Question 4 of 7: Op-Amp Low-Pass Filter Design and High-Impedance Practice

Question 4: Op-Amp Low-Pass Filter Design and High-Impedance Practice (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

a) Given. Required DC gain magnitude $A=5$; time constant $\tau=60$ s; candidate amplifiers LM741 ($I_b=8\times10^{-8}$ A, own $V_{os}=1$ mV) and LF13741 ($I_b=5\times10^{-11}$ A, own $V_{os}=5$ mV).

Find. Component values $R_i$, $R_f$, $C$; the resulting filter input impedance; and which amplifier the design should use.

Approach. Use the standard single-pole INVERTING active low-pass topology — capacitor $C$ in parallel with the feedback resistor $R_f$, input resistor $R_i$ — size $R_f$ and $C$ for a practical component pair, get $R_i$ from the gain requirement, then check the bias-current-induced DC output offset for each candidate op-amp to choose between them.

[Figure not reproduced: Fig. 2 — Single-pole active low-pass filter: $R_i$ sets the input, $R_f$ with $C$ in its feedback path sets both the DC gain and the pole. The summing (inverting) node is a virtual ground, so the circuit's input impedance as seen from the source is simply $R_i$. See the official exam paper.]

1. Write the filter's gain and time constant. For $C$ across $R_f$ in an inverting op-amp stage, $H(s)=-\dfrac{R_f}{R_i}\cdot\dfrac{1}{1+sR_fC}$, so the DC gain magnitude is $A=R_f/R_i$ and the time constant is $\tau=R_fC$.
2. Pick $R_f$, then solve $C$ from the time constant. Choose a large but still practical feedback resistor, $R_f=6\ \text{M}\Omega$, so that $C$ does not have to be an unwieldy value: $$C=\frac{\tau}{R_f}=\frac{60\ \text{s}}{6\times10^6\ \Omega}=10\ \mu\text{F}.$$
3. Solve $R_i$ from the required DC gain — this IS the filter's input impedance. $$R_i=\frac{R_f}{A}=\frac{6\times10^6\ \Omega}{5}=\boxed{1.2\ \text{M}\Omega.}$$ Because the op-amp's inverting input is a virtual ground, the impedance the source actually sees looking into the filter is exactly $R_i$.
4. Check the bias-current-induced DC output offset for each candidate, $V_{os,out}\approx I_b\,R_f$ (uncompensated inverting stage): $$V_{os,out}^{LM741}=(8\times10^{-8}\ \text{A})(6\times10^6\ \Omega)=0.48\ \text{V}=480\ \text{mV},$$ $$V_{os,out}^{LF13741}=(5\times10^{-11}\ \text{A})(6\times10^6\ \Omega)=3\times10^{-4}\ \text{V}=0.3\ \text{mV}.$$
5. Choose the amplifier. The LM741's bias-current offset (480 mV) swamps its own 1 mV input-offset-voltage spec by almost 500× and would push the output far off a low-level sensor signal's useful range; the LF13741's bias-current offset (0.3 mV) is smaller than even its own 5 mV offset-voltage spec, i.e. negligible. $$\boxed{\text{Use the LF13741 (JFET input) for this design.}}$$

b) Low-pass filters recur throughout measuring-instrument design: (1) as an ANTI-ALIASING filter ahead of an ADC, removing signal content above the Nyquist frequency before sampling; (2) as general NOISE REDUCTION/smoothing on a noisy sensor signal (e.g. a thermocouple or strain-gauge line) to improve SNR at the cost of bandwidth; (3) as EMI/mains-hum rejection, attenuating high-frequency interference picked up on signal wiring; (4) smoothing the output of an RMS-to-DC converter or an envelope/peak detector, which is inherently ripple-laden until filtered; and (5) band-limiting a signal ahead of a slow readout device (chart recorder, low-rate data logger) so fast transients are not aliased into the recorded trace.

c) A high input impedance minimizes LOADING of whatever source (a sensor or a previous stage) drives the amplifier: with negligible current drawn, the measured voltage is essentially undisturbed by the act of measuring it. This is exactly the same divider argument used for high-source-impedance sensors elsewhere in this paper (Question 3's gas sensor, Question 7's pH electrode) — if the amplifier's input impedance is not much larger than the source impedance, a significant fraction of the true signal is dropped across the source itself and lost before it ever reaches the amplifier.

d) High-value resistors bring several real disadvantages, several of which are visible directly in part (a): (1) they contribute more THERMAL (Johnson) noise, which scales with $\sqrt{R}$; (2) they make the circuit far more sensitive to BIAS-CURRENT-INDUCED offset errors, as the $R_f=6\ \text{M}\Omega$ design above demonstrated (480 mV vs. 0.3 mV, purely from the choice of op-amp); (3) at a high-value resistor's high-impedance node, small STRAY/PARASITIC capacitance becomes significant, forming an unintended additional RC time constant and making the node more susceptible to capacitively-coupled interference pickup; and (4) very high-value precision resistors are more expensive, have wider tolerance, and drift more with temperature than common mid-range values, making the design harder to manufacture repeatably.

e) Every real op-amp input draws a small bias current that must have a DC path to flow through; if that path is missing, the current instead charges up whatever capacitance is on that node, and the node's DC level drifts uncontrolled until the amplifier saturates against a supply rail. This is particularly critical in a NON-INVERTING configuration because the $(+)$ input is not held at a defined virtual ground by feedback the way the inverting input is — it directly sets the output's DC operating point, so an undefined, drifting $(+)$ input propagates straight through to an undefined, drifting output. A bias-current return resistor from the $(+)$ input to ground (or to a defined reference) supplies the missing DC path without materially loading a high-impedance source signal.