### Slide 1

```BASIC INSTRUMENTATION
ELECTRICITY
Voltage and Current
R
1.5V
R
I
Q=P/R
I=V/R
Electrical current
Liquid flow
Resistance
• Every substance has resistance
• Conductor is substance having low
resistance
• Isolator is substance having high
resistance
• 16 AWG wire resistance is ±12 Ω/km
• 18 AWG wire resistance is ± 20 Ω/km
• Question:
– What is the resistance of 600 m 16 AWG wire?
Voltage Drop
• When current flows across a wire the voltage will
drop
• Example
V=24 V
I=16 mA
length= 500 m
What is the voltage across the PT
PT
Problem
• The allowed voltage for a Pressure
Transmitter is 18V to 30 V. What is the
maximum wire length if the power supply
voltage in the control room is 24 V?
AC VOLTAGE AND CURRENT
v(t)
VAC
R
v(t) = Vmcos(ωt + )
Frequency , f = 50 Hz/ 60 Hz
T = 1/f = 1/50 = 0.02 s
ω = 2πf
 is the phase angle
Vm
0
2π
ωt
AC VOLTAGE and CURRENT in RESISTOR
v(t)
VAC
R
Vm
ωt
v(t) = Vmcos ωt
i(t) = v(t)/R
= Vm (cos ωt) /R
i(t)
Im
ωt
= Imcos ωt
Im = Vm/R
V and I in resistor are in phase
AC VOLTAGE, CURRENT and POWER
in RESISTOR
v(t)
v(t) = Vmcos ωt
i(t) = Imcos ωt
2π
i(t)
ωt
p(t) = v(t) i(t)
= VmImcos2ωt
p(t)
= VmIm {1+cos(2ωt )}/2
CAPACITOR
+
+
+
+
+
+
-
-
Q
Q  CV , C 
V
q (t )  Cv (t )
dq(t )
i (t ) 
dt
dv(t )
i (t )  C
dt
Unit of C is F (Farad)
1 Farad = 1 Coul/Volt = 1As/V
Real capacitor always have intrinsic capacitor and
resistor with it
VOLTAGE AND CURRENT IN CAPACITOR
v(t) = Vmcos(ωt +)
VAC
C
dv(t )
i (t )  C
dt
 CVm sin(t   )
 CVm cos(t     / 2)
 I m cos(t     / 2)
v(t)
i(t)
Im
0
π/2
Vm
2π
ωt
voltage
POWER IN CAPACITOR
v(t)
VAC
i(t)
2π
+
+
-
p(t)
-
C
v(t) = Vmcos(ωt +)
i(t )   I m sin(t   )
p(t) = v(t) i(t)
= VmImcos (ωt + )sin(ωt + )
inductor
di (t )
v(t )  L
dt
i(t)
coil
Unit of L is H(Henry)
1 H = 1 Vs/A
core
inductor
Equivalent Ckt of inductor
Inductor is made of coil and core.
Real inductors always have intrinsic
capacitor and resistor with it
VOLTAGE AND CURRENT IN INDUCTOR
v(t)
i(t)
di (t )
v(t )  L
dt
for
v(t)
Vm
i(t)
Im
2π
i(t )  I m cost
We have
v(t )  LI m sin t
ωt
POWER IN INDUCTOR
v(t)
v(t)
i(t)
2π
i(t )  I m cos(t   )
+
+
-
v(t) =  Vmsin(ωt +)
p(t)
-
p(t) = v(t) i(t)
= VmImcos (ωt + )sin(ωt + )
V and I in RL circuit
Ri(t )
di (t )
v(t )  Ri (t )  L
dt
v(t)
di (t )
L
dt
i(t)
v
v(t )  RIm cost  LI m sin t
v(t )  I m R 2  (L) 2 cos( t   )
i
di (t )
L
dt
for i(t )  I m cost
  tan
Ri(t )

ωt
1
L
R
Power in RL circuit
v(t)
v(t)
i(t)
i(t)

ωt
0
+
-
+
-
p(t)
VOLTAGE AND CURRENT RC CIRCUIT
v(t )  Ri(t )  VC (t )
v(t)
i(t)
C for
i(t )  I m cost
Im
v(t )  RI m cos t 
sin t
C
v(t )  I m
1 2
R  ( ) cos(t   )
C
2

1
  tan (
)
RC
1
v(t)
i(t)
ωt
Power in RC circuit
v(t)
v(t)
i(t)
C
i(t)

ωt
0
+
-
+
-
p(t)
AC VOLTAGE, CURRENT and POWER
in R, L, and C (summery )
v
v
v
i
i
i
ωt
ωt
ωt
p(t)
RESISTOR
CAPACITOR
INDUCTOR
Power in RC circuit
v(t)
v(t)
i(t)
i(t)

+
-

ωt
0
+
RC CIRCUIT
p(t)
ωt
0
+
-
+
-
RL CIRCUIT
p(t)
Phasors
• A phasor is a complex number that
represents the magnitude and phase of a
sinusoid:
v(t )  VM cost   
VM
V

2
Example for V and I phasor in resistor
VAC
v(t) = Vmcos(ωt + )
i(t) = Vm/R cos(ωt + )
R
Vm/√2
v(t)
i(t)
Im /√2

ωt
Example for V and I phasor in capacitor
v(t) = Vmcos (ωt+)
VAC
C
i(t )  CVm sin(t   )
 CVm cos(t     / 2)
 I m cos(t     / 2)
i(
t)
Im
0
Im /√2
v(t)
Vm
Vm/√2
2π ωt

Example for V and I phasor in capacitor
Im/√2
Vm/√2

We can set the angle 
arbitrarily. Usually we set
the voltage is set to be zero
phase abritrary
v(t) = Vmcos ωt
Im/√2
i(t )  CVm sin t
 CVm cos(t   / 2)
 I m cos(t   / 2)
Vm/√2
Example for V and I phasor in inductor
Vm/√2
i(t )  I m cost
v(t)
i(t)
v(t )  LI m sin t
Im/√2
Here we can set the voltage
to be zero phase, then the
phase of current will be 
Vm/√2
i(t )  I m cost  
v(t )  I m R 2  (L) 2 cos t
Im/√2
Impedance
• By definition impedance (Z) is
Z = V/I
• AC steady-state analysis using phasors
allows us to express the relationship
between current and voltage using a
formula that looks likes Ohm’s law:
V=IZ
Impedance (cont’d)
Impedance depends on the frequency .
Impedance is (often) a complex number.
Impedance is not a phasor (why?).
Impedance allows us to use the same solution
techniques for AC steady state as we use for DC
• Impedance in series/parallel can be combined
as resistors
•
•
•
•
Impedance of resistor
VAC
R
Vm/√2
Im /√2

v(t) = Vmcos(ωt + )
i(t) = Vm/R cos(ωt + )
Vm

V
2
Z 
R
Vm
I

R 2
ZR = R
Impedance of capacitor
v(t) = Vmcos (ωt+)
VAC
C
i(t )  CVm cos(t     / 2)
Vm / 2
V
Z 
I CVm / 2(   / 2)
Im /√2
Vm/√2

1
1

  /2 
C
jC
1
Zc 
jC
Impedance of capacitor inductor
ωLIm /√2
i(t )  I m cost
v(t)
i(t)
v(t )  LI m sin t
Im/√2
V LI m / 2 / 2
Z 
I
I m / 20
 L   / 2  j L
ZL = jωL
Impedance
ZR = R
1
Zc 
jC
ZL = jωL
Impedance Example:
+
1mF
-
f = 50Hz
Find ZC
Zc = 1/jC
2f =2 × 3.14 × 50 = 314 rad/s
Zc = 1/jC=1/(j × 314 × 106)
Zc = j3184.71
Symbol of Impedance
Z
Impedance in series
Z1
Z2
Z T = Z1+ Z2
Impedance in parallel
Z1
Z2
1
1
1


ZT Z1 Z 2
Impedance in series example
R = 1K2
ZT = ?
Zc = 1/jC=1/(j × 314 × 15 × 106)
Zc = j212.31
ZT = 1200 – j212.31
C = 15 mF
  314
Impedance in series example
R = 1K2
ZT = ?
ZL = jL=j × 314 × 5 × 103
ZL = j1.57
ZT = 1200 + j1.57
L = 5 mH
  314
Impedance in series example
R = 1K2
C = 15 mF
  314
ZT = ?
L = 5 mH
  314
ZT = 1200 – j212.31 + j1.57
= 1200 –j210.74
Zc = j212.31
ZL = j1.57
Impedance, Resistance, and Reactance
Generally impedance consist of:
The real part which is called Resistance, and
The imaginary part which is called reactance
Z = R + jX
impedance
Resistance
reactance
Example: Single Loop Circuit
10V  0
+
+
VC
1mF
-
 = 377 Find VC
20kW
-
Example (cont’d)
• How do we find VC?
• First compute impedances for resistor and
capacitor:
ZR = 20kW= 20kW  0
ZC = 1/j (377 1mF) = 2.65kW  -90
Impedance Example (cont’d)
20kW  0
Vi =10V  0
+
-
+
VC
-
2.65kW  -90
Then use the voltage divider to find VC:
Zc
Vc 
Vi
Z R  ZC
Impedance Example (cont’d)
20kW  0
Vi =10V  0
+
-
+
VC
-
2.65kW  -90
2.65kW  900
Vc 
2.65kW   / 2  20kW
2.65  90
Vc  10V 0
20.17  7.54
 1.31V   82.46
Complex Power
Complex power is defined as
S = VI*
The unit of complex power is Volt Ampere (VA)
S= VI* = I2Z = I2(R+jX) = I2R+jI2X
= I2Z cos  +jI2Z sin 
S = VI cos  +jVI sin  = P + jQ
S is called apparent power and the unit is va
P is called active power and the unit is watt and
Q is called reactive power and the unit is var
SIGNAL CONDITIONER
• Signals from sensors do not usually have
suitable characteristics for display,
recording, transmission, or further
processing.
• They may lack the amplitude, power, level,
or bandwidth required, or they may carry
desired information.
SIGNAL CONDITIONER
• Signal conditioners, including amplifiers, adapt
sensor signals to the requirements of the
receiver (circuit or equipment) to which they are
to be connected.
• The functions to be performed by the signal
conditioner derive from the nature of both the
requires a single-ended, low-frequency (dc)
voltage with low output impedance and
amplitude range close to its power-supply
voltage(s).
SIGNAL CONDITIONER
• A typical receiver here is an analog-to-digital converter
• Signals from sensors can be analog or digital. Digital
signals come from position encoders, switches, or
oscillator-based sensors connected to frequency
counters.
• The amplitude for digital signals must be compatible with
logic levels for the digital receiver, and their edges must
be fast enough to prevent any false triggering.
• Large voltages can be attenuated by a voltage divider
and slow edges can be accelerated by a Schmitt trigger.
Operational Amplifiers
The term operational amplifier or "op-amp" refers to a
class of high-gain DC coupled amplifiers with two inputs
and a single output. The modern integrated circuit version
is typified by the famous 741 op-amp. Some of the general
characteristics of the IC version are:
•High input impedance, low output
impedance
•High gain, on the order of a million
•Used with split supply, usually +/- 15V
•Used with feedback, with gain
determined by the feedback network.
Inverting Amplifier
Non-inverting Amplifier
For an ideal op-amp, the non-inverting amplifier gain is given by
Voltage Follower
The voltage follower with an ideal op amp gives simply
But this turns out to be a very useful service, because the input impedance of
the op amp is very high, giving effective isolation of the output from the signal
source. You draw very little power from the signal source, avoiding "loading"
effects. This circuit is a useful first stage.
The voltage follower is often used for the construction of buffer for logic
circuits.
Current to Voltage Amplifier
A circuit for converting small
current signals (>0.01
microamps) to a more easily
measured proportional
voltage.
so the output voltage is given
by the expression above.
Voltage-to-Current Amp
The current output through the load resistor is proportional
to the input voltage
Summing Amplifier
Integrator
Differentiator
Difference Amplifier
Differential Amplifier
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