E E 2315 Lecture 08 - Introduction to Operational Amplifiers.
E E 2315
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Calculating ResistanceWhen conductor has uniformcross-section
Area,A
l
RA
l
6
6
1.67 10
2.70 10
cu
al
cm
cm
Temperature Coefficient of Resistance
Metallic conductors have a linear increase of resistance with increased temperature.
To is the reference temperature (usually 20oC) and Ro is the resistance at the reference temperature. is the temperaturecoefficient of resistance for the material. At 20oC, some values for are:
Material Alpha @ 20oC
Aluminum 0.004308
Copper 0.004041
( ) 1o oR T R T T
Resistors in Series
By KCL: Is = I1= I2
By Ohm’s Law: V1 = R1·I1 and V2 = R2·I2
Combine: Vs = R1I1 + R2I2 = (R1 + R2) Is = ReqIs
In General: Req = R1 + R2 +···+ Rn
Vs
R1
R2
+ V1 -Is
I1
I2+V2-
+Vs-
Vs Req
Is
Resistors in Parallel (1/2)
By KVL: Vs = V1 = V2 By KCL: Is = I1 + I2
By Ohm’s Law: and
Combine:
Is R1 R2
+V1-
I1I2+
V2-
+Vs-
+Vs-
Is Req
11
1
VI
R 2
22
VI
R
Resistors in Parallel (2/2)
For two resistors:
For many resistors:
In terms of conductance:
1 2
1 1 1 1
eq nR R R R
1 2eq nG G G G
Voltage Divider Circuit
Vs
R1
R2
+ V1 -
+V2-
MeasureV2
I
1 2
sVIR R
22 2 2
1 2 1 2
ss
V RV I R R V
R R R R
Voltage Divider Equations
Unloaded:
Loaded:
If RL >> R2:
2
1 2o s
RV V
R R
2
21 21
o s
L
RV V
RR R
R
Current Divider Circuit (1/2)
Is G1 G2
+vo-
i1 i2
2 22
1 21 2
1
1 1s s
G Ri I I
G GR R
1 2
1 2 1 2
so
Ii iv
G G G G
Current Divider Circuit (2/2)
In general:
If there are onlytwo paths:
Is G1 G2
+vo-
i1 i2
2 1 22
1 21 2
1
1 1s
R R Ri I
R RR R
1 2
nn s
n
Gi I
G G G
D’Arsonval Meter Movement
• Permanent Magnet Frame• Torque on rotor proportional to coil
current• Restraint spring opposes electric torque• Angular deflection of indicator
proportional to rotor coil current
S N
D’Arsonval Voltmeter• Small voltage rating on movement (~50 mV)• Small current rating on movement (~1 mA)
• Must use voltage dropping resistor, Rv
Rv
+Vd'A
-
+ VRv -+Vx-
Id'A
Example: 1 Volt F.S. Voltmeter
Note: d’Arsonval movement has resistance of 50
Scale chosen for 1.0 volt full deflection.
950
+50 mV
-
+ 0.95 V -+1.0 V
-
1 mA
Example: 10V F.S. Voltmeter
Scale chosen for 10 volts full deflection.
9950
+50 mV
-
+ 9.95 V -+10 V
-
1 mA
D’Arsonval Ammeter• Small voltage rating on movement (~50 mV)• Small current rating on movement (~1 mA)
• Must use current bypass conductor, Ga
Ga
+Vd'A
-
IGa Id'AIx
Example: 1 Amp F.S. Ammeter
Note: d’Arsonval movement has conductance of 0.02 S
Scale chosen for 1.0 amp full deflection.
Ga = 19.98 S has ~50.050 m resistance.
19.98 S
+50 mV
-
999 mA 1 mA1.0 A
Example: 10 Amp F.S. Ammeter
Scale chosen for 10 amp full deflection.
Ga = 199.98 S has ~5.0005 m resistance.
199.98 S
+50 mV
-
9.999 A 1 mA10 A
Measurement Errors
• Inherent Instrument Error• Poor Calibration• Improper Use of Instrument• Application of Instrument Changes
What was to be Measured– Ideal Voltmeters have Infinite
Resistance– Ideal Ammeters have Zero Resistance
Example: Voltage Measurement
True Voltage:
(If voltmeter removed)
45 V
400
100 +Vo-
10 kvolt-
meter
10045 9
500oV V V
Another Voltage Measurement (1/2)
True Voltage:
(If voltmeter removed)
45 V
40 k
10 k+Vo-
10 kvolt-
meter
1045 9
50o
kV V V
k
Example: Current Measurement (1/2)
True Current:
(If ammeter replaced by short circuit)
5A
100
25 50 mAmmeter
Io
255 1.0
125oI A A
Example: Current Measurement (2/2)
Measured Current:
255 0.9996
125.05oI A A
0.9996% 1 100% .04%
1.0
AError
A
Another Current Measurement (1/2)
True Current:
(If ammeter replaced by short circuit)
5A
100 m
25 m 50 mAmmeter
Io
255 1.0
125o
mI A A
m
Measuring Resistance
• Indirect– Measure Voltage across Resistor– Measure Current through Resistor– Calculate Resistance (Inaccurate)
• d’Arsonval Ohmmeter– Very Simple– Inaccurate
• Wheatstone Bridge (Most Accurate)
Ohmmeter Example
10 mA Full Scale (Outer Numbers)
Rb+Radj+Rd’A=150 Vb=1.5 V
Inner (Nonlinear) Scale in Ohms
5
1002.5 7.5
050
150
450
8
Wheatstone Bridge Vab= 0 and Iab= 0
Vad = Vbd
R1I1=R2I2
R3I3=RxIx
Vg
Rg R1R2
R3 Rx
a b
c
d
+ Vab -
I1 I2
I3 Ix
Iab