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I am analyzing this continuous waveform amplifier circuit for a Microwave Motion Sensor. This high gain amplifier is recommended in the datasheet as the signal output of the sensor is extremely small (~1uV). I understand that they are using an op-amp integrator and an op-amp differentiator configurations, but what are the roles/purposes of the circled parts? Could someone please kindly explain it to me as my circuit analysis is a little bit rusty.

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    \$\begingroup\$ Please edit the drawing to include a designator for each part. Otherwise we have to talk about "the 100 kohm resistor in the top of the divider" instead of just "R1" and it gets tedious. \$\endgroup\$ – The Photon Oct 30 '17 at 16:30
  • \$\begingroup\$ Edited. Also, maybe I don't see it correctly but how #3 and #4 are filters when the RC components are in series? And For #2, shouldn't it be the other way round (capacitor first) for it to be a high pass filter? \$\endgroup\$ – Shibalicious Oct 30 '17 at 16:36
  • \$\begingroup\$ Filters can have different topologies.. A capacitor in series is used to remove the DC coupling. Probably the name "filter" was misleading. \$\endgroup\$ – Eugene Sh. Oct 30 '17 at 16:42
  • \$\begingroup\$ 1) is a voltage divider which creates a 2.5V bias voltage. \$\endgroup\$ – Janka Oct 30 '17 at 16:43
  • \$\begingroup\$ But if the capacitors are to remove the DC component, what's the purpose of R1 in #2, #3 and #4? Also, what's the purpose of C1 in #1? \$\endgroup\$ – Shibalicious Oct 30 '17 at 16:49
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Figure 1. Waveform amplifier.

  1. The resistors form a half-supply reference for the single-rail powered op-amps. With a 5 V supply this reference voltage will be 2.5 V. The 100 uF capacitor stabilises this voltage. As @Trevor points out in the comments, it also prevents any AC input signal from (3) affecting (2).
  2. The non-inverting input is held at 2.5 V. With no signal the output should go to 2.5 V as well.
  3. This stage is a non-inverting amplifier. Without the 330k resistor the bias current of the op-amp would charge or discharge the 4.7 uF capacitor until it reached +5 V or 0 V. Providing a DC path to the 2.5 V reference prevents this.
  4. Difficult to know what the 12k is for without a schematic of the motion sensor innards. The capacitor means that the rest of the circuit will only respond to rapid changes from the motion sensor.
  5. The op-amp is a non-inverting one. The gain is given by the standard formula \$ A = 1 + \frac {R_F}{R_G} = \frac {1M}{10k} = 100 \$. (Note the sloppy units on the schematic. It should be 'k' for kilo and 'M' for mega.) The capacitor blocks DC again as the DC path for the bias is provided by the 1M feedback resistor. It could probably have been omitted and (5) connected to the 2.5 V reference but it may be doing some high-pass filtering too.
  6. The capacitor again blocks DC from reaching the next stage. The 8k2 resistor is the input resistor of an inverting op-amp.
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  • \$\begingroup\$ Thank you for an a clear and detailed answer. One last question, is the 100uF capacitor at #1 a bypass capacitor (just like the one before +5V terminal for the motion sensor)? If so, isn't the 100uF capacity a little bit too big for that purpose? \$\endgroup\$ – Shibalicious Oct 30 '17 at 17:41
  • \$\begingroup\$ It's purpose is to hold the 2.5 V reference rock-steady in the medium term if the 5 V supply varies. If we look at the time constant for it and two 100k in parallel we get \$ \tau = RC = 50k \cdot 100\mu = 5 \ \mathrm s \$. That's the time it would take to fall 63% if the supply was switched off. You could calculate how well it would hold for a given dip in supply but I'd say the designer didn't bother and just made it big enough that it was never going to be an issue. The 0.1 µF decoupling capacitors are for very short switching transients in the motion sensor. \$\endgroup\$ – Transistor Oct 30 '17 at 17:51
  • \$\begingroup\$ @Hypomania additionally, the input signal also (accidentally) feeds directly back to that half rail "reference through that 330K. Without the cap it could couple some of that signal past the first stage to the input of the second... sloppy design if you ask me to just save one resistor. \$\endgroup\$ – Trevor_G Oct 30 '17 at 17:55
  • \$\begingroup\$ @Transistor thank you, perfect explanation, it's all clear to me now! \$\endgroup\$ – Shibalicious Oct 30 '17 at 18:03
  • \$\begingroup\$ @Transistor, I done some simulations and tried measuring the voltage across the 4.7uF cap. In both cases - with and without the 330k resistor the max it would ever reach was 2.5V, I honestly can't see how it could fluctuate from 0V to 5V without it. The only difference was that it charged slower with the 330k resistor added in. Could you please elaborate further on this one? \$\endgroup\$ – Shibalicious Oct 30 '17 at 19:13
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C1 AC-couples the signal from the motion sensor into the first amp.

R1 and Rx set the gain of the first stage. The other C1 makes the gain 1 at DC. This is because DC is irrelevant, and making its gain 1 prevents offset voltages from swamping the signal.

The other C1 AC-couples again.

Rblob and Rx set the gain of the second stage.

R9 and R2 make a bias voltage of half the supply, and C1 filters the result to remove supply noise.

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  • \$\begingroup\$ But for the bias voltage, why do we need a 100uF capacitor, isn't that a bit redundant? \$\endgroup\$ – Shibalicious Oct 30 '17 at 17:44
  • \$\begingroup\$ @Hyp: No, I don't see what a capacitor filtering the bias divider output is supposed to be redundant with. There is no other capacitance performing the same function. \$\endgroup\$ – Olin Lathrop Oct 30 '17 at 19:51

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