operational amplifiers (op-amps) · an operational amplifier, or op-amp, ... the difference voltage...

Module: Electronics II Module Number: 650321 Electronic Devices and Circuit Theory, 9th ed., Boylestad and Nashelsky

Philadelphia University Faculty of Engineering

Communication and Electronics Engineering

Amplifier Circuits-III

Operational Amplifiers (Op-Amps):

An operational amplifier, or op-amp, is a very high gain differential amplifier with high input impedance and low output impedance.

Typical uses of the operational amplifier are to provide voltage amplitude changes (amplitude and polarity), oscillators, filter circuits, and many types of instrumentation circuits.

An op-amp contains a number of differential amplifier stages to achieve a very high voltage gain.

- Symbol and Terminals:

The following Figure shows a basic op-amp with two inputs and one output as would result using a differential amplifier input stage. Each input results in either the same or an opposite polarity (or phase) output, depending on whether the signal is applied to the plus (+) or the minus (-) input. To illustrate what an op-amp is, let's consider its ideal characteristics. A practical op-amp, of course, falls short of these ideal standards, but it is much easier to understand and analyze the device from an ideal point of view. These two considerations are clearly shown in the Figure Below.

Firstly, the ideal op-amp has - infinite voltage gain, - infinite bandwidth, - it has an infinite input impedance (open) so that it does not load the driving

source, - a zero output impedance.

Lecturer: Dr. Omar Daoud Part III 1


The input voltage, Vin appears between the two input terminals, and the output voltage is AvVin as indicated by the internal voltage source symbol.

Secondly, characteristics of a practical op-amp are:

- very high voltage gain, - very high input impedance, - low output impedance, and - wide bandwidth.

- Internal Block Diagram:

A typical op-amp is made up of three types of amplifier circuit: a differential amplifier, a voltage amplifier, and a push-pull amplifier, as shown in the Figure below.

A differential amplifier is the input stage for the op-amp; it provides amplification of the difference voltage between the two inputs. The second stage is usually a class A amplifier that provides additional gain. Some op-amps may have more than one voltage amplifier stage. A push-pull class B amplifier is typically used for the output stage.



The Differential Amplifier Input Stage The term differential comes from the amplifier's ability to amplify the difference of two input signals applied to its inputs . Only the difference in the two signals is amplified; if there is no difference, the output is zero .

A basic differential amplifier circuit and its symbol are shown in the Figure. The transistors (Q1 and Q2) and the collector resistors (Rc1 and Rc2) are carefully matched to have identical characteristics. Notice that the two transistors share a single emitter resistor, RE.

The Differential Amplifier Operation:

- Assume both bases are connected to ground, - The emitter voltage will be -0.7 V because the voltage drops across both base-

emitter junctions are equa1. - The emitter currents are equal (IE1 = IE2) and each is one-half of the current

through RE. - The collector currents are both equal and are approximately equal to the emitter

currents. Because the collector currents are the same, the collector voltages are also the same, which reflects the zero difference in the input voltages (both bases are at 0V.

- If the base of Q1 is disconnected from ground and connected to a small positive voltage,

- Ql will conduct more current because the positive voltage on its base causes the emitter voltage to increase slightly.

- Although the emitter voltage is a little higher, the total current through RE is nearly the same as before.

- The emitter current is now divided so that more of it is in Ql and less in Q2.

- As a result, the collector voltage of Q1 will decrease and the collector voltage of Q2 will increase, reflecting the difference in the input voltages (one is 0 V and the other at a small positive value).



- If the base of Ql is placed back at ground and a small positive voltage is connected to the base of Q2,

- Q2 will conduct more current, - Ql will conduct less, - The emitter current is now divided so that more of it is in Q2 and less in

Q1, - As a result, the collector voltage of Q1 will increase and the collector

voltage of Q2 will decrease. - The differential amplifier exhibits three modes of operation based on the type of

input signals. These modes are single-ended, differential, and common. Since the differential amplifier is the input stage of the op-amp, the op-amp exhibits the same modes.

Op-Amp Input Modes and Parameters 1) Input Signal Modes Single-Ended Mode: one input is grounded and a signal voltage is applied

only to the other input. - In the case where the signal voltage is applied to the inverting input, an

inverted, amplified signal voltage appears at the output. - In the case where the signal is applied to the noninverting input with

the inverting input grounded, a non inverted, amplified signal voltage appears at the output.

Differential Input Mode: Two opposite-polarity (out-of-phase) signals are

applied to the inputs. This type of operation is also referred to as double-ended. The amplified difference between the two inputs appears on the output.



Common Mode Input: Two signal voltages of the same phase, frequency, and amplitude are applied to the two inputs. When equal input signals are applied to both inputs, they cancel, resulting in a zero output voltage. This action is called common-mode rejection. Its importance lies in the situation where an unwanted signal appears commonly on both op-amp inputs. 2) Input Parameters

Common Mode Rejection Ratio: Common-mode rejection means that this

unwanted signal will not appear on the output and distort the desired signal. Common-mode signals (noise) generally are the result of the pick-up of radiated energy on the input lines, from adjacent lines, the 60 Hz power line, or other sources.

The measure of an amplifier's ability to reject common-mode signals is a parameter called the CMRR (common-mode rejection ratio).

where Ad = differential gain of the amplifier Ac = common-mode gain of the amplifier Vd = difference voltage Vc = common voltage Ex. Calculate the CMRR for the circuit measurements shown in Fig.



Solution:

Ex. Determine the output voltage of an op-amp for input voltages of Vi1 = 150 V, Vi2 =140 V. The amplifier has a differential gain of Ad = 4000 and the value of CMRR is: (a) 100. (b) 105. Solution:



Common mode Input voltage range: It is the range of input voltages which,

when applied to both inputs, will not cause clipping or other output distortion. Many op-amps have common-mode input voltage ranges of ±10 V with dc supply voltages of ±15 V.

Input Impedance: Two basic ways of specifying the input impedance of an

op-amp are the differential and the common mode. The differential input impedance is the total resistance between the

inverting and the noninverting inputs (measured by determining the change in bias current for a given change in differential input voltage).

The common-mode input impedance is the resistance between each input and ground (measured by determining the change in bias current for a given change in common-mode input voltage).

Input Offset Voltage: The ideal op-amp produces zero volts out for zero volts

in. In a practical op-amp, however, a small de voltage. VOUT(error), appears at the output when no differential input voltage is applied (causes a slight mismatch of the base-emitter voltages of the differential amplifier input stage of an op-amp) . The input offset voltage, Vos, is the differential dc voltage required between

the inputs to force the output to zero volts (Typical values of in the range of 2 mV or less, while it is 0V in ideal cases).

Input Offset Current: Ideally, the two

input bias currents are equal, and thus their difference is zero. In a practical op-amp, however, the bias currents are not exactly equal. The input offset current, Ios, is the

difference of the input bias currents, expressed as an absolute value (Actual magnitudes of offset current are usually at least an order of magnitude (ten times) less than the bias current).

inosvOUT

inosos

inininos

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RIV

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)error(

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Slew Rate: The maximum rate of change of the output voltage in response to a

step input voltage is the slew rate of an op-amp. The slew rate is dependent upon the high-frequency response of the

amplifier stages within the op-amp. A pulse is applied to the input and the resulting ideal output voltage is

indicated as below. The width of the input pulse must be sufficient to allow the output to

"slew" from its lower limit to its upper limit. A certain time interval Δt is required for the output voltage to go from

its lower limit – Vmax to its upper limit +Vmax once the input step is applied.

Slew Rate= t

Vout

where ΔV =+V - (- V ), The unit of slew rate is volts per microsecond (V/s). out max max

Ex. The output voltage of a certain op-amp appears as shown in the given Figure in response to a step input. Determine the slew rate.

Solution:

Slew Rate= sV/μ18101

)9(96

Input Bias Current: The input terminals of a bipolar differential amplifier are

the transistor bases and, therefore, the input currents are the base currents. The input bias current is the dc current required by the inputs of the amplifier to properly operate the first stage (It is the average of both input currents).



Comparison between Op-amps with and without negative feedback:

Practical Op-Amps Circuits: 1) Inverting Op-Amp The signal input is applied to the inverting (–) input. The non-inverting input (+) is grounded. The resistor Rf is the feedback resistor. It is connected from the output to the negative (inverting) input. This is negative feedback.

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2) Noninverting Op-Amps



3) Summing Op-Amps Because the op-amp has high input impedance, the multiple inputs are treated as separate inputs.

4) Integrator Op-Amps The output is the integral of the input. Integration is the operation of summing the area under a waveform or curve over a period of time. This circuit is useful in low-pass filter circuits and sensor conditioning circuits.



5) Differentiator Op-Amps The differentiator takes the derivative of the input. This circuit is useful in high-pass filter circuits.

Ex. Calculate the output voltages V2 and V3 in the given circuit below.



Solution:

Ex. What range of output voltage is developed in the given circuit below.

Solution:



Ex. The differential amplifier shown below has a differential gain of 2500 and a CMRR of 30000.

In part (a), a single ended input of 500Vrms is applied and at the same time a 100mV, 60Hz common-mode interference signal appears on both inputs as a result of radiated pick-up from ac power system.

In part (b), differential input signals of 500Vrms each are applied to the inputs. The common-mode interference is the same as in part (a).

1) Determine the common-mode gain, 2) Express the CMRR in dB, 3) Determine the rms output signal for both parts a and b, 4) Determine the rms interference voltage on the output.

Solution:

1) CMRR= 083.0CMRR

dc

c

d AA

A

A

2) CMRRdB=20log(30000)=89.5dB 3) (a) rmsiddo vAv V25.150025001

(b) rmsiddo vAv V5.250050025001

4) rmsiccoc mmvAv V3.8100083.0

Ex. In the given op-amp, Ri=10k, Rf=220k, and from the datasheet Zin=2M, Zout=75, and Ad (Aol)=200000.

a) Determine the input and output impedances, b) Ac(Acl).

Solution:

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Ex. In the given op-amp, Ri=1k, Rf=100k, and from the datasheet Zin=4M, Zout=50, and Ad (Aol)=200000.

c) Determine the input and output impedances,

d) Ac(Acl). Solution:

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Ex. Determine the output for the circuit of the given Fig. with components Rf =1M, R1=100k, R2 =50k, and R3 =500k.

Solution:



Op-Amps Applications (Active Filters):

A filter circuit can be constructed using passive components: resistors and capacitors. An active filter additionally uses an amplifier to provide voltage amplification and signal isolation or buffering (The high input impedance of the op-amp prevents excessive loading of the driving source. and the low output impedance of the op-amp prevents the filter from being affected by the load that it is driving. Active filters are also easy to adjust over a wide frequency range without altering the desired response). A filter that provides a constant output from dc up to a cutoff frequency

fOH and then passes no signal above that frequency is called an ideal low-pass filter.

A filter that provides or passes signals above a cutoff frequency fOL is a

high-pass filter.

When the filter circuit passes signals that are above one ideal cutoff

frequency and below a second cutoff frequency, it is called a bandpass filter.



Active Low Pass Filters (LPF):

First Order LPF:

112

1

1

CRf

R

RA

OH

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Fv

Second Order LPF:

22112

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Active High Pass Filters (HPF):

Active Band Pass Filters (BPF):



d

Ex. For the given Active BPF circuit, if R1=R3=30k, R2=R4=51k, RA2=RB2=10k, RA1= 15k, RB1=30k, C1= C2= C3=0.01F and C2= 0.02 F then fin

a) The circuit BW, b) The geometric center frequency, and c) The Q value (Quality factor is defined as the peak energy stored in the

circuit divided by the average energy dissipated in it per cycle at resonance).

Solution:

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Multiple-Feedback Band-Pass Filter Another type of filter

configuration, shown in the Figure below, it is a multiple-feedback bandpass filter. The two feedback paths are through R2 and C1). Components R1 and C1 provide the low-pass response, and R2 and C2 provide the high-pass response. The maximum gain Ao occurs at the center frequency.

Assume that C1>C2 fC1<fC2 1) C2 is Open circuit as long as C1 is open circuit Vout=0V. 2) fC1<fin<fC2 C1 is short circuit and C2 is still open circuit Thus, it acts as inverting amplifier Vout exists. 3) fin≥fC2 Both C1 and C2 are short circuit Rf is shorted Vout =0V.

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Ex. Determine the center frequency, maximum gain, and bandwidth for the filter in the given Figure. Solution:




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Active Notch/Band-Stop Filters (NF):

1) fin<fC2 C1 and C2 is open circuit Vout=0V from the inverting

stage but there is a noninverting one. 2) fC1<fin<fC2 C1 is short circuit Vi is shown to be common

mode signal.

3) fin≥fC2 Both C1 and C2 are short circuit C2 is short out the signal at inverting input Vout comes from

the noninverting input (amplify the difference between the two inputs).

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1

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operational amplifiers (op-amps) · an operational amplifier, or op-amp, ... the difference voltage...

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