Files-3-Lesson Notes Lecture 15

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Lecture 15 Integrator and Differentiator Objective: To present an op amp integrator and differentiator. The circuit techniques applied for resistive op amp circuits can be extended to study op amp circuits with both resistors and capacitors. The most fundamental RC op amp circuits are the integrator and the differentiator. At the end of this class you should be able to: * Understand the operation of op amp integrator and differentiator * Analysis the integrator and differentiator circuits assuming ideal and non-ideal op amps. Inverting Integrator: In this configuration: Feedback resistor R2 in the inverting amplifier is replaced by capacitor C. The circuit uses frequency-dependent feedback. Fig. 1: Integrator: (a) circuit (b) response to step input In analyzing RC op amp circuit, you may use time domain or frequency domain method. Using time domain analysis: R s v s i = dt o dv C c i =

description

Integrator and differentiator

Transcript of Files-3-Lesson Notes Lecture 15

Page 1: Files-3-Lesson Notes Lecture 15

Lecture 15

Integrator and Differentiator

Objective: To present an op amp integrator and differentiator. The circuit techniques applied for resistive op amp circuits can be extended to study op amp circuits with both resistors and capacitors. The most fundamental RC op amp circuits are the integrator and the differentiator.

At the end of this class you should be able to:

* Understand the operation of op amp integrator and differentiator * Analysis the integrator and differentiator circuits assuming ideal and non-ideal op amps.

Inverting Integrator: In this configuration:

Feedback resistor R2 in the inverting amplifier is replaced by capacitor C. The circuit uses frequency-dependent feedback.

Fig. 1: Integrator: (a) circuit (b) response to step input

• In analyzing RC op amp circuit, you may use time domain or frequency domain method.

Using time domain analysis:

Rsv

si =

dtodv

Cci −=

Page 2: Files-3-Lesson Notes Lecture 15

Since ic= is

)0()0(

)0(0

)(1)(

1

cVov

ovdt

svRCtov

dsvRCodv

−=

+∫−=∴

∫−=∫

ττ

τ

* Voltage at the circuit’s output at time t is given by the initial capacitor voltage plus the integral of the input signal from start of integration interval, here, t=0. * Integration of an input step signal results in a ramp at the output as shown in Fig. 7(b) Example 1: Draw the output waveform of the integrator shown in Fig. 1(a) in response to the input shown in Fig. 2. Assume R=10kΩ, C=10nF and vo(0)=0.

Fig. 2: Step input

Solution:

tdCR

tvot

40 101

111)( −×−=×−= ∫ τ for mst 10 ≤≤

Thus, the output voltage will decrease linearly with time from 0V at t=0 to –10V at t=1ms as shown in Fig. 3.

Fig. 3: Response of the step input

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* Frequency domain analysis of the integrator is very simple. By writing the KCL equation at v- and solving for vo(s) results in:

sRCsvsv

i

o 1)()(

−=

For physical frequency ωjs → : RCjjv

jv

i

o

ωωω 1

)()(

−=

Therefore, the magnitude and phase responses are as follows:

Magnitude: CRv

v

i

o

ω1=

Phase: constant o9090180 =−=φ The Bode plot of the magnitude response of the integrator is shown in Fig. 4.

Fig. 4: Bode plot of the integrator transfer function

Problem: Note that this means any small dc component of vi(t) results in output. In

practice, the op amp will saturates at a voltage close to V∞

+ or V- depending on input voltage polarity.

Example 2: Determine the output voltage due to both offset voltage and biasing current for the integrator circuit shown in Fig. 5

Fig. 5: DC errors in the integrator

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At t<0, reset switch is closed, circuit becomes a voltage-follower

OSVoV =

At t=0, reset switch is opened, circuit starts integrating its own offset voltage and bias current. Using superposition analysis, output due to VOS can be found as follows. Assume the initial capacitor voltage is zero.

By KVL: ∫+=t

COSo dtiC

VtV0

1)(

But R

Vi OS

c =

tCRV

VdR

VC

VtV OSOS

tOS

OSo +=+= ∫0

1)( τ (1)

This means that Vo(t) increases linearly with time until op amp saturates. Therefore, the offset voltage is very dangerous for the integrator operation. When VOS is set to zero, the output due to IB2 will be:

tCB

Itov 2)( =

Thus the total output is given by:

tCB

It

RCOS

VOSVtTov 2)( ++=

This means that the output becomes ramp with slope determined by VOS and IB2 and saturates at one of the power supplies. Differentiator:

Fig. 6: Differentiator

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In this configuration:

Input resistor R1 in the inverting amplifier is replaced by capacitor C. Derivative operation emphasizes high-frequency components of input signal,

hence is less often used than the integrator. * Using time domain analysis:

Rov

Ri −=

dtsdv

Csi =

Since iR= is

dtsdv

RCov −=

Output is scaled version of derivative of input voltage. Problem: the differentiator is noise magnifier (i.e. spikes are produced at output due to sharp changing in vi(t). Therefore, it is usually avoided in practice. In frequency domain, writing KCL at v- and solving for transfer function yields:

90

)()(

)()(

−=⇒

=⇒

−=⇒−=

φ

ω

ωωω

CRvv

CRjjvjv

sCRsvsv

i

o

i

o

i

o

The Bode plot of the magnitude response of the differentiator is shown in Fig. 7.

1CR

Fig. 7: Bode plot of the differentiator transfer function