1
\$\begingroup\$

I have an input signal from -3.3V to 3.3V and the ADC I would like to use can only read from 0 to 3.3V.

I also have access to a stable source of power at 3.3V.

From previous visits, I have tried to read Carter's "Designing Gain and Offset Circuits" as best I can.

Based on:

  • Vref = 3.3
  • Vout fs = 3.3
  • Vout zs = 0
  • Vin fs = 3.3
  • Vin zs = -3.3

I believe: my m = 0.5 > 0 and my b = 1.65 > 0. Thus, I am in Carter's first scenario (Carter's Figure 1):

schematic

simulate this circuit – Schematic created using CircuitLab

For R1, I have selected 10 kOhm. I have computed R2 to be the same 10 kOhm.

HELP:

  1. (I think R1 = 10 k Ohm is OK, right?) [EDIT: Thanks for confirmation]
  2. I do not know how to find Rf. Will a 10 k Ohm be OK?
  3. When I try to compute Rg, I get 0 in the denominator! What resistance should I choose? [EDIT: Thank you @Michael ~ OK no resistor here]
  4. Is the LM358 [EDITED @Dwayne Reid Thank you.] NE5532 is a reasonable amplifier for this project? If not, please provide some guidance.

Thank you, in advance, for your time & assistance.

\$\endgroup\$
  • \$\begingroup\$ A 0 in denominator for R_g might suggest that the resistance should be infinite aka an open. If you analyze this for this situation, it's correct, you just need a voltage follower. R1 and R2 are a voltage divider that half the input voltage, giving you a value of 0.5 * (Vin + Vref), which in your case is exactly what you want. \$\endgroup\$ – Michael Mar 22 at 6:16
  • \$\begingroup\$ @Michael Thank you. \$\endgroup\$ – Old Guy Mar 22 at 8:39
0
\$\begingroup\$

Let's analyze what you need to do.

1) Input voltage is plus minus 3.3V, reference supply is 3.3V. Your choice of resistors R1 & R2 is good. Check: if input is 3.3V, input to op amp is 3.3V. If input is zero Volts, input to op-amp is 3.3V /2. If input is -3.3V, input to op-amp is zero Volts.

2) Determine the total gain of the input resistor network. Inspection tells us that it is 0.5.

3) Determine the gain required from the op-amp.

I'm going to digress a little bit here. The LM358 is an okay op-a,p. but lousy input offset voltage error and a major problem here: the highest output voltage is about (Vdd - 2 Volts). IF you run the LM358 from a higher voltage than your 3.3V rail, you can make it work. But if all you have is that 3.3V rail, you need a different op-amp.

Now let's look at what gain you need from the op-amp.

From statement (1) above, the output voltage range of the input divider network 3.3V through 0V. Since that is also the desired output voltage range, the op-amp need a gain of (1.0) or unity gain.

It's really that simple!

| improve this answer | |
\$\endgroup\$
  • \$\begingroup\$ thank you for the quick reply. Do you have any opinion about the NE5532 opamp for this "gain" and offset application? \$\endgroup\$ – Old Guy Mar 22 at 7:32
0
\$\begingroup\$

schematic

simulate this circuit – Schematic created using CircuitLab

Figure 1. Simplified solution.

You have worked out the offset / divider solution correctly and have a voltage follower with a gain of 1.

Your source impedance is the combination of the two 10 kΩ resistors in parallel - 5 kΩ. If you're ADC has an input impedance of 500 kΩ or more you can feed directly in without worrying about loading effects. No op-amp needed.

| improve this answer | |
\$\endgroup\$
  • \$\begingroup\$ Do note that 5 kΩ input impedance into a 500 kΩ ADC means 1% error, which is worse than 8-bit resolution. Of course, if the impedances are well controlled, you could do a correction in software. Lastly, an ADC I’m familiar with (the integrated one in STM32 microcontrollers) has a much lower impedance of only 50 kΩ, corresponding to a 10% error. So the use of an op amp may be indicated here. \$\endgroup\$ – swineone Mar 22 at 13:59

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.