transfer characteristic

8/8/2019 Transfer Characteristic

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Transfer Characteristics

Often define circuits by their "Transfer Characteristics"Apply an input voltage to one side of a circuitOutput voltage measured across some part of the circuitTransfer characteristics: Plots the output against inputThus that state what the output will be for any input


2/14

Op Amp Integrator

Recall resistor followed by a capacitor RC integratorIf the RC time constant is long relative to periodThe resistor dominates the voltage drop andThe voltage across the capacitor becomes the integralConsider an inverting op amp circuitBut replace Rfwith a capacitor CfSince summing point SP = a virtual ground.

00 == spsp IV

As with the regular inverting op amps

inisin

R

VIII ===

For the following capacitor then the current is

dt

dVCI

f

ff =


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Op Amp Integrator Cont'd

Since there can be no current through the op ampfs

II =

s

s

inf

ff IR

V

dt

dVCI ===

Thus the voltage across the output capacitor is= dtVCR

V infs

f

1

Sincefout VV =

Thus the op amp output voltage is= dtVCR

V infs

out

1

Where = RsCf= time constant of RC circuitHowever the op amp supplies the currentAnd the summing point is a groundThus RC need not be longer than the input period.


4/14

Op Amp Integrator Single Pulse Input

Consider an op amp integrator circuit for a single square pulse4 V for 10 ms durationWhat is the response?Assuming C is initially uncharged then

sec5105000 6 mCR fs ===

During the pulse; t


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Op Amp Integrator For a Single Pulse

Result: slope to a constant value of 8 VFalling edge of pulse does not matterOnly the period of voltage input


6/14

Op Amp Integrator for a Stream of Pulses

For a stream of pulses period TWhen period T < RC get a triangle wave outputNegative voltage gives positive rising edgeSlope of out wave is

==fs

inin

fs

outCR

VdtV

CRV

1

Input of positive voltage starts decreasing voltage portionCalled a sawtooth wave or triangular wave output


7/14

Op Amp Capacitive Differentiator

Can change the op amp circuit to a DifferentiatorExchange the resistor and capacitorHave the capacitor on the input, resistor as feedbackWant RC time constant short relative to period of any signalFor the feedback side

f

f

fR

VI =

Recall that for a capacitordt

dVCtI in

sin=)(

Since the summing point SP is a ground this equation is exact


8/14

Op Amp Capacitive Differentiator Output

Again for inverting op amp circuitsoutffs

VVII ==

Thus the output becomesdt

dVCRIRV in

sfinfout==

Where = RfCs is the time constant of the RC circuit.Note the response time of op amp limits the operationEven if RC is very small


9/14

Op Amp Capacitive Differentiator & Stream of Pulses

Thus for a string of pulses (square wave)Get a sudden change called an impulseDirection opposite to that of falling/rising edgeFollowed by an exponential decayDecay is as capacitor charges/dischargesDecay time set by the RC time constantOther waveforms integrated eg sin wave gives cos wave


10/14

Second Order Systems (EC 8)

Second Order circuits involve two energy storage systemsCreate second order Differential EquationsTransfer of energy from one storage to another and back againIn circuits Resistors, Inductors and CapacitorsCalled RLC circuitsL stores energy in Magnetic field from the currentC stores energy in Electric field from stored chargeAs L discharges energy from B field it is stored in CAs C discharges charge it is stored in LResistor is always loosing energyEg. series Voltage source, Resistor, Inductor and CapacitorAlso parallel RLC (equations different)


11/14

Damped Spring DE (EC 8)

Math often uses the damped spring with weight for 2nd order DE02

2

=++ kydt

dyc

dt

ydm

Where m=mass of weight

c=damping constant

k=spring constant

y=vertical displacement

Energy is stored in momentum of weightEnergy also in position of springEnergy lost in damping pot


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Solution of Second Order Systems

General solution to Second Order circuitsProceeds similar to First Order Circuits(1) Use Kirchhoff's laws for circuit equation

(2) Manipulate to get I or V in terms of derivatives in time

(3) Generate the "Differential Equation form"

also called "Homogeneous equation form"(4) Solve the Differential Equation:

Solution substitution method: assume a solutionFor step change assume exponential type solution.Second order equations generally have two solutionsResponse is combination of both solutions(5) Use initial or final conditions for constants of integrationConditions may include derivatives at those times


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Solution of Series RLC Second Order Systems

Consider a series RLC with voltage source suddenly appliedFor series RLC used KVLNote for parallel will use KCL(1) Using KVL to write the equations:

++=t

idtC

iRdt

diLV

0

0

1

(2) Want full differential equation

Differentiating with respect to timei

CR

dt

di

dt

idL

10

2

2 ++=

(3) This is the differential equation of second order

Second order equations involve 2nd order derivatives


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Comparison of RLC and Damped Spring DE (EC 8)

Looking at the damped spring with weight 2nd order DE02

2

=++ kydt

dyc

dt

ydm

Where m=mass of weight

c=damping constant

k=spring constant

y=vertical displacement

Energy is stored in momentum of weight and springFor the RLC series the DE is

01

2

2

=++ iCdt

diR

dt

idL

Current i is related to the displacement yL is equivalent to the momentum energy stored m1/C is equivalent to the spring constant kR is equivalent to the damping loss c

transfer characteristic

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