1
\$\begingroup\$

I am looking at the following article from Analog Devices: Use Synchronous Detection to Make Precision, Low Level Measurements, and I'm wondering if there is a way to build the following demodulator circuit (Figure 4 at the linked site) with a single supply op-amp:

Demodulator

Any ideas out there?

EDIT

This circuit appears to work just fine with a single supply op-amp. I think I had the switch direction backwards the first time I tried simulating it. The switch should be set up such that the gain is +1 when the sync signal is high, and -1 when the sync signal is low. Below are images of a spice simulation showing this working with the LT1498 on a single +3.3 V supply.

Circuit

Simulation result

\$\endgroup\$
  • 2
    \$\begingroup\$ Yes. Move switch gnd to Vcc/2 \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 May 10 at 20:12
  • \$\begingroup\$ I'm not sure that the "AC coupling" in the block diagram can just be a cap -- you may need something that enforces an average around VCC/2. \$\endgroup\$ – TimWescott May 10 at 21:33
  • \$\begingroup\$ Alternately, generate the excitation to the sensor from a microprocessor, and drive an ADC instead of an op-amp. Then do the demodulation in software by changing the sign of every other sample and averaging. This works great -- I've done it more than once. \$\endgroup\$ – TimWescott May 10 at 21:35
  • \$\begingroup\$ It can just be a cap... \$\endgroup\$ – Voltage Spike May 10 at 22:14
1
\$\begingroup\$

The problem is the square wave from the sensor is centered around zero, and the switch rectifies the signal back to DC. If the op amp's negative rail is tied to ground, the range is 0V to +Vcc which means that half of the square wave. Actually now that I think about it, this should be able to work, because the amplifier is essentially switching between a gain of 1 and -1 its kind of like applying an absolute value to the signal (as long as the signal is synchronous with the chopping signal, sometimes delays can create error in these types of circuits, and not like an absolute value in the sense that it can go negative in the circuit above). If I have time, I'll run some calculations to see if this holds true, it's been a few years since I have messed with a circuit like this.

However, this would affect the range of the sensor on the other end, which I would imagine that it would be restricted to above 0 as well.

The real power behind these circuits is to be able to restrict the noise to a certain bandwidth. Back in the day 1/f noise was prevalent on most amps, so circuits like this were used to modulate the signal out of that frequency domain, and a lock in amplifier like the one above to bring the chopped frequency back to DC. Another good tool is stick a bandpass filter before the lock in amplifier to restrict the bandwidth of noise even further.

\$\endgroup\$
  • \$\begingroup\$ I find it useful (sometimes) for getting rid of offset drift in the sensor -- you can buy nice chopper-stabilized amps, but that doesn't do any good if the sensor is biased. Of course, the sensor has to be amenable to the method. \$\endgroup\$ – TimWescott May 10 at 23:34
  • \$\begingroup\$ Works wonders on Wheatstone bridges. \$\endgroup\$ – Voltage Spike May 11 at 1:23
  • \$\begingroup\$ @laptop2d I think you're right. It appears to work just fine with a single supply op-amp as long as the switch is configured such that the gain is +1 when the sync signal is high and -1 when the sync signal is low. I added a spice simulation to my question to show this. \$\endgroup\$ – user181297 May 13 at 14:38
  • \$\begingroup\$ @user181297 Thats cool, another improvement on these circuits is switching to a single frequency sine wave (similar to a carrier frequency in AM) and changing C2 to a band pass filter, which gives you more control over bandwidth and noise as square waves have frequency content across many frequencies and should be capacitance coupled. \$\endgroup\$ – Voltage Spike May 13 at 15:29

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.