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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