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(1)

The Emitter Coupled Pair

The emitter coupled pair or differential pair is shown in figure below. The current source I

EE

is realized by current mirror. We assume Q

1

and Q

2

to be identical transistors and both the collector resistances are fabricated with equal values. Our objective here is to demonstrate that the differential pair can be utilized as an amplifier or a switch. To accomplish this we develop the dc transfer characteristics of the circuit.

−𝑉1 + 𝑉𝐵𝐸1 − 𝑉𝐵𝐸2 + 𝑉2 = 0 𝐼𝐶1 = 𝛼𝐼𝐸𝑆𝑒𝑉𝐵𝐸1 𝑉𝑇

𝐼𝐶2 = 𝛼𝐼𝐸𝑆𝑒𝑉𝐵𝐸2 𝑉𝑇 𝐼𝐶1

𝐼𝐶2 = 𝑒 𝑉𝐵𝐸1−𝑉𝐵𝐸2 𝑉𝑇

𝑉𝐵𝐸1 − 𝑉𝐵𝐸2 = 𝑉1 − 𝑉2 = 𝑉𝑑

𝐼𝐸1 + 𝐼𝐸2 = 𝐼𝐸𝐸 = 𝐼𝐶1

𝛼 + 𝐼𝐶2 𝛼

(2)

𝛼𝐼𝐸𝐸

𝐼𝐶2 = 𝐼𝐶1

𝐼𝐶2 + 1 = 1 + 𝑒𝑣𝑑 𝑉𝑇 𝐼𝐶2 = 𝛼𝐼𝐸𝐸

1 + 𝑒+𝑉𝑑 𝑉𝑇 A similar analysis gives

𝐼𝐶1 = 𝛼𝐼𝐸𝐸 1 + 𝑒−𝑉𝑑 𝑉𝑇 𝑉𝑜1 ≡ 𝑉𝐶𝐶 − 𝐼𝐶1𝑅𝐶 𝑉𝑜2 ≡ 𝑉𝐶𝐶 − 𝐼𝐶2𝑅𝐶

𝑉𝑜 = 𝑉𝑜1 − 𝑉𝑜2

Thus it acts as a switch and as an amplifier. When 𝐼𝐶2 ≈ 0 at 𝑉𝑑 ≥ 4𝑉𝑇,

𝑉𝑜2 = 𝑉𝐶𝐶 𝑎𝑛𝑑 𝑉𝑜1 = 𝑉𝐶𝐶 − 𝛼𝐼𝐸𝐸𝑅𝐶 can be made small by taking large 𝑅𝐶. Thus the output of Q1 acts as a closed switch and Q2 as an open switch. By applying 𝑉𝑑 ≤ −4𝑉𝑇, Q1 is open and Q2 is closed. Between −2𝑉𝑇𝑎𝑛𝑑 2𝑉𝑇 the circuit is linear and acts as an amplifier.

(3)

The Differential Amplifier

The differential amplifier, emitter-coupled pair, or differential pair, is essential building block in modern IC amplifiers. This circuit is displayed in Figure below. Both the transistors are matched, i.e., fabricated on a single chip. For a small difference voltage Vd (4VT > |Vd|), the differential pair behave as a linear amplifier. Included in the figure below is the output resistance RE of the current source bias network. We assume that the current across RE is negligible as compared to IEE. We assume that Rs=0.

Differential Mode

For V1 = V2 and assuming 𝛽𝐹 ≫ 1, the collector and emitter currents in each stage are equal 𝐼𝐶 ≈ |𝐼𝐸| . All of these currents have magnitudes equal to 𝐼𝐸𝐸 2 (approx.) because of the symmetry of the circuit and the negligible current in RE. Let us now increase V1 by ∆𝑣 2 and simultaneously decrease V2 by ∆𝑣 2. In effect, we are applying an incremental signal ∆𝑣 2 at B1 and applying -∆𝑣 2 to B2. The differential voltage 𝑉𝑑 = 𝑉1 − 𝑉2 increases by ∆𝑣. For ∆𝑣 < 4𝑉𝑇, the circuit behaves linearly. Thus IC1 increases by ∆𝐼𝐶, and IC2 decreases by the same amount.

As 𝐼𝐶 ≈ 𝐼𝐸 , the changes in IC1 and IC2 also appears at the emitters.

Consequently, the current in RE remains unchanged (the incremental current in RE is zero), causing the voltage VE to remain constant.

(4)

The situation just described is referred to as the differential mode because the input signals (∆𝑣 2) applied to Q1 and Q2 are equal and opposite and a difference signal Vd exists. For this mode, the incremental circuit can be redrawn as shown in Figure (a) below.

(a) The differential-mode small-signal equivalent. (b) The common-mode small signal equivalent of the differential pair

(5)

The Common Mode

Now let us consider that both V

1

and V

2

increase by ∆𝑣 2. The difference voltage V

d

remains zero, and I

C1

and I

C2

remain equal. However, because R

E

is present, both I

C1

and I

C2

exhibit a small increase 𝛿𝐼

𝐶

. Again changes in I

C

appear at the emitter, and hence the current in R

E

increases by 2𝛿𝐼

𝐶

. The voltage V

E

is no longer constant but must increase by 2𝛿𝐼

𝐶

𝑅

𝐸

. This situation, where equal signals are applied to Q

1

and Q

2

is called common mode.

The incremental equivalent circuit is displayed in Figure (b) above, in which it is implied that Q

1

and Q

2

are represented by their small signal models. We can write 2𝛿𝐼

𝐶

𝑅

𝐸

as 𝛿𝐼

𝐶

2𝑅

𝐸

which is shown in the Figure (b) above showing 2R

E

. The voltage across each is 2𝛿𝐼

𝐶

𝑅

𝐸

and equals the incremental change in V

E

; thus the two resistances are in parallel and 2𝑅

𝐸

||2𝑅

𝐸

= 𝑅

𝐸

.

It is evident in the Figures (a) and (b) above that depending on the input signal, the differential amplifier behaves as either common-emitter stage or a common-emitter stage with emitter resistance. Therefore, the gain of this stage is significantly higher for differential mode operation than for the common mode operation. Usually, differential amplifiers are designed so that, for practical purposes, only difference signals are amplified.

As shown in the Figure (a) above, the emitter is at ground for small signal for small signal analysis. Thus it

appears that R

E

is bypassed. Similarly, the voltage between the two collectors V

o1

– V

o2

is zero in common mode

and is twice the change in V

o1

(V

o2

) for the differential mode. Since the applied signal ∆𝑣 can be made positive or

negative, the voltage V

o1

– V

o2

can be positive or negative.

(6)

Differential Mode Gain ADM

Consider that a signal VDM is applied at the base of Q1 in Figure below and that –VDM is applied to B2.

For this condition, the circuit in Figure (a) above is valid (with ∆𝑣 2 replaced with VDM). Use of half-circuit concept, that is, analysis of only one-half of the circuit, results in small signal model in Figure (a) below.

𝐴𝐷𝑀 = 𝑉𝑜1

𝑉𝐷𝑀 = −𝛽𝑅𝐶

𝑟𝜋 = −𝑔𝑚𝑅𝐶

(7)

For VDM positive, Vo1 = ADMVDM and as seen in the Equation above, ADM is negative, so that Vo1 is 180o out of phase with VDM (Vo1 is inverted). Because Q2 is driven by –VDM, Vo2 = -ADMVDM and Vo2 is in phase with VDM (Vo2 is noninverting).

Common-Mode Gain ACM

When the signal VCM is applied to both bases in Figure on the left above, consider the Circuit in Figure (b) to the right above.

For this circuit, the gain ACM is 𝐴𝐶𝑀 = 𝑉𝑜1

𝑉𝐶𝑀 = −𝛽𝑅𝐶

2 𝛽 + 1 𝑅𝐸 + 𝑟𝜋 With 𝛽 ≫ 1 and the division by 𝑟𝜋, reduces to

𝐴𝐶𝑀 = −𝑔𝑚𝑅𝐶

1+2𝑔𝑚𝑅𝐸 ≈ − 𝑅𝐶

2𝑅𝐸 For 2𝑔𝑚𝑅𝐸 ≫ 1.

Because the same signal is applied to Q1 and Q2, both Vo1 and Vo2 are 180o out of phase with VCM.

(8)

The Common-Mode Rejection Ratio

The differential amplifier is primarily designed to amplify differential signals; hence we require 𝐴

𝐷𝑀

≫ 𝐴

𝐶𝑀

. A convenient measure of the differential amplifier performance is the common-mode rejection ratio or CMRR, defined as

𝐶𝑀𝑅𝑅 ≡

𝐴𝐷𝑀

𝐴𝐶𝑀

Combination of equation for ADM and ACM above and substituting in this equation yields 𝐶𝑀𝑅𝑅 = 1 + 2𝑔

𝑚

𝑅

𝐸

≈ 2𝑔

𝑚

𝑅

𝐸

As seen in this equation, large values of CMRR require large values of R

E

and often

necessitates the use of current sources having high output resistances. Note that if 𝑅

𝐸

→ ∞,

𝐶𝑀𝑅𝑅 → ∞, A

CM

= 0 and no common-mode component appears at the output.

(9)

Output for Arbitrary Input Signals

Consider signals V1 and V2 applied to Q1 and Q2, respectively. This pair of signals can be represented as the sum and difference of two signals VDM and VCM

𝑉1 = 𝑉𝐶𝑀 + 𝑉𝐷𝑀 𝑉2 = 𝑉𝐶𝑀 − 𝑉𝐷𝑀

Solving these equations for VDM and VCM gives

𝑉𝐷𝑀 = 𝑉1−𝑉2

2 = 𝑉𝑑

2 𝑉𝐶𝑀 = 𝑉1+𝑉2

2

The effect of this decomposition is illustrated in the Figure on the right. Because the circuit behaves linearly, superposition applies. The output consists of the two-components, one attributed to the pair of sources VDM and the other to the pair of sources VCM. Thus one component is due to differential input signal, and the second is due to common-mode input.

(10)

The output voltage V

o1

is given as

𝑉

𝑜1

= 𝐴

𝐷𝑀

𝑉

𝐷𝑀

+ 𝐴

𝐶𝑀

𝑉

𝐶𝑀

= 𝐴

𝐷𝑀

𝑉

𝐷𝑀

+

𝑉𝐶𝑀

𝐶𝑀𝑅𝑅

Equation above demonstrate the importance of CMRR if we are to amplify only the difference signals. As the CMRR is increased, the common-mode output component has diminished significance. The out put voltage V

o2

is expressed as

𝑉

𝑜2

= −𝐴

𝐷𝑀

𝑉

𝐷𝑀

+ 𝐴

𝐶𝑀

𝑉

𝐶𝑀

= −𝐴

𝐷𝑀

𝑉

𝐷𝑀

𝑉𝐶𝑀

𝐶𝑀𝑅𝑅

Further simplifying we get 𝑉

𝑜1

=

𝐴𝐷𝑀

2

𝑉

𝑑

+

𝑉1+𝑉2

𝐶𝑀𝑅𝑅

𝑉

𝑜2

=

−𝐴𝐷𝑀

2

𝑉

𝑑

𝑉1+𝑉2

𝐶𝑀𝑅𝑅

These equations are an alternative form for the output voltages that appear in the literature. The difference signal

V

d

appears explicitly.

(11)

OPERATIONAL AMPLIFIER CHARACTERISTICS

The OPAMP is a two-input voltage controlled voltage source whose output voltage is proportional to the difference between the two input voltages. It is a direct coupled device, high gain amplifier. The schematic diagram of the Op-Amp is displayed in Figures (a) and (b) below.

(12)

The schematic diagram of the OPAMP is displayed in Figure (a) above and its equivalent circuit in Figure (b) above. As seen in the Figure (b), the OPAMP is a voltage controlled voltage source. The output voltage vo is the amplified difference signal v

i

= v

1

– v

2

. The – and + symbols at the input of the OPAMP refer to the inverting and noninverting input terminals.

That is, if v

2

= 0, v

o

is 180

o

out of phased with respect to the input signal v

i

. When v

1

= 0, the output v

o

is in phase with v

i

.

The Ideal OPAMP

The ideal OPAMP has the following characteristics:

1. The input resistance 𝑅

𝑖

→ ∞ (open circuited). Consequently, no current enters either input terminal.

2. The output resistance R

o

= 0.

3. The voltage gain 𝐴

𝑣

→ ∞. The output voltage 𝑣

𝑜

= −𝐴

𝑣

𝑣

𝑖

is finite ( 𝑣

𝑜

< ∞); thus as 𝐴

𝑣

→ ∞, it is required that 𝑣

𝑖

= 0.

4. The amplifier responds equally at all frequencies. (i.e. the bandwidth is infinite).

5. When 𝑣

1

= 𝑣

2

, 𝑣

𝑜

= 0 and is independent of 𝑣

1

. The converse is also true.

(13)

(a)Inverting OPAMP stage

(b)Non-inverting Amplifier

(c)Non-Inverting Amplifier

(14)

The same symbol is used for ideal and practical OPAMP. To distinguish them, we indicate the finite gain A

v

in the triangle for practical OPAMPS and omit it in ideal case. This is illustrated in Figures (a) (b) & (c) above. The circuit in Figure (a) above is an inverting amplifier stage utilizing an ideal OPAMP. Because the input current is zero, the current I exists in both R

1

and R

2

. Furthermore, since V

i

= 0, it follows that

𝐼 =

𝑉𝑠

𝑅1

= −

𝑉𝑜

𝑅2

𝑓𝑟𝑜𝑚 𝑤ℎ𝑖𝑐ℎ 𝑤𝑒 𝑔𝑒𝑡 𝐴

𝑣

=

𝑉0

𝑉𝑠

= −

𝑅2

𝑅1

The OPAMP is used as noninverting amplifier stage in the circuit shown in Figures (b) & (c) above. Making 𝑉

𝑖

= 0 requires that

𝑉

𝑖

= 𝑉

1

− 𝑉

2

=

𝑅1

𝑅1+𝑅2

𝑉

𝑜

− 𝑉

𝑠

= 0 𝑆𝑜𝑙𝑣𝑖𝑛𝑔 𝑓𝑜𝑟 𝐴

𝑣

= 𝑉

𝑜

𝑉

𝑠

yields 𝐴

𝑣

=

𝑅1+𝑅2

𝑅1

= 1 +

𝑅2

𝑅1

The above equations indicates that the feedback provided by R

2

causes A

v

to depend only on the

resistance ratio R

2

/R

1

.

(15)

The Voltage Follower

The property of high impedance is very desirable feature of the noninverting configuration. It enables using this

circuit as a buffer amplifier to connect a source with a high impedance to a low-impedance load. In such cases we

make R

2

= 0 and R

1

= ∞ to obtain the unity gain amplifier shown in the Figure (a) below. This circuit is commonly

referred to as a voltage follower, since the output follows the input. Figure (b) is its equivalent circuit model.

(16)

Difference Amplifiers

A difference amplifier is one that responds to difference between the two signals applied at its input and ideally rejects signals that are common to the two inputs. The representation of signals in terms of their differential and common mode components are given in the Figure below.

For practical cases voltage Vo is given by 𝑣𝑜 = 𝐴𝑑𝑣𝐼𝑑 + 𝐴𝑐𝑚𝑣𝐼𝑐𝑚

Where Ad denotes the amplifier differential gain and Acm denotes the common-mode gain. The efficacy of the differential amplifier is measured by the degree of its rejection of common-mode signals in preference to differential signals. This is usually quantified by a measure known as Common-Mode Rejection Ratio (CMRR) defined as

𝐶𝑀𝑅𝑅 = 20𝑙𝑜𝑔 𝐴𝑑 𝐴𝑐𝑚

(17)

A Single Op-Amp Difference Amplifier

Our first attempt at designing a difference amplifier is motivated by the observation that the gain of the noninverting amplifier configuration is positive, 1 + 𝑅2 𝑅1 , while that of the inverting configuration is negative −𝑅2 𝑅1 . Combining the two configurations together is then a step in the right direction – namely, getting the difference between two input signals. We have to make the two gain magnitudes equal in order to reject common-mode signals. This, however, can be easily achieved by attenuating the positive input signal to reduce the gain of the positive path from 1 + 𝑅2 𝑅1 to 𝑅2 𝑅1 . The resulting circuit would like as shown in Figure below.

Hence from this Figure

𝑅4

𝑅3+𝑅4

1 +

𝑅2

𝑅1

=

𝑅2

𝑅1

Or

𝑅4

𝑅3+𝑅4

=

𝑅2

𝑅1+𝑅2

This condition is satisfied by selecting

𝑅4

𝑅3

=

𝑅2

𝑅1

(18)

To apply superposition, we first reduce vI2 to zero – that is, ground the terminal to which vI2 is applied – and then find the output voltage, which will be due entirely to vI1. Hence, it is illustrated in Figure (a) below.

𝑣𝑜1 = −𝑅2

𝑅1𝑣𝐼1

Next we reduce vI1 to zero and evaluate the corresponding output vo2. The circuit will take the form as in Figure (b) below.

(19)

The output voltage vo2 is given by 𝑣𝑜2 = 𝑣𝐼2 𝑅4

𝑅3+𝑅4 1 + 𝑅2

𝑅1 = 𝑅2

𝑅1𝑣𝐼2

The superposition principle tells us that the output voltage vo is equal to the sum of vo1 and vo2. Thus we have

𝑣𝑜 = 𝑅2

𝑅1 𝑣𝐼2 − 𝑣𝐼1 = 𝑅2

𝑅1𝑣𝐼𝑑

Thus the circuit acts as a differential amplifier with the differential gain Ad of 𝐴𝑑 = 𝑅2

𝑅1.

Let us now consider the circuit with only the common mode signal applied at the input, as shown in the Figure on the right.

𝑖1 = 1

𝑅1 𝑣𝐼𝑐𝑚𝑅4

𝑅4+𝑅3𝑣𝐼𝑐𝑚 = 𝑣𝐼𝑐𝑚 𝑅3

𝑅4+𝑅3 1 𝑅1

The output voltage can now be found as 𝑣𝑜 = 𝑅4

𝑅3+𝑅4𝑣𝐼𝑐𝑚 − 𝑖2𝑅2

𝑖2 = 1

𝑅2

𝑅4

𝑅4+𝑅3 𝑣𝐼𝑐𝑚1

𝑅2𝑣𝑜

(20)

Substituting i

2

= i

1

we get

𝑣

𝑜

= 𝑅

4

𝑅

4

+ 𝑅

3

𝑣

𝐼𝑐𝑚

− 𝑅

2

𝑅

1

𝑅

3

𝑅

3

+ 𝑅

4

𝑣

𝐼𝑐𝑚

= 𝑅

4

𝑅

3

+ 𝑅

4

1 − 𝑅

2

𝑅

1

𝑅

3

𝑅

4

𝑣

𝐼𝑐𝑚

Thus,

𝐴

𝑐𝑚

𝑣𝑜

𝑣𝐼𝑐𝑚

=

𝑅4

𝑅3+𝑅4

1 −

𝑅2

𝑅1 𝑅3

𝑅4

for this design if R

3

= R

1

and R

2

= R

4

we get A

cm

= 0.

To find the input resistance between the two input terminals, called the differential input resistance R

id

consider the figure above again.

𝑅

𝑖𝑑

𝑣𝐼𝑑

𝑖1

also writing the loop equation 𝑣

𝐼𝑑

= 𝑅

1

𝑖

1

+ 0 + 𝑅

1

𝑖

1

Thus

𝑅

𝑖𝑑

= 2𝑅

1

(21)

The Instrumentation Amplifier

It consists of two stages. The first stage is formed by opamps A1 and A2 and their associated resistors, and the second stage is the difference amplifier formed by opamp A3 and its four associated resistors. A1 and A2 are connected in noninverting configuration and thus realizes a gain of 1 + 𝑅2 𝑅1 . It follows that each of vI1 and vI2 is amplified by this factor, and the resulting amplified signal appear at the outputs of A1 and A2 respectively.

The difference amplifier in the second stage operates on the difference signal 1 + 𝑅2 𝑅1 𝑣𝐼2 − 𝑣𝐼1 =

1 + 𝑅2 𝑅1 𝑣𝐼𝑑 and provides its output as 𝑣𝑜 = 𝑅4

𝑅3 1 + 𝑅2

𝑅1 𝑣𝐼𝑑

Thus the differential gain realized is 𝐴𝑑 = 𝑅4

𝑅3 1 + 𝑅2

𝑅1

The common mode gain will be zero because of the differencing action of the second stage amplifier. This circuit has the advantage of very high (ideally infinite) input resistance and high differential gain.

(22)

Voltage-to-Current Converter (Transconductance Amplifier)

The circuit for this configuration is given in Figure below. Thus we get the following equation. Note that i

L

is independent of load Z

L

.

𝑖

𝐿

𝑡 =

𝑣𝑠(𝑡)

𝑅

(23)

Current –to-Voltage Converter (Transresistance Amplifier)

Photocells and Photomultiplier tubes give an output current which is independent of the load. The circuit in Figure below shows an Op-Amp used as a current-to-voltage converter. Due the virtual ground at the amplifier input, the current through Rs is zero and it flows through the feedback resistor R’. Thus the output vo = -isR’. It is common to parallel R’ with a Capacitance C’ to reduce high frequency noise and the possibility of oscillations. The current to voltage converter makes excellent current- measuring instrument since it is an ammeter with zero voltage across the meter.

(24)

Logarithmic Amplifiers Using Matched Transistors

Consider the circuit of logarithmic amplifier given below.

We now derive the logarithmic expression for vo. For the present discussion, ignore the high resistance balancing potentiometer arrangements. For matched transistors and with 𝑖𝐵 ≪ 𝑖𝐶, the positive input to A2 is at a voltage

𝑣 ≡ 𝑉𝐵𝐸2 − 𝑉𝐵𝐸1 = 𝑉𝑇 ln 𝑖𝐶2 − 𝑉𝑇𝑙𝑛 𝑖𝐶1 = −𝑉𝑇𝑙𝑛 𝑖𝐶1

𝑖𝐶2

Since v equals the small difference in the base-emitter voltages of Q2 and Q1, we neglect v compared with the reference voltage VR. Then since 𝑖𝐵2 ≪ 𝑖𝐶2 and because of the virtual ground at the input of A1, it follows that

𝑖𝐶2 = 𝑉𝑅

𝑅2 and 𝑖𝐶1 = 𝑣𝑠

𝑅1 Since A2 is noninverting Op- Amp,

𝑣𝑜 = 𝑣 𝑅3 + 𝑅4 /𝑅3, Combining these equations we get 𝑣𝑜 = −𝑅3 + 𝑅4

𝑅3 𝑙𝑛 𝑣𝑠 𝑅1

𝑅2 𝑉𝑅

Experimentally, it is found that the above equation is satisfied over a dynamic range of four decades., from input voltages of 2 mV to 20 V.

(25)

Exponential (Antilog) Amplifier

This amplifier is depicted in the given Figure. In the exponential amplifier the feedback current iC1 is constant and is derived from the reference voltage VR, whereas iC2 depends upon the input signal. In logarithmic amplifier the converse of this is true.

Because of the virtual ground at the inputs to A1 and A2, the collector and base of Q1 are at the same potential −𝑣 = 𝑉𝐵𝐸1 − 𝑉𝐵𝐸2. Neglecting v relative to VR, we obtain

𝑖𝐶1 = 𝑉𝑅

𝑅2 and 𝑖𝐶2 = 𝑣𝑜

𝑅1 From the input attenuator it is clear that −𝑣 = 𝑅3𝑣𝑠

𝑅3+𝑅4 = 𝑉𝑇𝑙𝑛 𝑖𝐶1

𝑖𝐶2 . Substituting the currents iC1 and iC2 from the above equations we obtain

𝑣𝑠 = −𝑉𝑇 𝑅3+𝑅4

𝑅3 𝑙𝑛 𝑣𝑜

𝑅1 𝑅2 𝑉𝑅

Hence, 𝑣0 = 𝑅1𝑉𝑅

𝑅2 𝑒𝑥𝑝 − 𝑣𝑠

𝑉𝑇 𝑅3 𝑅3+𝑅4

The system is calibrated for mismatch and offset voltages by setting vs = 0 and then adjusting the potentiometer P until

𝑣𝑜 = 𝑅1𝑉𝑅 𝑅2.

(26)

Comparator

An analog comparator, or simply a comparator, has two input voltages v1 and v2 and one output voltage vo. Often, the input (v2) is constant reference voltage VR, and the other is time-varying signal. The ideal comparator is shown in Figure below. The voltage transfer characteristic shown in Figures to the right below. It has the constant output voltage vo = V(0) if v1 – v2 = vi < 0 and a different constant voltage vo = V(1) if vi > 0. Hence, if v2 = VR, a reference voltage, vo = V(0) when v1 < VR and vo = V(1) if v1 > VR.

(27)

Square Wave Generator From a Sinusoid

The comparator performs highly nonlinear wave shaping because the output bears no resemblance to the input waveform. It is often used to transform a signal which varies slowly with time to another which exhibits an abrupt change. One such application is the generation of square wave from sinusoidal signal. If VR is set to zero, the output will change from one stage to the other very rapidly (limited by slew rate) every time the input passes through zero. Such a configuration is called zero- crossing detector. If the input to a comparator is a sine wave, the output is square wave. If a zero crossing detector is used, a symmetrical square wave results, as shown in the Figures below.

(28)

Regenerative Comparator (Schmitt Trigger)

The regenerative comparator of Figure (a) below is commonly referred to as a Schmitt trigger. The input voltage is applied to the inverting terminal 2 and the feedback voltage to the noninverting terminal 1. Assuming that the output resistance of the comparator is negligible compared with R1 + R, we obtain 𝑣1 = 𝑅2

𝑅1+𝑅2𝑣𝑜

(29)

Since v

1

= v

i

with v

2

= 0, v

o

= A

v

v

i

and use of small signal analysis gives the return ratio as 𝑇 =

−𝑅2𝐴𝑣

𝑅1+𝑅2

Clearly, with A

v

> 0, T < 0 and the feedback is positive (Regenerative). Let V

o

= V

Z

+ V

D

and assume that v

2

< v

1

so that v

o

= +V

o

. From the Figures above we find that the voltage at the noninverting terminal is given by

𝑣

1

= 𝑉

𝐴

+

𝑅2

𝑅1+𝑅2

𝑉

𝑜

− 𝑉

𝐴

≡ 𝑉

1

If v

2

is now increased, v

o

remains constant at V

o

, and v

1

= V

1

= constant until v

2

= V

1

. At this threshold, critical, or triggering voltage, the output regeneratively switches to v

o

= -V

o

and remains at this value as long as v

2

> V

1

. The transfer characteristics is indicated in Figure (b) above. The voltage at the noninverting terminal for v

2

> V

1

is

𝑣

1

= 𝑉

𝐴

𝑅2

𝑅1+𝑅2

𝑉

𝑜

+ 𝑉

𝐴

≡ 𝑉

2

Note that V

2

< V

1

, and the difference between theses two values is called the hysteresis V

H

𝑉

𝐻

= 𝑉

1

− 𝑉

2

=

2𝑅2𝑉𝑜

𝑅1+𝑅2

(30)

Square Wave Generator

The inverting Schmitt trigger can be used to obtain a free-running square wave generator by connecting an RC feedback network between the output and the inverting input. The circuit is displayed in Figure below.

(31)

In Figure (a) above the differential input voltage v

i

is given by 𝑣

𝑖

= 𝑣

𝑐

− 𝑣

1

= 𝑣

𝑐

𝑅2

𝑅1+𝑅2

𝑣

𝑜

= 𝑣

𝑐

− 𝛽𝑣

𝑜

From the ideal-comparator characteristics 𝑣

𝑜

= 𝑉

𝐷

+ 𝑉

𝑍

= 𝑉

𝑜

if v

i

< 0 and v

o

= - V

o

if v

i

> 0. Consider an instance of time when v

i

< 0 or v

c

< βv

o

= βV

o

. The capacitor C charges exponentially toward V

o

, through the RC combination. The output remains at V

o

until v

c

equals +βV

o

at which time the comparator output reverses to –V

o

. Now v

c

charges exponentially toward –V

o

. The output voltage v

o

and the capacitor voltage v

c

waveforms are shown in Figure (b) above. If we let t = 0 when v

c

= -βV

o

for the first half cycle, we have

𝑣

𝑐

𝑡 = 𝑉

𝑜

1 − (1 + 𝛽)𝑒

𝑡 𝑅𝐶

Since at t = T/2 v

c

(t) = +βV

o

, we find T solving equation above is given by 𝑇 = 2𝑅𝐶𝑙𝑛

1+𝛽

1−𝛽

= 2𝑅𝐶𝑙𝑛 1 +

2𝑅1

𝑅2

Note that T is independent of V

o

This square wave generator is particularly useful in the frequency range of 10 Hz to 10 KHz. At higher frequencies

the slew rate of the OPAMP limits the slope of the output square wave.

(32)

Triangular Wave Generator

The triangular wave generator is illustrated in the Figure (a) below and the triangular wave is illustrated in the Figure (b) below.

To find the maximum value of the triangular waveform assume that output voltage vo of the Schmitt trigger is at its negative value, -(VZ + VD) = -Vo. With a negative input, the output v’(t) of the integrator is at an increasing ramp.

(33)

The voltage at the noninverting comparator input v1 is obtained by the use of superposition and is 𝑣1 = − 𝑉𝑜𝑅2

𝑅1+𝑅2 + 𝑣𝑜𝑅1

𝑅1+𝑅2

When v1 rises to VR, the comparator changes state, vo = +Vo and v’o(t) starts decreasing linearly. Hence the peak Vmax of the triangular waveform occurs for v1 = VR. From the above equation

𝑉𝑚𝑎𝑥 = 𝑉𝑅 𝑅1+𝑅2

𝑅1 + 𝑉𝑜 𝑅2

𝑅1

By similar argument it is found that 𝑉𝑚𝑖𝑛 = 𝑉𝑅 𝑅1+𝑅2

𝑅1 − 𝑉𝑜 𝑅2

𝑅1

The peak to peak swing is 𝑉𝑚𝑎𝑥 − 𝑉𝑚𝑖𝑛 = 2𝑉𝑜 𝑅2

𝑅1

(34)

We now calculate the sweep time T

1

and T

2

for V

s

= 0. The capacitor charging current is given as

𝑖

𝑐

= 𝐶

𝑑𝑣𝑐

𝑑𝑡

= −𝐶

𝑑𝑣𝑜

𝑑𝑡

Where v

c

= 𝑣

𝑜

is the capacitor voltage. For v

o

= -V

o

, i = -V

o

/R, and the positive sweep speed is 𝑑𝑣

𝑜

𝑑𝑡 = 𝑉

𝑜

𝑅𝐶 . Hence

𝑇

1

=

𝑉𝑚𝑎𝑥−𝑉𝑚𝑖𝑛

𝑉𝑜 𝑅𝐶

=

2𝑅2𝑅𝐶

𝑅1

Since the negative sweep speed has the same magnitude as calculated above, T

2

= T

1

= T/2 = 1/2f, where the frequency f is given by

𝑓 =

𝑅1

4𝑅2𝑅𝐶

(35)

Important Problems (Sedra Smith 5 th Edition)

3.9 3.16 3.21 3.22 3.26 3.28 3.29 3.30 3.40 3.42 4.33 4.34 4.37 4.38 4.41 4.42 4.43 4.46 4.48 4.55

5.31 5.38 5.40 5.53 5.55 5.56 5.57 5.68 5.69 5.70

5.71 5.72 5.73 5.78 5.115 5.124 5.130 5.134

References

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