Operational Amplifier Applications
Most analog signal processing is done today with circuits using operational amplifiers because opamps are available with good data for little money. Most signals that must be processed in electronic circuits arise in analog form and are needed after proce
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Amplifier When connected as an inverting amplifier, an operational amplifier can be used for the addition of several voltages. As Fig. 11.1 indicates, the input voltages are connected via series resistors to the N-input of the operational amplifier. Since this node represents virtual ground, Kirchhoff’s current law (KCL) directly yields the relation for the output voltage: V1 V2 Vn Vo + + ··· + + = 0 R1 R2 Rn RN If a DC voltage is added to the signal voltage in the manner described, the inverting summing amplifier can also be used as an amplifier with a wide-range zero adjustment. Vn V2
Fig. 11.1. Inverting summing amplifier
V1
Output voltage: RN RN RN V1 + V2 + · · · + Vn −Vo = R1 R2 Rn
Vo
U. Tietze et al., Electronic Circuits © Springer 2008
726
11 Operational Amplifier Applications
11.2
Subtracting Circuits 11.2.1 Reduction to an Addition A subtraction operation can be reduced to the problem of an addition by inverting the signal to be subtracted. This requires the circuit shown in Fig. 11.2. The operational amplifier OA1 inverts the input voltage V2 ; the output voltage is then Vo = αP V2 − αN V1
(11.1)
If both resistance ratios are equal αP = αN = α then the output voltage is the amplified difference Vo = α(V2 − V1 ) = AD (V2 − V1 ). If the resistance ratios are not equal because of tolerances, we get an outpout voltage even if both input voltages are equal V1 = V2 = VCM
(11.2)
resulting in Vo = αP VCM − αN VCM = VCM (αP − αN ). With 1 αP = α + α 2
and
1 αN = α − α 2
we obtain Vo = VCM α = VCM ACM . From this we can calculate the common mode rejection G =
AD 1 α = . = α tolerance ACM
(11.3)
The common-mode rejection ratio thus equals the reciprocal of the relative matching of the individual gains. If resistors with a tolerance of 1% = 0.01 are used, a common-mode rejection of G = 1/0.01 = 100 can be expected. V1
RN/αN
V2
RN/αP V2 OA1
OA2
Fig. 11.2. Subtracting circuit using a summing amplifier
Output voltage: Condition for coefficients:
Vo = αD (V2 − V1 ) αN = αP = α
Vo
11.2 Subtracting Circuits
727
11.2.2 Subtraction Using a Single Operational Amplifier To calculate the output voltage of the subtracting amplifier in Fig. 11.3, we may use the principle of superposition. We therefore write Vo = k1 V1 + k2 V2 For V2 = 0, the circuit is an inverting amplifier, where Vo = −αN V1 . It follows k1 = −αN . For V1 = 0, the circuit represents a noninverting amplifier that has a voltage divider connected at its input. The potential VP =
RP αP V2 = V2 RP + RP /αP 1 + αP
is thus amplified by the factor (1 + αN ), and this results in the output voltage αP (1 + αN )V2 Vo = 1 + αP If both resistor ratios are the same – that is, if αN = αP = α – it follows: Vo = αV2 and that k2 = α. We now obtain the output voltage for the general case, in the form Vo = α(V2 − V1 ) Should the ratios of the resistors at the P- and N-inputs not be precisely equal to α, the circuit will not evaluate the precise difference between the input voltages. In this case, Vo =
1 + αN αP V 2 − α N V1 1 + αP
To calculate the common-mode
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