1

I'm using a controller that needs a 0 to +5V input to activate an actuator on the output side. Normally, a potentiometer of 10k is used to delver 0-5V on the input (left drawing). For my application, I need to control it with a sensor that acts as a variable resistor and has a range of 1.3k - 2.3k (right drawing). enter image description here So I need a thing in the middle ("X") that will deliver as close as possible to 0-5V on the input depending the sensor's resistance. I tried to work out how a DC amplifier for this would look like, but without success. Your help would be appreciated.

MarcC
  • 13
  • 4

2 Answers2

3

If you want to do this with an analog circuit you need to look at the math involved.

You want to get a voltage from the variable resistor whose resistance varies from 1.3k to 2.5k. You can use a voltage divider to do this. If you make the other leg of the divider 2.5k using an input of 5V you get an output voltage range of:

$$ 5V \cdot\frac{1300\Omega}{1300\Omega+2500\Omega}= 1.7105V$$ to $$ 5V \cdot\frac{2500\Omega}{2500\Omega+2500\Omega}= 2.5V$$

Now you need to scale this to 0 V to 5 V. So start by subtracting the lower range voltage. $$ 1.7105 V - 1.7105V = 0 V $$ $$ 2.5 V - 1.7105V = 0.7895 V $$

This gets the lower limit of 0 V, now you need to get the upper limit so you need to multiply by some factor. $$ \frac{5 V}{0.7895 V} = 6.333 $$

Now you need a circuit to do the math, you can do this with opamps. You'll need a 1.7105 V reference to be subtracted from your sensor output, this can be had with a divider using the high and low values of your sensor resistance, 2.5k and 1.3k. Then you need a summing amplifier with gain of 6.333.

Here's a circuit I threw together in LTspice as an example. R2 is the sensor.
The time scale at the bottom is the resistance of the sensor in kilohms. enter image description here enter image description here

This is just a basic example to demonstrate the idea and could probably be improved upon.

You need to use single supply rail-to-rail opamps if you only have the 5 V supply available, and since your output is going to be right up to the rails it's going to suffer a bit at the extremes. A higher supply voltage, and preferably a dual polarity supply would help with this.

A digital alternative to this might be to use a micro with an ADC and DAC to read the voltage from the divider, use software to scale it, and output the result through the DAC.

GodJihyo
  • 17,628
  • 1
  • 16
  • 45
  • Hi! Thanks for this. This simulation looks great. The LT1677 is a bit of an odd beast and is not supported in CircuitLab, so I can't get it verified. LTspice is way over my head. Any alternatives for the LT1677? – MarcC Oct 20 '22 at 10:04
  • @MarcC See if circuitlab has any single supply rail to rail opamps. I just picked one ltspice had. Like I said, if you can get better supply voltages the opamp choice becomes less critical. – GodJihyo Oct 20 '22 at 11:11
  • Works perfectly. – MarcC Nov 02 '22 at 08:16
  • @MarcC In CircuitLab or did you actually build it? And in either case what opamp did you end up using? – GodJihyo Nov 02 '22 at 13:40
  • I actually built it using the specified components. It produces close to 0.1 to 4.8V on the output. Now it works I know what to look for and could try to build it using other (cheaper) components. Thanks again. – MarcC Nov 04 '22 at 08:13
0

You can use a digital potentiometer and/or programmable ATTINY13A-SUR to control precise output voltage. For example, if input voltage == 3V, then output 3.5V. Depends on drop voltage and needed output.

https://en.wikipedia.org/wiki/Digital_potentiometer

Hayk
  • 3
  • 2
  • Thanks very much! You're thinking along the same line as I was: I asked the vendor of the controller if I could introduce this kind of logic on the in-port of the box. As an alternative, I was looking at some discrete components, but never thought of adding a processor outside of the box to do it. – MarcC Oct 20 '22 at 10:05