BASIC INSTRUMENTATION ELECTRICITY. Voltage and Current 1.5V R I I=V/R Q=P/R R Electrical current...
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Transcript of BASIC INSTRUMENTATION ELECTRICITY. Voltage and Current 1.5V R I I=V/R Q=P/R R Electrical current...
BASIC INSTRUMENTATION ELECTRICITY
Voltage and Current
1.5V
R
I
I=V/R
Q=P/R
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
PTI=16 mAV=24 V
length= 500 m
What is the voltage across the 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
RVAC
Frequency , f = 50 Hz/ 60 Hz T = 1/f = 1/50 = 0.02 sω = 2πf is the phase angle
v(t)
ωt
Vm
2π0v(t) = Vmcos(ωt + )
AC VOLTAGE and CURRENT in RESISTOR
v(t) = Vmcos ωt
i(t) = v(t)/R
= Vm (cos ωt) /R
= Imcos ωt
Im = Vm/R
v(t)
ωt
Vm
i(t)
ωt
Im
V and I in resistor are in phase
RVAC
v(t) = Vmcos ωt
i(t) = Imcos ωt
v(t)
ωt2πi(t) p(t) = v(t) i(t)
= VmImcos2ωt
= VmIm {1+cos(2ωt )}/2p(t)
AC VOLTAGE, CURRENT and POWER in RESISTOR
CAPACITOR
dt
tdvCti
dt
tdqti
tCvtqV
QCCVQ
)()(
)()(
)()(
,
----
++++
+ -+ - 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 +) CVAC
)2/cos(
)2/cos(
)sin(
)()(
tI
tCV
tCVdt
tdvCti
m
m
m
The current lead the voltage
v(t)
ωt
Vm
2π0
i(t)
Im
π/2
POWER IN CAPACITOR
v(t) = Vmcos(ωt +)
)sin()( tIti m
p(t) = v(t) i(t)
= VmImcos (ωt + )sin(ωt + )
v(t)
2π
i(t)
+ +
- -
p(t)
CVAC
inductor
dt
tdiLtv
)()( i(t)
Unit of L is H(Henry) 1 H = 1 Vs/A
Inductor is made of coil and core.Real inductors always have intrinsic capacitor and resistor with it
coil
inductor
core
Equivalent Ckt of inductor
VOLTAGE AND CURRENT IN INDUCTOR
dt
tdiLtv
)()(
v(t)
for tIti m cos)(
tLItv m sin)(
v(t)
ωt
Vm
2π
i(t)Im
The voltage lead the currentWe have
i(t)
POWER IN INDUCTOR
v(t) = Vmsin(ωt +)
v(t)
2π
i(t)
)cos()( tIti m
p(t) = v(t) i(t)
= VmImcos (ωt + )sin(ωt + )
+ +
- -
p(t)
v(t)
V and I in RL circuit
dt
tdiLtRitv
)()()(
tLItRItv mm sincos)(
)cos()()( 22 tLRItv m
R
L 1tan
for tIti m cos)(
ωt
vi
)(tRidt
tdiL
)(
v(t) i(t)
)(tRi
dt
tdiL
)(
Power in RL circuit
v(t)
ωt0
i(t)
+ +--
p(t)
v(t) i(t)
VOLTAGE AND CURRENT RC CIRCUIT
)()()( tVtRitv C
for tIti m cos)(
tC
ItRItv m
m
sincos)(
)cos()1
()( 22
tC
RItv m )1
(tan 1
RC
v(t) i(t) C
v(t)
ωt
i(t)
Power in RC circuit
v(t)
ωt0
i(t)
+ +--
p(t)
v(t) i(t) C
v
ωt
i
v
ωt
i
v
ωt
i
p(t)
AC VOLTAGE, CURRENT and POWER in R, L, and C (summery )
RESISTOR CAPACITOR INDUCTOR
Power in RC circuit
v(t)
ωt0
i(t)
+ +--
p(t)
v(t)
ωt0
i(t)
+ +--
p(t)
RC CIRCUIT RL CIRCUIT
Phasors
• A phasor is a complex number that represents the magnitude and phase of a sinusoid:
tVtv M cos)(
2MV
V
Example for V and I phasor in resistor
RVAC
v(t) = Vmcos(ωt + )i(t) = Vm/R cos(ωt + )
v(t)
ωt
i(t)Im /√2
Vm/√2
Example for V and I phasor in capacitor
v(t) = Vmcos (ωt+)
v(t)
ωt
Vm
2π0
i(t) Im
CVAC
)2/cos(
)2/cos(
)sin()(
tI
tCV
tCVti
m
m
m
Im /√2Vm/√2
Example for V and I phasor in capacitor
We can set the angle arbitrarily. Usually we set the voltage is set to be zero phase abritrary
v(t) = Vmcos ωt
)2/cos(
)2/cos(
sin)(
tI
tCV
tCVti
m
m
m
Vm/√2
Im/√2
Vm/√2
Im/√2
Example for V and I phasor in inductor
v(t) i(t)
Here we can set the voltage to be zero phase, then the phase of current will be
tIti m cos)(
tLRItv m cos)()( 22
Vm/√2
Im/√2
tIti m cos)(
tLItv m sin)(
Vm/√2
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 = I Z
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 steady state.
• Impedance in series/parallel can be combined as resistors
Impedance of resistor
RVAC
v(t) = Vmcos(ωt + )i(t) = Vm/R cos(ωt + )
Im /√2
Vm/√2
R
R
V
V
I
VZ
m
m
2
2
ZR = R
Impedance of capacitor
CjC
CV
V
I
VZ
m
m
12/
1
)2/(2/
2/
CVAC
Im /√2Vm/√2
v(t) = Vmcos (ωt+))2/cos()( tCVti m
CjZc
1
Impedance of capacitor inductor
v(t) i(t)
tIti m cos)(
tLItv m sin)(
ωLIm /√2
Im/√2
LjL
I
LI
I
VZ
m
m
2/
02/
2/2/
ZL = jωL
Impedance
ZL = jωL
ZR = R
CjZc
1
Impedance Example:
f = 50Hz
Find ZC
Answer:
Zc = 1/jC
f =2 × 3.14 × 50 = 314 rad/s
Zc = 1/jC=1/(j × 314 × 106)
Zc = j3184.71
1F+
-
Symbol of Impedance Z
Impedance in series
Impedance in parallel
Z1
Z2
Z1 Z2ZT = Z1+ Z2
21
111
ZZZT
Impedance in series example
C = 15 F
R = 1K2
ZT = ?
Answer:
Zc = 1/jC=1/(j × 314 × 15 × 106)
Zc = j212.31
ZT = 1200 – j212.31
Impedance in series example
L = 5 m
R = 1K2
ZT = ?
Answer:
ZL = jL=j × 314 × 5 × 103
ZL = j1.57
ZT = 1200 + j1.57
Impedance in series example
Answer:
ZT = 1200 – j212.31 + j1.57
= 1200 –j210.74
C = 15 F
R = 1K2
ZT = ?
L = 5 m
ZL = j1.57
Zc = j212.31
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
Resistanceimpedance reactance
Example: Single Loop Circuit
20k+
-1F10V 0 VC
+
-
= 377 Find VC
Example (cont’d)
• How do we find VC?
• First compute impedances for resistor and capacitor:
ZR = 20k= 20k 0
ZC = 1/j (377 1F) = 2.65k -90
Impedance Example (cont’d)
Then use the voltage divider to find VC:
20k 0
+
-2.65k -90Vi =10V 0 VC
+
-
iCR
cc V
ZZ
ZV
Impedance Example (cont’d)
46823115471720
90652010
.V ...
.V Vc
kk
kVc 202/65.2
9065.2 0
20k 0
+
-2.65k -90Vi =10V 0 VC
+
-
Complex PowerComplex 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 superimposed interference that masks the 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 signal and the receiver. Commonly, the receiver 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 (ADC).
• 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 AmplifiersThe 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 AmplifierFor an ideal op-amp, the non-inverting amplifier gain is given by
Voltage Follower
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.
The voltage follower with an ideal op amp gives simply
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