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Force sensitive resistor controlling output voltage

Discussion in 'Sensors and Actuators' started by shivasage, Apr 7, 2020.

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  1. shivasage


    Apr 7, 2020
    Hello everyone! I'm planning a project which aims to regulate an output voltage from 0 to +5V DC by using a force sensitive resistor. This is for a musical implementation: I own a guitar pedal which accepts 0 to +5V voltage to control certain parameters. I want to push on the sensor with my finger to control these parameters.

    Originally I had planned a circuit using an op-amp (schematic attached; the sensor is represented by the 68k resistor because it starts at several million ohms with zero pressure and gets down to ~68k ohms with maximum pressure). with a 1meg potentiometer in the feedback loop to fine-tune the response. The 50k load represents the input of the guitar pedal (this value sourced from the owner's manual). I planned the op-amp circuit because, according to datasheets, this would result in the most linear response.

    When attempting to breadboard the circuit, the op-amp was damaged and I didn't have a replacement. I didn't want to order parts just yet so I tested a 9V wall wart running straight through the sensor, and was surprised to find the response was quite linear. (Perhaps this can be explained by the non-linear nature of the FSR being cancelled out by the non-linear nature of increasing pressure with your fingers?). So I tried coming up with a simpler circuit (schematic attached).

    In this schematic, the 100k pot controls the incoming voltage to fine-tune the response. The 6.8k resistor in series is to lower the current running through the pot when it is at minimum resistance to avoid damaging the pot (I have since lowered this value to 2k). The 50k resistor in parallel has a similar effect of allowing the current to flow elsewhere besides through the pot, and it also smoothed out the taper of the output voltage. Again, the 50k load represents the guitar pedal. (Ignore the capacitor, I think it's unnecessary.)

    Do you guys think the second circuit will suffice? One issue that I'm aware of is that the load will have a significant effect on the output. However, because the load will always be the same (i.e. I'm always going to use the same guitar pedal), does this matter? I thought it was weird when I simulated the circuit online that there was zero current flowing, but an email from the manufacturer: "the input is 0-5V, current won’t matter because there’s a >50k input resistor and the current is self regulating and protected." He also told me the pedal has overvoltage protection, which alleviates another potential concern.

    Another person helping me warned that the FSR might be damaged by the high voltage. I'm unable to find any relevant information in the FSR's datasheet. That being said, when breadboarding I was running 9v (and later 12v) straight through the sensor, with no damage. But perhaps extended use will damage it?
  2. Fish4Fun

    Fish4Fun So long, and Thanks for all the Fish!

    Aug 27, 2013
    Hi shivasage, Welcome to EP!

    It might be better if you explain what it is you are trying to do. I am certain this seems self-evident to you; however, I have read through your post twice and I am clueless.

    While I don't think "regulating" is what you actually want to do, there is nothing in your circuit specifically related to either voltage or current regulation. I think you are simply looking for a 0-5V output from your pressure sensor? Essentially what you want is this:

    However, what you have is a sensor that exhibits variable resistance based on the force exerted on it.

    If I have this correctly, you want to use the FSR to make the OP amp output swing from 5V to 0V based on the resistance of the force sensor? (FSR?)

    But then you start in about a 1M pot, a pedal, a 50k load, a 100k pot and then you throw in some random resistors and a capacitor for good measure .... you change the source voltage from 5V to 9V and maybe 12V, (but your second schematic shows 18V) announce it seems to be working .... and then ask:

    Slow down tiger.

    In the menage you mention a 50k load ... Using a sensor with a 68k to 5?M range it is improbable that you will be able to reliably drive a 50k load without the use of active components (ie an OP amp). I suspect this is the reason you keep increasing your source voltage.

    If you truly want help with designing a reliable circuit, the first thing you need to do is define what the circuit is suppose to do. You need to clearly explain what the inputs and outputs need to be, and how they are related.

    I am confident you think you have already clearly stated the circuit's purpose as well as all of the input and output parameters; however, let me assure you that I have failed to glean much (if any) of the information requisite to help you from your post.

    To be clear on what we would need to help you:

    Clearly define the inputs and outputs of each of the components and how you want the components to interact.

    To be clear on what is NOT needed:

    Running dialog on your experiments, schematics of randomly assembled passive components, solicitations for approbation on said schematics.

    I am not trying to be mean, just clear on what is and is not requisite to help you.

    A very basic, but HUGELY IMPORTANT formula in electronics is Ohm's Law:

    e = iR

    The reason you keep increasing the source voltage, the reason increasing your source voltage is problematic AND the reason an OP amp will almost certainly be required to design a reliable circuit are rooted in Ohm's Law.

    Good Luck!

  3. shivasage


    Apr 7, 2020
    Hello Fish. Thanks for your response. Yes, that is exactly what I am looking for. I will attempt to explain it as best as I can, given my limited knowledge in the field of electronics.

    In the world of musical gadgets, there are certain devices that have a so called "control voltage" or "CV" input. It was invented to allow musicians to control a specific parameter on the device without having to physically reach over and turn the potentiometer controlling that parameter (common parameters being gain, filter frequency, pitch, etc.). You do this by taking a second, voltage-generating device and connecting its voltage output to the first device's CV input. The CV input sends this voltage to the potentiometer in question, which is acting as a voltage divider in the circuit. Depending on what the voltage-generating device is, it will have some way of adjusting this output voltage within some range (commonly 0V to 5V DC) either automatically (e.g. with an LFO) or manually (e.g. a foot pedal that you rock back and forth). Voila, you can now control a specific parameter of the first device without having to fiddle with a knob.

    I thought it'd be cool if the voltage-generating device was an FSR that I control with my fingers. So to sum up, I want to have a voltage source (from a standard 9V DC wall wart) running through the FSR and to send the output voltage to the first device's CV input. With no pressure, there will be 0V at the output because the FSR has >1M ohms resistance, but as I increase pressure with my fingers, the resistance decreases and the output voltage increases, thus changing the targeted parameter on the first device.

    Regarding the 50k 'load,' perhaps that was the totally incorrect term and perhaps it shouldn't even be in the schematic (originally it wasn't). The CV input is not required for the first device to function, so I'm not sure my output voltage has to 'drive' anything, but is merely connected to a potentiometer acting as a voltage divider. Without the 50k, the output current was zero, which confused me, so I emailed the manufacturer and was told "current won’t matter because there’s a >50k input resistor and the current is self regulating and protected." I was attempting to simulate the FSR device voltage output being connected to the first device's CV input, but I probably messed up.

    So yeah, I hope things are a little clearer now. Basically, I'm curious if an op-amp is even necessary considering I was able to get a nice, linear voltage output without ANY additional parts. As far as increasing voltage, that was just to fine-tune the voltage output as it relates to finger pressure. When running 9V through the FSR and measuring output with increasing pressure, I got a swing of 0V to 3V, but I want 5V max, hence increasing to 12V source. In both the opamp and non-opamp circuits, the additional potentiometer is to fine tune this response after the fact.
  4. Fish4Fun

    Fish4Fun So long, and Thanks for all the Fish!

    Aug 27, 2013
    Hi shivasage!

    Ok, lets do a quick review .... In this very simple schematic:

    We have:

    1) A 5V source voltage presumably with a negligible impedance.
    2) A reference to ground, again, presumably with negligible impedance.
    3) A 1 Mega Ohm potentiometer with the 5V source connected @ one end and Ground connected to the other.
    4) A "wiper" output that will read 0-5V depending on where the wiper is situated.

    Remeber "Ohm's Law" ? ...

    e = iR --> 5 = 1,000,000 * i --> i = 0.000005 Amps From the 5V terminal to GND while the wiper output is "floating".

    However, if you connect the Wiper to an input with 50k of impedance, then the schematic looks this:

    To get an idea of what is going on, let's calculate the output at some different wiper values ....

    Wiper = 0 Ohms From +5 ... in this case we ignore the 1M Pot all together and the output is very near 5V and the only impedance is the source impedance of the 5V (ie, if your supply is rated for 2A @ 5V then e = iR --> 5V = 2A * R ==> R = 2/5 Ohms = 0.4 ohms source impedance ... ) In the case of a regulated power supply the actual impedance of the source is a lot more complicated than the static value 0.4 ohms, but for clarity & consistency let's work it out as if it were in fact a fixed source impedance of 0.4 ohms ...
    5V = (50,000 + 0.4) * i ==> i = 5/(50,000 + 0.4) = 9.99992 * 10^-5 ....
    i = 0.0000999992 Amps
    R = 50,000
    e = 0.0000999992 * 50000 = 4.99996V
    And the corresponding voltage drop = 0.00004V

    Moving on to a more interesting case ...

    Wiper = 5,000 Ohms from +5V ... In this case the output current is a function of the 5,000 ohm wiper resistance and the 50,000 ohm resistor ... (we can now dispense with the "source impedance" as being trivial).

    5V = (50,000 + 5000) * i ==> i = 5/55,000 = 9.09 x 10^-5 ==> 0.00009090
    The output voltage =>> e = 0.00009090 * 50,000 = 4.5454V
    and the corresponding voltage drop = 0.4545V

    Let's slide the lever on over a bit to Wiper = 50,000 ....
    5V = (50,000 + 50,000) * i ==> i = 5/100,000 = 0.00005A
    Skipping the math, intuitively, both the output and the drop = 2.5V

    So right now you are thinking, "Yes, this is a very linear progression ... " **BUT** On a "linear" potentiometer, the physical wiper would only be 50,000/1,000,000 = 5% of full scale .... while the output voltage would be exactly halfway between 5V and 0V.

    If the wiper were "Floating" (ie, NOT attatched to a 50k load) then the wiper voltage would be:

    5V = 1,000,000 * R ==> 5/1,000,000 = 0.0000005
    e = 0.0000005 * (1,000,000 - 50,000) = 4.75V
    and the corresponding voltage drop would only be 0.25V

    Bottom line: a 50k load will have a dramatic influence on a 1Meg source ... While you might be able to experimentally compensate by increasing the voltage and using various passive attenuation schemes, the chances of such a set up working reliably if connected to any other load would be spotty at best.

    So far I have only discussed a standard potentiometer, which is a simpler case than what you are working with. I suspect the best solution would be to create something like a Wheatstone Bridge ... I will let Wikipedia explain that:

    and Google (my friend) led me to this:

    Before I go go any further, read through these and see where they take you ...

    (I haven't read the linked paper by Kostas Tarchanidis ; however, I feel confident from the abstract that he discusses in detail circuits similar to what you will need ... )

    I also suspect there are numerous open source schematics//documents detailing practical **proven** circuits for capturing data from devices similar to, or exactly like your "FSR" ... while most of these probably involve ADCs and micro-controllers, they would all need to use some type of active buffering to yield reliable//repeatable data, it would be a lot easier for me to simply copy the analog section of one of these designs than start from scratch.

    To be honest I don't work much with analog control signals, if we can narrow your schematic down to something reasonable there are others here who will almost certainly chime in to help clean up the details. But you have to continue to refrain from (randomly?) inserting components onto a schematic ;-)

    As a general rule, if you cannot demonstrate mathematically WHY you added a component, then the chances of that component positively impacting your circuit is virtually zero.

    We will call that "Fish's Law". LoL!

    Let me know what you find in your reading.

    Good Luck!

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