Electronics/Op-Amps/Linear Configurations

Inverting Op Amp

The closed loop gain "mon then" 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 $Z_{i n} = R_{i n}$ (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

An op-amp connected in the non-inverting amplifier configuration

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

.

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. It provides an input impedance even higher than a normal Non-Inverting amplifier 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.

Inverting 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 an understanding of Superposition.

Ideal integrator

The configuration is an Inverting Amplifier with the feedback resistor a Capacitor. The derivation proceeds 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 Real integrator much more stable.

Ideal 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 through 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.

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Electronics/Op-Amps/Linear Configurations

Contents

Inverting Op Amp

Non-Inverting Op Amp

Ideal Op Amp Derivation

Feedback Analysis Derivation

Voltage Follower

Difference amplifier

Inverting summing amplifier

Ideal integrator

Ideal differentiator

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Electronics/Op-Amps/Linear Configurations

Inverting Op Amp

Non-Inverting Op Amp

Ideal Op Amp Derivation

Feedback Analysis Derivation

Voltage Follower

Difference amplifier

Inverting summing amplifier

Ideal integrator

Ideal differentiator

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