slide.

Circuit diagrams

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Circuit diagrams Every circuit can be drawn in different ways without changing the physics of the circuit. In fact, the two circuits shown are identical !! Examine complicated diagrams and redraw them to clearly show how the elements are interconnected.

What is the voltage across R4?

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Circuit diagrams

R4 is shorted-out as if it doesn’t exist, therefore, no voltage across it; and no current passing through it.

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Simple circuit Analysis

Simple circuit can be analyzed simply by: • represent the circuit with a clear diagram

• identify parallel and series combinations of resistors

• replace such combinations by their equivalent resistors

• continue until the circuit is reduced to a straightforward parallel or series circuit

• finally use Ohm’s law, or the voltage and current divider rules, to find the individual resistor voltages and currents if required.

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1kΩ

1kΩ

2kΩ

1kΩ

2kΩ

1kΩ

+ – 10V

+

–

V1 +

–

V3 +

–

V2

Simple circuit analysis: an example

Find V1, V2, and V3

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1kΩ

1kΩ

2kΩ

1kΩ

2kΩ

1kΩ

10V +

–

V1 +

–

V3 +

–

V2

Find an equivalent resistance for the network with V1 across it, then find V1

using a voltage divider.

+ –


33

1kΩ

1kΩ

10V +

–

V1

V5k1k1

k1V101 =Ω+Ω

Ω=V

+ –


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1kΩ

1kΩ

2kΩ

1kΩ

2kΩ

1kΩ

10V +

–

5V +

–

V3 +

–

V2

Find an equivalent resistance for the network with V2 across it, then find V2.

+ –


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1kΩ

2kΩ

1kΩ

10V +

–

5V +

–

V2 1kΩ

V5.2k1k1

k1V52 =Ω+Ω

Ω=V

+ –


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1kΩ

1kΩ

2kΩ

1kΩ

2kΩ

1kΩ

10V +

–

5V +

–

V3 +

–

2.5V

V25.1k1k1

k1V5.23 =Ω+Ω

Ω=V

+ –


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1kΩ

1kΩ

2kΩ

1kΩ

2kΩ

1kΩ

+ – 10V

+

–

V1 +

–

V3 +

–

V2


Mark the currents I, I1, I2, I3 and I4

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I1= 5 A I2=2.5 A I3= 2.5 A I4 = 1.25 A

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• Simple circuit analysis can only be applied for a simple circuit. Often, circuit is a complicated network even after simplification.

• Simple circuit analysis cannot be applied systematically using computer program. • Therefore, network analysis based on Kirchoff’s current and voltage laws usually need to be applied.

Is simple circuit analysis applicable in all cases?

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Kirchoff’s current law (KCL) Current law: for any node of the circuit, the sum of the currents flowing into the node is equal to the sum of the currents flowing out of the node. Or the algebraic (net) sum of current flow is zero, if you assign inward and outward current with different sign.

∑∑ =outin

II

I1=I2+I3

KCL

think of water: what goes into a pipe junction must come out

Because charge cannot build up at a node.

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Kirchoff’s voltage law (KVL)

Voltage law: The algebraic sum of all the voltages across individual components around any loop of a circuit is zero.

v1+v2+v3+v4=0 KVL

think of gravity: what goes up must come down.

∑ =loop iV 0

Because electrical potential is conservative.

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Network analysis In a circuit diagram: • A branch is one path through which the same current flow through all the components in series in this path.

• A loop is a closed path in the circuit

• A node or junction is a point where three or more circuit components are joined.

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Network analysis Two methods for calculating currents and voltages in a circuit based on KCL and KVL: (1) Nodal method; (2) Mesh (or loop) current method.

• In all methods, usually it is a good idea to replace simple series and parallel combinations of resistances by their equivalent resistance to simplify the circuit. • Choose one node as a reference (ground) node, which is considered to have zero voltage. Make a convenient choice of ground to simplify the analysis. A good choice is the node with the most branches, or a node which can immediately give you another node voltage (e.g., directly connect to a voltage source).

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Nodal method

• Choose ground node, identify all nodes and assign node voltages to all nodes (other than reference node). They are the unknowns.

• First assume (arbitrarily) a current direction in each branch of the circuit (draw an arrow) and define a current symbol (e.g., I1). If there is a current source in the branch, the branch current is then specified by the current output from this source.

• A KCL equation is written for each node in terms of nodal voltages.

• Number of equations equals number of independent nodes (k-1 for k total nodes).

• Complete the analysis by working out branch currents

If a current in your solution comes out negative, it just means that it’s in the opposite direction from the one you initially assume. So don’t worry about the actual current directions before you start solving.

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Nodal method: an example

Assume: R1=R2=R3=1Ω; R4=R5=2Ω V1 = 1V V2 = 2V VA = 0.6 V and VB = 0.8 V

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Nodal method: another example with current sources

500Ω

500Ω I1

V1 V2

Ω+

Ω−

=500500

1211

VVVI

KCL at node 1

500Ω

500Ω

1kΩ

500Ω

500Ω I1

3 mA

1 2 3

V1 V2 V3

I2 4mA

i1

i2

i3

i4

i5

i1

i2

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Nodal method: another example with current sources

500Ω

1kΩ

500Ω V2 V3 V1

Ω−

+Ω

=Ω

−500k1500

32221 VVVVV

KCL at node 2

Ω=

Ω−

+500500

3322

VVVI500Ω

500Ω

I2

V2 V3 KCL at node 3

V1 = 1.33V, V2=1.17V, V3=1.58V

i2

i3

i4

i4

i5

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Mesh (loop) current method • Choose ground node and identify meshes (loops).

• Assign a current (imaginary loop current) to each mesh (they are unknowns). By convention, define all mesh currents in the clockwise direction.

• Write a KVL equation in terms of mesh currents for each mesh.

• Complete the analysis by working out the branch currents. (actual branch current = algebraic sum of relevant imaginary loop currents at a given branch).

1kΩ

1kΩ

1kΩ

V1 V2 I1 I2 + –

+ –

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Mesh (loop) current method: voltages across resistors

R

I1

+ – VR

VR = I1 R

R

I1

+ – VR I2

VR = (I1 - I2 ) R When R is shared by two loops: Actual current through R = algebraic sum of relevant imaginary loop currents following through R 50

Mesh (loop) current method: an example

If, Vbb=1V, All Ri = 1 ohm, I1 = 0.5 A I2 = I3 = 0.25A Actual current passing R2 is I1-I2=0.25 A (downwards direction)

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1kΩ

2kΩ

2kΩ

12V 4mA

2mA

I0

+ –

Mesh (loop) current method: another example with current sources

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• The current sources will have whatever voltage necessary produce the nominal current which cannot be determined immediately. We can’t use KVL around the loop because we don’t know the voltages across the current sources.


I1 I2

I3 1kΩ

2kΩ

2kΩ

12V 4mA

2mA

I0

+ –

What to do?

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• The 4mA current source sets I2: I2 = -4 mA

• The 2mA current source sets a constraint on I1 and I3:

I1 - I3 = 2 mA • We have two equations and three

unknowns. Where is the third equation?


Notice the negative sign. Why?

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1kΩ

2kΩ

2kΩ

12V

4mA

2mA

I0 I1 I2

I3

+ –

Apply KVL to supermesh

12V - I3*2kΩ - (I3 - I2)*1kΩ - (I1 - I2)*2kΩ = 0

ground node

Supermesh is highlighted by the dashed yellow line and excludes current sources.

Solution: I1 = 1.2 mA; I2 = -4 mA; I3 = -0.8 mA; I0 = I1 -I2 = 5.2 mA 55

Linear superposition principle

For any linear circuit containing more than one independent voltage source or current source, the total current in any part of the circuit equals the algebraic sum of the currents produced by each source separately. To isolate the current from a particular source

• replace all other voltage sources by short circuits • replace all other current sources by open circuits

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Linear superposition principle: example 1

ARR

Vi s

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483'

212 =

+=

+=

Find I2 using superposition

ARR

Rii s

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4882''

21

12 =

+⋅

=+

=

''' 222 iii +=57

2kΩ 1kΩ

2kΩ 12V

I0

2mA

4mA – +


Solve for I0?

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2kΩ 1kΩ

2kΩ

I’0

2mA

I’0 = -4/3 mA why?

• 2mA source contribution


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2kΩ 1kΩ

2kΩ

I’’0

4mA

I’’0 = 0 why?

• 4mA source contribution


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2kΩ 1kΩ

2kΩ 12V

I’’’0

– +

I’’’0 = -4 mA why?

• 12V source contribution


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I’0 = -4/3 mA I’’0 = 0

I’’’0 = -4 mA

I0 = I’0+ I’’0+ I’’’0 = -16/3 mA

• Final result


As a practice, can you analyze this circuit using nodal or mesh current analysis? 62

1kΩ

1kΩ

2kΩ

1kΩ

2kΩ

1kΩ

+ – 10V

+

–

V1 +

–

V3 +

–

V2

Varying load

Estimation of load currents for changing loads

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Thevenin equivalent circuit

Thévenin's theorem for electrical networks states that any combination of sources and resistors with two terminals is electrically equivalent to a single voltage source Vth in series with a single resistor Rth. For single frequency AC systems the theorem can also be applied to general impedances (AC resistance).

• Vth = open circuit voltage at the output port • Rth = Vth / (short circuit current at the output port)

Rth also equals to the equivalent resistance looking into the circuit while replace voltage sources with shorts and current sources with open circuits. 64

The Thevenin idea is most useful when one considers two circuits or circuit elements, with the first circuit’s output providing the input for the second circuit. In the figure, the output of the first circuit (A), consisting of Vth and Rth, is fed to the second circuit element (B), which consists simply of a load resistance (RL) to ground. This simple configuration represents, in a general way, a very broad range of analog electronics.

Thevenin equivalent circuit

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Circuit loading

)/(1 Lth

th

Lth

Lthout RR

VRR

RVV+

=

+

=

Thus, we should try to keep the ratio Rth/RL small in order to avoid the loss in voltage due to loading the circuit . A maximum ratio of 1/10 is often used as a rule of thumb in circuit design. 66

1kΩ

1kΩ

2kΩ

1kΩ

2kΩ

1kΩ

+ – 10V

+

–

V1 +

–

V3 +

–

V2

Varying load

Draw the Thevenin equivalent circuit for the below circuit

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1kΩ

2kΩ

1kΩ

2kΩ

1kΩ

+ – 10V

+

–

+

–

+

–

To find Thevenin voltage

?

3.64V

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1kΩ

2kΩ

1kΩ

2kΩ

1kΩ

+

–

+

–

+

–

Draw the Thevenin equivalent circuit for the below circuit

For finding Thevenin resistance

1.91 kΩ

69

To find the load current through 1kΩ

3.64V

1.91 kΩ

1 kΩ

Load current = 3.64 V /(1.91+1) kΩ =1.25 mA Voltage across 1 kΩ resistor = 1.25V Note that you have obtained the same result using simple circuit analysis! (Refer slide number 36) But, Thevenin circuit is useful for computing load current in a variable load condition; its only a one time effort! 70

Norton equivalent circuit

Norton's theorem for electrical networks states that any network of sources and resistors can be replaced by a single current source in parallel with a single resistor. For single frequency AC systems the theorem can also be applied to general impedances, not just resistors.

• Int = short circuit current at the output port • Rnt = (open circuit voltage at the output port) / Int Rnt also equals to the equivalent resistance looking into the circuit while replace voltage sources with shorts and current sources with open circuits. (Rnt= Rth).

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• Short A and B, calculate the current: Int=Vin/R1 • Now replace voltage sources with shorts and current sources with open circuits. • Evaluate the resistance across points A and B "looking back" into the circuit. Rnt=R1*R2/(R1+R2)

A

B

A

B

nt nt

Example: Norton equivalent circuit

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nteqth IRV =eq

thnt R

VI =

Vth

Rth

+ –

Int Rnt

nttheq RRR ==

Thevenin equivalent vs Norton equivalent

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Revisiting types of loads

1. Resistive 2. Capacitive 3. Inductive 4. Combination (practical)

Passive

Active Amplifiers

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Capacitor • Stores charge on electrodes (parallel conductive plates separated by insulator)

Q=VC

• Capacitance measure in units of farads (F = 1 C/volt)

• Range of typical values (1pF to 1000 μF)

• Capacitor types ceramic (pF) mylar (nF) electrolytic (μF) 75

Capacitor

For DC voltages, no current passes through a capacitor. It “blocks DC”. When a time varying potential is applied, the current passing through the capacitor is:

The impedance (or AC resistance) of a capacitor is:

fCjCjZC πω 2/1/1 ==

Where ω is the frequency. ZC 0 when ω ∞

dtdVCI =

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in series

Equivalent Capacitance

in parallel

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Inductor

• Energy stored in magnetic field

• Measure in units of henries (H)

• Typical range (1μH to 1H)

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Inductor

LjZL ω=

It passes DC current, and resist AC current. The impedance of an inductor is:

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in series

Equivalent Inductance

in parallel

neq LLL

L 111

21

+++=

neq LLLL +++= 21

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Semiconductor Devices

The semiconductor devices we will focus in this course behave very differently from resistors, capacitors, and inductors that are familiar to you. Semiconductors all regulate, in some manner, the amount of current that can pass through them in a given direction.

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Basic Circuit Analysis