Single Loop Circuits

* with a current source* with a voltage source* with multiple sources* voltage divider circuits * Equivalent resistance

Superposition method

* Principle* Procedures* How to apply

Lecture 3. Single Loop Circuits & Superposition Method

1

2

Single Loop Circuits

• The same current flows through each element of the circuit—the elements are in series.

+– VS

R1

R2

Rn

I V1

V2

IS

R1

R2

Rn

V1I

V2

V3V3

With an independent voltage source With an independent current source

3

• What is I?

Single Loop Circuits – with a Current Source

sII

• In terms of I, what is the voltage across each resistor?

111 RIIRV s

…

IS

R1

R2

Rn

V1I

V2

V3

222 RIIRV s

nsnn RIIRV

4

• To solve for I, apply KVL around the loop.

+– VS

R1

R2

Rn

I + –

I R2

+

–

I R1

I Rn

+

–IR1 + IR2 + … + IRn – VS = 0

Single Loop Circuits – with a Voltage Source

i

s

n

s

R

V

RRR

VI

...21

• In terms of I, what is the voltage across each resistor?

11 IRV 22 IRV …

nn IRV

5

With Multiple Voltage Sources

• The current i(t) is:

• Resistors in series

sresistanceofsumsourcesvoltageofsum

RV

tij

Si )(

jNequivalent RRRRR 21

6

Voltage Division

• Consider two resistors in series with a voltage v(t) across them:

R1

R2

– v1(t)+

+

–

v2(t)

+

–

v(t)

21

11 )()(

RR

Rtvtv

21

22 )()(

RR

Rtvtv

• If n resistors in series:

j

iSR R

Rtvtv

ki)()(

7

Voltage Divider: A Practical Example

Electrochemical Fabrication of

Quantum Point Contactor Atomic-scale wire

Molecular Junction

Anode: Etching delocalized, but Cathode: Deposition localized at sharpest point,

due to: • Self-focusing – directional growth

Decreasing Gap!

+

+

+

++

+

--+++

+

+

+

+

+

++ --

E

Voltage Divider: An Example

+

+

+

++

+

Vgap

A

Rext

Vext

V0

0VRR

RV

extgap

gapgap

• Initially, Rgap >> Rext, Vgap ~ V0 full speed deposition.

• Finally, Rgap << Rext, Vgap ~ 0 deposition terminates.

• The gap resistance is determined by Rext.

Voltage Divider: An Example

2

• Growth starts after applying 1.5 V

1

• Two electrodes with10 m initial separation

3

• Self-terminates after forming a tunneling gap

Voltage Divider: An Example

Ohmic behavior

Time (sec.)

G (

2e2 /

h)

Stepwise increase in Conductance

Voltage Divider: An Example

12

Example: Two Resistors in Parallel

How do you find I1 and I2?

I R1 R2 V

+

–

I1 I2

13

212121

11

RRV

R

V

R

VIII

I R1 R2 V

+

–

I1 I2

21

21

21

111

RR

RRI

RR

IV

• Apply KCL with Ohm’s Law

Example: Two Resistors in Parallel

Equivalent Resistance of Parallel Resistors

• Two parallel resistors is often equivalent to a single resistor with resistance value of:

21

21

RR

RRReq

inpar RRRRR

11111

21

• n-Resistors in parallel:

15

21

2

1

21

21

11 RR

RI

R

RRRR

I

R

VI

What are I1 and I2 ?

• This is the current divider formula

• It tells us how to divide the current through parallel resistors

21

1

2

21

21

22 RR

RI

R

RRRR

I

R

VI

16

Circuits with More Than One Source

How do we find I1 or I2?

Is1 Is2 VR1 R2

+

–

I1 I2

17

21212121

11

RRV

R

V

R

VIIII ss

21

2121 RR

RRIIV ss

Is1 Is2 VR1 R2

+

–

I1 I2

• Apply KCL at the Top Node

What if More Than One Source?

18

Class Examples

• Example: P1-33 (page 43).

• Drill Problem P1-34 (page 43).

19

Superposition Method – A More General Approach to Multiple Sources

“In any linear circuit containing multiple independent sources, the current or voltage at any point in the circuit may be calculated as the algebraic sum of the individual contributions of each source acting alone.”

How to Apply Superposition

• To find the contribution due to an individual independent source, zero out the other independent sources in the circuit– Voltage source short circuit– Current source open circuit

• Solve the resulting circuit using your favorite technique(s)

21

Superposition of Summing Circuit

+

–Vou

t

1k

1k

1k

V

1

V

2

+–

+–

+

–V’

ou

t

1k

1k

1k

V

1

+

–V’’

out

1k

1k

1k

V

2

++–

+–

22

V’out = V1/3

V’’out = V2/3

Vout = V’out + V’’out = V1/3 + V2/3

+

–

V’out

1k

1k

1k

V1

+

–

V’’out

1k

1k

1k

V2++–

+–

Superposition of Summing Circuit (cont’d)

23

Superposition Procedure

1. For each independent voltage and current source (repeat the following): a) Replace the other independent voltage sources with

a short circuit (i.e., V = 0). b) Replace the other independent current sources with

an open circuit (i.e., I = 0). Note: Dependent sources are not changed!

c) Calculate the contribution of this particular voltage or current source to the desired output parameter.

2. Algebraically sum the individual contributions (current and/or voltage) from each independent source.

24

Class Examples

• Example 2-9 (page 70).

• Drill Problem 2.7.