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# Operational Amplifiers

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This discussion focuses on Amplifiers, Operational Amplifiers in particular.

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### Operational Amplifiers

1. 1. • The Ideal Op Amp • The Inverting and Non-Inverting Configurations • The Voltage follower For aid and reference only
2. 2. INTRODUCTION • This discussion focuses on Amplifiers, Operational Amplifiers in particular. • Signal Amplification- A fundamental signal processing task is employed in some form in almost every electronic system. • Need for amplification arises because transducers provide signals that are said to be “weak,” that is, in the microvolt (µV) or millivolt (mV) range and possessing little energy.
3. 3. INTRODUCTION • Such signals are too small for normal processing. Processing can be made easier if signal magnitude is increased. The functional block that accomplishes this task is the amplifier. • It is equally important to understand the need for linearity in amplifiers. While amplifying, one must keep in mind that the information contained in the signal is not changed. • Thus when feeding the signal/waveform to an amplifier, we want the output waveform to be an exact replica of that of the input except having larger magnitude.
4. 4. THE IDEAL OP AMP The Op-Amp terminals • The Op-Amp has three terminals: Terminal 1 and 2 are input terminals and terminal 3 is the output terminal. 1 3 2 • Two terminals 4 and 5 are brought out of the op-amp and connected to a positive voltage VCC and a negative voltage –VEE ,respectively. We assume that these two terminals are implicitly present in the op-amp device.
5. 5. THE IDEAL OP AMP Function of the Op-Amp • The Op-Amp senses the difference between the voltage signals applied at its two input terminals 1 & 2, multiplies this by a value A , and causes the resulting voltage A(v2 – v1) to appear at output terminal 3. Key features of an ideal op-amp 1. The input impedance of an ideal op amp is supposed to be infinite. The ideal op amp is not supposed to draw any current through its input terminals. 2. The output impedance of an ideal op amp is supposed to be zero. Voltage between terminal 3 and ground is independent of the current drawn from terminal 3.
6. 6. THE IDEAL OP AMP 3. From the expression of the output, note that the output is in phase with v2 and is out of phase with v1.Hence,terminal 1 is called inverting input terminal and terminal 2 is called non-inverting input terminal . 4. Common-mode rejection if v1=v2,the output will ideally be zero. From this, we conclude that an ideal op amp has zero common-mode gain or infinite common-mode rejection. - +
7. 7. THE IDEAL OP AMP 5.Ideal op amps will amplify signals of any frequency with equal gain A, and thus are said to have infinite bandwidth. 6.The ideal op amp should have a gain A whose value is very large and ideally infinite. Why so? This will be justified in the later sections. Characteristics of the Ideal Op Amp (in short) Infinite input impedance Zero output impedance Zero common-mode gain or infinite common-mode rejection Infinite open-loop gain A Infinite bandwidth
8. 8. THE INVERTING CONFIRUGATION • Op amps are not used alone, rather, the op amp is connected to passive components (resistors) in a feedback circuit. There are two such op amp circuit configurations employing 2 resistorsinverting & non-inverting .
9. 9. THE INVERTING CONFIRUGATION • Figure below depicts the inverting configuration. It consists of two resistorsResistor R2 is connected from the output terminal 3,back to the inverting or negative input terminal, terminal 1.R2 is seen as applying negative feedback because it is connected to the negative terminal. Terminal 2 is grounded and R1 is connected between terminal 1 and input voltage source v1.
10. 10. THE INVERTING CONFIGURATION • Closed loop gain G, is defined asG= vo /vi Assuming the op amp to be ideal and gain A very large(infinite), then by definition, v2 - v1 = vo /A = 0 vo being the output voltage. • From the result we may conclude that because A is very large, voltage v1 approaches and ideally equals v2 .Hence a virtual short circuit appears between the terminals 1 & 2. • Since terminal 2 is grounded thus, v1 =0 & v2 =0.
11. 11. THE INVERTING CONFIGURATION • On applying ohm’s law across R1 ,we geti1 = (vi -v1)/R1 = (vi -0)/R1 =vi /R1 • This current cannot flow through the op amp because an ideal op amp draws zero current. It follows that i1 will have to flow through R2 to low-impedance terminal 3. Thus, vo =v1 - i1R2 = 0-(vi /R1)R2 vo /vi = - R2 / R1 which is the closed loop gain of the inverting configuration.(Refer to the figure in the next slide).
12. 12. THE INVERTING CONFIGURATION
13. 13. THE NON-INVERTING CONFIGURATION vi In the non-inverting configuration, the input signal is applied directly to the positive input terminal of the op amp while one terminal of R1 is connected to the ground.
14. 14. THE NON-INVERTING CONFIGURATION • The closed-loop gain – Assuming that the op amp has infinite gain A, a virtual short circuit exists between its two input terminals. Hence the difference input signal isvid =v2 - v1 = vo /A = 0 • The current through R1 can be determined as v1 /R1 .
15. 15. THE NON-INVERTING CONFIGURATION • Because of the infinite input impedance of the op amp, the current vi /R1 will flow through R2 as shown in previous figure. Now the output voltage can be determined from vo = vi +(vi /R1)R2 which yields vo /vi =1+ R2/R1 which is the open loop gain of the noninverting configuration.
16. 16. THE VOLTAGE FOLLOWER • The property of high input impedance is a very desirable feature of the non-inverting configuration. • We can make R1 =∞ and R2 =0 to obtain the unity gain amplifier shown in the figure. This circuit is known as the voltage follower, since the output “follows” the input. In the ideal case, vo =vi ,Rin =∞,Rout =0. The equivalent circuit of the follower is also shown above.
17. 17. Thanks. For feedback, drop a mail at – f2011263@hyderabad.bits-pilani.ac.in Mahesh Naidu B.E. Electrical & Electronics, BITS-Pilani Hyderabad Campus For aid and reference only
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