Chapter 7   Operational Amplifier Circuits.

Example 7.1.

In the circuit of Figure 7.2 R₁ = 2 k ohms, R₂ = 99 k ohms, R₃ = 10 k ohms and R₄ = 100 k ohms. If the voltage at the top input is +2.0 mV and the voltage at the bottom input is -1.0 mV, what is the output voltage?

Solution:

The best way to keep track of voltages is to make a chart of all three terminals of all three op amps as shown. The voltages at the noninverting inputs of OA₁

Terminal	OA1	OA2	OA3
noninverting input	+2 mV	-1 mV	136.81818 mV
inverting input	+2 mV	-1 mV	136.81818 mV
Output	150.5 mV	-149.5 mV	3.00 V

and OA₂ are given and so can be filled in immediately. The op amps will make the voltage at the inverting input be the same as that at the noninverting input and so these values can also be filled in by inspection. The voltage across R₁ is 2 mV -(-1 mV) = 3 mV. The current in R₁ is I = 3 mV / 2 k ohms = 3/2 uA. The voltage drop across either R₂ is V = 3/2 uA x 99 k ohms = 148.5 mV with the upper end positive. The lower end of the top R₂ is at a potential of +2 mV; therefore, the upper end of the top R₂ is at 2 mV + 148.5 mV = 150.5 mV. The upper end of the bottom R₂ is at -1 mV; therefore, the lower end is at (-1 mV) + (-148.5 mV) = -149.5 mV. Notice that the difference is 150.5 mV - (-149.5 mV) = 300 mV. This confirms a first-stage gain of 100. The voltage at the noninverting input of OA₃ is found by simply applying the voltage divider equation to R₃ and R₄. The noninverting input of OA₃ does not load the voltage divider. V₊₃ = V_O1 R₄ / (R₃ + R₄) = 150.5 mV x 100 k ohms / 110 k ohms = 136.81818 mV. The entries for both the noninverting input and the inverting input of OA3 can now be filled in. The voltage drop across the bottom R₃ is 136.81818 mV - (-149.5 mV) = 286.31818 mV. The current through the lower R₃ is I = 286.31818 mV / 10 k ohms = 28.631818 uA. The voltage drop across the lower R₄ is 28.631818 uA x 100 k ohms = 2863.1818 mV with the right end positive. The voltage at the output of OA₃ is its inverting input voltage + the drop across the lower R₄, which is 136.81818 mV + 2863.1818 mV = 2999.99998 mV which, when rounded off, gives 3.00 volts.

The results of this example confirm that the gain of the second (OA₃) stage is given by

The input to the circuit of Figure 7.2 is a differential input because neither side of the input needs to be connected to ground. The output of the first stage (OA₁ and OA2) is a differential output because neither side is connected to ground. The input to the second (OA3) stage is also differential. The output from OA₃ is said to be single-ended because one side is connected to ground. Thus a circuit consisting of two R₃s, two R₄s and an op amp can be used to convert a differential source to a single-ended load. The problem with using such a circuit is that the input resistances at its two inputs are not equal. The input resistance at the noninverting input is R3 + R4 while the input impedance at the inverting input is R3. The two noninverting amplifiers serve to balance the input impedances and at the same time make them very high.

Connecting R₁ between the two amplifiers instead of each amplifier having its own R₁ makes the amplifiers self-balancing, which increases the common mode rejection ratio. Note in the example that even though the original inputs had a 2 to 1 imbalance, the outputs had less than a 1% imbalance. The OA₃ stage can easily accommodate the small imbalance. If this circuit is to have a high CMRR the values of the two R₃s and the two R₄s must be very well matched.

The resistors R_S are used to prevent serious imbalance due to input bias currents in the event that the source has a high Thevenin resistance. If input bias current is no problem the two resistors may be omitted (replaced by pieces of wire).

Electronic voltmeters use some electronic tricks to make the response of the rectifiers linear. The circuit of a linear AC voltmeter is shown in Figure 7.3.

The value of R2 sets the full-scale range of the meter. With the value given the range will be approximately 1 volt if a 1 mA meter is used. The 100 pf capacitor is to prevent the op amp from oscillation at high frequencies. This capacitor will cause the meter to lose accuracy at high frequencies and if you want to measure voltages of high frequency waves the capacitor must be as small as possible while still preventing oscillation. The capacitor at the input is necessary to block any DC which may be mixed with the AC to be measured.

When a positive voltage is applied to the input the op amp will adjust the brightness of the lamp so that the voltage across the photocell on the left is equal to the input voltage. If the two photocells are closely matched the voltage across the photocell on the right will also be equal to the input voltage. Because there is no electrical connection between the two parts of the circuit, the input can be at any arbitrarily high potential and allow the output to be grounded.

In this circuit the batteries are meant to be batteries. The circuit should be constructed in an insulated box to prevent accidental electric shock. The value of the resistor and pot which are in series with the photocells depend on the illuminated resistance of the photocells. The values of these resistors will have to be determined after the photocells are obtained. The purpose of the two diodes in series with the lamp is to permit the lamp voltage to go all the way to zero. The output of an op amp will have a minimum voltage of 2 P-N junctions from the negative power supply lead. The circuit only works for positive input voltages. If a negative input is applied, the lamp will attempt to get darker than dark and nothing will happen, except in the twilight zone.

When the input goes negative the output of OA₂ goes negative, which reverse biases D₄, disconnecting the output of OA₂ from the output node of the circuit. D₃ is forward biased. For a negative input, the output of OA₁ will go positive. D₂ is forward biased and D₁ is reverse biased. Resistor R₂ takes feedback from the output node and if R₁ and R₂ are of equal value the output node will have the same magnitude and opposite sign as the input node. Thus for a negative input the output is positive but has the same magnitude as the input.

If a capacitor is placed in parallel with R₄ the circuit becomes a peak detector. If an RC low-pass filter is connected to the output the circuit becomes an average value detector. In the latter case the resistor in the low-pass filter should be greater than 1 M ohms for an error of less than 1%. In either case the value of the capacitor will depend on the response time required.

If it is necessary to take the log of data which is being collected in an experiment, the method of choice is to digitize the data and let a computer do the log taking. In very well-funded laboratories the computer is on-line, (connected to the experimental equipment). In less well-funded experiments the data must be hand carried - on disk, tape or paper - to the computer.

It is always desirable and often necessary to see the log of the data as the experiment is running. Seeing the log of the data may be necessary in order to properly adjust the equipment. If the experimenter is able to see the data come out as the experiment runs, it is possible to tell if something has gone wrong. A strip-chart recorder is often used for this purpose.

Many experiments deliver the data in analog form and most strip-chart recorders accept data in analog form. For the purpose of so-called "quick look" data reduction an analog device which will take the log of analog data may be helpful.

The circuit of a working logarithmic converter is shown in Figure 7.6. This circuit is designed to be used with a strip-chart recorder having a Y axis input of 10 mV full- scale. With the values given the unit may be calibrated to 1 mV = 10 dB. The converter operates from two 9 volt transistor radio batteries.

The batteries are connected as shown in Figure 6.1. A double pole switch is required to interrupt the current from both batteries.

The input signal is applied through R₁ to the inverting input of OA₁ which is nothing more than a unity-gain inverter. R₅ is a zero adjust control. The op amp used has no offset adjustment pin connections and so the somewhat elaborate network of R₄ through R₉ is required to provide offset adjustment for the op amp.

Transistor Q₁ is the element which does the log taking. The I-V curve of a P-N junction is an exponential or a logarithmic function depending on how you look at it. In this circuit we are looking at it as a log function. If a positive input signal is applied to the input, it will be inverted and appear at the output of OA₁ as a negative voltage. This will cause current to flow to the left in R₁₀. The output of OA₂ will adjust itself so that the collector current of Q₁ is equal to the current in R₁₀. The voltage at the op amp's output will be adjusted so as to forward bias the base-emitter junction of Q₁. If the input voltage changes, the collector current of Q₁ will change linearly and the base-emitter voltage will change logarithmically.

Transistor Q₂ serves to temperature compensate the circuit. For example, if the temperature goes up, the base- emitter voltage of Q₁ will decrease, which will lower the output voltage. At the same time the temperature of Q₂ will increase and its base-emitter voltage will also decrease. The decreased drop across Q₂ will increase the output voltage which will compensate for the decrease caused by Q₁. Q₁ and Q₂ should be in physical contact to insure that they are at the same temperature. If the transistors are in metal packages, they should be electrically insulated from each other because the case is often connected electrically to one of the transistor elements.

The network of R₁₂ through R₁₄ reduces the output voltage to the proper level for the 10 mV chart recorder. R₁₃ should be adjusted for a span of 10 dB per mV at the upper end of the working range. R₅ may be used to correct errors at the low end of the range. The unit will not have a working range of 100 dB as implied by 10 / mV over a range of 10 mV. The best achievable is about 50 dB. Below -50 dB the response tends to become linear rather than logarithmic.

If the input voltage is negative, OA2 will go to full negative saturation which will slam the chart recorder's pen down hard against the zero stop. The log of a negative number is undefined in the analog world as well as in the world of numbers. The only capacitor in the circuit is to prevent OA2 from oscillating at ultrasonic frequencies.

Figure 7.7 shows a high-pass filter circuit. In this circuit if C = 0.01 uf and R = 10 k ohms the cutoff frequency is 2.8 kHz and the attenuation at 60 Hz is 50 dB. The cutoff frequency can be calculated from,

Figure 7.8 shows a low-pass filter circuit. In this circuit if C = 0.1 uf and R = 150 k ohms the cutoff frequency is 4.5 Hz and the attenuation at 60 Hz is 36 dB. The cutoff frequency can be calculated from,

Example 7.2.

It is desired to construct a low-pass filter with a cutoff frequency of 0.1 Hz. The capacitors to be used have a value of 1.0 uf. (a) What is the calculated value of R? (b) What standard values should be used for the two resistors in the filter?

Solution:

(a) Solving equation 7.6 for R gives R = 1 / (15 fC C) =
1 / (15 x 0.1 x 1 x 10-6) = 667 k ohms

(b) The closest value for R is 680 k ohms and the closest value for 2R is 1.3 M ohms.

1. In the circuit of Figure 7.2 R₁ = 200 ohms, R₂ = 99.9 k ohms, R₃ = 10 k ohms and R₄ = 1 M ohm. If the voltage at the top input is -20 uV and the voltage at the bottom input is -70 uV, (a) what are the voltages at each of the op amp terminals? Make a chart similar to that in Example 7.1. (b> What is the gain of the amplifier?

2. Using the circuits given in this chapter draw the schematic diagram of a circuit which will take an AC signal and rectify it without introducing significant errors. Then the circuit must average the rectified AC and convert it to dB.

3. Design a high-pass filter which has a cutoff frequency of 10 kHz. The capacitors have a value of 0.0033 uf. Give the standard values for the resistors to be used in the circuit. Draw the circuit.