This circuit amplifies the potential difference between the two input terminals, with the gain being related to the values of the four resistors.

It can be shown that the output voltage V_{out} is related to the input voltages V_{1} and V_{2}, according to the expression:

It is not immediately obvious how to apply this expression, but the following section shows a useful simplification which makes the circuit both easier to design and much more useful.

## Simplified Difference Amplifier

Sensible choice of component values can greatly reduce the complexity of the above circuit calculations. If R_{1} is chosen to be equal to R_{3} and R_{2} equal to R_{4}, the circuit may be redrawn as:

The equation for V_{out} simplifies to:

The circuit now takes the difference between V_{2} and V_{1}, and amplifies this by the gain of the circuit which is given by R_{2} / R_{1}.

#### Example 1

- Design a simple difference amplifier to give a gain of 10.
- If V
_{2}= 2 V and V_{1}= 1.5 V, calculate the expected value of V_{out}.

- R
_{2}and R_{1}are both unknown so simply pick a ‘reasonable’ value for one resistor and then calculate the other. If we assume that R_{1}= 1 kΩ then R_{2}will need to equal 10 kΩ to give the required gain of 10. - V
_{out}= (V_{2}− V_{1}) × R_{2}/ R_{1}= (2 − 1.5) × 10,000 / 1,000 = 5 V

#### Example 2

Derive the expression for the output voltage of the above diferential amplifier circuit.

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Considering the inverting input, begin by assuming the input impedance of the op-amp is very high so no current flows into the inverting input. The current flowing through R_{1} will be equal to that flowing through R_{2}. Hence:

(V_{1} − V_{A}) / R_{1} = (V_{A} − V_{out}) / R_{2}

Regrouping and rearranging terms to make V_{out} the subject gives:

V_{out} = V_{A} (R_{1} + R_{2}) / R_{1} − V_{1} R_{2} / R_{1} · · · · · · · (1)

Equation 1 above contains terms with V_{out} and V_{1}, but not V_{2}. It also contains an unwanted intermediate term V_{A} which will need to be eliminated. The next step will be to analyse the non-inverting input branch of the circuit in order to obtain an expression which will be substituted into Equation 1.

Now, considering the non-inverting input, we once again assume the input impedance of the op-amp is very high, so the input current is zero. A second assumption is made that voltages V_{A} and V_{B} will be equal, provided the output voltage is not saturated. This is because the difference V_{B} − V_{A} is multiplied by the (extremely high) open loop gain of the op-amp to give V_{out}. Working backwards from any given output voltage, the difference between these input voltages is unlikely to be more than a few microvolts.

(V_{2} − V_{B}) / R_{3} = V_{B} / R_{4}

Regrouping and rearranging terms to make V_{B} the subject gives:

V_{B} = V_{2} R_{4} / (R_{3} + R_{4}) = V_{A} · · · · · · · (2)

Now substitute Equation 2 into Equation 1 in order to eliminate V_{A}:

V_{out} = ( V_{2} R_{4} / (R_{3} + R_{4}) ) × ( (R_{1} + R_{2}) / R_{1} ) − V_{1} R_{2} / R_{1}

= V_{2} (R_{1} + R_{2}) R_{4} / ( (R_{3} + R_{4}) R_{1} ) − V_{1} R_{2} / R_{1}

#### Example 3

Derive the simplified expression for the output voltage by assuming that R_{3} = R_{1} and R_{4} = R_{2}.

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Begin by replacing all occurences of R_{3} and R_{4} in the unsimplified expression:

V_{out} = V_{2} (R_{1} + R_{2}) R_{2} / ( (R_{1} + R_{2}) R_{1} ) − V_{1} R_{2} / R_{1}

Cancelling identical terms gives:

V_{out} = V_{2} R_{2} / R_{1} − V_{1} R_{2} / R_{1}

= (V_{2} − V_{1}) R_{2} / R_{1}