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 Vout is related to the input voltages V1 and V2, 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 R1 is chosen to be equal to R3 and R2 equal to R4, the circuit may be redrawn as:
The equation for Vout simplifies to:
The circuit now takes the difference between V2 and V1, and amplifies this by the gain of the circuit which is given by R2 / R1.
- Design a simple difference amplifier to give a gain of 10.
- If V2 = 2 V and V1 = 1.5 V, calculate the expected value of Vout.
- R2 and R1 are both unknown so simply pick a ‘reasonable’ value for one resistor and then calculate the other. If we assume that R1 = 1 kΩ then R2 will need to equal 10 kΩ to give the required gain of 10.
- Vout = (V2 − V1) × R2 / R1 = (2 − 1.5) × 10,000 / 1,000 = 5 V
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 R1 will be equal to that flowing through R2. Hence:
(V1 − VA) / R1 = (VA − Vout) / R2
Regrouping and rearranging terms to make Vout the subject gives:
Vout = VA (R1 + R2) / R1 − V1 R2 / R1 · · · · · · · (1)
Equation 1 above contains terms with Vout and V1, but not V2. It also contains an unwanted intermediate term VA 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 VA and VB will be equal, provided the output voltage is not saturated. This is because the difference VB − VA is multiplied by the (extremely high) open loop gain of the op-amp to give Vout. Working backwards from any given output voltage, the difference between these input voltages is unlikely to be more than a few microvolts.
(V2 − VB) / R3 = VB / R4
Regrouping and rearranging terms to make VB the subject gives:
VB = V2 R4 / (R3 + R4) = VA · · · · · · · (2)
Now substitute Equation 2 into Equation 1 in order to eliminate VA:
Vout = ( V2 R4 / (R3 + R4) ) × ( (R1 + R2) / R1 ) − V1 R2 / R1
= V2 (R1 + R2) R4 / ( (R3 + R4) R1 ) − V1 R2 / R1
Derive the simplified expression for the output voltage by assuming that R3 = R1 and R4 = R2.
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Begin by replacing all occurences of R3 and R4 in the unsimplified expression:
Vout = V2 (R1 + R2) R2 / ( (R1 + R2) R1 ) − V1 R2 / R1
Cancelling identical terms gives:
Vout = V2 R2 / R1 − V1 R2 / R1
= (V2 − V1) R2 / R1