em312 chap 4(b) amplifier

Engineering Experimentation and Measurements (EM312) School Of E

Unless otherwise specified, all materials and diagrams are adapted from the following sources: 1. Principles of Measurement Systems (3rd Edition), by John P. Bentley, Pearson/Prentice Hall 1995 2. Electronic Instrumentation and Measurement Techniques (2nd Edition), by William David Cooper, Prentice Hall 1978

005 nts: 1/19

9.2: Amplifiers - Amplifiers are used to amplify low-level signals, to a level which enables them to be further

processed. 9.2.1: The ideal operational amplifier and its applications - The operational amplifiers (op-amp) is the basic building block for modern amplifiers. - It is capable of amplifying signals from d.c. up to many kHz.

Figure 9.6: Circuit symbol and simplifiers equivalent circuit for operational amplifier [1]

Table 9.1: Ideal and typical operational amplifier characteristics [1]

- The transfer function of an operational amplifier is derived based on the virtual short

concept:

V+ = V and i+ = i = 0



005 nts: 2/19

The Basic Op-Amp (adapted from Analog Filter Design, by M.E. Van Valkenburg, Saunders College Publishing)

[van 19] - The negative input terminal is aka the inverting terminal - The positive input terminal is aka the non-inverting terminal - In general, depending on which pin (the ve or +ve terminal) you connect to ground, the

configuration will be either inverting or non-inverting. Idealized Characteristics:

[van 19]

- The op-amp is only linear when | V+ V | < AVcc requires input to be ~ V

- The ideal characteristics are (assumed to be): Ri = Ro = 0 A = - These assumptions will imply that v+ = v and i+ = i = 0 * virtual short



005 nts: 3/19

Typical Operational Amplifier Circuits Inverting Amplifier

Figure 9.7(a): Operational amplifier circuit used in measurement systems: Inverting amplifier [1]

- This amplifier is mainly used for gain adjustments in devices that are not phase sensitive. - Note that Vin is applied at the inverting terminal (V). - Assuming that R = 0 (hence V = V+ = 0), from KCL at node V , ii = iF

1RVVin =

F

out

RVV

1R

Vin = F

out

RV

in

out

VV =

1RRF



005 nts: 4/19

Non-inverting Amplifier

Figure 9.7(b): Operational amplifier circuit used in measurement systems: Non-Inverting amplifier [1]

- This amplifier can be used for gain adjustments in devices that are not phase sensitive.

However, the gain is always greater than unity. - Note that Vin is applied at the non-inverting terminal (V+). - Assuming the virtual short, (hence V = V+ = 0), from KCL at node V , ii = iF

1

0R

V = F

out

RVV

1R

V = F

out

RVV

1R

Vin = F

outin

RVV

Vin RF = R1 Vin R1 Vout Rearranging R1 Vout = (R1 + RF)Vin

in

out

VV =

1

1

RRR F = 1 +

1RRF



005 nts: 5/19

Voltage Follower

Figure 9.7(c): Operational amplifier circuit used in measurement systems: Voltage Amplifier [1]

- The voltage follower has unity gain and high input impedance like the other amplifiers. The

us of a voltage follower is to act as a buffer circuit. - Assuming the virtual short, (hence V = V+ = 0), from KCL at node V , Vin = V+ = V = Vout

in

out

VV = 1

Differential Amplifier

Figure 9.7(d): Operational amplifier circuit used in measurement systems: Differential amplifier [1]

- The differential amplifier boosts the bridge out-of-balance voltage Eth, which is the

difference between the voltages of V2 and V1.



005 nts: 6/19

- Applying voltage divider at node V+,

V+ = 23

3

RRR V2

- Assuming the virtual short, (hence V = V+), from KCL at node V,

F

out

RVV

RVV

1

1

- Substituting V = V+ = 23

3

RRR V2

F

out

R

VVRR

R

R

VRR

RV 2

23

3

1

223

31

- Multiplying both sizes with R1RF

outF

F VRVRRRRV

RRRRVR 12

23

312

23

31

- Rearranging

1223

31

23

31 VRVRR

RRRR

RRVR FFout

1223

131

)( VRVRR

RRRVR FFout

11

2231

13

)()( V

RRV

RRRRRRV FFout

Special case of differential amplifier - If R2 = R1, and R3 = RF

11

211

13

)()( V

RRV

RRRRRRV F

F

Fout

= 11

21

3 VRRV

RR F

Vout = 1R

RF (V2 V1)



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A.C Amplifier (lead-lag) (Future! Refer to section 9.3 of Bentley)

Figure 9.7(e): Operational amplifier circuit used in measurement systems: A.C amplifier (lead-lag) [1]

- The A.C. amplifier is mainly used in an A.C. carrier system which rejects drift and

interference voltages. Voltage Summer (Future! Refer to section 10.1 of Bentley)

Figure 9.7(f): Operational amplifier circuit used in measurement systems: Voltage summer [1]

- The voltage summer forms the basis of a digital-to-analog converter which is in turn used in

an analogue-to-digital converter.



005 nts: 8/19

Instrumentation Amplifiers

Figure 15-1: The Basic Instrumentation Amplifier [Flyod electronic devices]

- An instrumentation amplifier is a differential voltage-gain device that amplifies the

difference between the voltages existing at its two input terminals. - The main purpose is to amplify small signals that are riding on large common-mode voltages. - Op-amps 1 & 2 are non-inverting configurations that provide the high Zin and voltage gain.

Op-amp 3 is used as a unity-gain differential amplifier.

Figure 15-2: The Instrumentation Amplifier with the external gain-setting resistor RG.

Differential and common-mode signals are indicated. [Flyod electronic devices]



005 nts: 9/19

- Applying KCL at node V of op-amp 1

1

1,1,

RVVout =

GRVV 2,1,

Applying the virtual short

1

1,

1

1,

RV

RV inout =

G

in

RV 1,

G

in

RV 2,

1

1,

RVout =

G

in

RV 1, +

1

1,

RVin

G

in

RV 2,

Vout,1 = 1,1 inG

VRR + 1,

1

1inVR

R 2,1 inG

VRR

Vout,1 = 1,11 inG

VRR

2,1 in

G

VRR

Note that

Gin

out

RR

VV 1

1,

1, 1

- Applying KCL at node V of op-amp 2

GRVV 2,1, =

2

2,2,

RVV out

Applying the virtual short

G

in

RV 1,

G

in

RV 2, =

2

2,

2

2,

RV

RV outin

2

2,

RVout =

2

2,

RVin +

G

in

RV 2,

G

in

RV 1,

Vout,2 = 2,2

2inVR

R + 2,2 inG

VRR 1,2 in

G

VRR

Vout,2 = 2,21 inG

VRR

1,2 in

G

VRR

Note that

Gin

out

RR

VV 2

2,

2, 1



005 nts: 10/19

- Next we can apply the derived Vout,1 and Vout,2 into differential amplifier of op-amp 3. The output of the differential amplifier is

1,3

52,

543

356

)()(

outoutout VRRV

RRRRRRV

- Typically, R3 = R4 = R5 = R6, hence Vout = Vout,2 Vout,1

Vout = 2,21 inG

VRR

1,2 in

G

VRR 1,11 in

G

VRR

+ 2,1 in

G

VRR

= 2,121 inGG

VRR

RR

1,211 in

GG

VRR

RR

If we let R1 = R2 = R

Vout = 2,1 inGG

VRR

RR

1,1 in

GG

VRR

RR

=

GRR21 1,2, inin VV

- It is observed that any common mode voltage embedded within Vin,2 and Vin,1 will be

cancelled out (subtracted out) in the Vout. - The overall (closed-loop_ gain of the instrumentation amplifier is

A = 1,2, inin

out

VVV =

GRR21



005 nts: 11/19

Figure 15-3: Illustration of the rejection of large common-mode voltages and the amplification of smaller signal

voltages by an instrumentation amplifier [Flyod electronic devices] Example: Determine the value of the external gain-setting for resistor RG for a certain instrumentation amplifier with R1 = R2 = 25k. The closed-loop voltage gain is to be 500. Solution:

From A = 1,2, inin

out

VVV =

GRR21

500 =

GRk2521

RG = 100



005 nts: 12/19

Log and Antilog Amplifiers Log Amplifiers - A logarithmic (log) amplifier produces an output that is proportional to the logarithm of the

input. - Log amplifiers are used in applications that require compression of analog input date,

linearization of transducers that have exponential outputs, and analog multiplication and division.

- The output voltage of a log amplifier is given as Vout = K ln(Vin) Where the natural logarithm to the base e can be converted to base 10 using ln x = 2.3 log10 x - The logarithmic characteristic is usually obtained from a semiconductor pn junction in the

form of either a diode or the base-eitter junction of a bipolar transistor placed in the feedback loop of an op-amp circuit.



005 nts: 13/19

Signal Compression with Log Amplifiers - In certain applications, a signal may have portions that are too large in magnitude for a

particular system to handle. - If a linear signal compression circuit is used, the lower voltages are reduced by the same

percentage as the higher voltages, resulting in the lower voltages being overwhelmed by noise.

- If a log amplifier is used instead, the higher voltages are reduced by a greater percentage than

the lower voltages, thus keeping the lower voltage signals from being lost in noise.

Figure 15-33: The basic concept of signal compression with a logarithmic amplifier [Flyod electronic devices]



005 nts: 14/19

Log Amplifier with a Diode

- A typical diode has a forward diode current of IF defined as kTqVRF FeII

/ where q is the charge on an electron VF is the forward diode voltage k is Boltzmanns constant IR is the reverse leakage current T is the absolute temperature in Kelvin - The forward diode voltage can be derived as shown Applying the natural log on both side of the equation kTqVRF FeII

/lnln kTqVRF FeII /lnlnln kT

qVI FR ln Rearranging

kT

qVF = RF II lnln =

R

F

IIln

VF =

qkT

R

F

II

ln



005 nts: 15/19

Figure 15-29: A basic log amplifier using a diode as the feedback element [Flyod electronic devices]

- Applying the virtual short

Vout = VF and IF = Iin = 1R

Vin

- Substituting VF =

qkT

R

F

II

ln and eventually IF = Iin = 1R

Vin into Vout = VF

Vout =

qkT

R

F

IIln =

qkT

1

lnRI

V

R

in

At the usual operating temperature of 25C, kT/q 25 mV

Vout = (0.025V)

1

lnRI

V

R

in

Example: Determine the output voltage for the log amplifier below given IR = 50nA.

Figure 15-29 [Floyd electronic devices]

Solution

Vout = (0.025V)

1

lnRI

V

R

in = (0.025)

)10100)(1050(2ln 39 = -0.15V



005 nts: 16/19

Log Amplifier with a BJT - A typical bipolar junction transistor, BJT has a collector current of IC defined as kTqVEBOC BEeII

/ where q is the charge on an electron IEBO is the emitter-to-base leakage current k is Boltzmanns constant VBE is the base-to-emitter (threshold) voltage T is the absolute temperature in Kelvin - Following the same steps as the diode, the base-to-emitter voltage can be determined kTqVEBOC BEeII

/lnln kTqVEBOC BEeII /lnlnln kT

qVI BEEBO ln Rearranging

kT

qVBE = EBOC II lnln =

EBO

C

IIln

VBE =

qkT

EBO

C

IIln



005 nts: 17/19

Figure 15-30: A basic log amplifier using a transistor as the feedback element [Floyd electronic devices]

- Applying the virtual short

Vout = VBE and IC = Iin = 1R

Vin

- Substituting VBE =

qkT

EBO

C

IIln and eventually IC = Iin =

1RVin into Vout = VBE

Vout =

qkT

EBO

C

IIln =

qkT

1

lnRI

V

EBO

in


Vout = (0.025V)

1

lnRI

V

EBO

in

Example: Determine the output voltage for a transistor log amplifier with Vin = 3V, R1 = 68k, and assume IEBO = 40nA. Solution

Vout = (0.025V)

1

lnRI

V

EBO

in = (0.025)

)1068)(1040(3ln 39 = -0.175V

Vin



005 nts: 18/19

Antilog Amplifier with a BJT

Figure 15-30: A basic antilog amplifier [Floyd electronic devices]

- Applying the virtual short Vout = RF IC and kTqVEBOC BEeII

/ - Merging the two gives Vout = RF IC = RF kTqVEBO BEeI

/ - The exponential term can be expressed as an antilogarithm as follows:

Vout = RF IEBO antilog

kT

qVin



V

Vin025.0

Note: antilog

V

Vin025.0

=

V

Vin

e 025.0



005 nts: 19/19

Example: For the antilog amplifier below, determine the output voltage. Assume IEBO = 40nA.

Figure 15-32 [Floyd electronic devices]

Solution:


V

Vin025.0

= (68 103)(40 109) antilog

025.0101.175 3 = 3V

o. o2. 2 o. o2. 2

em312 chap 4(b) amplifier

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