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I have a receiver circuit that is magnetically coupled via a parallel resonant circuit with the sender (similar to RFID but just one directional). I want to feed the signal into an ADC which has, according to the datasheet, an input impedance of around 5 kOhm.

I measured the voltage across the resonant circuit with an amplitude of about 60mV when using an oscilloscope that has an input resistance of 1 MOhm. Obviously, the measured voltage is going to be a lot less when using the low input resistance of the ADC. This is why I decided to use an impedance converter before the ADC:

schematic

simulate this circuit – Schematic created using CircuitLab Note: V1 isn't an actual voltage source, it's just there to point out that there is a voltage induced there. The opamp is powered from +5V.

When measuring the output voltage of the impedance converter, I get a signal that is very chopped up like so:

Measurements

I'm using a MAX492 Opamp (https://datasheets.maximintegrated.com/en/ds/MAX492-MAX495.pdf). The parameters I took into account were the Gain Bandwidth Product as well as Slew Rate:

  • Gain/BW Product = 500 kHz, so it should be able to amplify the 125 kHz Signal by a factor of 4. In the impedance converter setup, the gain is 1 so this shouldn't be the problem.
  • Slew Rate = 0.2 V/us, the 125 kHz sine wave has a rise time of \$\frac{1}{4} * \frac{1}{125 kHz} = 2 us\$. So the Opamp should be able to change the Voltage by 400 mV in that time.

Is there something I'm missing with this particular opamp or is my approach just wrong in the first place?

I'd appreciate any help.

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  • \$\begingroup\$ If V1 is a true voltage source, C1 and L1 will have no effect other than consuming and supplying reactive power. \$\endgroup\$ – winny Jul 17 '16 at 21:49
  • \$\begingroup\$ Also, do you have a bipolar supply for your opamp? Does it have less than 60 mV of offset? \$\endgroup\$ – winny Jul 17 '16 at 21:52
  • \$\begingroup\$ V1 doesn't really exist, I just wanted to point out that there there is a voltage induced there. I'm powering the opamp from a positive supply only, so I'd expect only the top half of the signal to be correct. The offset Voltage is 200uV according to the datasheet. \$\endgroup\$ – Felix S Jul 17 '16 at 21:56
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    \$\begingroup\$ You don't have a negative supply, so during the negative cycles of the sinewave you are driving the amplifier hard into saturation. I don't know if this is the case for this amplifier, but some op-amps take a very long time to come out of saturation, so you might want to try using a negative supply to see if it works then. \$\endgroup\$ – John D Jul 17 '16 at 22:17
  • \$\begingroup\$ Does it have rail to rail input and output range? How is the response when driven below the negative rail? \$\endgroup\$ – winny Jul 17 '16 at 22:31
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I believe you simply do not have the MAX492 biased correctly. Look at the "Common Mode Input Voltage Range" ("CMIVR") in the MAX492 data sheet. It goes from Vee-0.25 to Vcc+.25. This is the legitimate range of voltage you should be applying to either input on the MAX492.

Since your L-C circuit is grounded to the power supply common, its sinusoidal voltage output will swing above and below ground. On the positive cycle this is OK, but on the negative cycle you are violating the CMIVR of the MAX492 by dipping to far below ground.

You will want to bias the output of the L-C to 1/2 Vcc by make capacitively coupling the L-C output to 2.5 volts and connect this signal to the NI input of the op-amp. This will position the L-C's output squarely in the middle of the legitimate CMIVR.

You can make the 2.5 volt bias point with a simple resistive divider between 5 VDC and GND. The coupling capacitor needs to have a low reactance at the resonant frequency of the L-C. You have to choose the resistor values (both are equal) so they don't swamp out the L-C's output. I'd start with 20KOhms each and compare results with the unloaded output of the L-C. If the loaded output is significantly swamped, go higher on the resistors.

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