Electronics/Op-Amps

Amplifiers

Op-Amp (operational amplifier)

Remember when we talked about "ideal" capacitors and inductors, but then discussed all the imperfections of real capacitors and inductors? We're going to do the same thing here, except 3 times instead of twice.

Designing op-amp circuits
Ideal op-amps
technical details about non-ideal op-amps

The idea of an "ideal" op-amp makes it easier to design circuits. Later we will discuss the various ways that real op-amps fail to work as we expected.

An operational amplifier or op-amp is an electronic circuit module, which has a non-inverting input (+), an inverting input (-) and one output.

Originally, op-amps were so named because they were used to model the basic mathematical operations addition, subtraction, integration, differentiation etc in electronic analog computers. In this sense a true operational amplifier is an ideal circuit element.

An example

See Wikipedia:Operational amplifier and look for the "Internal circuitry of 741 type op-amp" section. See also the data in the Linear Databook published in 1982 by the National Semiconductor Corporation.

This particular op-amp requires either two supply-voltage sources of never more than (18 volts(V) for 741C), (22V for 741, 741A and 741E) each, or a center/centre tapped equivalent voltage source with each half supplying never more than the same voltage.
The maximum permitted power dissipation is 500 milliWatt(mW).
The maximum input voltage must never be more than 15V, with the maximum permitted differential voltage being 30V.
The maximum storage temperature range permitted for all these op-amps is minus 65 degrees Celsius to plus 150 degrees Celsius.
The actual operating ambient temperature for 741C and 741E is zero to 70 degrees Celsius, and it is minus 55 degrees Celsius to plus 125 degrees Celsius for 741 and 741A.
For the many additional data see the Linear Databook.

Another example

The Linear Databook, among data for also many other types, lists information, for example, about nearly equal types 124, 124A, 224, 224A, 324, and 324A op-amps from page 3-172 to 3-180. Their special advantage is that only one power supply is needed, and that there are 4 op-amps in a single dual-in-line package.

Electrical characteristics are listed from pages 3-173 to 3-174, mainly for a more usual single supply voltage of plus 5 V.

Notation

A typical circuit symbol for an op-amp looks like this:

Its terminals are:

V₊: non-inverting input
V₋: inverting input
V_out: output
V_S+: positive power supply
V_S−: negative power supply

Often these pins are left out of the diagram for clarity, and the power configuration is described or assumed from the circuit.

The input pin polarity is often reversed in diagrams for clarity. In this case, the power supply pins remain in the same position; the more positive power pin is always on the top, and the more negative on the bottom. The entire symbol is not flipped; just the inputs.

Quick Design Process

Imagine we have a (weak) input signal, and we want to amplify it to generate a strong output signal; or we have several different voltages, and we want to add them together. For these purposes, we need op-amp amplifying and summing circuits. It seems that there is nothing new to say about these bare (usually single op-amp) circuits. Only, the innovator Dieter Knollman has suggested a new simpler design procedure in an EDN's article ^[1]. He has also placed his work on the web^[2] (see also a circuit story^[3] written after Dieter's work).

We assume:

All inputs are ideal voltage sources.
Gain is defined as the gain from the ideal voltage source to the op-amp output.
The feedback resistor, Rf, connects from the op-amp output to the inverting input.
The input resistor, Ri, connects from the ideal voltage source input to the op-amp input. (This *includes* any source resistance).
- positive-gain inputs connect (through that resistor) to V₊
- negative-gain inputs connect (through that resistor) to V₋

Design procedure:

First, specify the desired circuit gains for each input.
Then calculate the ground gain, as follows:

Daisy's theorem:

The sum of the gains = +1 in a properly-designed op amp circuit.

Therefore, choose

Ground gain = 1 - ( the sum of the desired positive and negative gains).

Choose a feedback-resistor value. For example, let Rf=100 kiloOhms.

Next, calculate the resistor values for each input, including the ground-gain resistor, using

R i = \frac{R f}{| g a i n |}

.

where |gain| is the absolute value of the desired gain.

Some op-amp circuits need a resistor to ground from the op amp's inverting input. Others need a resistor to ground on the noninverting input. The sign of the ground gain determines where to place the ground resistor.

If the desired gains add up to one, a resistor to ground is unnecessary.

You will see below some examples that illustrate using the quick design procedure. First of all, answer the question, "What's the highest and lowest output voltage you want?" Make sure the power terminals of the op-amp are at least that high and low. (If you want it to swing from +5V to -5V, then an op-amp connected to 0V and +10V won't work).

Ideal Op Amps

The output voltage is the difference between the + and - inputs multiplied by the open-loop gain: V_out = (V₊ − V₋) * A_vo.

The ideal op-amp has an infinite open-loop gain, infinite input impedances and zero output impedance. The input impendance is the impedance seen between the non-inverting and inverting inputs.

The model for the Op Amp is shown in Figure 1. Where V₊ − V₋ is equal to v_d; R_in is the input impedance; R_out is the output impendance; A_vo is the open loop gain; and R_s is the source impedance.

Figure 1: The model of the Op Amp.

Using voltage divider rule, we can determine the voltage v_d.

v_{d} = \frac{V_{i n} R_{i n}}{R_{i n} + R_{s}}

(1)

Since the dependent voltage source amplifies the voltage v_d by A_vo. We can work out the output voltage across R_L once again using voltage divider rule.

V_{o u t} = \frac{A_{v o} v_{d} R_{L}}{R_{L} + R_{o u t}}

(2)

Substituting equation 1 into 2.

V_{o u t} = \frac{A_{v o} V_{i n} R_{i n} R_{L}}{(R_{L} + R_{o u t}) (R_{i n} + R_{s})}

Now if the properties of an Ideal Op Amp are applied. The ideal properties of an Op Amp are infinite input impedance and zero output impendance. Since R_in >>, much greater than, R_s, R_in/(R_in+R_s) $\approx$ 1.

V_{o u t} = A_{v o} V_{i n}

Which is basically the definition of an Op Amp. But if the input impendance was not infinite and the output impendance nonzero then this would not be so.

Basic Op-Amp Configurations

All the more complicated Op Amp configurations are based on two basic ones; the Inverting and Non Inverting configurations. An understanding of these two configurations, makes it much simpler to understand the more advanced configurations.

Inverting Op Amp

The closed loop gain of an Inverting Op Amp is

R_{1} = R_{i n}

A_{f} = - \frac{R_{f}}{R_{1}}

(i)

Input Impedance of this configuration is Zin = Rin (because V− is a virtual ground, no current flows into the Op Amp ideally.)

To get formula (i) we take a KVL loop with $V_{i n}$ , $R_{1}$ and the inputs of the Op Amp. This gives

V_{i n} = i_{i n} R_{1} + v_{d}

Where $v_{d}$ is $v_{+} - v_{-}$ the voltage between the non-inverting and inverting inputs. But for ideal Op Amps $v_{d}$ is approximately zero. $v_{d}$ is zero because the input impedance is infinite, which means the current through the impedance must be zero by Ohms law. The zero current means that there is no voltage drop across the impedance. This gives:

i_{i n} = \frac{V_{i n}}{R_{1}}

(5)

Using this idea.

i_{f} = \frac{V_{o u t}}{R_{f}}

(6)

If we take KCL at the inverting input then

i_{i n} = i_{d} - i_{f}

For an ideal Op Amp there is no input current because there is infinite resistance. So using equations 5 and 6.

- \frac{V_{i n}}{R_{1}} = \frac{V_{o u t}}{R_{f}}

Since

A_{f} = \frac{V_{o u t}}{V_{i n}} = - \frac{R_{f}}{R_{1}}

Example 1

Design an Inverting Amplifier to amplify a 100mV signal to 1V signal. Assume: No source impedance.

Solution:

Step 1: Work out required gain.

| A_{f} | = \frac{1}{0.1} = 10

Step 2: Select an $R_{f}$ value.

Choose

R_{f} = 100 k Ω

.

Step 3: Work out the required value of $R_{1}$ using equation (i).

R_{1} = \frac{100 k}{10} = 10 k Ω

Example 2

Design an Inverting Amplifier to amplify a 10mV signal to a 1V signal. The signal has a 100 $Ω$ source impedance. The amplifier must not invert the signal.

Solution:

Since the voltage cannot be inverted then there must be an even number of stages. For simplicity let us choose two stages. Assume an ideal Op Amp.

Step 1: Work out the required gain.

A_{f_{t o t}} = \frac{1}{0.01} = 100

Step 2: Choose the required gain for each stage.

The gain will be 10 for both stages. But in the first stage we must worry about loading.

Step 3: Choose a value of input impedance i.e choose $R_{1}$ .

Choose

10 k Ω

. We can now calculate the voltage that input to the Op Amp by voltage divider.

V_{i n} = \frac{R_{1} V_{s}}{R_{1} + R_{s}} = \frac{(10 k) (10 m V)}{10 k + 100} = 9.9 m V

We want the output of this stage to be 100mV.

A_{f 1} = \frac{100}{9.9} = 10.1

Step 4: Use $A_{f 1}$ to work out $R_{f 1}$ .

Using equation (i)

R_{f} = A_{f 1} R_{1} = 101 k Ω

Step 5: Choose a $R_{f}$ for the second stage.

Choose 100k.

Step 6: Calculate $R_{1}$ using equation (i).

R_{1} = 10 k Ω

Non-Inverting Op Amp

The closed loop gain of a Non Inverting Op Amp is

A_{f} = 1 + \frac{R_{2}}{R_{1}}

(ii)

The input impedance of this configuration is Zin = ∞ (realistically, the input impedance of the op-amp itself, 1 MΩ to 10T Ω).

Ideal Op Amp Derivation

Take a KVL with the inputs of the Op Amp and R₁.

V_{i n} = v_{d} + V_{R 1}

But $v_{d}$ is zero since the Op Amp is ideal. Therefore

V_{i n} = V_{R 1}

(3)

According to voltage divider rule

V_{R 1} = \frac{V_{o u t} R_{1}}{R_{1} + R_{2}}

(4)

Substitute equation 4 into 3.

V_{i n} = \frac{V_{o u t} R_{1}}{R_{1} + R_{2}}

Thus

A_{f} = \frac{V_{o u t}}{V_{i n}} = 1 + \frac{R_{2}}{R_{1}}

Feedback Analysis Derivation

If the output is connected to the inverting input, after being scaled by a voltage divider K = R₁ / (R₁ + R₂), then:

V₊ = V_in

V₋ = K V_out

V_out = G(V_in − K V_out)

Solving for V_out / V_in, we see that the result is a linear amplifier with gain:

V_out / V_in = G / (1 + G K)

If G is very large, V_out / V_in comes close to 1 / K, which equals 1 + (R2 / R1).

This negative feedback connection is the most typical use of an op-amp, but many different configurations are possible, making it one of the most versatile of all electronic building blocks.

When connected in a negative feedback configuration, the op-amp will tend to output whatever voltage is necessary to make the input voltages equal. This, and the high input impedance, are sometimes called the two "golden rules" of op-amp design (for circuits that use feedback):

No current will flow into the inputs
The input voltages will be equal to each other

The exception is if the voltage required is greater than the op-amp's supply, in which case the output signal stops near the power supply rails, V_S+ or V_S−.

Example 3

Design a Non-Inverting Amplifier to amplify a 100mV signal to 1V signal. Assume: No source impedance.

Step 1: Work out required gain.

A_{f} = \frac{1}{0.1} = 10

Step 2: Select an $R_{2}$ value.

Choose

R_{2} = 90 k Ω

.

Step 3: Work out the required value of $R_{1}$ using equation (ii).

10 = 1 + \frac{90 k}{R_{1}}

R_{1} = \frac{90 k}{9} = 10 k Ω

Example 4 (a quick design procedure):

I want to amplify a signal A with a gain of 8. We want an output swing of at least -3 to +3 V.

We have a 5V and -5V power supply handy, so we can use that.

1. gain of 8 for A.

2. Ground gain = 1 - (8) = -7.

3. feedback-resistor value: Rf = 100 kiloOhms.

4. resistor values for each input:

R_A = 100 kΩ / 8 = 12.5 kΩ
R_ground = 100 kΩ / | -7 | = 14.3 kΩ

Since A has a positive gain, connect its resistor to V₊. Since ground has a negative gain, connect its resistor to V₋

Example 5

Design a two stage Non-Inverting Amplifier to amplify a 10mV signal to 1V signal. The signal has a source impedance of 100 Ω.

Solution Assuming an ideal Op Amp. Since this configuration has the input impendance of the Op Amp itself. We do not have to worry about loading since the input impedance is infinite.

Step 1: Work out the required gain.

A_{f} = \frac{1}{0.01} = 100

Step 2: Choose the gain for each stage.

Choose 10 for both stages.

Step 3: Choose a value for the $R_{2}$ resistors in both stages.

Choose 90kΩ for both.

Step 4: Work out the value of $R_{1}$ .

Using equation (ii)

R_{1} = 10 k Ω

.

Advanced Op-Amp Configurations

The following configurations are all combinations of the Inverting and Non-Inverting Configurations or are special cases of them.

Voltage Follower

This configuration is also known as the unity gain Buffer. Since it can be used to counter the effects of loading of the source.

This configuration provides an input impedance even higher than a normal Non-Inverting since the gain reduces that input impedance. The gain is given by equation (ii). But R_2 is short circuited and R_1 is an open circuit.

A_{f} = 1 + \frac{R_{2}}{R_{1}} = 1 + \frac{0}{\infty} = 1

Difference amplifier

This configuration is just an Inverting and a Non-Inverting configuration connected simultaneously. Resistors R₂ and R₄ are a voltage divider. Consider the situation when R₄ is open circuited and R₂ is short circuited. Now from equations (i) and (ii) we know that the gain of V₁ is

A_{f 1} = - \frac{R_{3}}{R_{1}}

and the gain of V₂ is

A_{f 2} = 1 + \frac{R_{3}}{R_{1}}

Now if we set $A_{f 1}$ to -10 then $A_{f 2}$ will be 11. This means that V_out will be

V_{o u t} = V_{1} A_{f 1} + V_{2} A_{f 2} = V_{2} (11) - (10) V_{1}

This means that if $V_{1} = V_{2}$ that V_out will be V₂. This is not very useful for the most because mathematically we would like the answer to be zero. But if we make the voltage at the non-inverting input equal to 10/11 then when the voltages are equal we will have zero.

When R₄ and R₂ are connected the gain of V₂ is

A_{f 2} = \frac{(1 + \frac{R_{3}}{R_{1}}) R_{4}}{R_{4} + R_{2}}

(x)

The gain of V₁ is

A_{f 1} = - \frac{R_{3}}{R_{1}}

(y)

If we want $A_{f 2} = | A_{f 1} |$

\frac{R_{3}}{R_{1}} = \frac{(R_{1} + R_{3}) R_{4}}{(R_{4} + R_{2}) R_{1}}

We just set $R_{4} = R_{3}$ and $R_{2} = R_{1}$ .

This configuration has a low input impedance. The input impedance seen by V₁ is R₁ as in the Inverting Amplifier. The input impedance seen by V₂ is R₂ + R₄.

Example 6 (a quick design procedure):

We often want to subtract one signal A from another signal B, and amplify the difference by 10. We want an output swing at least -6V to +6V. Oh, and for safety reasons, A and B each have a source impedance of 8 kΩ.

Our 5V and -5V power supply isn't adequate, so we pick a +12V and -12V power supply.

1.

gain of +10 for B.
gain of -10 for A.

2. Ground gain = 1 - ( +10 + -10 ) = +1.

3. feedback-resistor value: Rf = 100 kiloOhms.

4. resistor values for each input:

R_A = 100 kΩ / | -10 | = 10 kΩ
R_B = 100 kΩ / | +10 | = 10 kΩ
R_ground = 100 kΩ / | +1 | = 100 kΩ

So we connect the 100 kΩ Rf from V_out to V₋, another 100 kΩ from ground to V₊. Then we connect a 2 kΩ from R_A to V₋, and another 2 kΩ from R_B to V₊.

If the input has a source impedance, the source impedance is part of the circuit. The 8 kΩ source impedance, plus the 2 kΩ physical resistors we added, give us a total of 10 kΩ between the ideal voltage source and the op amp input.

Summing amplifier

This is merely an Inverting Amplifier with extra inputs. The analysis is nearly identical but we have many currents equal to the feedback current. If we take a KCL at the Inverting input.

i_{n} + \dots + i_{2} + i_{1} = - i_{f}

The value of the currents can be determined by Ohm's Law using the fact that v_d is zero for an ideal Op Amp.

\frac{V_{n}}{R_{n}} + \dots + \frac{V_{2}}{R_{2}} + \frac{V_{1}}{R_{1}} = - \frac{V_{o u t}}{R_{f}}

If $R_{n} = \dots = R_{2} = R_{1}$ then

\frac{1}{R_{1}} (V_{n} + \dots + V_{2} + V_{1}) = - \frac{V_{o u t}}{R_{f}}

A_{f} = - \frac{V_{o u t}}{V_{s u m}} = - \frac{R_{f}}{R_{1}}

Just as it is for the Inverting Amplifier.

Note: There is also a Summing Amplifier made using the Non-Inverting Amplifier configuration. The configuration is a bit more complicated and harder to use. Since it requires and understanding of Superposition.

Integrator

The configuration is an Inverting Amplifier with the feedback resistor a Capacitor. The derivation procedes the same.

i = - i_{f}

\frac{V_{i n}}{R} = - C \frac{d V_{o u t}}{d t}

\frac{V_{i n}}{R C} = - \frac{d V_{o u t}}{d t}

Integrate both sides with respect to

V_{o u t} = \int_{t_{0}}^{0} - \frac{V_{i n}}{R C} d t + V_{i n i t i a l}

Practically a Resistor is often connected in parallel with the feedback capacitor. This means that there is not infinite gain at very low frequencies, which makes the Integrator much more stable.

Differentiator

The configuration is an Inverting Amplifier with a Capacitor as Resistor one so the derivation proceeds the same as before.

i = - i_{f}

C \frac{d V_{i n}}{d t} = - \frac{V_{o u t}}{R}

V_{o u t} = - R C \frac{d V_{i n}}{d t}

This configuration is unstable for several reasons. The higher frequency inputs are going to have higher derivatives. Which means that circuit acts like a low pass filter, but more importantly this means that it will just saturate if a high frequency signal is put into the differentiator. This is also seen throught the gain.

A_{f} = - \frac{R}{1 / j ω C}

This means high frequencies mean high gain and thus saturation.

Practically a Resistor is often connected in series with the capacitor.

Real Op Amps

Real Op Amps are (normally) built as an integrated circuit, but occasionally with discrete transistors or vacuum tubes. A Real Op Amp is an approximation of an Ideal Op Amp. This means that they do not have infinite open loop gain, infinite input impedance nor zero output impedance. Real Op Amps also create noise in the circuit, have an offset voltage, thermal drift and finite bandwidth.

An offset voltage means that there exists a voltage v_d when both inputs are grounded. This offset is called an input offset because the voltage v_d is offset from its ideal value of zero volts. The input offset voltage is multiplied by the open loop gain to create an output offset voltage.

Thermal drift means that the characteristics of the Op Amp change with temperature. That is the open loop gain, input and output impedances, offset voltages and bandwidth change as the temperature changes.

Op Amps are made up of transistors. Transistors can only respond at a certain rate because of certain capacitances that they have. This means that the Op Amps cannot respond fast enough to frequencies above a certain level. This level is the bandwidth.

Modern integrated circuit MOSFET op-amps approximate closer and closer to these ideals in limited-bandwidth, large-signal applications at room temperature. When the approximation is reasonably close, we go ahead and call the practical device an 'op-amp', forget its limitations and use the thinking and formulae given in this article.

DC Behaviour

Open-loop gain is defined as the amplification from input to output without any feedback applied. For most practical calculations, the open-loop gain is assumed to be infinite; in reality, however, it is limited by the amount of voltage applied to power the operational amplifier, i.e. Vs+ and Vs- in the above diagram. Typical devices exhibit open loop DC gain ranging from 100,000 to over 1 million. This allows the gain in the application to be set simply and exactly by using negative feedback. Of course theory and practice differ, since op-amps have limits that the designer must keep in mind and sometimes work around.

AC Behaviour

The op-amp gain calculated at DC does not apply at higher frequencies. This effect is due to limitations within the op-amp itself, such as its finite bandwidth, and to the AC characteristics of the circuit in which it is placed. The best known stumbling-block in designing with op-amps is the tendency for the device to resonate at high frequencies, where negative feedback changes to positive feedback due to parasitic lowpasses.

Typical low cost, general purpose op-amps exhibit a gain bandwidth product of a few MHz. Specialty and high speed op-amps can achieve gain bandwidth products of 100s of MHz.

Applications

audio and video pre-amplifiers and buffers
voltage comparators
differential amplifiers
differentiators and integrators
Filters
Precision rectifiers
voltage and current regulators
analogue calculators
analogue-to-digital converters
digital-to-analogue converters

The generic op-amp has two inputs and one output. (Some are made with floating, differential outputs.) The output voltage is a multiple of the difference between the two inputs:

floating outputs? Are you perhaps misunderstanding "fully differential amplifier" ?

V_out = G(V₊ − V₋)

G is the open-loop gain of the op-amp. The inputs are assumed to have very high impedance; negligible current will flow into or out of the inputs. Op-amp outputs have very low source impedance.

If an Operational Amplifier is used with positive feedback it can act as an oscillator.

Other Notation

The power supply pins (V_S+ and V_S−) can be labeled many different ways. For FET based op-amps, the positive, common drain supply is labeled V_DD and the negative, common source supply is labeled V_SS. For BJT based op-amps, the V_S+ pin becomes V_CC and V_S− becomes V_EE. They are also sometimes labeled V_CC+ and V_CC−, or even V₊ and V₋, in which case the inputs would be labeled differently. The function remains the same.

Oscillators

When used with specific positive feedback, an Op Amp can act as an oscillator. Instances of this are:

Fully Differential Amplifier

Integrated Fully Differential Amplifier

... insert drawing of internal structure here ...

The generic op-amp has two inputs and one output.

Like an instrumentation amplifier, a fully differential amplifier (FDA) has 3 inputs.

Unlike any of the previous single-output amplifiers mentioned before, a fully differential amplifier has 2 outputs. This makes it ideal for driving w:differential signaling wire pairs and the differential inputs of some ADCs. (ADCs with 16 or more bits of precision almost always have differential inputs).

Basic Applications

... insert drawings of basic FDA amplifying circuits here ...

Fully Differential Amplifier versus Instrumentation Amplifier

An instrumentation amplifier has extremely high input impedance-- much higher than the input impedance of a fuly differential amplifier (once all the feedback resistors are in place). So an instrumentation amplifier is better for measuring voltage inputs with unknown (and possibly time-varying) output resistance.

A fully differential amplifier is better than an instrumentation amplifier for precisely generating differential output voltages, with good rejection of differential noise on the input, output, and power lines.

Applying a Quick Design Procedure

Can we apply Dieter's procedure to a fully differential amplifier? We can, if we revise it a little:

Daisy's theorem revised:

The sum of the gains = 0 in a properly-designed fully differential amplifier circuit.

Here is the FDA quick design procedure:

Use equal feedback resistors Rf, one from the + output to the - input, and one from the - output to the + input. Connect the "common" input to the appropriate voltage source.
The sum of the gains = zero in a properly-designed fully differential amplifier circuit (note that this new formulation differs from Daisy's theorem, which claims that the sum is always 1).
Calculate resistor values for each input, using Ri = Rf / |desired gain|.
For inputs with positive gain, connect the calculated resistor value between that input and the + input on the amplifier.
For inputs with negative gain, connect the calculated resistor value between that input and the - input on the amplifier.
If all your inputs are differential pairs, then the sum of all the gains is now zero. Done. Otherwise add a ground resistor to bring the total gain to zero.

Example 7 (a classic design procedure):

Example 8 (a quick design procedure):

References

↑ Single-formula technique keeps it simple - EDN's article by Dieter Knollman, PhD, Lucent Technologies
↑ K9 analysis make analog circuit design and analysis doggone simple
↑ How to Simplify the Design of the Mixed Op-amp Voltage Summer (after EDN's Single-formula technique keeps it simple)

Electronics/Op-Amps

Contents

Amplifiers

Op-Amp (operational amplifier)

An example

Another example

Notation

Quick Design Process

Ideal Op Amps

Basic Op-Amp Configurations

Inverting Op Amp

Non-Inverting Op Amp

Ideal Op Amp Derivation

Feedback Analysis Derivation

Advanced Op-Amp Configurations

Voltage Follower

Difference amplifier

Summing amplifier

Integrator

Differentiator

Real Op Amps

DC Behaviour

AC Behaviour

Applications

Other Notation

Oscillators

Fully Differential Amplifier

Integrated Fully Differential Amplifier

Basic Applications

Fully Differential Amplifier versus Instrumentation Amplifier

Applying a Quick Design Procedure

References

See also

Further reading

Navigation menu

Electronics/Op-Amps

Amplifiers

Op-Amp (operational amplifier)

An example

Another example

Notation

Quick Design Process

Ideal Op Amps

Basic Op-Amp Configurations

Inverting Op Amp

Non-Inverting Op Amp

Ideal Op Amp Derivation

Feedback Analysis Derivation

Advanced Op-Amp Configurations

Voltage Follower

Difference amplifier

Summing amplifier

Integrator

Differentiator

Real Op Amps

DC Behaviour

AC Behaviour

Applications

Other Notation

Oscillators

Fully Differential Amplifier

Integrated Fully Differential Amplifier

Basic Applications

Fully Differential Amplifier versus Instrumentation Amplifier

Applying a Quick Design Procedure

References

See also

Further reading

Navigation menu

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