Where to add a potentiometer to vary input voltage?

[Jim Kailey]

The voltage from the FSR varies between 2.5 V and 3.6 V as the FSR resistance varies from 10 k to 26 k. When the voltage is less than 2.5 V, the lower comparator holds the transistor base at ground; when the voltage is greater than 3.6 V, the upper comparator holds the transistor base at ground. In between these values, the 1 k resistor turns on the transistor, turning on the buzzer.



Where can I add a pot to vary the 10K ohm/2.5v value that's input into the TLV1702? Thanks.

 

[danadak]

If you just want to maintain the 3.6 while varying the 2.5, I would use
two seperate R dividers, 2 R's to set the 3.6, and one pot ground to 5V,
wiper into lower comparator, to set/change the 2.5.


Regards, Dana.

 

[Jim Kailey]

Thanks much for the reply, that's exactly what I want to do.

So would I replace the fixed 10K resistor at the bottom of the diagram that feeds into the 2.5v point with a pot?

 

[Tony Stewart]

You can make the adjustment finer using a smaller value pot and adjust the min-max range of 2.5V by adjust the fixed values to like 2.5 to 2.6V with a pot that is symmetrical 2.5V +/- 2.5% for a 5% span using 2 fixed 10k (up and down) and 1 trimmer in the middle = 1k for +/-125 mV

You decide what ratios and limits of control you need for the tolerances on the FSR and resistors you expect.

Here you can change the fixed R values with the mouse wheel or left-click on it to edit.

https://tinyurl.com/yu4qduz2

 

[danadak]

The 5.6K 5.4K divider in post 4 does not give you the 3.6 you want, you
should recalculate that.

This will pick standard R's for you - http://howardtechnical.com/voltage-divider-calculator/


Regards, Dana.

 

[Jim Kailey]

I think I'm seeing that I have to split the line going to the comparator inputs, I'm sure that's what you said Dana but I didn't know what you meant.

Those are neat little tools, thanks. It looks like using a 10K pot with 10K fixed resistors on either side would give me a range of 1.7 to 3.3v which might work, or a 5K pot to give 2-3v, I'll have to experiment a little.

Using your voltage divider calculator Dana I'm getting only 13 & 34 ohms with 100ma, can that be right?

With this one I'm getting like 2K and 5.1K:

Voltage Divider Calculator

Try our easy to use Voltage Divider Calculator. Enter any three known values and press Calculate to solve for the other.

ohmslawcalculator.com

Thanks again for the help guys, I'm quite circuit challenged as you can see, at least I haven't let out the magic smoke yet...

 

[danadak]

Using your voltage divider calculator Dana I'm getting only 13 & 34 ohms with 100ma, can that be right?

Yes, forcing design to pull 100 mA thru the divider, standard values, we get 100 mA x 47 ohms =
4.7V, closest we can get with standard values, eg. the 5V supply. But the divider ratio does not seem
to give us the 2.5 we are seeking, ratio is ~.72, where we need .5. Looks like tool has an issue.

100 mA thru a simple divider with a near infinite load on divider wack. Waste of power. 1 mA
more sane. Even less would be OK.

Your 2K, 5.1K ratio is also ~ .72, in your tool site.

If we have 5V and want 2.5 the ratio has to be .5, resistors same value.

Regards, Dana.

 

[Jim Kailey]

Great, thanks. I had no idea what current would be used by the circuit, I'll play around with the values and report back.

 

[Tony Stewart]

Using your voltage divider calculator Dana I'm getting only 13 & 34 ohms with 100ma, can that be right?

Jim, your calculator lets you choose;
{ohms kilohms, kΩ or megaohms mΩ. }
( should be MΩ, mΩ is milliohms !)

So it just computes the ratio values.

If you choose a standard value for either R, then it calculates the other more precisely where you look at the nearest 1% value from Digikey or Mouser etc.

If you want a variable threshold pot from 0 to 100% for force, this may work for you for one sensor, so if you plan to make more, this must be trimmed.

Did you really want the buzzer ON inside the window? or outside?

This old circuit is called a Window Comparator.

The resistance is logarithmic. We don't know how accurate you need this, as they aren't very accurate (15~25% ?) but they are very dynamic.

https://cdn-learn.adafruit.com/assets/assets/000/010/126/original/fsrguide.pdf

R in kohms

 

[Jim Kailey]

Ahh, did a little reading up on divider circuits and see it's the ratio that matters, thanks.

Yes, I want the buzzer on inside the window, like this:

It works perfectly I just want to be able to adjust the point it turns off as mentioned. In your mod to the original circuit you cut the line feeding the two comparators into separate lines, I got a reply on another forum suggesting it could be done with the through line but have no idea how they came up with those values, any idea?

Example B:

 

[Tony Stewart]

If you want a pot for each level, define how much % of the voltage you want and the nominal values.

I did this thinking you wanted to reduce the max force only from 0 to 100% of the min. fixed force threshold of your initial values posted. It can be done with 2 pots, but you must define the ranges.


Notice here I reduced the design to use all the same 10k resistors. The force slider is harder than pushing the sensor to use but at least it is interactive with the LED.

Two sliders, one for the RV1 pot and one for the FSR..

https://tinyurl.com/ymx6leuu

 

[Jim Kailey]

Thanks again for your time on this!

I just want to vary the output of the 2.5v line but I'm not sure yet what sort of range I'll need. I also have 5K and 10K pots "in stock" so was hoping to use one of those.

At this point I could use your split divider line but am really wondering how to do it as in the B example below. How are those resistances determined?

 

[Nigel Goodwin]

Apply ohms law V = I x R and rearrange it as required.

So for the first example the total resistance is 10K + 4.4K + 5.6K = 20K

So the total resistance (R) is 20K, voltage is 5V (V), rearranging gives I = V / R

So I = 5/20000 (which equals 0.00025A or 0.25mA) - this is the current through the entire chain of resistors.

So now to find the voltage at each point, go back to V = I x R

So for the bottom tap R = 10000, so V = 0.00025 x 10000 (which equals 2.5V)

Next tap R is 10000+4400 ohms, so V = 0.00025 x 14400 (which equals 3.6V)

It's simple primary school maths.

 

[Jim Kailey]

Great, thanks so much, I figured it was ohm's law but didn't know how the circuit "saw" the different resistances nor did I know what sort of current was involved.

 

[Tony Stewart]

using my simulation https://tinyurl.com/ymx6leuu

V1= 5V * . . . . (Rv1 + R2) . . .V2 = from V1 to = 5V* . . . . . ..R2_
. . . . . . . . (R1 + RV1 + R2) . . . . . . . . . . . . . . . . . . . (R1 + RV1 + R2)

 

[Jim Kailey]

Cool simulation, thanks a bunch!