A signal that swings between -1V and +1V must be amplified with a gain of 2 using MCP602 and then biased by +2.5V, so it eventually swings between 0.5V and 4.5V.

The amplifier only connected to 5V, GND and an input of 2.000V gives an output of 4.034V which is OK. However when it's VDD and VSS pins are connected to 5V and the middle of the voltage divider (+2.5V when not connected to opamp), its output becomes 3.611V and the middle of the voltage divider becomes 3.609.

Is the opamp being biased the incorrect way? If not, how can the circuit be changed?

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  • 2
    \$\begingroup\$ You can't create an offset by tying the Vss (ground) pin of the op-amp to 2.5v. The output will never go below 2.5v. \$\endgroup\$ – tcrosley Apr 3 '14 at 23:28
  • \$\begingroup\$ @tcrosley Oh, how should I bias it in this case? \$\endgroup\$ – Nyxynyx Apr 4 '14 at 1:37
  • \$\begingroup\$ The datasheet doesn't specify whether an MCP602 can work as an amplifier with -1 Volt at any of the analog inputs while Vss is grounded. It says such condition doesn't damage the device. Please tell us more about your signal source. Is your signal AC or DC? What is the output impedance order? \$\endgroup\$ – motoprogger Apr 4 '14 at 2:22
  • \$\begingroup\$ @motoprogger The signal has an AC waveform. What is meant by the output impedance? The signal source uses a 62 ohm resistor across its output leads to provide the voltage waveform. In the question's diagram, I connect one end of the signal source to GND and the other end to + non-inverting input of the op amp. \$\endgroup\$ – Nyxynyx Apr 4 '14 at 2:40
  • \$\begingroup\$ Is the pinout for the MCP602 correct? It doesn't seem to match the datasheet. \$\endgroup\$ – gsills Apr 5 '14 at 4:18

The most easy way to convert a 1V to -1V signal to a 4.5V to 0.5V output signal through a non-inverting gain of 2 amplifier is to AC couple the input signal through a capacitor. To IC1-1 connect the input via the capacitor and +2.5V reference through a 50kOhm. The feedback divider (R3 and R4) gets referenced to +2.5V as drawn. The +2.5 needs to be a relatively low impedance, like from a MCP1525 for example,a resistor divider from +5V won't cut it (can you see why?). Capacitor value needs to be able to pass the lowest frequency of the signal while not being loaded down by the 50kOhm. And, as tcrosley says IC1-5 goes to +5V and IC1-3 goes to ground.


Here is a schematic to help illustrate:

enter image description here

R5 is there to give a DC offset to Vin, so that on the amplifier side of things Vin will be centered around 2.5V (1.5V to 2.5V instead of -1V to 1V). Now instead of ground being the reference voltage it is +2.5V and the difference between Vin(offset) and Vref gets amplified by 2. 50kOhm is chosen for R5 to be equivalent to R3||R4. This balances amplifier input bias currents (although since bias currents here are in the pA it isn't critical).

Gain of the amp is set by the ratio of R4/R3 (plus 1 if the reference is ground), so having the +2.5V reference from a 33kOhm divider (effectively ~16kOhm) adds resistance to R3 causing gain error. Also using a high impedance reference means that the DC offset of Vin wouldn't work right. Whatever the impedance of the reference voltage, it must be unaffected by currents through R3.

A generalized expression for Vout would be:

\$\frac{\text{R3} \text{Vin}+\text{R4} \text{Vin}-\text{R4} \text{Vref}}{\text{R3}}\$

You can see that if Vref is zero (ground) that this is just the usual non-inverting gain. And you can also see that if Vin operates about +2.5V as offset across the capacitor that the output would be between 0.5V and 4.5V.

Finally for the offset cap to work, it's impedance must be much less than 50 kOhm (really more like 500 Ohm) at the lowest frequency of interest. You say that is 60 Hz, so set the impedance of the capacitance to be no greater than 500 Ohm at 60 Hz (it will be about 5uF, which is a lot).

For more information about OpAmp analysis and use, a good reference is "OpAmps for Everyone". This reference covers Thevenin's equivalent circuits and why it makes sense to talk about R3||R4 for equivalent impedance into the OpAmp inverting input in section 2.5.

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  • \$\begingroup\$ I've added a second diagram, is that what you mean it should be? Why did you choose 50kOhm and how would you select the capacitor value for a 60Hz signal. Why does the +2.5V reference require a low impedance? Sorry for the many basic questions... \$\endgroup\$ – Nyxynyx Apr 4 '14 at 17:25
  • \$\begingroup\$ Not quite what I meant. I should have added a schematic. I'll edit for clarity. \$\endgroup\$ – gsills Apr 5 '14 at 3:42
  • \$\begingroup\$ Thanks for the edit! A few more basic questions: 1. Can you help me identify how R3 and R4 are connected in parallel? 2. How do you get the values to be 33kOhm and effective resistance 16kOhm? \$\endgroup\$ – Nyxynyx Apr 5 '14 at 23:52
  • \$\begingroup\$ @Nyxynyx ... These 2 questions are really the same. If you think about the circuit as a small signal AC model you'll see that the impedance seen by the OpAmp inverting input is R3||R4. Same is true for the common node of your R1 & R2. More rigorous, is that R3||R4 and R1||R2 is the Thevenin equivalent resistance at those nodes. \$\endgroup\$ – gsills Apr 6 '14 at 23:43

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