Educational Objectives

        Pre-lab Assignment

Bipolar Supply - The Op-Amp has a single output terminal plus two input terminals, one of which is inverted (-) and one which is not. The symbol on the right is also used. The two power terminals labeled ±V highlight the fact that the op-amp is an active device requiring an independent power source in addition to the two inputs. Notice that the op-amp requires two DC supply voltages, one at +V and one at -V. Both voltages can be delivered by using a bipolar power supply or by using a voltage divider. This use of bipolar supply voltages is to allow the amplified output voltage to swing in both positive and negative directions (as would be required in the amplification of AC signals.)

Pin Connections - In this lab we will be using an 8-pin 741 op-amp as shown below.

                              Figure 2 – Pin connections of op-amp 741 (top-view)

Notice that the two inputs are connected to pins 2 and 3. Pin 2 is the inverted input and pin 3 is the non-inverting input. The output is at pin 6. The bipolar power leads are connected to pins 4 and 7 as indicated.  Pins 1 and 5 can be used to offset any null voltage which can arise due to variations between the transistors contained within the op-amp chip.  Pin 8 is not connected (NC) to the op-amp electrically, but can be used as a heat sink.

The Ideal Op-Amp

When analyzing op-amp circuits, one usually begins by treating it as an ideal op-amp.

1. The ideal op-amp has infinite input impedance (open) and so does not draw any power or current from the driving source.

2. The ideal op-amp has zero output impedance.

    Thus the terminal output voltage is the same as that generated after the amplification stages.

3. The ideal op-amp has infinite voltage gain.

4. The ideal op-amp has infinite bandwidth.

In the real world, op-amps have limitations. The practical op-amp has high (but not infinite) input impedance, low output impedance, high voltage gain and wide bandwidth. The output voltage can never exceed the bipolar supply voltage ±V. The maximum positive and negative output voltages are called the saturation limits and tend to be very near ±V.  A simplified graph of output voltage against differential input voltage is shown below.  The straight portion between the saturation limits is called the linear region and its slope gives the gain of the amplifier.

                        Figure 3 – Linear and saturation regions of the input-output relation of an op-amp

Op-Amp Circuits

Below, we will look at a few of the many circuits in which op-amps have found applications. Three of the more important circuits include:

    1. Inverting Amplifier

    2. Non-inverting Amplifier

    3. Differential Amplifier

1. Inverting Amplifier

The circuit below shows a simple inverting amplifier made using an Op-Amp.

                                   Figure 4 – Circuit of a simple inverting amplifier

The positive end of the input voltage V_in is connected through a resistor R₁to the inverting input pin (-) on the op-amp (the negative end of V_in is connected to ground). The non-inverting input pin(+) is connected to ground. The gain G of an amplifier is defined as the ratio of the output to input voltages.

When the input voltage is a DC signal, V_out and V_in in the above equation represent the actual values of the two voltages. When the input is a sinusoidal AC signal, the output will also be a sinusoidal AC signal. In this case, V_out and V_in represent the magnitudes of the input and output signals.

For the above circuit, we can assume that the same current flows through both resistors because of the extremely high input impedance and that the voltage at the inverting pin is nearly the same as ground (virtual short). Thus

so that the gain is:

When the input and output are sinusoidal AC signals, the negative sign in the above equation indicates a 180-degree phase shift between the output and input.

Example - If R₁ = 1000 Ohms and R₂ = 10,000 Ohms, then the gain is - 10. An input signal of 3 mV (either a 3 mV DC voltage or a sine wave with 3 mV magnitude) will be inverted and multiplied to result in an output signal of V_out = -30 mV (either a -30 mV DC voltage or a sine wave with 30 mV magnitude and a 180-degree phase shift from the input signal).

2. Non-inverting Amplifier

The circuit below shows a simple design for a non-inverting amplifier. The input voltage V_in is applied directly to the non-inverting terminal of the op-amp. Negative feedback is provided by the two external resistors R₁ and R₂ which form a voltage divider and apply a fixed fraction of the output voltage to the inverting input terminal of the op-amp.

                                  Figure 5 – Circuit of a simple noninverting amplifier

The gain of the amplifier is determined by the external resistors R₁ and R₂ according to the equation:

Example - If R₂ = 10,000 Ohms and R₁ = 1000 Ohms, the gain is 11 so that an input voltage of V_in = 3 mV is multiplied to an output voltage of V_out = 33 mV. (The output is not-inverted for the circuit shown above.)

Example - A non-inverting amplifier is assembled using the values R₁ = 1 Kilo-Ohm and R₂ = 9 Kilo-Ohms and is powered by a ± 10 Volt bipolar supply.

a. Find the gain factor in the linear region (no saturation).

b. What are the ideal saturation voltages?

c. Find the maximum positive and negative voltages which can be provided before the output saturates.

Solution

a. The gain is G=1+9/1 = 10.

b. The ideal saturation voltages are determined by the bipolar supply voltages, that is ± 10 Volts.

c. The amplifier saturates when the input differential voltage multiplied by the gain equals ± 10 volts. That is the input voltage must be smaller than + 1 volt but larger than - 1V.

3. Differential Amplifier

In the previous two circuits (inverting and non-inverting amplifiers), there is only one input signal. A differential amplifier is used to deal with two input signals, and only the difference between the two input signals is amplified. The circuit below shows a simple differential amplifier.

                                                Figure 6 – Circuit of a simple differential amplifier

Notice the two resistors connected to V₁ and V₂ should have the same value (well-matched resistors), and the two resistors, one connected to V_out and one connected to ground, should also have the same value. The output signal is related to the two input signals according to the following equation:

The gain of the differential amplifier is defined as the ratio between V_out and (V₂-V₁):

The measurement of ECG signal provides an excellent example to demonstrate the need for using a differential amplifier. As we know, the magnitude of the ECG signal on the body surface is very small: often less than 1 mV. On the other hand, due to the surrounding power supply lines, there is a strong 60 Hz noise signal on the body surface, and the magnitude of this noise is usually 1000 times larger than that of the ECG signal. This can be demonstrated by touching Channel 1 input of an oscilloscope with your finger and observing the distorted 60 Hz signal shown on the screen. If we either use an inverting amplifier or a non-inverting amplifier to amplify the ECG signal, what we will get is the large 60 Hz noise: the 60 Hz noise is so large that most likely the amplifier will be saturated.  Now, if we connect the electrode on the right arm to the V₁ input, connect the electrode on the left leg to the V₂ input, and connect the electrode on the right leg to ground (remember the previous lab?), the large 60 Hz noise in both V₁ and V₂  (which is called the common mode signal) will be cancelled out, and the ECG signal ¾ the voltage drop from left leg to right arm (which is called the differential signal) is amplified.

Part A – Experimenting on the Inverting Amplifier 

(1)   The circuit that you will be using for this part of experiment is shown below.

                                  Figure 7 – Circuit for the experimental with an inverting amplifier

The +10V and -10V power supply will be obtained from the ±25V group of VHP E3631A Triple Output DC Power Supply. Turn on the power supply by pressing 'Power' button. While keeping output disabled (output 'Off'), turn the knob to adjust +25 V supply to +10 V and the current limit to 0.3 A, and adjust -25 V supply to -10V and the current limit to 0.3A. The terminal 'COM' should be connected to the ground of your circuit. 

V_out from pin 6 should be connected to Channel 1 input of the oscilloscope. The input signal V_in is a sine waveform generated by the HP33120A Function Generator. The frequency of the signal is 1 kHz, and the peak-to-peak (p-p) amplitude is initially 0.2V. In addition to connecting V_in to R₁ of the circuit, you should also connect V_in to Channel 2 input of the oscilloscope so that the input and output of the circuit can be displayed simultaneously on the screen. Use the upper half of the screen to display V_out and use the lower half of the screen to display V_in.

(2)   After double checking the circuit connection first by yourself and then by the TA, enable the output of the power supply. Press the Autoscale key of the oscilloscope. If everything is right, you should see two sine waveforms displayed on the screen. Notice, there is a 180-degree phase difference between the two waveforms. Measure the p-p amplitudes of both input and output by the oscilloscope. If the amplitude of the input waveform is significantly different from 0.2V, adjust the Function Generator to make the p-p amplitude of V_in as close to 0.2 V as possible.

(3)   Record on your lab notebook the following set of values of the input and output peak-to-center amplitudes:

Input (p-p):                0.2V    0.6V    1.0V    1.4V    1.8V

Output (p-p):              xxV    xxV     xxV     xxV     xxV

a)     Graph your data and fit the data with a straight line.

b)    Determine the gain of the amplifier based on your graph (slope of the line).

c)     Compare this gain with the gain determined by Eq. (3).

(4) Increase the p-p amplitude of V_in to 3V, observe the clipping (due to saturation) of V_out. Sketch the waveforms (with the correct vertical scale) of both V_in and V_out on your lab notebook. Indicate the voltage of V_out where clipping occurs.

Part B – Experimenting on the Non-Inverting Amplifier 

(1) The circuit that you will be using for this part of experiment is shown below.

                       Figure 8 – Circuit for the experimental with a noninverting amplifier

(2)   The input signal V_in is the same as used in Part A. Again, use Channel 1 of the oscilloscope to display V_out and Channel 2 to display V_in.  Notice, now the two signals are in-phase.

(3)   Record on your lab notebook the following set of values of the input and output peak-to-center amplitudes:

Input (p-p):                0.2V    0.6V    1.0V   1.4V   1.8V

Output (p-p):              xxV    xxV     xxV    xxV    xxV

a)     Graph your data and fit the data with a straight line.

b)    Determine the gain of the amplifier based on your graph (slope of the line).

c)     Compare this gain with the gain determined by Eq. (4).

(4) Increase the p-p amplitude of V_in to 3V, observe the clipping (due to saturation) of V_out. Sketch the waveforms (with correct vertical scale) of V_in and V_out on your lab notebook. Indicate the voltage of V_out where clipping occurs.

Pre-lab Assignment

1. Calculate the gain of the inverting amplifier shown in Fig. 7.

2. Based on the power supply shown in Fig. 7 and the gain of the amplifier, determine the maximum  peak-to-peak amplitude of V_in that can still produce a V_out without clipping (i.e. further increase of the amplitude of V_in will produce a V_out with some clipping).

3. Calculate the gain of the non-inverting amplifier shown in Fig. 8.

4. Based on the power supply shown in Fig. 8 and the gain of the amplifier, determine the maximum  peak-to-peak amplitude of V_in that can still produce a V_out without clipping.