# Electronics/Op-Amps/Linear Configurations

## Inverting Op AmpEdit

The closed loop gain "mon then" of an Inverting Op Amp is

- (i)

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

To get formula (i) we take a KVL loop with , and the inputs of the Op Amp. This gives

Where is the voltage between the non-inverting and inverting inputs. But for ideal Op Amps is approximately zero. 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:

- (5)

Using this idea.

- (6)

If we take KCL at the inverting input then

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

Since

**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.

Step 2: Select an value.

- Choose .

Step 3: Work out the required value of using equation (i).

**Example 2**

Design an Inverting Amplifier to amplify a 10mV signal to a 1V signal. The signal has a 100 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.

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 .

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

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

Step 4: Use to work out .

- Using equation (i)

Step 5: Choose a for the second stage.

- Choose 100k.

Step 6: Calculate using equation (i).

## Non-Inverting Op AmpEdit

The closed loop gain of a Non Inverting Op Amp is

- (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 DerivationEdit

Take a KVL with the inputs of the Op Amp and R_{1}.

But is zero since the Op Amp is ideal. Therefore

- (3)

According to voltage divider rule

- (4)

Substitute equation 4 into 3.

Thus

#### Feedback Analysis DerivationEdit

If the output is connected to the inverting input, after being scaled by a voltage divider K = R_{1} / (R_{1} + R_{2}), 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.

Step 2: Select an value.

- Choose .

Step 3: Work out the required value of using equation (ii).

**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.

Step 2: Choose the gain for each stage.

- Choose 10 for both stages.

Step 3: Choose a value for the resistors in both stages.

- Choose 90kΩ for both.

Step 4: Work out the value of .

- Using equation (ii) .

## Voltage FollowerEdit

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.

## Difference amplifierEdit

This configuration is just an Inverting and a Non-Inverting configuration connected simultaneously. Resistors R_{2} and R_{4} are a voltage divider. Consider the situation when R_{4} is open circuited and R_{2} is short circuited. Now from equations (i) and (ii) we know that the gain of V_{1} is

and the gain of V_{2} is

Now if we set to -10 then will be 11. This means that V_{out} will be

This means that if that V_{out} will be V_{2}. 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 R_{4} and R_{2} are connected the gain of V_{2} is

- (x)

The gain of V_{1} is

- (y)

If we want

We just set and .

This configuration has a low input impedance. The input impedance seen by V_{1} is R_{1} as in the Inverting Amplifier. The input impedance seen by V_{2} is R_{2} + R_{4}.

**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 amplifierEdit

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.

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.

If then

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 integratorEdit

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

Integrate both sides with respect to

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 differentiatorEdit

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

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.

This means high frequencies mean high gain and thus saturation.

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