The signal conditioner -- changes the voltage Amplify Attenuate Filter

The signal conditioner -- changes the voltage

AmplifyAttenuateFilter

Electrical Drawings

• Symbols• Wires are straight lines usually horizontal and

vertical• Connection points are shown as circles on the end

of a wire:• Ground is a common connection point from which

most voltages are measured:• Shown as either a small triangle:• or as a set of lines forming a triangle:

• Resistors are shown as zigzag lines, vertical or horizontal:

Operational Amplifiers -- a.k.a. “op-amps”

• Practical signal amplifiers are frequently constructed from inexpensive, integrated circuit “chips” called operational amplifiers.

• The circuit symbol for an op-amp is a triangle (see Figure 3.10a).

+Vp

Vo

Vn

V+

V


• A circuit containing an op-amp can be used to amplify a weak signal from a transducer.

• Can we get something for nothing?

• No! There are two power supply connections, marked V+ and V.• These connections are often not shown on circuit diagrams.

+Vp

Vo

Vn

V+

V


• The common connection point (ground) at the bottom of the diagram can also be shown as a wire running from left to right.

+Vp

Vo

Vn


• The common connection point (ground) at the bottom of the diagram can also be shown as a wire running from left to right.

+Vp

Vo

Vn

• The input voltages (Vn and Vp) are applied between two input terminals (labeled + and ) and ground.

• The output voltage (Vo) appears between a single output terminal and ground.


• Properties• The op-amp is sometimes called a differential amplifier

because its output equals its internal gain times the difference between the voltages at the + and terminals.

• The internal gain is denoted by the lower case g.

• Vo = g(Vp Vn) Eq. (3.10)

+Vp

Vo

Vn


• Properties (continued)• The internal voltage gain is is very high

(usually g > 100,000).

• As a result, Vn Vp.

• Stated another way, the voltage between the + and input terminals 0.

+Vp

Vo

Vn

V


• Properties (continued)• The resistance between the input terminals (the input

resistance) is very high, usually 1M• As a result, the current entering the input In 0.

• Also, the current entering the input Ip 0.

+Vp

Vo

Vn

In 0

Ip 0

Practical Amplifier Circuits Using Op-Amps

• Practical amplifier circuits can be constructed by connecting other components (e.g., resistors) to an op-amp.

• Recall that the power supply connections are usually not shown in circuit diagrams.

• The gain of a practical amplifier circuit can be calculated by using the previously described properties of an “ideal” op-amp. Voltage between the + and input terminals 0. Current into (or out of) the + and input terminals

0.

Practical Amplifier Circuits Using Op-Amps

• A simple noninverting amplifier using an op-amp can be constructed as follows:

+Vi

Vo

R1

R2

• We will now analyze this circuit (Figure 3.11).

• Objective: Find gain G in terms of R1 and R2.

i

o

V

VG

Noninverting Amplifier Using an Op-Amp

• Apply KCL at junction B:

• I1 = I2 + In

• But In 0, so …

I1 = I2

+Vi

Vo

In 0

Ip 0

R1

R2

I1

I2

B


• Apply KVL around loop A:

• I1R1 0 + Vi = 0, so …

VI1

I2

+

A

1

i1 R

VI

+Vi

Vo

R1

R2

+


• Apply KVL around the outer loop:

0RIRIV 2211o

+Vi

Vo

R1

R2

I1

I2

+

+


• We now have three equations:

0RIRIV 2211o 1

i1 R

VI 21 II

• Solve these for Vo in terms of Vi , R1 and R2:

0RRV

RRV

V 21

i1

1

io

0VRR

VV i1

2io


0VRR

VV i1

2io

• Continuing ...

1

2ii

1

2io R

R1VV

RR

VV

• So the gain of this noninverting amplifier is ...

1

21

1

2

i

o

RRR

RR

1V

VG

… a Positive Number!


• Example:

1

21

1

2

i

o

RRR

RR

1V

VG

+Vi

Vo

R1 = 1000

R2 = 9000

101000

90001000V

VG

i

o

Inverting Amplifier Using an Op-Amp

• As in Figure 3.13 (and your homework):

1

2

i

o

RR

V

VG

Vo

R2R1

Vi

• For this circuit:

… a Negative Number!

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The signal conditioner -- changes the voltage Amplify Attenuate Filter