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Investing op amp gain formulas

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Op amps usually have three terminals: two high-impedance inputs and a low-impedance output port. Operational amplifiers work to amplify the voltage differential between the inputs, which is useful for a variety of analog functions including signal chain, power, and control applications. Because most op amps are used for voltage amplification, this article will focus on voltage amplifiers. There are many different important characteristics and parameters related to op amps see Figure 1. These characteristics are described in greater detail below.

This means the feedback path, or loop, is open. Voltage comparators compare the input terminal voltages. Even with small voltage differentials, voltage comparators can drive the output to either the positive or negative rails. High open-loop gains are beneficial in closed-loop configurations, as they enable stable circuit behaviors across temperature, process, and signal variations.

Input impedance is measured between the negative and positive input terminals, and its ideal value is infinity, which minimizes loading of the source. In reality, there is a small current leakage. Arranging the circuitry around an operational amplifier may significantly alter the effective input impedance for the source, so external components and feedback loops must be carefully configured. It is important to note that input impedance is not solely determined by the input DC resistance.

Input capacitance can also influence circuit behavior, so that must be taken into consideration as well. However, the output impedance typically has a small value, which determines the amount of current it can drive, and how well it can operate as a voltage buffer.

An ideal op amp would have an infinite bandwidth BW , and would be able to maintain a high gain regardless of signal frequency. Op amps with a higher BW have improved performance because they maintain higher gains at higher frequencies; however, this higher gain results in larger power consumption or increased cost. GBP is a constant value across the curve, and can be calculated with Equation 1 :. These are the major parameters to consider when selecting an operational amplifier in your design, but there are many other considerations that may influence your design, depending on the application and performance needs.

Other common parameters include input offset voltage, noise, quiescent current, and supply voltages. In an operational amplifier, negative feedback is implemented by feeding a portion of the output signal through an external feedback resistor and back to the inverting input see Figure 3. Negative feedback is used to stabilize the gain. This is because the internal op amp components may vary substantially due to process shifts, temperature changes, voltage changes, and other factors.

The closed-loop gain can be calculated with Equation 2 :. There are many advantages to using an operational amplifier. Op amps have a broad range of usages, and as such are a key building block in many analog applications — including filter designs, voltage buffers, comparator circuits, and many others. In addition, most companies provide simulation support, such as PSPICE models, for designers to validate their operational amplifier designs before building real designs.

The limitations to using operational amplifiers include the fact they are analog circuits, and require a designer that understands analog fundamentals such as loading, frequency response, and stability. It is not uncommon to design a seemingly simple op amp circuit, only to turn it on and find that it is oscillating. Due to some of the key parameters discussed earlier, the designer must understand how those parameters play into their design, which typically means the designer must have a moderate to high level of analog design experience.

There are several different op amp circuits, each differing in function. The most common topologies are described below. The most basic operational amplifier circuit is a voltage follower see Figure 4. This circuit does not generally require external components, and provides high input impedance and low output impedance, which makes it a useful buffer. Because the voltage input and output are equal, changes to the input produce equivalent changes to the output voltage.

The most common op amp used in electronic devices are voltage amplifiers, which increase the output voltage magnitude. Inverting and non-inverting configurations are the two most common amplifier configurations. Both of these topologies are closed-loop meaning that there is feedback from the output back to the input terminals , and thus voltage gain is set by a ratio of the two resistors. In inverting operational amplifiers, the op amp forces the negative terminal to equal the positive terminal, which is commonly ground.

In this configuration, the same current flows through R2 to the output. The current flowing from the negative terminal through R2 creates an inverted voltage polarity with respect to V IN. This is why these op amps are labeled with an inverting configuration. V OUT can be calculated with Equation 3 :. The operational amplifier forces the inverting - terminal voltage to equal the input voltage, which creates a current flow through the feedback resistors.

The output voltage is always in phase with the input voltage, which is why this topology is known as non-inverting. Note that with a non-inverting amplifier, the voltage gain is always greater than 1, which is not always the case with the inverting configurations. VOUT can be calculated with Equation 4 :. The gain of the operational amplifier is very high, this means that for outputs within the rail voltage, which it is for an analogue amplifier, the voltage difference between the inverting and non-inverting inputs must be very small.

As the non-inverting input is at ground, the inverting input must be virtually at ground. It is for this reason that the circuit is sometimes referred to as a virtual earth amplifier. The op amp inverting amplifier is very easy to design, but as with any design there are a few hints and tips that can be of use. These are only three tips for the circuit design of an op amp inverting amplifier that have been found useful over the years.

The main concept is to keep an open mind as to things that might happen in the circuit under unusual circumstances. It also helps not to stretch the circuit design too far, expecting too much from a single stage. Having the tips and these points in mind when designing the circuit can help avoid issues later.

Typically an op amp circuit will be operated from differential supplies, e. This is quite acceptable in many applications, but in many electronic circuit designs only one supply may be available. Under these circumstances it is relatively easy to implement what is termed a single ended version of the inverting amplifier op amp circuit - this uses only one supply and ground. The single voltage supply version of the op amp circuit for the inverting amplifier circuit uses more components when compared to the dual rail version, but the design of the amplifier elements remains the same.

Effectively a half way point is created for the non-inverting input. And in this way the operational amplifier sees the same conditions it would as if it were operating from a dual supply. The single ended rail version of the op amp circuit finds applications where only one voltage supply rail is available. Often circuits running from battery supplies will only have one supply and this solution is often employed in these applications.

There are some op amps that are designed to operate in a single ended mode, but this approach can be adopted for op amps that are available. An operational amplifier is a differential amplifier, and therefore there are two inputs: for the inverting amplifier, the negative feedback from the output and the input signal are both applied to the inverting input, whilst the non-inverting input is taken to ground.

The op amp circuit for the inverting amplifier offers many advantages including relatively low input impedance, a low output impedance and the level of gain that is required within the limits of the op amp and the gain required from the overall circuit. It also requires very few electronic components to produce a high performance circuit. Shopping on Electronics Notes Electronics Notes offers a host of products are very good prices from our shopping pages in association with Amazon.

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A consideration to keep in mind, though, is common-mode gain in differential op-amp circuits such as instrumentation amplifiers. We should expect to see no change in output voltage as the common-mode voltage changes:. Aside from very small deviations actually due to quirks of SPICE rather than real behavior of the circuit , the output remains stable where it should be: at 0 volts, with zero input voltage differential. Our input voltage differential is still zero volts, yet the output voltage changes significantly as the common-mode voltage is changed.

More than that, its a common-mode gain of our own making, having nothing to do with imperfections in the op-amps themselves. With a much-tempered differential gain actually equal to 3 in this particular circuit and no negative feedback outside the circuit, this common-mode gain will go unchecked in an instrument signal application.

There is only one way to correct this common-mode gain, and that is to balance all the resistor values. Suppose that all resistor values are exactly as they should be, but a common-mode gain exists due to an imperfection in one of the op-amps. With the adjustment provision, the resistance could be trimmed to compensate for this unwanted gain. One quirk of some op-amp models is that of output latch-up , usually caused by the common-mode input voltage exceeding allowable limits.

In JFET-input operational amplifiers, latch-up may occur if the common-mode input voltage approaches too closely to the negative power supply rail voltage. On the TL op-amp, for example, this occurs when the common-mode input voltage comes within about 0. Such a situation may easily occur in a single-supply circuit, where the negative power supply rail is ground 0 volts , and the input signal is free to swing to 0 volts.

Latch-up may also be triggered by the common-mode input voltage exceeding power supply rail voltages, negative or positive. As a rule, you should never allow either input voltage to rise above the positive power supply rail voltage, or sink below the negative power supply rail voltage, even if the op-amp in question is protected against latch-up as are the and op-amp models. At worst, the kind of latch-up triggered by input voltages exceeding power supply voltages may be destructive to the op-amp.

While this problem may seem easy to avoid, its possibility is more likely than you might think. Consider the case of an operational amplifier circuit during power-up. If the circuit receives full input signal voltage before its own power supply has had time enough to charge the filter capacitors, the common-mode input voltage may easily exceed the power supply rail voltages for a short time. If the op-amp receives signal voltage from a circuit supplied by a different power source, and its own power source fails, the signal voltage s may exceed the power supply rail voltages for an indefinite amount of time!

Another practical concern for op-amp performance is voltage offset. That is, effect of having the output voltage something other than zero volts when the two input terminals are shorted together. When that input voltage difference is exactly zero volts, we would ideally expect to have exactly zero volts present on the output. However, in the real world this rarely happens. Even if the op-amp in question has zero common-mode gain infinite CMRR , the output voltage may not be at zero when both inputs are shorted together.

This deviation from zero is called offset. A perfect op-amp would output exactly zero volts with both its inputs shorted together and grounded. However, most op-amps off the shelf will drive their outputs to a saturated level, either negative or positive. In the example shown above, the output voltage is saturated at a value of positive For this reason, offset voltage is usually expressed in terms of the equivalent amount of input voltage differential producing this effect.

In other words, we imagine that the op-amp is perfect no offset whatsoever , and a small voltage is being applied in series with one of the inputs to force the output voltage one way or the other away from zero. Offset voltage will tend to introduce slight errors in any op-amp circuit. So how do we compensate for it? Unlike common-mode gain, there are usually provisions made by the manufacturer to trim the offset of a packaged op-amp.

These connection points are labeled offset null and are used in this general way:. On single op-amps such as the and , the offset null connection points are pins 1 and 5 on the 8-pin DIP package. Inputs on an op-amp have extremely high input impedances. We analyze the circuit as though there was absolutely zero current entering or exiting the input connections. This idyllic picture, however, is not entirely true.

Op-amps, especially those op-amps with bipolar transistor inputs, have to have some amount of current through their input connections in order for their internal circuits to be properly biased. These currents, logically, are called bias currents. Under certain conditions, op-amp bias currents may be problematic. The following circuit illustrates one of those problem conditions:.

At first glance, we see no apparent problems with this circuit. In other words, this is a kind of comparator circuit , comparing the temperature between the end thermocouple junction and the reference junction near the op-amp. The problem is this: the wire loop formed by the thermocouple does not provide a path for both input bias currents, because both bias currents are trying to go the same way either into the op-amp or out of it.

In order for this circuit to work properly, we must ground one of the input wires, thus providing a path to or from ground for both currents:. Another way input bias currents may cause trouble is by dropping unwanted voltages across circuit resistances. Take this circuit for example:. We expect a voltage follower circuit such as the one above to reproduce the input voltage precisely at the output.

But what about the resistance in series with the input voltage source? But even then, what slight bias currents may remain can cause measurement errors to occur, so we have to find some way to mitigate them through good design. One way to do so is based on the assumption that the two input bias currents will be the same. In reality, they are often close to being the same, the difference between them referred to as the input offset current.

If they are the same, then we should be able to cancel out the effects of input resistance voltage drop by inserting an equal amount of resistance in series with the other input, like this:. With the additional resistance added to the circuit, the output voltage will be closer to V in than before, even if there is some offset between the two input currents. In either case, the compensating resistor value is determined by calculating the parallel resistance value of R 1 and R 2.

Why is the value equal to the parallel equivalent of R 1 and R 2? This gives two parallel paths for bias current through R 1 and through R 2 , both to ground. A related problem, occasionally experienced by students just learning to build operational amplifier circuits, is caused by a lack of a common ground connection to the power supply. This provides a complete path for the bias currents, feedback current s , and for the load output current.

Take this circuit illustration, for instance, showing a properly grounded power supply:. The effect of doing this is profound:. Thus, no electrons flow through the ground connection to the left of R 1 , neither through the feedback loop. This effectively renders the op-amp useless: it can neither sustain current through the feedback loop, nor through a grounded load, since there is no connection from any point of the power supply to ground.

The bias currents are also stopped, because they rely on a path to the power supply and back to the input source through ground. The following diagram shows the bias currents only , as they go through the input terminals of the op-amp, through the base terminals of the input transistors, and eventually through the power supply terminal s and back to ground. Without a ground reference on the power supply, the bias currents will have no complete path for a circuit, and they will halt.

Since bipolar junction transistors are current-controlled devices, this renders the input stage of the op-amp useless as well, as both input transistors will be forced into cutoff by the complete lack of base current. Bias currents are small in the microamp range , but large enough to cause problems in some applications. It is not enough to just have a conductive path from one input to the other.

You can learn more about Op-amps by following our Op-amp circuits section. An op-amp has two differential input pins and an output pin along with power pins. Those two differential input pins are inverting pin or Negative and Non-inverting pin or Positive. An op-amp amplifies the difference in voltage between this two input pins and provides the amplified output across its Vout or output pin.

Depending on the input type, op-amp can be classified as Inverting Amplifier or Non-inverting Amplifier. In previous Non-inverting op-amp tutorial , we have seen how to use the amplifier in a non-inverting configuration. In this tutorial, we will learn how to use op-amp in inverting configuration. It is called Inverting Amplifier because the op-amp changes the phase angle of the output signal exactly degrees out of phase with respect to input signal. Same as like before, we use two external resistors to create feedback circuit and make a closed loop circuit across the amplifier.

In the Non-inverting configuration , we provided positive feedback across the amplifier, but for inverting configuration, we produce negative feedback across the op-amp circuit. In the above inverting op-amp, we can see R1 and R2 are providing the necessary feedback across the op-amp circuit.

The R2 Resistor is the signal input resistor, and the R1 resistor is the feedback resistor. This feedback circuit forces the differential input voltage to almost zero. The voltage potential across inverting input is the same as the voltage potential of non-inverting input. So, across the non-inverting input, a Virtual Earth summing point is created, which is in the same potential as the ground or Earth.

The op-amp will act as a differential amplifier. So, In case of inverting op-amp, there are no current flows into the input terminal, also the input Voltage is equal to the feedback voltage across two resistors as they both share one common virtual ground source. Due to the virtual ground, the input resistance of the op-amp is equal to the input resistor of the op-amp which is R2.

This R2 has a relationship with closed loop gain and the gain can be set by the ratio of the external resistors used as feedback. As there are no current flow in the input terminal and the differential input voltage is zero, We can calculate the closed loop gain of op amp. Learn more about Op-amp consturction and its working by following the link. In the above image, two resistors R2 and R1 are shown, which are the voltage divider feedback resistors used along with inverting op-amp.

R1 is the Feedback resistor Rf and R2 is the input resistor Rin. If we calculate the current flowing through the resistor then-. So, the inverting amplifier formula for closed loop gain will be. So, from this formula, we get any of the four variables when the other three variables are available. Op-amp Gain calculator can be used to calculate the gain of an inverting op-amp. In the above image, an op-amp configuration is shown, where two feedback resistors are providing necessary feedback in the op-amp.

The resistor R2 which is the input resistor and R1 is the feedback resistor. The input resistor R2 which has a resistance value 1K ohms and the feedback resistor R1 has a resistance value of 10k ohms. We will calculate the inverting gain of the op-amp.

The feedback is provided in the negative terminal and the positive terminal is connected with ground. So the gain will be times and the output will be degrees out of phase. Now, if we increase the gain of the op-amp to times, what will be the feedback resistor value if the input resistor will be the same? So, if we increase the 10k value to 20k, the gain of the op-amp will be times. As the lower value of the resistance lowers the input impedance and create a load to the input signal.

In typical cases value from 4.

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When that input voltage difference is exactly zero volts, we would ideally expect to have exactly zero volts present on the output. However, in the real world this rarely happens. Even if the op-amp in question has zero common-mode gain infinite CMRR , the output voltage may not be at zero when both inputs are shorted together. This deviation from zero is called offset. A perfect op-amp would output exactly zero volts with both its inputs shorted together and grounded.

However, most op-amps off the shelf will drive their outputs to a saturated level, either negative or positive. In the example shown above, the output voltage is saturated at a value of positive For this reason, offset voltage is usually expressed in terms of the equivalent amount of input voltage differential producing this effect.

In other words, we imagine that the op-amp is perfect no offset whatsoever , and a small voltage is being applied in series with one of the inputs to force the output voltage one way or the other away from zero. Offset voltage will tend to introduce slight errors in any op-amp circuit. So how do we compensate for it? Unlike common-mode gain, there are usually provisions made by the manufacturer to trim the offset of a packaged op-amp.

These connection points are labeled offset null and are used in this general way:. On single op-amps such as the and , the offset null connection points are pins 1 and 5 on the 8-pin DIP package. Inputs on an op-amp have extremely high input impedances. We analyze the circuit as though there was absolutely zero current entering or exiting the input connections.

This idyllic picture, however, is not entirely true. Op-amps, especially those op-amps with bipolar transistor inputs, have to have some amount of current through their input connections in order for their internal circuits to be properly biased. These currents, logically, are called bias currents. Under certain conditions, op-amp bias currents may be problematic. The following circuit illustrates one of those problem conditions:. At first glance, we see no apparent problems with this circuit.

In other words, this is a kind of comparator circuit , comparing the temperature between the end thermocouple junction and the reference junction near the op-amp. The problem is this: the wire loop formed by the thermocouple does not provide a path for both input bias currents, because both bias currents are trying to go the same way either into the op-amp or out of it. In order for this circuit to work properly, we must ground one of the input wires, thus providing a path to or from ground for both currents:.

Another way input bias currents may cause trouble is by dropping unwanted voltages across circuit resistances. Take this circuit for example:. We expect a voltage follower circuit such as the one above to reproduce the input voltage precisely at the output. But what about the resistance in series with the input voltage source? But even then, what slight bias currents may remain can cause measurement errors to occur, so we have to find some way to mitigate them through good design.

One way to do so is based on the assumption that the two input bias currents will be the same. In reality, they are often close to being the same, the difference between them referred to as the input offset current. If they are the same, then we should be able to cancel out the effects of input resistance voltage drop by inserting an equal amount of resistance in series with the other input, like this:.

With the additional resistance added to the circuit, the output voltage will be closer to V in than before, even if there is some offset between the two input currents. In either case, the compensating resistor value is determined by calculating the parallel resistance value of R 1 and R 2.

Why is the value equal to the parallel equivalent of R 1 and R 2? This gives two parallel paths for bias current through R 1 and through R 2 , both to ground. A related problem, occasionally experienced by students just learning to build operational amplifier circuits, is caused by a lack of a common ground connection to the power supply. This provides a complete path for the bias currents, feedback current s , and for the load output current. Take this circuit illustration, for instance, showing a properly grounded power supply:.

The effect of doing this is profound:. Thus, no electrons flow through the ground connection to the left of R 1 , neither through the feedback loop. This effectively renders the op-amp useless: it can neither sustain current through the feedback loop, nor through a grounded load, since there is no connection from any point of the power supply to ground.

The bias currents are also stopped, because they rely on a path to the power supply and back to the input source through ground. The following diagram shows the bias currents only , as they go through the input terminals of the op-amp, through the base terminals of the input transistors, and eventually through the power supply terminal s and back to ground.

Without a ground reference on the power supply, the bias currents will have no complete path for a circuit, and they will halt. Since bipolar junction transistors are current-controlled devices, this renders the input stage of the op-amp useless as well, as both input transistors will be forced into cutoff by the complete lack of base current. Bias currents are small in the microamp range , but large enough to cause problems in some applications.

It is not enough to just have a conductive path from one input to the other. To cancel any offset voltages caused by bias current flowing through resistances, just add an equivalent resistance in series with the other op-amp input called a compensating resistor. This corrective measure is based on the assumption that the two input bias currents will be equal.

Any inequality between bias currents in an op-amp constitutes what is called an input offset current. It is essential for proper op-amp operation that there be a ground reference on some terminal of the power supply, to form complete paths for bias currents, feedback current s , and load current.

Being semiconductor devices, op-amps are subject to slight changes in behavior with changes in operating temperature. Any changes in op-amp performance with temperature fall under the category of op-amp drift. Drift parameters can be specified for bias currents, offset voltage, and the like. The latter action may involve providing some form of temperature control for the inside of the equipment housing the op-amp s.

This is not as strange as it may first seem. If extremely high accuracy is desired over the usual factors of cost and flexibility, this may be an option worth looking at. Op-amps, being semiconductor devices, are susceptible to variations in temperature.

Any variations in amplifier performance resulting from changes in temperature is known as drift. Drift is best minimized with environmental temperature control. With their incredibly high differential voltage gains, op-amps are prime candidates for a phenomenon known as feedback oscillation.

An op-amp circuit can manifest this same effect, with the feedback happening electrically rather than audibly. A case example of this is seen in the op-amp, if it is connected as a voltage follower with the bare minimum of wiring connections the two inputs, output, and the power supply connections.

The output of this op-amp will self-oscillate due to its high gain, no matter what the input voltage. To combat this, a small compensation capacitor must be connected to two specially-provided terminals on the op-amp.

If the op-amp is being used to amplify high-frequency signals, this compensation capacitor may not be needed, but it is absolutely essential for DC or low-frequency AC signal operation. Some op-amps, such as the model , have a compensation capacitor built in to minimize the need for external components. Op-amp manufacturers will publish the frequency response curves for their products.

The circuit designer must take this into account if good performance is to be maintained over the required range of signal frequencies. Those two differential input pins are inverting pin or Negative and Non-inverting pin or Positive. An op-amp amplifies the difference in voltage between this two input pins and provides the amplified output across its Vout or output pin.

Depending on the input type, op-amp can be classified as Inverting Amplifier or Non-inverting Amplifier. In previous Non-inverting op-amp tutorial , we have seen how to use the amplifier in a non-inverting configuration. In this tutorial, we will learn how to use op-amp in inverting configuration. It is called Inverting Amplifier because the op-amp changes the phase angle of the output signal exactly degrees out of phase with respect to input signal.

Same as like before, we use two external resistors to create feedback circuit and make a closed loop circuit across the amplifier. In the Non-inverting configuration , we provided positive feedback across the amplifier, but for inverting configuration, we produce negative feedback across the op-amp circuit. In the above inverting op-amp, we can see R1 and R2 are providing the necessary feedback across the op-amp circuit.

The R2 Resistor is the signal input resistor, and the R1 resistor is the feedback resistor. This feedback circuit forces the differential input voltage to almost zero. The voltage potential across inverting input is the same as the voltage potential of non-inverting input. So, across the non-inverting input, a Virtual Earth summing point is created, which is in the same potential as the ground or Earth. The op-amp will act as a differential amplifier. So, In case of inverting op-amp, there are no current flows into the input terminal, also the input Voltage is equal to the feedback voltage across two resistors as they both share one common virtual ground source.

Due to the virtual ground, the input resistance of the op-amp is equal to the input resistor of the op-amp which is R2. This R2 has a relationship with closed loop gain and the gain can be set by the ratio of the external resistors used as feedback. As there are no current flow in the input terminal and the differential input voltage is zero, We can calculate the closed loop gain of op amp. Learn more about Op-amp consturction and its working by following the link.

In the above image, two resistors R2 and R1 are shown, which are the voltage divider feedback resistors used along with inverting op-amp. R1 is the Feedback resistor Rf and R2 is the input resistor Rin. If we calculate the current flowing through the resistor then-. So, the inverting amplifier formula for closed loop gain will be. So, from this formula, we get any of the four variables when the other three variables are available.

Op-amp Gain calculator can be used to calculate the gain of an inverting op-amp. In the above image, an op-amp configuration is shown, where two feedback resistors are providing necessary feedback in the op-amp. The resistor R2 which is the input resistor and R1 is the feedback resistor. The input resistor R2 which has a resistance value 1K ohms and the feedback resistor R1 has a resistance value of 10k ohms. We will calculate the inverting gain of the op-amp. The feedback is provided in the negative terminal and the positive terminal is connected with ground.

So the gain will be times and the output will be degrees out of phase. Now, if we increase the gain of the op-amp to times, what will be the feedback resistor value if the input resistor will be the same? So, if we increase the 10k value to 20k, the gain of the op-amp will be times. As the lower value of the resistance lowers the input impedance and create a load to the input signal.

In typical cases value from 4. When high gain requires and we should ensure high impedance in the input, we must increase the value of feedback resistors. But it is also not advisable to use very high-value resistor across Rf.

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Inverting and unity-gain op-amp with virtual ground

The equation for the output voltage Vout also shows that the circuit is linear in nature for a fixed amplifier gain as Vout = Vin x Gain. This property can be. Inverting Op-amp is called Inverting because the op-amp changes the phase angle of The formula for inverting gain of the op-amp circuit-. The logarithm of the amplification factor (multiplied by 20) is expressed in units of decibels (dB). For example, for an opamp with an open gain of ,x.