Chapter 6   The Operational Amplifier.

The circuits are quite complex, the 741 op amp has 20 transistors in it. The curious student is referred to instructional books and data books devoted exclusively to op amps. There are many such books available.

Op amps are usually treated as 5 terminal devices. In the following pages they will be treated as just another component the same as a transistor is a component.

Electrical Connections.

Except for section 6.8 you will not again see the power supply connections to an op amp. The power supply connections are always there; they are usually not shown on schematic diagrams. The op amp will not work without power supply connections. If you see a diagram which does not show power supply connections and you assume that it is not necessary to connect a power supply, you will be in error. The op amp will not work without power supply connections. They are always assumed if not shown.

Many op amps have inputs for adjusting the offset voltage to zero. When the voltage at the two inputs is exactly equal the output should be zero but it is not. The amount by which the input voltage must be different to bring the output to zero is the offset voltage.

The Differential Inputs.

Now we will disconnect the inverting input from ground and connect the noninverting input to ground. The variable DC voltage will be applied to the inverting input. When the voltage at the inverting input is positive, the output voltage is negative. When the voltage at the inverting input is negative, the output voltage is positive. Thus you can see why the inputs have been named as they have.

Theoretically, if both inputs are connected together, the output voltage is zero. The offset adjustments can be used to adjust the output to zero, but there is a little more to it than that. If the two inputs (tied together) are made positive or negative, the output voltage will not change. But that's only in theory.

In practical op amps the output voltage will change a little when the two inputs have the same voltage applied to them. A voltage which is applied to both inputs simultaneously is called a common mode voltage. A voltage which is applied between the two inputs is called the normal mode voltage or differential voltage.

The normal mode gain is defined as the ratio of output voltage to the differential input voltage.

Example 6.1.

An op amp has had its offset adjusted to zero. Its inverting input is at +3.001 v and its noninverting input is at +3.000 v. Will its output be positive or negative?

Solution:

The noninverting input is negative with respect to the inverting input; therefore, the output will be negative.

Example 6.2.

An op amp has had its offset adjusted to zero. Its inverting input is at -5.001 v and its noninverting input is at -5.000 v. Will its output be positive or negative?

Solution:

The noninverting input is positive with respect to the inverting input; therefore, the output will be positive.

Example 6.3.

When the normal mode voltage is changed by 25 uv the output changes by 5 volts. When the common mode voltage is changed by 1 volt the output voltage changes by 6.325 volts. (a) What is the CMRR? (b) Express the CMRR in dB.

Solution:

The normal mode gain is 5 v / 25 uv = 200,000. The common mode gain is 6.325 v / 1 v = 6.325. (a) The CMRR is 200,000 / 6.325 = 31620. (b) The CMRR expressed in dB is 20 LOG (31620) = 90 dB.

Example 6.4.

An op amp has a DC gain of 100,000 and a unity gain bandwidth of 1 MHz. If feedback is applied to the op amp with a beta value of 0.05, what is the exact closed-loop gain at (a) DC, (b) 1 kHz, (c) 20 kHz, and (d) 100 kHz?

Solution:

From equation 5.15.5 the corner frequency is,

f_c2 = f_hu / A_Max

Where f_c2 is the corner frequency, f_hu is the upper unity gain frequency, and A_Max is the DC gain.

f_c2 = 1 MHz / 100,000 = 10 Hz.

(a) Using equation 6.4,

A' = A / (1 + A beta) = 100,000 / (1 + 100,000 x 0.05) = 19.9960

(b) Using equation 5.4.1

A' = A_DC / sqrt([f / f_c2]² + [1 + A_DC beta]²)

A' = 100,000 / sqrt([1,000 / 10]² + [1 + 100,000 x 0.05]²) = 19.9920

(c) and (d) also use equation 5.4.1.

(c) A' = 18.5663

(d) A' = 8.94391

Example 6.5.

For the conditions of example 6.4 calculate the percent error of gain using 1/beta as the standard of comparisons at (a) DC, (b) 1 kHz, (c) 20 kHz, and (d) 100 kHz.

Solution:

Since 1/beta = 20.00000 the errors are as follows:

(a) -0.020%, (b) -0.040%, (c) -7.1685%, (d) -55.2805%.

A unity gain buffer has a voltage gain of unity as the name implies. The value of beta is set at unity by feeding back all of the output voltage to the inverting input. The circuit of Figure 6.2 is of a unity gain buffer.

The voltage at the output cannot become 2.000000000000 volts. If it did the differential input voltage would be zero and the output would be zero. Since the output voltage is finite, there must be a finite difference in voltage between the noninverting input and the inverting input. That small but finite difference is caused by the finite open-loop gain of the op amp. Because the open-loop gain of the op amp is very large, the error is very small.

The percent error is,

Example 6.6.

Calculate the gain error of a unity-gain buffer if the op amp has an open-loop gain of (a) 200,000; (b) 20,000; (c) 1,000; and (d) 20?

Solution:

The % error is (a) -100% / (1 + 200,000) = -0.0004999975%, (b) -0.00499975%, (c) -0.0999%, (d) -4.7619%.

Often it is desired to calculate the input and/or output impedance at higher frequencies where the gain is less than the DC gain. Equations 6.10 and 6.11 may be used but the value of A must be the gain at the frequency being considered, not the gain at DC. This gain can be obtained from equation 5.15.1 which is repeated below.

Example 6.7.

Determine the minimum input impedance and the maximum output impedance of a unity-gain buffer employing a type 741 op amp. The data book gives the minimum input resistance of the 741 as 0.3 M ohms, a typical output resistance of 75 ohms and a minimum gain of 20,000. The typical gain is given as 200,000.

Solution:

The minimum gain is given as 20,000. The typical gain is given as 200,000 but as the question asks for the minimum input impedance and maximum output impedance, the minimum gain should be used.

Z_IN = 0.3 M ohms x 20,000 = 6 G ohms,

The output impedance is, Z_O = 75 / 20,000 = 3.75 milli ohms.

This circuit works because the op amp will always adjust its output until both inputs have the same potential. If a given potential is placed between the input terminals in Figure 6.3 the noninverting input of the op amp will be placed at that potential. The op amp will adjust its output until the inverting input is at the same potential. Because of the voltage divider consisting of R₁ and R₂ the output voltage must be greater than the input voltage by the amount of the divider ratio. The conditions of equal potential at both inputs require that they have the same sign. Because resistors can't change the sign of a voltage, the output voltage must also have the same sign as the input voltage.

Example 6.8.

A noninverting amplifier has the following values: R₁ = 4.7 k ohms and R₂ = 2.7 k ohms. The open-loop gain is 200,000. What is the closed-loop gain?

Solution:

When the open-loop gain is very high the closed-loop gain is 1/beta. Therefore, the closed-loop gain is given by A' = (R₁ + R₂) / R₁ = (4.7 k ohms + 2.7 k ohms) / 4.7 k ohms = 1.57.

Example 6.9.

A noninverting amplifier is to have a gain of 20 dB and R₂ is to be a 10 k ohm resistor. What is the necessary value of R₁?

Solution:

The first step is to convert dB into gain.

A' = 10^dB/20 = 10^20/20 = 10¹ = 10.

Since A' = 1/beta to obtain a gain of 10, beta must equal 0.1. Solving equation 6.12 for R₁ gives,

R₁ = beta R₂ / (1 - beta)

R₁ = 0.1 x 10 k ohms / (1 - 0.1) = 1.11 k ohms.

This circuit also works because the op amp will always adjust its output until both inputs have the same potential. If a voltage of a given sign and magnitude is placed between the input terminals in Figure 6.4 the inverting input of the op amp will be placed at a potential which is determined by R₁ and R₂. The op amp will adjust its output until the inverting input is at the same potential as the noninverting input. Because the noninverting input is grounded, the inverting input will always be adjusted to zero potential. Figure 6.5 shows an equivalent circuit to the inverting amplifier.

Example 6.10.

An inverting amplifier has the following values: R₁ = 4.7 k ohms and R₂ = 2.7 k ohms. The open-loop gain is 200,000. What is the closed-loop gain?

Solution:

When the open-loop gain is very high the closed-loop gain is 1/beta. Therefore, the closed-loop gain is given by,

A' = R₂ / R₁ = 2.7 k ohms / 4.7 k ohms = -0.574.

That's right. It's possible for an inverting amplifier to have a gain of less than unity.

Example 6.11.

An inverting amplifier is to have a gain of 30 dB and R₂ is to be a 10 k ohm resistor. What is the necessary value of R₁?

Solution:

The first step is to convert dB into gain.

A' = 10^dB/20 = 10^30/20 = 10^3/2 = 31.62.

Since A' = 1/beta, to obtain a gain of -31.62, beta must equal 0.03163. Solving equation 6.13 for R₁ gives,

R₁ = beta R₂ = 0.03163 x 10 k ohms = 316 ohms.

Example 6.12.

In a circuit similar to that of Figure 6.6 the values are as follows: R_1A = 10 k ohms, R_1B = 7.5 k ohms, R_1C = 5 k ohms, R₂ = 10 k ohms, V_INA = +15 v, V_INB = -9 v, and V_INC = -8 v. What is the output voltage of the amplifier?

Solution:

We will sum the currents at the node where the 4 resistors are joined. This node is at zero potential.

I_A + I_B + I_C = I₂.

Let us define current into the node as positive.

I_A = V_INA / R_1A = 15 v / 10 k ohms = 1.5 mA.

I_B = -9 v / 7.5 k ohms = -1.2 mA.

I_C = -8 v / 5 k ohms = -1.6 mA.

I₂ = 1.5 mA -1.2 mA -1.6 mA = -1.3 mA.

The net current flows out of the node; therefore, I₂ must flow into the node to prevent a charge imbalance in the conductors. The current in R₂ is flowing to the left, which means that the right end is positive with respect to the summing node. The voltage across R₂ is I₂ R₂ = 1.3 mA 10 k ohms = +13 volts. Equation 6.17 will give the same answer.

Example 6.12 illustrates that a summing amplifier gives the negative of the true sum of the input voltages. This was one of the original uses of an operational amplifier.

Although we are not sound engineers, most of us own stereo systems and examples from that field are easy to relate to and understand. Suppose it is desired to add together the left and right channels of a stereo signal to form a monaural signal for a monaural amplifier. At the same time it is necessary to maintain the individual integrity of the left and right signals to feed a stereo amplifier. A circuit similar to Figure 6.6 but with two equal value resistors in the inputs will do the job nicely. The output will be a true L + R sum which is required for a monaural image. Because the summing node is a virtual ground, the individual L and R signals will not get mixed up in the lines before the summing amplifier. Audio engineers call this circuit a mixer but radio engineers mean something completely different when they say "mixer". Our best course is to avoid the term all together.

The op amp will adjust its output to keep its two inputs at the same potential. Since the noninverting input is grounded, the inverting input will also be at ground potential. The current in R is

Example 6.13.

In Figure 6.7 the capacitor starts out with zero initial charge. The input voltage is as shown in Figure 6.8. Draw the output voltage versus time, giving scale values. R = 2 M ohms and C = 1 uf.

Solution:

The integrator's output is shown in Figure 6.9. For the first 1/2 second there is no input and the output does not change. From 0.5 to 1.5 seconds the input voltage is 1.0 volts. That is an area of 1.0. If the gain of the integrator were unity, the output would be -1 after 1.5 seconds but this one has a 2 M ohm resistor which makes the 1/RC term in front of the integral to be 1/2. The input is 1 volt for 1 second or 1 volt second. The output will change by -0.5 volts/second for 1 second to give a value of -0.5 volt. From 1.5 seconds to 2.0 seconds the input is zero and the output will not change. Notice that the output is not zero when the input is zero. When the input is zero there is no additional area being added to the integral and so the output simply stays where it is. The input is -0.5 volts from 2 to 3.5 seconds which is an enclosed area of 0.75. Because of the gain of the integrator the output will change by +0.375 volts during the 1.5 seconds when the input is -0.5 volts. The output voltage will change at a rate of 0.25 volts/second for 1.5 seconds. At the end of 3.5 seconds the output will be -0.125 volts. The output will remain at this level from 3.5 to 4.0 seconds when the input is zero. When the input goes to -1.5 volts the output will change at a rate of 0.75 volts/second for a time of 0.5 seconds. The enclosed area is 0.75 and so the output will change by an amount of 0.375 volts. The output will end up at a voltage of +0.25 volts. The output will remain at this level until there is additional input or the capacitor is discharged. There is no requirement that the output come back to zero.

In some applications it may not be necessary to have a bipolar output or the signals to be processed may be AC only. In such a case it is best to power an op amp from a single power supply. One power supply is less costly, smaller, lighter and generates less heat than two power supplies.

There are some special purpose op amps which are designed for single supply operation. Their inputs may be placed at or even a little below ground even though the op amp is operating on a single positive power supply. However, the output of such an op amp can only have potentials of ground + 1.2 volts to VCC -1.2 volts. For best operation an offset should be added to the input to place the output at 1/2 of VCC.

Input Offset Voltage is the voltage which must be applied to the input terminals to bring the output exactly to zero. The typical and maximum values are given here. The untrimmed value is given, which means that no trimmer pot was connected to the pins provided for that purpose. The units are usually mV (the units are always stated in the table).

Input Bias Current is the current which flows into each input of an op amp. If the input stage is a pair of BJTs, the base current of the transistors is the bias current. If the input is a pair of FETs, the bias current is the leakage current of the P-N junction which makes up the gates of the FETs. The units may be uA, nA or pA depending on the op amp. This specification is only significant when the resistance in series with the inputs is large. If the source resistance is such that the voltage drop due to the input offset current is significant compared to the input offset voltage, the resistance in the two inputs must be equal.

Input Offset Current is the difference between the bias currents in the two inputs. If the input bias current is producing a significant drop across the source resistances, the offset current will be converted to an offset voltage by the IR drop across the source resistances.

Input Resistance is the resistance the op amp inputs present to an input signal without any feedback applied to the amp.

Input Capacitance is the effective capacitance from each input to ground including Miller capacitance.

Offset Voltage Adjustment Range is only given on op amps which have pins for an offset voltage trimmer. It indicates how large the offset can be and still be adjusted to zero.

Input Voltage Range tells how far the inputs may be moved around in potential. This is usually the voltage which the amplifier will tolerate without damage. The input voltage range for proper operation of the amplifier may be given separately as a common mode voltage rating. It is usually 2 or 3 volts less than the maximum input voltage.

Common Mode Rejection Ratio (CMRR) is the ratio of normal mode gain to common mode gain. CMRR is most often given in dB. It is defined in detail on pages 244 and 245 of this text.

Supply Voltage Rejection Ratio states how much the power supply voltage will affect the output voltage of the op amp. It is the change in power supply voltage in uV divided by the change in output voltage in volts. This parameter is measured under open-loop conditions.

Large Signal Voltage Gain is the open-loop gain.

Output Voltage Swing states over what range the output voltage can vary.

Output Resistance is the Thevenin resistance of the amplifier.

Output Short Circuit Current states how much current will flow if the output of the amplifier is shorted to ground. Most op amps have built-in current limiting to protect them from damage if a short occurs.

Supply Current is how much current the op amp will draw from the power supply.

Power Consumption is the supply current multiplied by the total power supply voltage.

Transient Response (Unity Gain) states the risetime and overshoot for a unity-gain amplifier with a step input voltage.

Slew Rate is very important because it does more to limit the upper frequency limit than does the unity gain bandwidth. It states how fast the output voltage can change and is usually given in V/us.

The Gain Bandwidth Product or Unity Gain Frequency may be given in a graph rather than in the data table. It is the frequency at which the gain is unity.

Op amp data usually contains a large number of graphs. The majority of these graphs show how various parameters are effected by temperature. If you are going to use an op amp under conditions of extreme temperature, you should not ignore these graphs. If you are looking for a particular specification and it is not in the table, it may be in the graphs.