sensor recommended circuit

In the circuit above the inverting input feeds to ground through a capacitor.

The output should be a constant voltage used for reading into an ADC/taking a measurement.

A negative feedback op-amp circuit, usually I thought, balances the current over the feedback resistor with the current feeding into it.

I am puzzled as current should not flow through a capacitor. Could someone explain how this circuit works?

  • \$\begingroup\$ Is this circuit just a gain X 1 buffer with low pass filtering? \$\endgroup\$
    – Robin
    Jan 11, 2023 at 11:30
  • \$\begingroup\$ @Robin not for AC \$\endgroup\$
    – Designalog
    Jan 11, 2023 at 11:31
  • \$\begingroup\$ So its a gain 1 for d.c. and removes a.c. from the output signal and acts as a buffer to not electrically load the sensor? \$\endgroup\$
    – Robin
    Jan 11, 2023 at 11:33
  • \$\begingroup\$ @Robin if your sensor can be modeled accurately with a voltage source, then it is not loaded just because it's connected to the (+) input, which we can assume has a large input impedance. \$\endgroup\$
    – Designalog
    Jan 11, 2023 at 11:43
  • 2
    \$\begingroup\$ @Robin this amp let's DC through with a gain of 1, while amplifying by 20x frequencies up to the pole set by the parallel combination of 3.3nF and 1Mohm. \$\endgroup\$
    – Designalog
    Jan 11, 2023 at 11:46

5 Answers 5


If you're referring to the 2.2 μF cap, it was probably placed for the purpose of not amplifying a DC signal (including the op-amp's own offset).

Think about it: at DC, that capacitor is an open, therefore the amplifier behaves as buffer. It lets DC through with a 1x factor only. The capacitor becomes, ideally, a short at your frequency of interest; your amplifier gain at AC is defined by the resistors, as you probably know.

The op-amp is still perfectly biased, as there is a DC path between output and the (-) input.


It's been a long time since I worked in the analog world, but I'm pretty sure this is nothing more than a power buffer. A low-power signal is detected and mirrored to the output of the op amp with plenty of power for whatever is next in line. The RC circuits touching the negative input are nothing more than debouncing circuits that attenuate any ringing on the op amp output.

Looking only at this schematic, it doesn't seem like that would be needed, but how the output of that circuit is terminated is incredibly important. Pretending that it's driving 500 feet of coax cable, the termination reflection would be nasty. Those circuits would filter most if not all of that out.

In short, it's not the signal that's being amplified, but the power delivered to the input of the next circuit that's being amplified.

But, like I say, it's been a long time since I worked with op amps. I could be wrong. This circuit would make a lot of sense if the sensor output was only high or low (+5 or GND) with the possibility of a slow transition. Thus, the circuit would also act to minimize the effects of hysteresis.


DC current won't flow though a capacitor but AC will quite happily.

We can do a rough analysis by looking at the points where the capacitors have similar impedances to the resistors, and looking at what the dominant components in the gain are in each region.

At 0.07 Hz the 2.2uF capacitor's impedance is around 1MΩ At 1.4 Hz the 2.2uF capacitor's impedance is around 50kΩ At 48 Hz the 3.3nF capacitors's impednace is around 1MΩ At 960 Hz the 3.3nF capacitor's impednace is around 50kΩ

We can then look at the dominant compoenents in each frequency regime.

From DC to 0.07 Hz the 2.2uF cap has an impedance greater than 1MΩ so the dominant factor is the "1+" in the gain equation for a non-inverting amplifier.

From 0.07 Hz to 1.4 Hz the dominant factors are the 2.2uF capacitor and the 1 MΩ resistor. The gain increases with frequency.

From 1.4 Hz to 48 Hz the dominant factors are the resistors and we have a gain of around 20.

From 48Hz to 960Hz the dominant components are the 50K resistor and the 3.3nF capacitor. The gain decreases with frequency.

Above 960Hz the "1+" gain dominates again and we once again have a gain of around 1.

So this appears to be a "band boost" filter.

  • \$\begingroup\$ Looks like pulsed IR sources are used and the gain at the pulsing frequency X20 brings the signal out of the noise \$\endgroup\$
    – Robin
    Jan 12, 2023 at 8:38
  1. The DC gain is 1 because the 2.2uF capacitor does not pass DC.
  2. The mid-frequency gain (around 9Hz) is 1M/50k= 20 times.
  3. The gain is reduced -3dB at 1.5Hz by the 2.2uF capacitor and 50k resistor and lower frequencies are reduced almost 6dB per octave below 1.5Hz until +3dB about 0.23Hz.
  4. The high frequency gain is reduced to +17dB at 48.5Hz and higher frequencies are reduced -6dB per octave.

Thanks for all above.enter image description here

I got my old "Analogue Filter design" Van Valkenburg book out and used some s transforms on it via python.

import cmath

for i in range (1,200000,10):
  TWO_PI = (3.142*2.0)
  omega = i/100.0;
  s = complex(0,1) * omega
  R1 = 56000
  C1 = 2.2E-6
  R2 = 1e6
  C2 = 3.3E-9
  # Z1 is R + C to ground
  Z1 = R1 + 1.0 / ( C1 * s )

  # fb from output to minus on opamp
  Z2 = 1.0 / ( (1.0/R2) + (C2 * s))

  # output of the opamp means the middle
  # of the volatge divider is at V1 input voltage.
  # thus v1 = v2 ( Z2 / (Z1 + Z2))
  # re-arranging V2 = V1 (Z1 + Z2)(Z1)
  Z = (Z1+Z2)/Z1
  print ("f",omega/TWO_PI, "Z", Z, "|Z|", abs(Z))

Exactly as people posted above. Graph of magnitude made in gnuplot using 2:6 to plot against.


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