Termpaper Of Ele

TERMPAPER OF ELE

TOPIC:-INVERTING OF OP AMPLIFIER

SUBMITTED TO:-MR.UDAY SINGH

SUBMITTED BY:-SUJEET PANDEY RES.NO.:-10802494 E7801B43

INTRODUCTION

The Inverter - A Simple Operational Amplifier Circuit Operational amplifiers can be used to perform mathematical operations on voltage signals such as inversion, addition, subtraction, integration, differentiation, and multiplication b

Inverting Amplifier

Figure:- inverting op amplifier The Open Loop Gain of an ideal Operational Amplifier can be very high, up to about 1,000,000 (120dB) or more. However, this very high gain is of no real use to us as it

makes the amplifier both unstable and hard to control as the smallest of input signals, just a few micro-volts, would be enough to cause the output to saturate and swing towards one or the other of the voltage supply rails losing control. As the open loop DC gain of an operational amplifier is extremely high we can afford to lose some of this gain by connecting a suitable resistor across the amplifier from the output terminal back to the inverting input terminal to both reduce and control the overall gain of the amplifier. This then produces and effect known commonly as Negative Feedback, and thus produces a very stable Operational Amplifier system. Negative Feedback is the process of "feeding back" some of the output signal back to the input, but to make the feedback negative we must feed it back to the "Negative input" terminal using an external Feedback Resistor called Rf. This feedback connection between the output and the inverting input terminal produces a closed loop circuit to the amplifier resulting in the gain of the amplifier now being called its Closed-loop Gain. This results in the inverting input terminal having a different signal on it than the actual input voltage as it will be the sum of the input voltage plus the negative feedback voltage giving it the label or term of a Summing Point. We must therefore separate the real input signal from the inverting input by using an Input Resistor, Rin. As we are not using the positive non-inverting input this is connected to a common ground or zero voltage terminal as shown below, but the effect of this closed loop feedback circuit results in the voltage potential at the inverting input being equal to that at the noninverting input producing a Virtual Earth summing point because it will be at the same potential as the grounded reference input. Inverting Amplifier Circuit

In this Inverting Amplifier circuit the operational amplifier is connected with feedback to produce a closed loop operation. There are two very important rules to remember about inverting amplifiers is that, "no current flows into the input terminal" and that "V1 equals V2". This is because the junction of the input and feedback signal (X) is at the same potential as the positive (+) input which is at zero volts or ground then, the junction is a "Virtual Earth". Because of this virtual earth node the input resistance of the amplifier is equal to the value of the input resistor, Rin and the closed loop gain of the inverting amplifier can be set by the ratio of the two external resistors.

We said above that there are two very important rules to remember about Inverting Amplifiers or any operational amplifier for that matter and they are. •

1. No Current Flows into the Input Terminals

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2. The Differential Input Voltage is Zero as V1 = V2 = 0 (Virtual Earth) Then by using these two rules we can find the equation for calculating the gain of an inverting amplifier, using first principles. Current ( i ) flows through the resistor network as shown.

Then, the Closed-Loop Voltage Gain of an Inverting Amplifier is given as.

and this can be transposed to give:

The negative sign in the equation indicates an inversion of the output signal with respect to the input as it is 180o out of phase. This is due to the feedback being negative in value. Example No1 Find the closed loop gain of the following inverting amplifier circuit.

Using the previously found formula for the gain of the circuit

we can now substitute the values of the resistors in the circuit as follows, Rin = 10kΩ and Rf = 100kΩ. and the gain of the circuit is calculated as

-Rf/Rin = 100k/10k = 10.

therefore, the closed loop gain of the inverting amplifier circuit above is given 10 or 20dB. One final point to note about Inverting Amplifiers, if the two resistors are of equal value, Rin = Rf then the gain of the amplifier will be -1 producing a complementary form of the input voltage at its output as Vout = -Vin. This type of inverting amplifier configuration is generally called a Unity Gain Inverter of simply an Inverting Buffer.

Analysis Of The Inverting Amplifier To analyze the inverting amplifier, we start by making an assumption that the output voltage, Vout, is some "reasonable" value - a value somewhere between the values of the positive and negative power supply voltages. Notice what happens here. This assumption that the difference between the inverting and noninverting input voltages is just a consequence of the very high gain of the operational amplifier. It's not special to this circuit. It's a general idea that we can make use of in other amplifier circuits. For all practical purposes that voltage is close enough to zero that we will call it zero when we calculate how the circuit behaves. We know it isn't zero, but it has such a small value that it will not affect any of our calculations. You'll need to remember the logic here. •

If the output of the operational amplifier is some reasonable value (usually ten volts or less, either positive or negative, as long as the amplifier isn't saturated. Then all bets are off.),

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And, if the operational amplifier has a really high gain (and remember 100,000 is probably a low value of the gain for typical amplifiers),  Then, the voltage at the input of the amplifier is zero for all practical purposes. And, note the following:

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For the voltage to be zero the gain of the amplifier would have to be infinite. When people discuss this assumption, they often refer to it as the infinite gain assumption.

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What the infinite gain assumption reduces to is that we can consider the voltage difference between the two inputs to be zero.

For all practical purposes that voltage is close enough to zero that we will call it zero when we calculate how the circuit behaves. We know it isn't zero, but it has such a small value that it will not affect any of our calculations. Since the difference between the operational amplifier input voltages are practially zero and the internal input resistance is very large, we can make the assumption that the current flowing into the amplifier through either of the input terminals is so small as to be negligible. Most of the time that's a good assumption because: •

The input voltage is small.

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The input resistance of the op-amp is large.

Here is a modified circuit diagram that shows the input resistance of the operational amplifier. You can visualize the input resistance as a resistor connected between the input terminals of the operational amplifier.

Let's take a minute to summarize the few assumptions we have made so far. •

The output voltage, Vout, is within the value between the positive and negative voltage supply. It's a "reasonable" value.

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The input difference, (V+ - V- ) is small enough that we can consider the value to be approximately zero. This is due to large gain of the amplifier - the infinite gain assumption. We will assume that the input voltage difference is zero.

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Since we will treat the input difference as zero, and assume input resistance (the resistance between the non-inverting and inverting inputs) is infinte, then the current flowing through both of the inputs of the amplifier will also be so small that it is negligible. We will assume that no current enters the input terminals of the op-amp.

Assuming that the input difference is small, we can write KCL at the inverting node: (Notice the little red dot at the inverting node in the circuit diagram.) (Note also, that we have defined two voltages, V1 and Vout that are both measured with respect to the ground.)

Here's the KCL equation using the assumption that the voltage at the amplifier input - at the input node - is zero. I1 + I0 = 0 Technically, we can write KCL in terms of all the voltages involved (taking V+ and V- as the voltages - with respect to ground - at the "+" and "-" terminals respectively). Doing that we obtain: ( V1 - V- )/ R1 + ( Vout - V- )/ R0 = 0 However, since we assume that there is no voltage difference between V+ and V- , we can replace V- with V+ and we have the inverting input terminal connected to ground, so V- = 0. That means we get:

V1 / R1 + Vout / R0 = 0 •

Note that the situation where V+ ~= 0 happens so often that it has a common name. The non-inverting terminal in a connection like this - where the inverting input terminal is connected to ground - is called a virtual ground.

After all is said and done, we can solve for the output voltage, and doing that we find: Vout = - V1 R0 / R1 There are two things to note about this expression for the output voltage. •

The input voltage is multiplied by a constant that depends only upon the two resistors, R0 and R1.

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No property of the amplifier shows up in the final expression.

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Properties of the amplifier that are used in the argument for this expression are: ○ Very large gain (approaching infinity) ○ Very large input resistance between the two input terminals.

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Finally, note that pesky minus sign.

And we reiterate the conditions/assumptions under which this result is true. •

The Input Voltage Difference, (V+ - V-), is very small because the gain is large and the output is not overly large.

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The current flowing into the input terminals is negligible because the input resistance is small.

The result above - for the output voltage - is the result we wanted. It gives the output voltage in terms of the input voltage and the two resistors in the circuit. Amazingly, no property of the OpAmp shows up in this expression although the presence of the OpAmp in the circuit is what makes it work the way it does. The opamp doesn't show up in the final expression for the output, but the circuit wouldn't work without the op-amp. Finally, you might want to read the note on gain - especially since it relates to this circuit.

Problems & Questions Q1. In this circuit, what is the expression for the ratio of output voltage to input voltage (Vout/V1)?

Top of Form Wiring The Op-Amp Inverter Circuit Next, we are going to show you how an inverter circuit is connected. Please read through the steps carefully and we will show you how the components look as you insert them into a circuit board. Before we do that, you need to understand how the chip is wired internally. Here is the pin-out for a typical 741 op-amp in a DIP (Dual In-line Package).

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Insert the op-amp into the circuit board. Put the chip on a circuit board. Insert the chip so that it "straddles" the groove down the middle between two sets of pin connector holes. It should look like the picture below. Note that if you didn't straddle the groove, you'd connect two pins together. Here's the amplifier in the circuit board. Notice that the notch is toward the "top".

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Connect the feedback resistor, R0. Connect R0 between pin 2, the inverting input, and pin 6, the output pin. Often, you can "bridge" the operational amplifier, that is you can just place the resistor above the operational amplifier between pins 2 and 6 as shown on the right below.

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Connect the output signal lead. Connect a lead to the output of the operational amplifier. This lead is where you can see the output of the circuit, with an oscilloscope, for example.

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Connect the input resistor, R1.

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Connect power supply and ground leads.

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Remember that you have to ground the non-inverting input. That's the next connection.

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Connect the input signal lead.

○ Finally, if you want to, you may proceed directly to the section on testing the circuit. Click here if you want to go to the section on testing the circuit. Otherwise, you should be ready to go. Check all the connections made in the first four steps. When you are sure you have it correct, then you can turn on the power supplies and begin testing your circuit. Below there's a photo of a completed circuit, and a hotlink to take you to the section on testing the amplifier.

Refrence:- www.facstaff.bucknell.edu www.ecircuitcenter.com/

www.williamson-labs.com www.technologystudent.com www.wisc-online.com