BASIC INSTRUMENTATION ELECTRICITY

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BASIC INSTRUMENTATION ELECTRICITY. 1.5V. Voltage and Current. R. R. I. Q=P/R. I=V/R. Liquid flow. Electrical current. Resistance. Every substance has resistance Conductor is substance having low resistance Isolator is substance having high resistance - PowerPoint PPT Presentation

Transcript of BASIC INSTRUMENTATION ELECTRICITY

  • BASIC INSTRUMENTATION ELECTRICITY

  • Voltage and CurrentRII=V/RQ=P/RRElectrical currentLiquid flow

  • ResistanceEvery substance has resistanceConductor is substance having low resistanceIsolator is substance having high resistance16 AWG wire resistance is 12 /km18 AWG wire resistance is 20 /kmQuestion:What is the resistance of 600 m 16 AWG wire?

  • Voltage DropWhen current flows across a wire the voltage will dropExamplePTI=16 mAV=24 Vlength= 500 mWhat is the voltage across the PT

  • ProblemThe 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 CURRENTFrequency , f = 50 Hz/ 60 Hz T = 1/f = 1/50 = 0.02 s = 2f is the phase angle v(t)t20v(t) = Vmcos(t + )

  • AC VOLTAGE and CURRENT in RESISTORv(t) = Vmcos t i(t) = v(t)/R = Vm (cos t) /R = Imcos tIm = Vm/R

  • v(t) = Vmcos ti(t) = Imcos t p(t) = v(t) i(t) = VmImcos2t = VmIm {1+cos(2t )}/2p(t)AC VOLTAGE, CURRENT and POWER in RESISTOR

  • CAPACITORUnit 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 CAPACITORv(t) = Vmcos(t +) CVACThe current lead the voltage

  • POWER IN CAPACITORv(t) = Vmcos(t +)p(t) = v(t) i(t) = VmImcos (t + )sin(t + ) v(t)2i(t)++--

  • inductorUnit 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

  • VOLTAGE AND CURRENT IN INDUCTORforThe voltage lead the currentWe havei(t)

  • POWER IN INDUCTORv(t) = Vmsin(t +)v(t)2i(t)p(t) = v(t) i(t) = VmImcos (t + )sin(t + ) ++--

  • V and I in RL circuitt

  • Power in RL circuit

  • VOLTAGE AND CURRENT RC CIRCUIT

  • Power in RC circuit

  • vtivtivtip(t)AC VOLTAGE, CURRENT and POWER in R, L, and C (summery ) RESISTORCAPACITORINDUCTOR

  • Power in RC circuitRC CIRCUITRL CIRCUIT

  • PhasorsA phasor is a complex number that represents the magnitude and phase of a sinusoid:

  • Example for V and I phasor in resistorv(t) = Vmcos(t + )i(t) = Vm/R cos(t + )

  • Example for V and I phasor in capacitorv(t) = Vmcos (t+)

  • Example for V and I phasor in capacitorWe can set the angle arbitrarily. Usually we set the voltage is set to be zero phase abritrary v(t) = Vmcos tVm/2Im/2

  • Example for V and I phasor in inductorv(t)i(t)Here we can set the voltage to be zero phase, then the phase of current will be

  • ImpedanceBy definition impedance (Z) is Z = V/IAC steady-state analysis using phasors allows us to express the relationship between current and voltage using a formula that looks likes Ohms law:V = I Z

  • Impedance (contd)Impedance depends on the frequency w.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 resistorv(t) = Vmcos(t + )i(t) = Vm/R cos(t + )ZR = R

  • Impedance of capacitorv(t) = Vmcos (t+)

  • Impedance of capacitor inductorv(t)i(t)LIm /2Im/2ZL = jL

  • ImpedanceZL = jLZR = R

  • Impedance Example:f = 50HzFind ZCAnswer:Zc = 1/jwCw=2pf =2 3.14 50 = 314 rad/sZc = 1/jwC=1/(j 314 106) Zc = j3184.71

  • Symbol of ImpedanceImpedance in seriesImpedance in parallel

  • Impedance in series example C = 15 mF w = 314 R = 1K2 ZT = ?Answer:Zc = 1/jwC=1/(j 314 15 106) Zc = j212.31ZT = 1200 j212.31

  • Impedance in series example L = 5 mH w = 314 R = 1K2 ZT = ?Answer:ZL = jwL=j 314 5 103ZL = j1.57ZT = 1200 + j1.57

  • Impedance in series exampleAnswer:ZT = 1200 j212.31 + j1.57 = 1200 j210.74

  • Impedance, Resistance, and ReactanceGenerally impedance consist of: The real part which is called Resistance, andThe imaginary part which is called reactanceZ = R + jXResistanceimpedancereactance

  • Example: Single Loop Circuitw = 377 Find VC

  • Example (contd)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 (contd)Then use the voltage divider to find VC:

  • Impedance Example (contd)20kW 02.65kW -90Vi =10V 0VC+-

  • 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 + jQS is called apparent power and the unit is vaP is called active power and the unit is watt andQ 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 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 AmplifierFor an ideal op-amp, the non-inverting amplifier gain is given by

  • Voltage FollowerBut 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 AmplifierA 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 AmpThe current output through the load resistor is proportional to the input voltage

  • Summing Amplifier

  • Integrator

  • Differentiator

  • Difference Amplifier

  • Differential Amplifier