Capacitor in force sensor circuit

Thread Starter

shivasage

Joined Jan 12, 2020
5
(Apologies if this is the wrong forum)

Hello. I'm planning a project which uses a force sensitive resistor to control voltage. I'm using this schematic. I understand it for the most part, but I'm confused by the capacitor C1. What is its function?

MOD: Added image clip.
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pmd34

Joined Feb 22, 2014
503
Hi! C1 is a common element on amplifiers, it basically shorts out any high frequency "noise". Effectively it is like having a very low value feedback resistor for high frequencies.
 

Reloadron

Joined Jan 15, 2015
4,873
Along with Rfeedback it becomes a part of a low pass filter. What pmd34 just said.

Since the operational amplifier is single supply if R1 & R2 are the same value they create a 1/2 Vsupply bias (offset). That's about it.

Ron
 

Thread Starter

shivasage

Joined Jan 12, 2020
5
Hi! C1 is a common element on amplifiers, it basically shorts out any high frequency "noise". Effectively it is like having a very low value feedback resistor for high frequencies.
Along with Rfeedback it becomes a part of a low pass filter. What pmd34 just said.

Since the operational amplifier is single supply if R1 & R2 are the same value they create a 1/2 Vsupply bias (offset). That's about it.

Ron
Ok, that makes sense! I play guitar and capacitors are used as low-pass filters for the tone control, so that is not unfamiliar to me.

I have another question regarding current. I understand that R1 and R2 are creating a voltage divider that is dependent on their ratio. That means I can have higher or lower value resistors but still have the same ratio. My understanding is that with higher values I will have lower current. I read that for the force sensor I am using, I should keep current under 1mA to avoid overheating.

My question is, am I trying to get close to 1mA, but under it? What's stopping me from having an incredibly small current? For example, if I choose R1 = 800k and R2 = 100k, with Vsupply = 9V for a desired Vref of 1V, I will have 9 / (800k + 100k) = .01mA current. But I could also use R1 = 80k and R2 = 10k which would yield a current of .1mA. What is the difference?
 

Reloadron

Joined Jan 15, 2015
4,873
Here is what is going on keeping it simple. Using your schematic and R1 and R2 as a divider network. Let's make Vsupply 10 volts. Now with R1 and R2 the same value then Vout at their junction will be 5 volts. There are a few caveats to this. The higher the combined R1 and R2 resistance (Rtotal) the lower the circuit current since voltage and current are inversely proportional. So if I make each resistor 5 Ohms then Rtotal = 10 Ohms and my applied voltage is 10 volts my current would be 10 volts / 10 Ohms = 1.0 amp so a 5 Ohm resistor with 1 amp current flow = 1 * 5 = 5 Watts power to dissipate, I need a heck of a high power resistor. How about we use 5,000 Ohms and 5,000 Ohms so Rtotal is now 10,000 Ohms and now 10 Volts / 10,000 Ohms = 0.001 Amp or 1 mA. The higher the current the more heat generated to a point where the resistors begin to have a "self heating" effect their values change and things get ugly. Another consideration is in this case the operational amplifier input impedance is in parallel with R2. However, since the operational amplifier input impedance is very, very high it is of no consequence.

Yes, the divider output is ratio metric. If R1 were 9.0K and R2 were 1.0K the Vout would be 1.0 volt with 10 volts applied. Voltage Dividers is a more in depth read on the subject. :)

Ron
 

Analog Ground

Joined Apr 24, 2019
271
The selection of the range of R1 and R2 resistance values will have no effect on the voltage on the sensor. You are correct that it is the ratio of R1 to R2 which sets the voltage. Specifically, the voltage on the sensor will be R2/(R1+R2) * Vsupply. In other words, the op amp will make the voltage on the sensor be the same as the voltage at the connect between R1 and R2. The current through the sensor is then the voltage divided by the sensor resistance.

You are also correct that R1 and R2 can be much larger values to save power. How big? One factor is the current through R1 and R2 should be a lot greater than the "bias" current which goes into the input of the op amp. For your op amp, this current is very, very low. Like 1 picoamp or 1E-12 amps. R1 and R2 can be in the megaohm range with no problem. Note, going much larger than the sensor resistance does not gain much. All your power consumption will then be in the sensor.

That said, going larger does increase the amount of internal noise in R1 and R2. This is a whole other topic but you probably don't care too much about it.
 

Analog Ground

Joined Apr 24, 2019
271
An improvement to the circuit would be addition of a capacitor across R2. This forms a low pass filter for any noise from Vsupply which will get into your circuit. The filter will also reduce internal noise from R1 and R2. The frequency response of the filter can be low and will not effect the frequency response of the circuit for the sensor signal. The cap across R2 is a simple win.
 

Thread Starter

shivasage

Joined Jan 12, 2020
5
How does the current through R1+R2 relate to the current through the sensor? Is the current on the sensor entirely independent, so Vout divided by Rsensor? I'm confused because this:
All your power consumption will then be in the sensor.
...makes it sound like the current through R1+R2 is related to the current through the sensor, but this:
The current through the sensor is then the voltage divided by the sensor resistance.
...makes it sound like it's independent.

I'm also wondering how Rfeedback affects the circuit. I found an almost identical circuit where it seems the only difference (besides no capacitor) is that Rfeedback (in this case labeled "gain control resistor") is in series with Vout and the sensor, whereas in the first circuit it's in parallel. How does this affect things?
 
How does the current through R1+R2 relate to the current through the sensor? Is the current on the sensor entirely independent, so Vout divided by Rsensor? I'm confused because this:...makes it sound like the current through R1+R2 is related to the current through the sensor, but this:...makes it sound like it's independent.

I'm also wondering how Rfeedback affects the circuit. I found an almost identical circuit where it seems the only difference (besides no capacitor) is that Rfeedback (in this case labeled "gain control resistor") is in series with Vout and the sensor, whereas in the first circuit it's in parallel. How does this affect things?
What your asking is difficult to explain without an understanding of the basics of how op amps circuits work. Be careful when trying to "compare" one circuit with another. It is just not possible to answer your questions without getting into op amp basics which is too involved for a forum answer. I will provide one hint. Most op amp circuits will try to keep the voltage on the + and - inputs the same. So, the voltage on the + input, Vref, will "appear" on the - input. This is what is meant by the voltage in the resistor divider is the same as the voltage on the sensor. The current through R1 and R2 is not relevant. It is the voltages at the op amp inputs.
 

MisterBill2

Joined Jan 23, 2018
4,846
Hi! C1 is a common element on amplifiers, it basically shorts out any high frequency "noise". Effectively it is like having a very low value feedback resistor for high frequencies.
The addition of the capacitor in the negative feedback loop is to reduce the gain as the signal frequency increases. The AC gain will be lower than the DC gain.
 

Sensacell

Joined Jun 19, 2012
2,544
Force-sensitive-resistors are horrible unstable, non-repeatable monsters.

You almost never see them used in commercial applications- just FYI before you go too deep with the details.
 
This is the type of circuit I would tell a student new into electronics to hook it up to a scope and observe the results with and without a capacitor. Try different sizes of capacitors as well.

Senseacell - "Force-sensitive-resistors are horrible unstable, non-repeatable monsters " - Agreed
 

Thread Starter

shivasage

Joined Jan 12, 2020
5
The addition of the capacitor in the negative feedback loop is to reduce the gain as the signal frequency increases. The AC gain will be lower than the DC gain.
What if the signal is DC? (Though I think that schematic is assuming AC; it gives two options).
Force-sensitive-resistors are horrible unstable, non-repeatable monsters.

You almost never see them used in commercial applications- just FYI before you go too deep with the details.
This is a musical application, so there's room for error. That being said, could an external potentiometer be used at one point in the circuit to adjust for things 'drifting' later on? Or perhaps two potentiometers, one to account for drift and the other to fine-tune the sensitivity/response of the sensor? I guess Rfeedback is used to fine tune the sensor, so perhaps a pot in place of R1 and/or R2 to account for drift?
 
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Y
Force-sensitive-resistors are horrible unstable, non-repeatable monsters.

You almost never see them used in commercial applications- just FYI before you go too deep with the details.
Could you be more specific? What do you mean by "unstable"? For example, after removing a load, the resistance does not return to the same value? Or, does the application of the same load not produce the same change in resistance? Or, is it sensitive to where the load is applied on the surface? Does it cause instability in a circuit because it has a large capacitance? Etc.....

Here is a spec table for one of the sensors. Is there a particular spec which is incorrect?

FlexiSpec.jpg
 
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Sensacell

Joined Jun 19, 2012
2,544
Y

Could you be more specific? What do you mean by "unstable"? For example, after removing a load, the resistance does not return to the same value? Or, does the application of the same load not produce the same change in resistance? Or, is it sensitive to where the load is applied on the surface? Does it cause instability in a circuit because it has a large capacitance? Etc.....
All except the last one.

It will never settle back to the previous starting value.
It will never produce the same output, even with the same load.
It very much depends on where you apply the force.
It will age poorly, all parameters drift.


Just think of how many times you have seen these used in real-world products and systems.
Low expectations are in order.
 
All except the last one.

It will never settle back to the previous starting value.
It will never produce the same output, even with the same load.
It very much depends on where you apply the force.
It will age poorly, all parameters drift.


Just think of how many times you have seen these used in real-world products and systems.
Low expectations are in order.
But is it out of spec? The spec sheet makes it pretty clear they have accuracy and repeatability in the 5% range and low matching between units (requiring calibration). Does your experience disagree with the specs?

"It will never settle back to the previous starting value". - Hysteresis: In the spec
"It will never produce the same output, even with the same load". - Repeatability: In the spec.
"It very much depends on where you apply the force". - Not spec'd. Contact manufacturer. Sensitivity to mechanics seems reasonable given it is a "sheet effect" device.
"It will age poorly, all parameters drift". Temperature drift: In the spec. Aging: Not in the spec, how much? Contact manufacturer. Does it delaminate? What is "poorly".

IMHO, not fair to trash a part unless something is out of whack compared to the specs. "Sensacell" makes a contact sensitive product but just because it did not work for your application does not mean it is not suitable for another product. We are engineers, not marketing. We need data, not opinions.
 
You almost never see them used in commercial applications- just FYI before you go too deep with the details.
"Located in the Innovation District of Boston, Tekscan has been in business since 1987. During that time, we've perfected our product, our team and our process to ensure we meet our customers' expectations. Today, the team is over 100 employees strong." - Tekscan "About Us"

Somebody must buy their products or there are over 100 people working for free.
 

Sensacell

Joined Jun 19, 2012
2,544
IMHO, not fair to trash a part unless something is out of whack compared to the specs. "Sensacell" makes a contact sensitive product but just because it did not work for your application does not mean it is not suitable for another product. We are engineers, not marketing. We need data, not opinions.
Your point is taken. I did not look at the datasheet- its performance is better than I expected.

It's not my intention to "trash" other products, but to encourage realistic expectations.
People tend to have rosy ideas about what sensors can do, especially when you read a "sell sheet" describing application scenarios for some new piece of kit.
I always approach any piece of technology with a healthy level of skepticism, the edge conditions and shortcomings are always hidden in the datasheet for you to trip over when you get the actual thing working.
 

MisterBill2

Joined Jan 23, 2018
4,846
In the application identified by the TS as a musical device, exact repeatability is certainly not a requirement. Mostly without exceptions musical instruments are played by ear to some extent. That is why there are no deaf violinists. Certainly the force dependent resistor is not suitable for an accurate scale to measure weights precisely, but it will be totally satisfactory for "louder" and "softer" control where the operator is in the feedback loop. Using the right tool for the job, or in this case, using the correct component for the job.
 
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