Chapter 3

Applications of Resistive Circuits

Objectives Learn the properties and the model of a real source Learn the characteristics of an amplifier Learn the major features of an operational amplifier

3.1 REAL SOURCES AND POWER TRANSFER

Real sources are …… physical devices, such as batteries and generators, capable of supplying electric

energy different from ideal sources in two aspects :

they cannot deliver unlimited amounts of power they dissipate power internally

Source Models and Loading

A real voltage source can be characterized as the following v-i curve in Figure 3.1(a):

Figure 3.1 terminal voltage measured under open-circuit conditions :sv as increasing current i , v might typically decrease nonlinearly modeling : with sufficiently small current, its v-i curve can be approximated by a

straight line, which slope is , that is, the linearized v-i relationship becomes

sR−

s s sv v R i i v R= − s

the lumped-element model is shown in Figure 3.1(b) source voltage :sv : series source resistance sR

the real voltage source have internal resistance that cause an internal voltage drop

1

and power dissipation when 0i ≠ A real current source can be characterized as the following v-i curve in Figure 3.2(a):

Figure 3.2

modeling : with sufficiently small voltage, its v-i curve can be approximated by a

straight line, which slope is 1 sR− , that is, the linearized i-v relationship

becomes

s s s sv R v R i= −i i

the lumped-element model is shown in Figure 3.2(b) : source current, short-circuit current si : parallel source resistance sR

From the concept of source conversion introduced in Section 2.5, there is no theoretical distinction between the two source networks in Figure 3.1(b) and Figure 3.2(b). However, the practical distinction emerges when the loading effect is considered.

Loading effect : a load resistance R is attached to a real voltage

source, then we have a voltage divider with

L

Ls s

s L

RsR i v

R R= − =

+v v

causes the load voltage sR i sv v< when s s LR Rv v≈

a load resistance R is attached to a real current source, then we have a current divider with

L

ss s

s s L

v Ri i iR R R

= − =+

sv R causes the load current si i<

when s s LR Ri i≈

2

From the above, the key distinction between practical sources is the value of relative to

sR

LR A “good” voltage source should have small internal resistance

compared to , so sR

LR sv v≈ A “good” current source should have large internal resistance

compared to , so sR

LR si i≈

Example 3.1 Model of a Battery Example 3.1 A Paradox Resolved

Figure 3.4

3

Power Transfer and Efficiency

For a real source, internal resistance limits the amount of power that can be transferred to a load resistance.

If a load resistance is connected to a real voltage source with internal resistance , then

power delivered to the load via

LR sv

sR

Lp ( )s si v R R= + L

is

( ) ( )

22 2

2 2 ,1

L sL L s

s ss L

R vp R i vR RR R

α αα

= = = =++

LR .

v.s. Lp L sR R

Figure 3.6 is at its maximum value

Lp2

max 4s

s

vpR

=

when 1, L sL sR R tha is Rt R==

is called the maximum available power, since a source with fixed values of cannot deliver more power than

maxp

sv maxp If a load resistance is connected to a real current source with R in

parallel, then

power delivered to the load via

LR si s

Lp s

s L

RsR R

=+

i i

is

( ) ( )

22 2 2

2 2 ,1

L s LL L s s s

ss L

R R Rp R i i R iRR R

α αα

= = = =++

.

the maximum power is Lp2

max 4s sR ip = , when 1, L sL sR R tha is Rt R==

4

maximum power transfer theorem : If a source has fixed nonzero , then maximum power transfer to a load

resistance requires sR

L sR R= when , the load is said to be matched to the source for maximum power

transfer LR R= s

this theorem applies to any linear source network with Thévenin resistance

t sR R=

The power dissipated by internal (source) resistance : sR is ohmic heating is equal to

( )

22

ss s

s L

Rp R i vR R

= =+

2s for a voltage source

( )

22

2s L

s ss L

R Rp R i iR R

= =+

2s

L

for a current source

steadily decreases as increases and when

sp LR sp p= L sR R=

power-transfer efficiency :

L

L s

pp p+

Eff

the total power generated by the source is , so it is desired that the wasted internal power should be small compared to , and thus Eff should be large

Lp p+ s

L

sp Lp

for maximum power transfer ( ), , and thus Eff=0.5=50% L sR R= sp p=

at this moment, the terminal voltage drops to 2sv v=

Maximum power transfer is often not desired since the power-transfer efficiency is only 50%. Instead, higher power-transfer efficiency is sought by making as small as possible

sp

When do we want maximum power transfer? Primarily in applications where the source are used to convey

information rather than to deliver large amounts of power

5

Figure 3.7

Example 3.3 Calculating Power Transfer and Efficiency

6

3.2 AMPLIFIER MODELS

amplifiers: generally “enlarge” the variations of an voltage or current waveform is achieved with the help of electronic devices, primarily transistors

the large voltage or current being controlled must come from something other than the input source, so every amplifier needs an power supply

Here, we are not concerned with the details of electronic amplification, and instead we focus on the equivalent models of amplifiers and how they are used to determine the input-output relationships

voltage amplifier : ideally, ( ) ( )out v sv t A v t= , : the overall voltage amplification vA

if , then such amplifier inverts the output waveform 1vA <− the plot of versus is called a transfer curve outv inv

Figure 3.9

model of a voltage amplifier :

Figure 3.10 : input resistance, to reflect the property that the amplifier may draw current

from the applied source iR

thus, the terminal voltage differs from the source voltage inv sv

7

: voltage gain of the VCVS µ for a noninverting amplifier

for an inverting amplifier 0µ >0µ <

the open-circuit output voltage c iv vµ= n

: output resistance, to account for power dissipation within the amplifier oR The output side thus takes the form of a Thévenin network, and equals the

Thévenin equivalent resistance seen looking back into the output terminals with the input source suppressed

oR

The practical amplification v out iA v v= n

v

the loadings ( and ) at input and output results in and

sR LR in sv ≠out cv v≠

the internal resistances in the amplifier results in vA µ≠ calculation of :

vA

out in c outv

s s in

v v v vAv v v v

= = × ×c

since

in i

s s

v Rv R R=

+ i

, c

in

vv

µ= , out L

c o

v Rv R R=

+ L

so, in i L

vs s i o

v R RAv R R R R

µ= = × ×+ + L

vA µ< , which results from the loading effects For a “good” voltage amplifier, in order to let , large input

resistance ( ) and small output resistance ( ) are required

vA µ≈

iR Rs LoR R

model of a current amplifier :

Figure 3.11

: current gain of the CCCS β the short-circuit output current c ii iβ= n

The output side takes the form of a Norton network consisting of the CCCS in parallel with the Thévenin equivalent resistance oR

the overall current amplification i outA i i= s

8

out s oi

s s i o

i R RAi R R R R

β= = × ×+ + L

a “good” current amplifier with has small input resistance ( ) and large output resistance ( )

iA β≈ i sR R

o LR R in some applications, in order to have more voltage or current gain than can be

achieved with a single amplifier, two or more amplifiers are connected in cascade

Example 3.4 Cascading Voltage Amplifier

Figure 3.12

9

3.3 OP-AMPS

op-amp : operational amplifier first appeared as vacuum-tube circuits, and then as the integrated version

Operational Amplifiers

The name “operational amplifier” actually refers to a large family of general-purpose and special-purpose units having the distinctive feature that ….

Op-amps provide high-gain amplification of the difference between two input voltages

The op-amp schematic diagram and its symbol :

Figure 3.13

two input voltages : v and , one output voltage : v , measured with

respect to a single reference point identified by the ground emblem

p n out

p n

n

v

the ground point is established external to the op-amp by the power-supply connections

two equal dc supply voltages : and , which allows the output voltage to be in the range to .

PSV+ PSV−

PSV− PSV+

since only signal voltage (v and v ) is concerned, the two dc supply

voltages are often not shown in the op-amp symbol, as in Figure 3.13(b) Two inputs :

the noninverting input : the input terminal marked + in the op-amp symbol the inverting input : the input terminal marked − in the op-amp symbol

the difference voltage :

which entirely controls the output voltage d pv v v= −

10

The transfer curve relating to shown in Figure 3.14 (a): outv dv

Figure 3.14

linear region : maxdv v≤ , ( )out d p nv Av A v v= = − , out PSv V≤

increasing at the noninverting terminal causes v to move in the

positive direction

p outv

increasing at the inverting terminal causes to move in the negative direction

nv outv

saturation region : maxdv v> , out PSv V≈ , the output waveform is distorted as shown in Figure 3.14(b)

When operating within in linear region, an op-amp behaves like the simplified model in Figure 3.15

Figure 3.15

open circuit at the input terminals, so 0i i= =p n

a grounded VCVS produces at the output terminal, usually

dAv10, 000A ≥

Due to the gigantic value of and relatively small value of , to operate the A PSV

11

op-amp in the linear region, the difference voltage must be very small dv Analysis :

max max

max

PS out PS

d

PS

V v Vv v v

v V A

− < <+− < <

→ ≈

for and V,

10, 000A ≥ 30PSV ≤

max 3 mVdv v< ≤

In view of the limitation on , an op-amp alone is not a practical amplifier because any signal voltage with variations more than a few millivolts would drive it into saturation. Consequently, Linear op-amp circuits always include negative feedback connecting the

output terminal to the inverting terminal

dv

The feedback results in a greatly reduced difference voltage to prevent saturation.

The amplification with feedback becomes nearly independent of the gain , which is an important consideration since A tends to fluctuate unpredictably, sometimes increasing or decreasing by a factor of 2 or greater.

A

Noninverting Op-Amp Circuits

noninverting voltage amplifier : has the same polarity as outv inv

Figure 3.16 , 0p in in pv v i i= = = the resistors and constitute a feedback connection from the output to

the inverting terminal and then to the ground point FR 1R

Analysis (from Figure 3.16(b)):

12

1

1

(1) 0,

: :

(2) ,

1 , 1 1

: from back to

n n outF

n

n d in out d

d in n in out in d

d in out in

n outd n

d

Ri v Bv BR R

v B

v v v v Av

v v v v Bv v ABv

Av v v vAB AB

v BvAB v vv v

= ∴ = =+

+ = =

∴ = − = − = −

→ = =+ +

=

feedback voltage feedback factor

loop gain

∵

∵

∵

: , the overal amplification 1

d

out

in

AB

A vAB v

=

+close - loop gain

for most op-amps, the gain is very large, thus the loop gain

, and then the input-output relation simplifies to

A1AB

1 AB AB+ ≈

1

1

1 Fout in in

R Rv vB R

+≈ = v

the close-loop gain depends almost entirely upon the feedback factor associated with the resistive voltage divider, as distinguished

from the unreliable gain of the op-amp itself B

A the difference voltage

1d inv v

AB≈ inv

feedback indeed reduces as required for linear amplification of dv inv an equivalent circuit for the noninverting amplifier when 1AB

Figure 3.17

there are no loading effects at input or output, and so and

in sv v=

1

1out Fv

s F

v RAv R R

= = =+ B

By picking appropriate values of and , you can design for any FR 1R

13

moderate amplification . The op-amp gain is unimportant, provided that to satisfy the loop-gain condition.

vA

vA A In practical applications, the external resistances should be in the range

from 1kΩ to 100kΩ, corresponding to the amplification 101A ≤v

Larger values of can be obtained by connecting two op-amp circuits in cascade.

vA

When adjustable amplification is needed, a potentiometer may be used for voltage divider 1 FR R−

voltage follower : a special type of noninverting amplifier

Figure 3.18 , thus

10, FR R→ → ∞

1

11

n

out

out

in

vBv

v Av A

= =

= ≈+

Since , the output voltage “follows” the input. out inv v≈ From Figure 3.18(b), where regardless of and , a

voltage follower therefore functions as a buffer that eliminates loading effects between a high-resistance ( ) voltage source and a low-resistance ( ) load.

out in sv v≈ ≈ v sR LR

sR LR Example 3.5 Design a Noninverting Amplifier

14

Ideal Op-Amps

four significant properties of an inverting amplifier circuit with loop gain : 1AB the magnitude of the difference voltage is very small dv the value of the output voltage is essentially independent of the op-amp

gain outv

the currents into the inverting and noninverting terminals are negligible the equivalent output resistance is negligible

these four effects lead to the handy concept of the ideal op-amp

ideal op-amp : a fictitious device having infinite gain

A =∞

the negative feedback forces as , and so

and v remains finite

nv v→ p

d p n out

A → ∞

0v v v= − →

when an ideal op-amp has a negative-feedback connections the input terminals act as a virtual short (like a short circuit)

while (like an open circuit)

0dv→ =

0p ni i= =

Figure 3.19

the symbol “ ” at the input stands for v p nv= 0p ni i= =

the value of will be whatever is needed to satisfy the conditions of

the virtual short outv

The procedure for analyzing a circuit with an ideal op-amp is in p.104 of the text

Example 3.6 A Noninverting Current Amplifier

15

Figure 3.20

Inverting and Summing Op-Amp Circuits

op-amp circuits in an inverting mode : the input being connected to the inverting terminal

an inverting voltage amplifier :

Figure 3.21 Analysis

16

1 1

1 1

(1) 0

(2) ,

in f f in

in n in out n outin f

F F

out in Fout in

F

i i i iv v v v v vi i

R R R Rv v Rv vR R R

+ = → =−− −= = = =

∴ =− → =−

→ The circuit amplifies the input voltage by the factor 1

FRR

and inverts the

output. When , the circuit becomes a unit-gain inverter with 1FR R= out inv v=−

The inverting amplifier circuit has finite input resistance (unlike a noninverting amplifier, which has infinite input resistance) because it draws input current through :

1R

1in

iin

vR Ri

= =

source loading : the effect of source internal resistance from Figure 3.21(c), the source has open-circuit voltage and internal resistance , then

sv

sR

1

1 1

out F Fv

s s s

v R R RAv R R R R R

⎛ ⎞⎟⎜ ⎟= = − =−⎜ ⎟⎜ ⎟⎜+ +⎝ ⎠ 1

to minimize the source loading,

[1] let , and then 1 sR R 1v FA R≈− R , or

[2] use a voltage follower to provide buffering between the source and the inverting amplifier

inverting summing amplifier :

created by connecting two voltage sources plus input resistors to the inverting terminal

Figure 3.22 find by invoking superposition : outv

( )2 10, out Fv v R R−= =− 1 1v

17

( )1 20, out Fv v R R−= =− 2 2v

( )( )

1 2 11 2

1 2 1 1if , then

F Fout out out

out F

R Rv v v v vR R

R R v R R v v

− −

⎛ ⎞⎟⎜ ⎟∴ = + =− +⎜ ⎟⎜ ⎟⎜⎝ ⎠

= =− +

2

2

Hence the circuit sums, amplifies and inverts the two input signals. It can be generalized to apply for three or more inputs.

The name “operational” amplifier :

the op-amp circuits are able to perform many operations such as amplification, inversion, and summation of voltage signals. Additionally, the operation of subtraction can also be performed by a system with two or more op-amps.

Example 3.7 Design of an Op-Amp System

Figure 3.23

18

Figure 3.24

Example 3.8 Difference Amplifier

19