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