Analog Electrical Devices and Measurements

2141-375 Measurement and Instrumentation

Analog Devices: Current Measurements

θsin ILBF

BLIF

=

×=rrr

Force on a conductor

F B

I A conductor is placed in a uniform magnetic field B T, at an angle of θθθθ. The current flow in the conductor is IA. Force exerted on the conductor can be calculated from

θθθθ

Analog Devices: Current Measurements

ILBkx =

Force on a conductor

B

I

With no current flowing through the conductor, the spring will be at its unstretched length. As current flows through the conductor, the spring will stretch and developed force required to balance the electromagnetic force.

IFs = kx

F = IBL

k = spring constant, x = the total distance moved by the spring and θθθθ is 90 o

xBL

kI =

D’Arsonval or PMMC Instrument

Important parts of PMMC Instrument

•Permanent magnet with two soft-iron poles• Moving coil • Controlling or restoring spring

PMMC = Permanent Magnet Moving Coil

Torque Equation and Scale

When a current I flows through a one-turn coil in a magnetic field , a force exerted on each side of the coil

L = the length of coil perpendicular to the paper

F = IBLF = IBLF = IBL

Since the force acts on each side of the coil, the total force for a coil of N turns is

D

F = NIBL

The force on each side acts at a coil diameter D, producing a deflecting torque

TD = NIBLD

Torque Equation and Scale

The controlling torque exerted by the spiral springs is proportional to the angle of deflection of the pointer:

Since all quantities except θθθθ and I are constant for any given instrument, the deflection angle is

θθθθ = CI

TC = Kθθθθ

Where K = the spring constant. For a given deflection , the controlling and deflecting torques are equal

BLIND = Kθθθθ

Therefore the pointer deflection is always proportional to the coil current. Consequently, the scale of the instrument is linear.

Galvanometer

• Galvanometer is essentially a PMMC instrument designed to be sensitive to extremely current levels.

• The simplest galvanometer is a very sensitive instrum ent with the type of center-zero scale, therefore the pointer can be deflected to either right or left of the zero position.

• The current sensitivity is stated in µµµµA/mm

• Galvanometers are often used to detect zero current or vo ltage in a circuit rather than to measure the actual level of curre nt or voltage. In this situation, the instrument is referred as a null detector.

DC Ammeter

PMMC instrument

Ammetershunt

Rs

Coil resistance

Rm

Im VmI = Is + Im Is

Rs

I

Shunt resistance

s

mms I

RIR =

m

mms II

RIR

−=

ssmm RIRI =sm VV =

• An ammeter is always connected in series with a circuit.

• The internal resistance should be very low

• The pointer can be deflected by a very small current

• Extension of ranges of ammeter can be achieved by connecting a very low shunt resistor

Example: An ammeter has a PMMC instrument with a coil resistance of Rm = 99 Ω and FSD current of 0.1 mA. Shunt resistance Rs = 1 Ω. Determine the total current passing through the ammeter at (a) FSD, (b) 0.5 FSD, and (c) 0.25 FSD.

Known: FSD of Im Rm and Rs

Solution:

DC Ammeter

0.25FSD0.5FSDFSD

2.5510I = Im + Is (mA)

2.4754.959.9Is (mA)

0.0250.050.1Im (mA)

Example: A PMMC instrument has FSD of 100 µA and a coil resistance of 1kΩ. Calculate the required shunt resistance value to convert the instrument into an ammeter with (a) FSD = 100 mAand (b) FSD = 1 A.

Known: FSD of Im Rm

Solution:

DC Ammeter

(a) FSD = 100 mA: Rs = 1.001 Ω

(b) FSD = 1 A: Rs = 0.10001 Ω

DC Ammeter: Multirange

B

AE

C

D

Rs1

Rs2

Rs3

Rs4

Rm

• A make-before-break must be used so that instrument is not left without a shunt in parallel to prevent a large current flow t hrough ammeter.

Multirange ammeter using switch shunts

Make-before break switch

B

A

E

C

D

DC Ammeter: Ayrton shunt

Rm

R1 R2 R3

VSIm

I Is Is

Im

B

C

D

AI

+

-

An Ayrton shunt used with an ammeter consists of several series-connected resistors all connected in parallel with the PMMC instrument. Range change is effected by switch between resistor junctions

Rm

R1 R2 R3

VSIm

I Is Is

Im

I

B

C

D

A

+

-

R1 + R2 + R3 in parallel with Rm

R1 + R2 in parallel with Rm + R3

Example: A PMMC instrument has a three-resistor Ayrton shunt connected across it to make an ammeter. The resistance values are R1 = 0.05 Ω, R2 = 0.45 Ω, and R3 = 4.5 Ω . The meter has Rm = 1 kΩ and FSD = 50 µA. Calculate the three ranges of the ammeter.

Known: FSD of Im Rm R1 R2 and R3

Solution:

DC Ammeter

DCBPosition

1000.05100.0510.05IFSD (mA)

Rm

R1 R2 R3

VSIm

I Is Is

Im

B

C

D

AI

+

-

DC Voltmeter

PMMC instrument

Series resistance or “multiplier”

mm

s RI

VR −=mmsm RIRIV +=

Given V = Range

V

RmRs

Im

V

Coil resistance

Multiplierresistance

mm

s RI

RangeR −=

The reciprocal of full scale current is the voltmeter sensitivity (k ΩΩΩΩ/V)

The total voltmeter resistance = Sensitivity X Range

• An ammeter is always connected across or parallel with the points in a circuit at which the voltage is to be measured.• The internal resistance should be very high

DC Voltmeter: Multirange

( )RRIV mm +=

• Multirange voltmeter using switched multiplier resistors

Meter resistance

Rm

R1

R2

R3

V

Multiplierresistors

R1 R2 R3Rm

V

• Multirange voltmeter using series-connected multiplier resistor

Where R can be R1, R2, or R3

( )RRIV mm +=

Where R can be R1, R1 + R2, or R1 + R2 + R3

Example: A PMMC instrument with FSD of 50 µA and a coil resistance of 1700 Ω is to be used as a voltmeter with ranges of 10 V, 50 V, and 100 V. Calculate the required values of multiplier resistor for the circuit (a) and (b)

Known: FSD of Im Rm

Solution:

DC Ammeter

Meter resistance

Rm

R1

R2

R3

V

Multiplierresistors

R1 R2 R3Rm

V

(a) (b)

R3R2R1

1.9983 MΩΩΩΩ998.3 kΩΩΩΩ198.3 kΩΩΩΩ

R3R2R1

1 MΩΩΩΩ800 kΩΩΩΩ198.3 kΩΩΩΩ

Ohmmeter: Voltmeter-ammeter method

V

A

VS

+

-

Rx

I

IVIx

V

+

-

VVS

+

-

Rx

I

A

+ -VA

+

-

VxV

-

Pro and con:

•Simple and theoretical oriented•Requires two meter and calculations•Subject to error: Voltage drop in ammeter (Fig. (a))

Current in voltmeter (Fig. (b))

Fig. (a) Fig. (b)

Measured Rx:

meas xR R≈

measx A A

x

V VV VR R

I I I

+= = = +

if Vx>>VA

Therefore this circuit is suitable for measure large resistance

Measured Rx: meas 1 /x

x V V x

RV VR

I I I I I= = =

+ +

meas xR R≈if Ix>>IV

Therefore this circuit is suitable for measure small resistance

Ohmmeter: Series Connection

•Voltmeter-ammeter method is rarely used in practical a pplications (mostly used in Laboratory)•Ohmmeter uses only one meter by keeping one parameter co nstant

Example: series ohmmeter

15k

0

∞

2550

75

100 µA

45k 5k

0

Infinity

Rx

Resistance tobe measured

Standardresistance

RmMeter

resistance

Meter

Battery

VS

R1

Basic series ohmmeter consisting of a PMMC and a series-connected standard resistor (R1). When the ohmmeter terminals are shorted (Rx = 0) meter full scale defection occurs. At half scale defection Rx = R1 + Rm, and at zero defection the terminals are open-circuited.

Basic series ohmmeter Ohmmeter scale

Nonlinear scale

1s

x m

VR R R

I= − −

Loading Effect: Voltage Measurement

LinearCircuit V

a

b

Vab RmVVab Rm

Rth

Vth

a

bUndisturbed condition: Rm = ∞

ab u thV V V= =

Measured condition: Rm ≠ ∞m

ab m thm th

RV V V

R R= =

+

Measurement error: 100m u

u

V Verror

V

−= ×

1

1 /m uth m

V VR R

=+General equation:

%100/1

1%100 ×

+−=×

+−=

thmthm

th

RRRR

R

Therefore, in practice, to get the acceptable results, we must have Rm ≥ 10 Rth (error ~ 9%)

Loading Effect

Circuit before measurement

10 V

R1100kΩΩΩΩ

R2100kΩΩΩΩ

5 V

5 V V

V

10 V

100kΩΩΩΩ

100kΩΩΩΩ

6.7 V

3.3 V 100kΩΩΩΩ

10 V

100kΩΩΩΩ

100kΩΩΩΩ

6 V

4 V 200kΩΩΩΩ

Circuit under measurement

V 3.3V 10100//100100

100//100=

+=measV

V 0.4V 10100//200100

100//200=

+=measV

V

10 V

100kΩΩΩΩ

100kΩΩΩΩ

5.2 V

4.8 V 1000kΩΩΩΩ

V 8.4V 10100//1000100

100//1000=

+=measV

Example Find the voltage reading and % error of each reading obtained with a voltmeter on (i) 5 V range, (ii) 10 V range and (iii) 30 V range, if the instrument has a 20 kΩ/V sensitivity, an accuracy 1% of full scale deflection and the meter is connected across Rb

Loading Effect

SOLUTION The voltage drop across Rb without the voltmeter connection

On the 5 V range

V 550k 5k 45

k 5=×

+=

+= V

RR

RV

ba

bb

kΩ 100 5V /20range =×Ω=×= VkSRm

kΩ 4.76 k 5 k 100

k 5 k 100=

+×

=+

=bm

bmeq RR

RRR

The voltmeter reading is

V 4.782 50k 76.4 k 54

k .764=

+=

+= V

RR

RV

eqa

eqb

50 V

Ra45kΩΩΩΩ

Rb5kΩΩΩΩ

45 V

5 V

Loading Effect

Error of the measurement is the combination of the loading effect and the meter error

The loading error = 4.782 - 5 = -0. 218 V

The meter error = ± 5 x = ± 0.05 V1100

∴% of error on the 5 V range:

%36.5100V 5

V 05.0V 218.0±=×

±−=

± 6.10± 0.35± 0.3-0.054.9530

± 4.40± 0.22± 0.1-0.124.8810

± 5.36± 0.27± 0.05-0.224.785

% errorTotal error (V)

Meter error (V)

Loading error (V)

Vb .

(V)Range

(V)

Loading Effect: Current Measurement

LinearCircuit A

a

b

Vab Rm

I

AVab Rm

Rth

Vth

a

b

I

Undisturbed condition: Rm = 0

Measured condition: Rm ≠ 0

Measurement error:

General equation:

Therefore, in practice, to get the acceptable results, we must have Rm ≤≤≤≤ Rth /10 (error ~ 9%)

ththu RVII /==

( )mththm RRVII +== /

( )thmum RRII /1/ +=

100×−

=u

um

I

IIerror

%100/1

1%100 ×

+−=×

+−=

mththm

m

RRRR

R

AC Voltmeter: PMMC Based

D W

Waveform Amplitude Average RMS

A

A

A

A

A

A

0

πA

πA2

AWD

D

+

0

0

AWD

D

+

A

2

A

2

A

2

A

3

A

AC Voltmeter: PMMC Based• Basic PMMC instrument is polarized, therefore its t erminals must be identified as + and -.• PMMC instrument can not response quite well with th e frequency 50 Hz or higher, So the pointer will settle at the average v alue of the current flowing through the moving coil: average-responding meter.

Full-wave Rectifier Voltmeter

RmD1

D2

D3

D4

Rs

Multiplierresistors

VpVrms

Vav

•Using 4 diodes

•On positive cycle, D1 and D4 are forward-biased, while D2 and D3 are reverse-biased

•On negative cycle, D2 and D3 are forward-biased, while D1 and D4 are reverse-biased

•The scale is calibrated for pure sine with the scale factor of 1.11 (A/ √√√√2 / 2A/ππππ)•Actually voltage to be indicated in ac

measurements is normally the rms quantity

Example: A PMMC instrument has FSD of 100 µA and a coil resistance of 1kΩ is to be employed as an ac voltmeter with FSD = 100 V (rms). Silicon diodes are used in the full-bridge rectifier circuit (a) calculate the multiplier resistance value required, (b) the position of the pointer when the rmsinput is 75 V and (c) the sensitivity of the voltmeter

Known: FSD of Im Rm

Solution:

AC Voltmeter: PMMC Based

RmD1

D2

D3

D4

Rs

Multiplierresistors

VpVrms

Vav

(a) Rs = 890.7 kΩ(b) 0.75 of FSD(c) 9 kΩ/V

AC Voltmeter: PMMC BasedHalf-wave Rectifier Voltmeter

Rs

RSH

D1

D2

RmVp

Vrms

Vav

•On positive cycle, D1 is forward-biased, while D2 i s reverse-biased

•On negative cycle, D2 is forward-biased, while D1 i s reverse-biased

•The shunt resistor R SH is connected to be able to measure the relative lar ge current.

•The scale is calibrated for pure sine with the scal e factor of 2.22 (A/ √√√√2 / A/ππππ)

Example: A PMMC instrument has FSD of 50 µA and a coil resistance of 1700 Ω is used in the half-wave rectifier voltmeter. The silicon diode (D1) must have a minimum (peak) forward current of 100 µA. When the measured voltage is 20% of FSD. The voltmeter is to indicate 50 Vrms at full scale Calculate the values of RS and RSH.

Known: FSD of Im Rm

Solution:

AC Voltmeter: PMMC Based

RS = 139.5 kΩ RSH = 778 Ω

Rs

RSH

D1

D2

RmVp

Vrms

Vav

Example The symmetrical square-wave voltage is applied to an average-responding ac voltmeter with a scale calibrated in terms of the rms value of a sine wave. If the voltmeter is the full-wave rectified configuration. Calculate the error in the meter indication. Neglect all voltage drop in all diodes.

T

Em

E

t

AC Voltmeter: PMMC Based

Solution 11%

Bridge Circuit

DC Bridge(Resistance)

AC Bridge

Inductance Capacitance Frequency

Schering Bridge Wien BridgeMaxwell BridgeHay BridgeOwen BridgeEtc.

Wheatstone BridgeKelvin BridgeMegaohm Bridge

Bridge Circuit

Bridge Circuit is a null method, operates on the princ iple of comparison. That is a known (standard) value is adjus ted until it is equal to the unknown value.

Wheatstone Bridge and Balance Condition

V

R1

R3

R2

R4

I1 I2

I3I4

The standard resistor R3 can be adjusted to null or balance the circuit.

A

B

C

D

Balance condition:

No potential difference across the galvanometer (there is no current through the galvanometer)

Under this condition: VAD = VAB

1 1 2 2I R I R=And also VDC = VBC

3 3 4 4I R I R=

where I1, I2, I3, and I4 are current in resistance arms respectively, since I1 = I3 and I2 = I4

1 2

3 4

R R

R R= or 2

4 31

x

RR R R

R= =

1 Ω

2 Ω 2 Ω

1 Ω

12 V

1 Ω

2 Ω 20 Ω

10 Ω

12 V

1 Ω

2 Ω 10 Ω

10 Ω

12 V

1 Ω

1 Ω 1 Ω

1 Ω

12 V

(a) Equal resistance (b) Proportional resistance

(c) Proportional resistance (d) 2-Volt unbalance

Example

Sensitivity of Galvanometer

A galvanometer is use to detect an unbalance conditi on in Wheatstone bridge. Its sensitivity is governed by: Current sensitivity (currents per unit defection) and internal resistance.

Thévenin Equivalent Circuit

G

A

B

C D

R3

R1 R2

R4

I1 I2

VS

Thévenin Voltage (VTH)

1 1 2 2CD AC ADV V V I R I R= − = −

where 1 21 3 2 4

and V V

I IR R R R

= =+ +

Therefore 1 2

1 3 2 4TH CD

R RV V V

R R R R

= = − + +

consider a bridge circuit under a small unbalance condition, and apply circuit analysis to solve the current through galvanometer

Sensitivity of Galvanometer (continued)

Thévenin Resistance (RTH)

R3

R1 R2

R4

A

B

C D 1 3 2 4// //THR R R R R= +

Completed Circuit

VTH

RTH

G

C

D

Ig=VTH

RTH+RgTH

gTH g

VI

R R=

+

where Ig = the galvanometer currentRg = the galvanometer resistance

Example 1 Figure below show the schematic diagram of a Wheatstone bridge with values of the bridge elements. The battery voltage is 5 V and its internal resistance negligible. The galvanometer has a current sensitivity of 10 mm/µA and an internal resistance of 100 Ω. Calculate the deflection of the galvanometer caused by the 5-Ω unbalance in arm BC

SOLUTION The bridge circuit is in the small unbalance condition since the value of resistance in arm BC is 2,005 Ω.

G

A

C

B

R3

R1 R2

R4

100 Ω Ω Ω Ω 1000 ΩΩΩΩ

2005 ΩΩΩΩ200 ΩΩΩΩ

5 V

(a)

100 ΩΩΩΩ A

B

C D200 Ω Ω Ω Ω 2005 Ω Ω Ω Ω

1000 Ω Ω Ω Ω

(b)

RTH= 734 ΩΩΩΩ

G

C

D

Ig=3.34 µµµµA

Rg= 100 ΩΩΩΩVTH

2.77 mV

(c)

D

Thévenin Voltage (VTH)

Thévenin Resistance (RTH)

100 // 200 1000 // 2005 734 THR = + = Ω

The galvanometer current2.77 mV

3.32 A734 100

THg

TH g

VI

R Rµ= = =

+ Ω + Ω

Galvanometer deflection10 mm

3.32 A 33.2 mmA

d µµ

= × =

mV 2.77

20051000

1000

200100

100 V 5

≈

+

−+

×=−= ACADTH VVV

Example 2 The galvanometer in the previous example is replaced by one with an internal resistance of 500 Ω and a current sensitivity of 1mm/µA. Assuming that a deflection of 1 mm can be observed on the galvanometer scale, determine if this new galvanometer is capable of detecting the 5-Ω unbalance in arm BC

Example 3 If all resistances in the Example 1 increase by 10 times, and we use the galvanometer in the Example 2. Assuming that a deflection of 1 mm can be observed on the galvanometer scale, determine if this new setting can be detected (the 50-Ω unbalance in arm BC)

Deflection Method

V

A

B

C D

R

R R

R+∆∆∆∆R

V

Consider a bridge circuit which have identical resistors, R in three arms, and the last arm has the resistance of R +∆R. if ∆R/R <<1

Small unbalance occur by the external environment

Thévenin Voltage (VTH)

Thévenin Resistance (RTH)

THR R≈

This kind of bridge circuit can be found in sensor applications, where the resistance in one arm is sensitive to a physical q uantity such as pressure, temperature, strain etc.

RR

RRVVV CDTH /24

/

∆+∆

==

In an unbalanced condition, the magnitude of the cu rrent or voltage drop for the meter or galvanometer portion of a bridge circuit i s a direct indication of the change in resistance in one arm.

Please correct

5 kΩΩΩΩ

Rv

6 V

Outputsignal5 kΩΩΩΩ

5 kΩΩΩΩ

(a)

0123456

0 20 40 60 80 100 120

Temp (oC)

Rv (

k ΩΩ ΩΩ)) ))

(b)

Example Circuit in Figure (a) below consists of a resistor Rv which is sensitive to the temperature change. The plot of R VS Temp. is also shown in Figure (b). Find (a) the temperature at which the bridge is balance and (b) The output signal at Temperature of 60oC.

4.5 kΩ

AC Bridge: Balance Condition

D

Z1Z2

Z4Z3

A C

D

B

I1 I2

all four arms are considered as impedance (frequency dependent components)The detector is an ac responding device: headphone, ac meterSource: an ac voltage at desired frequency

General Form of the ac Bridge

Z1, Z2, Z3 and Z4 are the impedance of bridge arms

At balance point: orBA BC 1 1 2 2E = E I Z = I Z

and1 21 3 2 4

V V I = I =

Z + Z Z + Z

V

1 4 2 3Z Z = Z Z

( ) ( )1 4 1 4 2 3 2 3Z Z =Z Zθ θ θ θ∠ + ∠ ∠ + ∠

Complex Form:

Polar Form:Magnitude balance:

Phase balance:

1 4 2 3Z Z =Z Z

1 4 2 3=θ θ θ θ∠ + ∠ ∠ + ∠

Example The impedance of the basic ac bridge are given as follows:

o1

2

100 80 (inductive impedance)

250 (pure resistance)

= Ω ∠

= Ω

Z

Z

Determine the constants of the unknown arm.

o3

4

400 30 (inductive impedance)

unknown

= ∠ Ω

=

Z

Z

Example an ac bridge is in balance with the following constants: arm AB, R = 200 Ωin series with L = 15.9 mH R; arm BC, R = 300 Ω in series with C = 0.265 µF; arm CD, unknown; arm DA, = 450 Ω. The oscillator frequency is 1 kHz. Find the constants of arm CD.

SOLUTION

The general equation for bridge balance states that

D

Z1Z2

Z4Z3

A C

D

B

I1 I2V1

2

3

4

200 100

1/ 300 600

450

unknown

R j L j

R j C j

R

ω

ω

= + = + Ω

= + = − Ω

= = Ω

=

Z

Z

Z

Z

1 4 2 3Z Z = Z Z

_

+

InvertingInput

VEE(-)

VCC(+)

Non-invertingInput

Output

_

+

Operational Amplifier: Op Amp

(a) Electrical Symbol for the op amp (b) Minimum connections to an op amp

Ideal Op Amp Rules:1. No current flows in to either input terminal2. There is no voltage difference between the two input terminals

I1

I2

V-V+

Rule 1: I1 = I2 = 0; R+/- = ∞∞∞∞Rule 2: V+ = V-; Virtually shorted

Vout

Inverting Amplifier

_

+

Rf

R1

vout

+

-vin

KCL

A

Use KCL at point A and apply Rule 1:

1

0A in A out

f

v v v v

R R

− −+ =

(no current flows into the inverting input)

Rearrange

1 1

1 10in out

Af f

v vv

R R R R

+ − + =

Apply Rule 2: (no voltage difference between inverting and non-inverting inputs)

Since V+ at zero volts, therefore V- is also at zero volts too. 0Av =

1

0in out

f

v v

R R+ =

1

fout

in

Rv

v R= −

Inverting Amplifier: another approach

Given vin = 5sin3t, R1=4.7 kΩ and Rf =47 kΩ

vout = -10vin = -50 sin 3t mV

_

+

Rf

R1

i

i

vout

+

-vin

No current flows into op amp

Since there is no current into op amp (Rule 1)

0f outV iR v−− + + =

From Rule 2: we know that V- = V+ = 0, and therefore

1

invi

R=

1 0inv iR V −− + − =0

0out fv iR= −

Combine the results, we get

1

fout

in

Rv

v R= −

1 2 3 4 5 6time

-60

-40

-20

20

40

60

mV

vin

vout

Non-inverting Amplifier

Rf

R1

vout

+

-vin

KCL

A

Use KCL at point A and apply Rule 1:

1

0A outA

f

v vv

R R

−+ =

Apply Rule 2:

_

+in Av v=

1

1 fout

in

Rv

v R= +

Given vin = 5sin3t, R1=4.7 kΩ and Rf =47 kΩ

vout = 11vin = 55 sin 3t mV 1 2 3 4 5 6time

-60

-40

-20

20

40

60

mV

vin

vout

Summing Amplifier: Mathematic Operation

Use KCL and apply Rule 1:

31 2 0A A outA A

f

v v v vv v v v

R R R R

− −− −+ + + =

_

+

Rf

Ri1

vout

+

-v1

v2

v3

i2

i3

R

R

i

vA

vB

1 2 3i i i i= + +

Since vA = 0 (Rule 2)

( )1 2 3f

out

Rv v v v

R= − + +

Sum of v1, v2 and v3

_

+

Difference Amplifier: Mathematic Operation

Use KCL and apply Rule 1:

1

1 4

0A outA v vv v

R R

−−+ =

R1

vout

+

-v1v2

R2

vA

vBSince vA = vB (Rule 2) and

32

2 3A B

Rv v v

R R

= = +

Substitute eq. (2) into eq. (1), we get

R3

R4

(1)

(2)

31 4 12

4 1 4 2 3 1

outv RR R vv

R R R R R R

+= − +

If R1 = R2 = R and R3 = R4 = Rf ( )2 1f

out

Rv v v

R= −

Difference of v1and v2

Differentiator and Integrator: Mathematic Operation

_

+

_

+

Differentiator

Integrator

R

vin

C

C

R

vin

i

vout

+

-

vout

+

-

i

i

i

vc+ -

outv iR= −

But Cdvi C

dt= in Cv v=and

inout

dvv RC

dt= −

out Cv v= −

But0

1( ) (0)

t

C Cv t idt vC

= +∫ inv iR=and

0

1(0)

t

out in Cv v dt vRC

= − +∫