I am trying to build a level shifting circuit for my project in order to make sure that the input will be appropriate for a 0-3.3V ADC.

I am using two 9V batteries connected in series so my thought is that I have to shift the incoming signal by around 1.6V in order to be in range for ADC.

I have tried two ways to create a voltage reference using 4 1n5818 diodes in series that gave me a reference of around 1.8V and a LM317l voltage regulator giving me around 1.66V.

The problem in my schematic is that the when I connect the reference circuitry to the opamp, it doesn't work as expected. I am not sure what I am doing wrong here.

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  • \$\begingroup\$ When you say "it doesn't work as expected", what exactly are you seeing? (Also, what op amp are you using?) \$\endgroup\$ Jan 14, 2021 at 0:22
  • \$\begingroup\$ @MathKeepsMeBusy hello, when i connect the diode circuit as the voltage refence, there is an no offset at all and the output waveform is the same with the input. however when I connect the voltage regulator, there is an offset but the output is a straight line instead of a waveform. I am using a LMC6484 \$\endgroup\$ Jan 14, 2021 at 0:27
  • \$\begingroup\$ The behavior, as you say is not what is expected. My first guess would be mis-wiring somewhere. Are you doing this on a bread-board? Can you post a picture? \$\endgroup\$ Jan 14, 2021 at 0:33
  • \$\begingroup\$ @MathKeepsMeBusy yes, i am using 100kΩ instead of 75k here and 10kΩ instead of 50k \$\endgroup\$ Jan 14, 2021 at 1:15
  • 2
    \$\begingroup\$ The resistor from 22 to 25, bottom side in image, seems to be misplaced? shouldn't one end be connected to op-amp input? \$\endgroup\$ Jan 14, 2021 at 1:59

1 Answer 1

  1. Where you indicate you are connecting to the diode string is not the right place. It should be between the anode of the top diode and the current limiting resistor.

  2. What is the signal source? Specifically, what is its output impedance? This appears as an additional resistor in series with the 75 K going off to the left (please add reference designators to your drawing), and affect both the overall circuit gain and the value of the DC voltage added.

  3. The non-inverting input of your opamp circuit is not a virtual ground, so there is interaction between the input and offset signals. The two 75 K resistors form a voltage divider for both. This will attenuate both the input signal and the DC offset voltage (assuming the input signal reference potential is GND).

  • \$\begingroup\$ re: (3) -- If the source impedance isn't too high, it seems like an okay summation circuit. Positive input is (Vin + Vref)/2 , and output is 2x that. \$\endgroup\$
    – Pete W
    Jan 14, 2021 at 1:55
  • \$\begingroup\$ thank you! could you explain how I can fix the gain? Do I just take the second 75k resistor out? \$\endgroup\$ Jan 14, 2021 at 17:11
  • \$\begingroup\$ a) which one is the "second" one? b) Doesn't matter; removing either one will not solve the problem. c) Nowhere do you way how much gain you need. \$\endgroup\$
    – AnalogKid
    Jan 14, 2021 at 18:46

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