Data Acquisition ET 228 Op –Amp Applications Subjects Covered Overview of OP Amp Applications High...

Data Acquisition ET 228Op –Amp Applications

• Subjects Covered• Overview of OP Amp Applications• High Resistance Voltmeters• Phase Shifter Circuit• Integrators• Diferentiators• Example use – Servo amplifiers


• Overview of Op-Amp Applications• Chapter five covers a varied set of OP-Amp

applications • Measuring short circuit current• Measure output from photo detectors • Equalize audio tones of different amplitudes• Control high currents• Allow matching of semiconductor characteristics• High resistance DC/AC voltmeter• A phase shift circuit• Integrators and differentiators• Servo Amplifiers


• High Resistance DC Voltmeter• See Figure 5.1 on page 119

• Voltage to be measured is input to the + pin• Ed = 0• Thus I = Ei /Ri

• Limits per design• Input voltage range: -1.0 to 1 V• Do Example problem 5-1 on page 120• Two simple steps to change the voltmeter into one

with an input range of -10.0V to 10V• _____________• _____________

• Do Example problem 5-2 on page 120


• Universal High Resistance Voltmeter• See Figure 5.2 on page 121

• Measures DC, rms, Peek, and Peak-Peak voltages• Sector switch for the Ranges• Bridge circuit around meter – current only flows

through it in one direction • Limits per design – Full Scale reading per

selected Scale• Max DC Input voltage: 5 VDC

• Max rms Input voltage: 5 Vrms

• Max Peak Input voltage: 5 VPeak

• Max Peak-Peak Input voltage: 5 VPeak-Peak


• Universal High Resistance Voltmeter• Design considerations

• Meter has full scale rating of 50 micro-amps it only reads average voltage

• On the DC setting it has a 100K ohm resistor to Common• 50 micro-amps X 100k ohms = 5VDC

• On the rms scale it has a 90 k ohm resistor and an average voltage on the input of 4.5V will yield a full scale deflection

PeakrmsPeakrms VxVxVV 07.7707.05707.0

PeakPeakPeakPeak VVxxVrectifierbridgefromV 50.407.7636.0636.007.7


• Universal High Resistance Voltmeter• Design considerations

• On the Peak setting it has a 63.6K ohm resistor to Common

• 50 microamps times 63.6k ohms = 3.18VAve

• On the Peak-Peak setting it has a 31.8K ohm resistor to Common

AvePeakAvePeak VxVxVV 18.3636.0

15

636.0

1

AvePeakPeakAvePeakPeak VxVxVV 59.1318.0

15

318.0

1


• Phase Shifter Circuit• See figure 5-13 on page 137

• Low pass passive filter on Op-Amp + input• Voltage on Op-Amp – input = Voltage on + input (VCap)

• Voltage across other Input resistor = Ei – VCap = I • Voltage out has a phase shift equal to twice the phase shift

on the input Cap • Output amplitude =Input amplitude

• Design procedure• Phase angle shift only depends upon Ri, Ci, and frequency

of the input signal OR


• Phase Shifter Circuit• Design procedure

• Steps• Determine the desired phase shift and frequency of input

signal

• Set Ci to an appropriate value for the input frequency

• Determine the value of Ri using one of the equations above

• Example problems 5-13 and 5-14

• Proof of concepts• • Voltage on pin 2 = • Voltage across other input resistor = -


• Phase Shifter Circuit• Proof of concepts

Note; phase shift across the resistor from the input = the phase shift from the LP filter. There will be the same phase shift across the feedback resistor THUS double the phase shift on the Cap • With f=1kHz and Ri = 15.9 k ohms and A=1V


• Phase Shifter Circuit• Proof of concepts

With f=1kHz and Ri = 15.9 k ohms and A=1V


• Integrators• Reference Circuit:

• Fig 5-15 on page 141• Inverting Op-Amp circuit with the feedback resistor

replaced with a capacitor• The case of the input voltage being a step function is

shown • VO is shown as a ramp voltage

• Analysis• Iin or input current with respect to time {i(t)}

• The voltage difference between the two Op-Amp input terminals is assumed to be 0V DC (Ed = 0V)

Data Acquisition ET 228 Op –Amp Applications

• Integrator Circuits• Analysis

• Iin or input current with respect to time {i(t)}• Voltage across a capacitor isand when i(t) replaces I in the voltage equation and realize the output of an inverting amplifier is a negative voltage we have

YIELDS

For t greater than the start time of the input voltage step

i

in

R

eti )(

idtC

vc1

dteCR

v infi

O

1

fi

ino CR

tev

t

0

Note: If Ei is a sine wave the output will be a negative cosine wave.


• Integrator Circuits• Analysis

• When t = 0, VO = 0V (initial equilibrium voltage)

• When t = 1/10 of RiCf, VO = - 0.1 x ein V

• When t = 1/2 of RiCf, VO = - 0.5 x ein V

• When t = 0.9 of RiCf, VO = - 0.9 x ein V

• When t = RiCf, VO = - ein V

• Conclusion• For input step functions to an integrator the output is a ramp

style output voltage • Goes from the output before the input and ramps to the

saturation voltage with the opposite polarity

fi

ino CR

tev


• Integrator Circuits• Conclusion

• For other input functions over a range of time find the indefinite integral answer

• Solve using the value of time for the end of the range• Solve using the value of time for the start of the range• Subtract the second answer from the first

fi

ino CR

tev

Notes:ES = 0V, IS = 0AIin = ein /Rin, If = Iin


• Servoamplifier• Uses an Integrator to delay the full effect of an input

voltage change to the output voltage • See Figure 5-16 on page 143

• The first Op-amp is the integrator• The second Op-amp is an inverter with a gain of “-1”• The positive feedback voltage (VF)causes the integrator’s output

to stop at “-2” times the input voltage» VF is set at ½ the output voltage

» How would you set VF at a different fraction of the output, thus changing the steady-state gain?

» VC must climb to 3 times Vin and the circuits time constant is 3 times that of a simple series RC circuit. Thus τ = 3RiC

• Work Example problem 5-15 Starting on bottom of page 143


• Servoamplifier• How long of a delay

• Stability in 5 τ • In addition noise during the transition is zeroed out• The time delay is caused by the time required for the Cap to

charge to a new value required by the change in the input voltage

• The voltage on the cap in figure 15-6 changes per the following:

t starts at 0 when Ei changes and

It stops when ΔVC = 3 ΔEi CR

tEV

i

iC


• Servoamplifier• Why does VC need to be 3 time Ein -- Why does τ equal

3 time RiC• Iin shrinks to zero as VF approaches Ein and becomes zero

when they are equal• When the voltages are stable, VF = ½ VO and the output

voltage of the integrator = - VO = -2 VF = -2 Vin

• By inspection then VC must = 3 times (note polarity shown in figure 5-16)

• Also, τ is 3 times RiC since the Cap must charge to 3 times the input voltage to reach stability

• Other cases• If VF = VO , then VC would equal 2Ei and τ would = 2RiC

• If VF = 1/4VO , then what would VC equal??


• Differentiators• Looks like an Integrator with the Cap and resistor

swapped• Operation

• Performs the mathematical operation of differentiation• VO equals the negative of the derivative of Ei (see text book on

page 145 in figure 5-17 for the equation)

OR

t

CREV infiO

dt

deCRV iifO


• Differentiators• Operation

Notes:Ed = 0V, Id = 0AIin = {Δ Ein CinRf}/Δt, If = Iin


• Differentiators• Problems

• Unstable – may oscillate • However it is a plus factor when making a multivibrator by

adding positive feedback• Gain increases with frequency

• XC decreases with increased frequency, thus the gain will increase with increased frequency

• Solution• Add a parallel Cap to the Feedback Resister and a series

resistor to the input Cap

• Design procedure • See bottom of page 145


• Differentiators• Design procedure

• Work example problem 5-18

• Device recommendations • Op Amps with high slew rates

Data Acquisition ET 228 Op –Amp Applications Subjects Covered Overview of OP Amp Applications High...

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