How to generate -1v from +3.3V Supply

Hi,

Hopefully someone will be able to help as analogue electronics is not my area, I would like to know if it's possible to generate -1v from a 3.3v supply? Suggested component or circuit would be helpful.

Thanks Rob

10 years, 10 months ago.

Not with passive components. You would need to generate an AC voltage and subsequently a chargepump or inductor to get the negative voltage. You could use an SMPS IC to do this but the easiest way is probably to use a MAX3232 which generates plus and minus 6V which you can divide down to -1V. Beware of limited currents however. What do you need it for?

I'm trying to put together the suggested circuit for the following sensor

http://proto-pic.co.uk/flexiforce-pressure-sensor-25lbs-1-area/

the circuit is part of the datasheet.

posted by Rob Cawsey 29 Aug 2013

-1V does not really seem to be required. One option is to replace the -1V with +5V, and the grounded input by a 3.3V input. This inverts the behavior, but that shouldn't be a problem, and the advantage is that those are probably voltages you already have, and the output is still between 0V and 3.3V, so can be directly put on the ADC of your mbed.

The sensor modulates its resistance. What the circuit does is put a constant voltage over the sensor. Depending on the resistance a current will flow, which is again converted to a voltage by the resistor. But it shouldn't matter if those voltages is -1V and 0V, or 5V and 3.3V. Only the current flows the other direction.

posted by Erik - 29 Aug 2013

Thanks for the detailed help :)

posted by Rob Cawsey 30 Aug 2013

Even a simple resistive divider will do: Vout = Vref * R1 / (R1+R2); So the voltage is not directly proportional (or inversely proportional) to the resistance but maybe that's not required.

posted by Ad van der Weiden 30 Aug 2013

Taking your original suggestion could I do the following:

1) Connect the +ve of the op-amp to 0V. 2) Connect the -ve of the op-amp to 3.3V. 3) Use a simple resistor voltage divider to achieve a voltage of approximately 1v on the sensor.

Reason I suggest 3.3V is that I don't have a +5v source on my board.

posted by Rob Cawsey 30 Aug 2013

I need to maintain a linear relationship between the pressure applied, so I don't think a resistive voltage divider would work in this case.

posted by Rob Cawsey 30 Aug 2013

You can always calculate it digitally what the corresponding pressure would be. Unless you are very busy with other calculations that shouldn't be much of an issue, especially on an 1768.

The output voltage you get is at no pressure equal to the voltage at the positive input of the opamp. So that needs to be between 0V and 3.3V. If the other terminal of your sensor has a higher voltage, the output will decrease at increased pressures, if it has a lower voltage, it will increase. So if you have 0V on the + input, you need a lower one on the pressure sensor, which is negative and a problem.

You can use a resistive divider to put 0.5-1V on the positive input, and ground the other side of your sensor. The downside is you have an offset, the output starts at 0.5V-1V, so you cannot use your complete range of the ADC. But it means you don't need other voltages.

Don't you have any other stable voltage above 3.3V? Doesn't need to be necesarily 5V.

posted by Erik - 30 Aug 2013

Thanks for taking the time to explain I now understand, at the moment there isn't a +5v supply but I can add one so no problems. You see this is why I only do software and some digital electronic design.

posted by Rob Cawsey 30 Aug 2013
10 years, 10 months ago.

If the device needing -1v doesn't need the same ground (0v) level than the mbed, I would suggest a voltage divider and just switch the pins on the device needing -1v. The divider works for low current. If higher is needed, use a regulator chip.

10 years, 10 months ago.

You could do a simple voltage doubler circuit. Using two capacitors and 2 diodes, forming a negative voltage generator. If you need some kind of feedback, this would be another story. If you just need it as a reference, I suggest a Zener across the last capacitor, just to clamp the voltage a -1V.

Lerche

Can you explain exactly how this works maybe with a diagram?

posted by Rob Cawsey 30 Aug 2013
10 years, 10 months ago.

if you use a PAM Output, connected to a capacitor, then diode, with another capacitor to gnd. It will generate a negative voltage. have a look at

http://en.m.wikipedia.org/wiki/Cockcroft%E2%80%93Walton_generator

Ceri