Feedback Negative and Divided | Analog and Digital Electronics - Electrical Engineering (EE) PDF Download

If we connect the output of an op-amp to its inverting input and apply a voltage signal to the noninverting input, we find that the output voltage of the op-amp closely follows that input voltage (I’ve neglected to draw in the power supply, +V/-V wires, and ground symbol for simplicity):

As V_in increases, V_out will increase in accordance with the differential gain. However, as V_out increases, that output voltage is fed back to the inverting input, thereby acting to decrease the voltage differential between inputs, which acts to bring the output down. What will happen for any given voltage input is that the op-amp will output a voltage very nearly equal to V_in, but just low enough so that there’s enough voltage difference left between V_in and the (-) input to be amplified to generate the output voltage.

The circuit will quickly reach a point of stability (known as equilibrium in physics), where the output voltage is just the right amount to maintain the right amount of differential. Taking the op-amp’s output voltage and coupling it to the inverting input is a technique known as negative feedback, and it is the key to having a self-stabilizing system (this is true not only of op-amps, but of any dynamic system in general). This stability gives the op-amp the capacity to work in its linear (active) mode, as opposed to merely being saturated fully “on” or “off” as it was when used as a comparator, with no feedback at all.

Because the op-amp’s gain is so high, the voltage on the inverting input can be maintained almost equal to V_in. We can write an equation relating the output voltage to the input voltage and the gain, G:

Vout=G⋅(Vin−Vout)$Vout=G·(Vin-Vout)$

Then, solving for the output voltage, we get the following:

Vout=Vin1+(1G)$Vout=\frac{V}{i}$

Let’s say that our op-amp has a differential voltage gain of 200,000 and V_in equals 6 V, we can calculate the output voltage using our equation:

Vout=61+(120,000)=5.999700015V$Vout=\frac{6}{1}=5.999700015V$

This creates just enough differential voltage (6 V - 5.99997000015 V = 29.99985 µV) to cause 5.99997000015 volts to be manifested at the output terminal, and the system holds there in balance. As you can see, 29.99985 µV is not a lot of differential, so for practical calculations, we can assume that the differential voltage between the two input wires is held by negative feedback exactly at 0 volts.

Advantage of Negative Feedback in Op-Amps

One great advantage in using an op-amp with negative feedback is that the actual voltage gain of the op-amp doesn’t matter, so long as its very large. If the op-amp’s differential gain were 250,000 instead of 200,000, all it would mean is that the output voltage would hold just a little closer to V_in (less differential voltage needed between inputs to generate the required output). In the circuit just illustrated, the output voltage would still be (for all practical purposes) equal to the non-inverting input voltage. Op-amp gains, therefore, do not have to be precisely set by the factory in order for the circuit designer to build an amplifier circuit with precise gain. Negative feedback makes the system self-correcting. The above circuit as a whole will simply follow the input voltage with a stable gain of 1.

How Does the Circuit in the Op-Amp Works?

Going back to our differential amplifier model, we can think of the operational amplifier as being a variable voltage source controlled by an extremely sensitive null detector, the kind of meter movement or other sensitive measurement device used in bridge circuits to detect a condition of balance (zero volts). The “potentiometer” inside the op-amp creating the variable voltage will move to whatever position it must to “balance” the inverting and non-inverting input voltages so that the “null detector” has zero voltage across it:

As the “potentiometer” will move to provide an output voltage necessary to satisfy the “null detector” at an “indication” of zero volts, the output voltage becomes equal to the input voltage: in this case, 6 volts. If the input voltage changes at all, the “potentiometer” inside the op-amp will change position to hold the “null detector” in balance (indicating zero volts), resulting in an output voltage approximately equal to the input voltage at all times.

This will hold true within the range of voltages that the op-amp can output. With a power supply of +15V/-15V, and an ideal amplifier that can swing its output voltage just as far, it will faithfully “follow” the input voltage between the limits of +15 volts and -15 volts. For this reason, the above circuit is known as a voltage follower. Like its one-transistor counterpart, the common-collector (“emitter-follower”) amplifier, it has a voltage gain of 1, a high input impedance, a low output impedance, and a high current gain. Voltage followers are also known as voltage buffers, and are used to boost the current-sourcing ability of voltage signals too weak (too high of source impedance) to directly drive a load. The op-amp model shown in the last illustration depicts how the output voltage is essentially isolated from the input voltage, so that current on the output pin is not supplied by the input voltage source at all, but rather from the power supply powering the op-amp.

It should be mentioned that many op-amps cannot swing their output voltages exactly to +V/-V power supply rail voltages. The model 741 is one of those that cannot: when saturated, its output voltage peaks within about one volt of the +V power supply voltage and within about 2 volts of the -V power supply voltage. Therefore, with a split power supply of +15/-15 volts, a 741 op-amp’s output may go as high as +14 volts or as low as -13 volts (approximately), but no further. This is due to its bipolar transistor design. These two voltage limits are known as the positive saturation voltage and negative saturation voltage, respectively. Other op-amps, such as the model 3130 with field-effect transistors in the final output stage, have the ability to swing their output voltages within millivolts of either power supply rail voltage. Consequently, their positive and negative saturation voltages are practically equal to the supply voltages.

Divided Feedback

If we add a voltage divider to the negative feedback wiring so that only a fraction of the output voltage is fed back to the inverting input instead of the full amount, the output voltage will be a multiple of the input voltage (please bear in mind that the power supply connections to the op-amp have been omitted once again for simplicity’s sake):

If R₁ and R₂ are both equal and V_in is 6 volts, the op-amp will output whatever voltage is needed to drop 6 volts across R1 (to make the inverting input voltage equal to 6 volts, as well, keeping the voltage difference between the two inputs equal to zero). With the 2:1 voltage divider of R1 and R2, this will take 12 volts at the output of the op-amp to accomplish.

Another way of analyzing this circuit is to start by calculating the magnitude and direction of current through R₁, knowing the voltage on either side (and therefore, by subtraction, the voltage across R₁), and R₁s resistance. Since the left-hand side of R₁ is connected to ground (0 volts) and the right-hand side is at a potential of 6 volts (due to the negative feedback holding that point equal to V_in), we can see that we have 6 volts across R₁. This gives us 6 mA of current through R₁ from right to left. Because we know that both inputs of the op-amp have extremely high impedance, we can safely assume they won’t add or subtract any current through the divider. In other words, we can treat R₁ and R₂ as being in series with each other: the current  flowing through R₁ must be the same with R₂. Knowing the current through R₂ and the resistance of R₂, we can calculate the voltage across R₂ (6 volts), and its polarity. Counting up voltages from ground (0 volts) to the right-hand side of R₂, we arrive at 12 volts on the output.

Upon examining the last illustration, one might wonder, “where does that 6 mA of current go?” Since the output voltage is positive, current flows from the positive side of the DC power supply, through the output pin of the op amp, through R₂, through R₁, to ground. Using the null detector/potentiometer model of the op-amp, the current path looks like this:

The 6 volt signal source does not have to supply any current for the circuit: it merely commands the op-amp to balance voltage between the inverting (-) and noninverting (+) input pins, and in so doing produce an output voltage that is twice the input due to the dividing effect of the two 1 kΩ resistors.

We can change the voltage gain of this circuit, overall, just by adjusting the values of R₁ and R₂ (changing the ratio of output voltage that is fed back to the inverting input). Gain can be calculated by the following formula:

Note that the voltage gain for this design of amplifier circuit can never be less than 1. If we were to lower R₂ to a value of zero ohms, our circuit would be essentially identical to the voltage follower, with the output directly connected to the inverting input. Since the voltage follower has a gain of 1, this sets the lower gain limit of the noninverting amplifier. However, the gain can be increased far beyond 1, by increasing R₂ in proportion to R₁.

Also note that the polarity of the output matches that of the input, just as with a voltage follower. A positive input voltage results in a positive output voltage, and vice versa (with respect to ground). For this reason, this circuit is referred to as a noninverting amplifier.

Relevance of Differential Gain of an Op-Amp to the Voltages and Currents in the Circuit

Just as with the voltage follower, we see that the differential gain of the op-amp is irrelevant, so long as its very high. The voltages and currents in this circuit would hardly change at all if the op-amp’s voltage gain were 250,000 instead of 200,000. This stands as a stark contrast to single-transistor amplifier circuit designs, where the Beta of the individual transistor greatly influenced the overall gains of the amplifier. With negative feedback, we have a self-correcting system that amplifies voltage according to the ratios set by the feedback resistors, not the gains internal to the op-amp.

Resulting Output Voltage and Gain with Input Voltage at the Inverting Input

Let’s see what happens if we retain negative feedback through a voltage divider, but apply the input voltage at a different location:

By grounding the noninverting input, the negative feedback from the output seeks to hold the inverting input’s voltage at 0 volts, as well. For this reason, the inverting input is referred to in this circuit as a virtual ground, being held at ground potential (0 volts) by the feedback, yet not directly connected to (electrically common with) ground. The input voltage this time is applied to the left-hand end of the voltage divider (R₁ = R₂ = 1 kΩ again), so the output voltage must swing to -6 volts in order to balance the middle at ground potential (0 volts). Using the same techniques as with the noninverting amplifier, we can analyze this circuit’s operation by determining current magnitudes and directions, starting with R₁, and continuing on to determining the output voltage.

We can change the overall voltage gain of this circuit, overall, just by adjusting the values of R₁ and R₂ (changing the ratio of output voltage that is fed back to the inverting input). Gain can be calculated by the following formula:

Note that this circuit’s voltage gain can be less than 1, depending solely on the ratio of R₂ to R₁. Also note that the output voltage is always the opposite polarity of the input voltage. A positive input voltage results in a negative output voltage, and vice versa (with respect to ground). For this reason, this circuit is referred to as an inverting amplifier. Sometimes, the gain formula contains a negative sign (before the R₂/R₁ fraction) to reflect this reversal of polarities.

These two amplifier circuits we’ve just investigated serve the purpose of multiplying or dividing the magnitude of the input voltage signal. This is exactly how the mathematical operations of multiplication and division are typically handled in analog computer circuitry.