force + meter

[pic16]

Hi guys,
I would like to do a DIY meter which could used to measure the force.

The meter i use is copy from the website below.
http://www.circuitstoday.com/wp-content ... ircuit.png
It actually was voltmeter built from ICL7107.

So, now i try to build a circuit which will give me "with every 1kg(force) it product 1 volt"
The sensor i'm used now are SN-FRC-0.5(from cytron), i have build the circuit, but it give me weird output
i'm using 741 op-amp as FSR voltage divider(follow datasheet) and i'm have no idea what i did is right or wrong

Could any expert here give me some idea to correct it?

[ABSF]

Hi pic16, its been a long time. Welcome back.

I read the datasheet of the FSR and i was somehow confused. When a weight of 1 Kg is applied, what is the resistance you get? One fig says 750 ohm while another fig says 1.2K ohm. So what do you get?

Show us your 741 circuit. Which figure did you use?

Allen

[pic16]

Hi pic16, its been a long time. Welcome back.

I read the datasheet of the FSR and i was somehow confused. When a weight of 1 Kg is applied, what is the resistance you get? One fig says 750 ohm while another fig says 1.2K ohm. So what do you get?

Show us your 741 circuit. Which figure did you use?

Allen

Hi ,
i didn't have the weight of 1kg, so i use 1.5kg.
It show me 12.2kohm


i'm refer the circuit at page 18
http://www.freeimagehosting.net/pb9g4

[ABSF]

I use the simulator to emulate your circuit and I got 0.55V from the output.

Lets calll the resistance Rfsr, the resistor that you added as R1 and the output voltage Vout.
Vout=[R1 / (R1+Rfsr)]* 5V = [1.5k/13.7K]*5 = 0.547V.
Since you want to get 1V per kg, and the weight now is 1.5Kg, you should be getting 1.5V to satisfy your requirement, right?
So how to get 1.5V from the above circuit? From the formula,
Vout=[R1 / (R1+Rfsr)]* 5V = 0.55V

Vout=[R1 / (R1+12.2K)]* 5V
If Vout=1.5 then
1.5=[R1 / (R1+12.2)]*5
1.5/5=R1 / {R1+12.2)
0.3(R1 + 12.2) = R1
R1=0.3R1+3.66
R1-0.3R1=3.66
0.7R1=3.66
R1=5.228K so using 5.1K as R1 would get you 1.5V

Allen

[pic16]

FSR.JPG

I use the simulator to emulate your circuit and I got 0.55V from the output.

Lets calll the resistance Rfsr, the resistor that you added as R1 and the output voltage Vout.
Vout=[R1 / (R1+Rfsr)]* 5V = [1.5k/13.7K]*5 = 0.547V.
Since you want to get 1V per kg, and the weight now is 1.5Kg, you should be getting 1.5V to satisfy your requirement, right?
So how to get 1.5V from the above circuit? From the formula,
Vout=[R1 / (R1+Rfsr)]* 5V = 0.55V

Vout=[R1 / (R1+12.2K)]* 5V
If Vout=1.5 then
1.5=[R1 / (R1+12.2)]*5
1.5/5=R1 / {R1+12.2)
0.3(R1 + 12.2) = R1
R1=0.3R1+3.66
R1-0.3R1=3.66
0.7R1=3.66
R1=5.228K so using 5.1K as R1 would get you 1.5V

Allen

i adjust to the resistance you have calculated, but when i test for 200g, it give me 0.84 volt

[ABSF]

Based on the figure inside the datasheet above. Using 3K for RM, you'll get 2V for 1Kg. But for 500g, it was about 1.5V and 200g is 1V. So you see the FSR is not a linear device. You can use it as a rough estimate of the weight you're going to measure but you cant use it as a scale. And I now know how you get the 1.5K resistor from.

You thought that 3K gave you 2V, then 1.5k should give you 1V, right? But if you look at the graph more carefully. Putting 10K (3.3 times 3K) only gives you 3.5V not 6V with a weight of 1Kg as you expected.

Allen

[pic16]

FSR Fig 9.JPG

Based on the figure inside the datasheet above. Using 3K for RM, you'll get 2V for 1Kg. But for 500g, it was about 1.5V and 200g is 1V. So you see the FSR is not a linear device. You can use it as a rough estimate of the weight you're going to measure but you cant use it as a scale. And I now know how you get the 1.5K resistor from.

You thought that 3K gave you 2V, then 1.5k should give you 1V, right? But if you look at the graph more carefully. Putting 10K (3.3 times 3K) only gives you 3.5V not 6V with a weight of 1Kg as you expected.

Allen

Too bad... i thought i could make a mini weight scale by combine this two circuits.
Then, is that any idea to create some circuit could make it become linear?

[ABSF]

Too bad... i thought i could make a mini weight scale by combine this two circuits.
Then, is there any way to create some circuits that could make it to become linear?

Yes, there is. But it wont be easy. See the discussion here..
http://forum.allaboutcircuits.com/archi ... -3961.html

You can google with "analog linearization technique" if you are really interested..

Another way is to use the "digital linearization technique using ADC, microcontroller(with look up table in EPROM) and DAC". This technique can be used with PIC that has an ADC channel. The output can be displayed directly after processing on LCD instead of sending to a DAC.

Allen

[Idris]

I have try that sensors..
This is the video
http://mytechinvention.blogspot.com/201 ... cator.html
This is the graph relationship between Force VS Resistance
http://3.bp.blogspot.com/-SJje21qKq8s/T ... nceFSR.png

[pic16]

I have try that sensors..
This is the video
http://mytechinvention.blogspot.com/201 ... cator.html
This is the graph relationship between Force VS Resistance
http://3.bp.blogspot.com/-SJje21qKq8s/T ... nceFSR.png

using microcontroller?