Op-amp - Converting a negative ramp to a positive ramp

Question

I have an analog input that I need to measure. The analog input is single-ended and ranges from -10V to +10V.

I want to try and measure that output with an ADC, which has a 0-3.3 V range.

I want -10 V to be 0 V, +10 V to be 2.5 V, and 0 V to be 1.25 V.

Right now, I have -10 V = 5 V, 0 V = 2.5 V, 10 V = 0 V (I can handle just a simple voltage divider to halve the range)

This is the circuit I have now:

^{simulate this circuit – Schematic created using CircuitLab}

I figure that there's some difference amplifier circuit I can use, but my knowledge of these analog circuits is lacking.

On the board, I have spare op-amps, as well as a 1.25 V reference to use.

Here is an easily editable simulation of the circuit.

Answer 1

This resistor divider gets you close:

^{simulate this circuit – Schematic created using CircuitLab}

A sweep of V1 from −10V to +10V outputs 0V to +2.22V:

If you want the full 0V to +2.5V, you'll need some gain \$A\$:

$$ A = \frac{2.5}{2.22} = 1.125 $$

In this next design, the op-amp is configured as a non-inverting amplifier with gain 1.125. The overall response to inputs from −10V to +10V is an output going from 0V to +2.5V:

^{simulate this circuit}

Sadly, that would require a negative supply for the op-amp, because it cannot get all the way to 0V output without one. If you want to use 0V for the negative supply, you'll have to sacrifice a few tens of millivolts at the bottom end.

^{simulate this circuit}

Note that I've increased R2 slightly, to raise the minimum potential at X to a little over 0V. Since this also increases maximum potential there, I've also had to reduce gain a little. For inputs between -10V and +10V, this last design will output +30mV to +2.5V:

Answer 2

As per RusselH's (user319836) suggestion below, here's the improved version:

^{simulate this circuit – Schematic created using CircuitLab}
The output is no longer reversed, but follows the input, except scaled & shifted from -10V ... +10V to 0V ... +2.5V (roughly).

Original Answer below

If you can live with the ramp being reversed, so -10V gets translated into +2.5V and +10V gets translated into 0V, then you can do it with a single opamp.
You could easily reverse it back again in your MCU's code.

^{simulate this circuit}

The inverting gain of 0.12 isn't perfect, but it gets you to within about 50mV at each end.

The output offset is set by Vref, scaled by R1 & R2, and multiplied by the non-inverting gain: \$1 + \frac{R_3}{R_4}\$, so:

so:

$$ \def\V{{\rm\, V}} \def\k{{\rm\, k\Omega}} % \begin{aligned} V_{ref} \cdot \frac{R_1}{R_1 + R_2} \cdot \left(1 + \frac{R_3}{R_4} \right) &= \\ &= 1.25\V \cdot \frac{10\k}{10\k + 1.2\k} \cdot \left(1 + \frac{1.2\k}{10\k}\right) \\ &= 1.116\V \cdot 1.12 \\ &= 1.25\V \end{aligned}$$

Answer 3

Here is a way that uses only two standard values of resistors and your present op-amps.

Note the need for a negative supply for the op-amps, however -5V is adequate. The worst-case output voltage of U1 is -1.25V at +10V input. With your idea of buffering it first, you'd need something like -12V to handle a -10V input. If you don't have a negative rail, you could use something like a 7660 charge pump to generate it from +5V.

Answer 4

I'd start with reducing the gain using a simple potential divider like this: -

• negative 10 volts in maps to -1.25 volts out
• positive 10 volts in maps to +1.25 volts out

So, all you have to do is add an offset like this: -

And, you get what you want.

Of course this task is made easier by using a standard voltage reference such as +2.5 volts and attenuating it to +1.42857 then adding an op-amp buffer before connecting it to R1.

Answer 5

I want -10 V to be 0 V, +10 V to be 2.5 V, and 0 V to be 1.25 V.

The function you want to implement is

\$V_{out} = (0.125\times V_{in}) + 1.25\$

Since amplification is positive, but less than 1, this can be accomplished completely with resistors and a voltage reference. An op-amp would only be necessary if one wants or needs to counter loading effects. In that case an op-amp configured as a voltage follower can be used.

To solve this problem, we first find a voltage \$V_x\$ that, when given as the input voltage is itself the output voltage. That is,

$$V_x = \frac{1}{8}V_x + 1.25$$

We find this value to be \$\frac{10}{7}\;\text{V}\$.

We also wish to find the voltage \$V_0\$ that, when given as the input voltage, results in a output voltage of \$0\$. That is,

$$\frac{1}{8}V_0 + 1.25 = 0$$.

By inspection, \$V_0 = -10 V\$.

The circuit we wish to implement is of the form:

^{simulate this circuit – Schematic created using CircuitLab}

To implement the appropriate function we choose resistors with the following ratios:

$$R_{in}:R_{ref}:R_0 \;\;=\;\; V_0(V_x-V_{ref})\;:\;V_{ref}(V_{ref}-V_x)\;:\;V_{ref}V_x$$

We choose the \$V_{ref}=2.5\;\text{V}\$

Substituting and simplifying gives us:

$$R_{in}:R_{ref}:R_0 \;\;=\;\; 12:3:4$$

^{simulate this circuit}

Then we verify the behavior with a simulation.

The behavior meets the spec.