CAI_Support
Senior Member
Ross, I am glad this new feature is useful for you.
Weighing Sensor for WebControl
- Thread starter ChubRock1
- Start date
I found a LM324 Quad OP Amp , would I be able to use that with the 5KG load cell I noted from Amazon? And if so, how would I wire it up? I know the load cell has the + and - Excite Volts (which hook to the 5V supply) and the + and - Sense Volts which need to go to the amp then to Webcontrol, just not sure how that should look?
LM2902 mentioned is really cheap on eBay, like this one 10 for $3.99 and free shipping:
http://www.ebay.com/itm/160599808101
or if you don't mind to wait little longer, from oversea shipping two for $0.99:
http://www.ebay.com/itm/221052730736
I did little search on web, here is a load cell wiring PDF:
http://loadcelltheor...ages/EleWir.pdf
Oh well, I did some search and found a lot of people using INA125 operation amplifier for this purpose and design can be really simple.
http://www.ebay.com/itm/160599808101
or if you don't mind to wait little longer, from oversea shipping two for $0.99:
http://www.ebay.com/itm/221052730736
I did little search on web, here is a load cell wiring PDF:
http://loadcelltheor...ages/EleWir.pdf
Oh well, I did some search and found a lot of people using INA125 operation amplifier for this purpose and design can be really simple.
Attachments
Here is the full datasheet for INA125:
http://www.ti.com/li...link/ina125.pdf
http://www.ti.com/li...link/ina125.pdf
I think that Rg is probably 60 Ohm based on table in datasheet FIG1
Suppose the output is 50mV with 10V power supply to INA125, to gain 1000 times to 5V, you will have accuracy about 1g. But deduct A/D accuracy, you still have about 1% accuracy or better. I am interested to see your actual result.
If power supply is 5V. you get max output 3.8V or so. To get bigger output, you will need to use higher supply voltage to this chip. I think the higher the voltage, the higher the accuracy.
This also work for thermal couple temperature measurement. I am also learn something today.
Suppose the output is 50mV with 10V power supply to INA125, to gain 1000 times to 5V, you will have accuracy about 1g. But deduct A/D accuracy, you still have about 1% accuracy or better. I am interested to see your actual result.
If power supply is 5V. you get max output 3.8V or so. To get bigger output, you will need to use higher supply voltage to this chip. I think the higher the voltage, the higher the accuracy.
This also work for thermal couple temperature measurement. I am also learn something today.
This load cell installation guide may be useful.
http://contactsystems.co.nz/files/Loadcell%20Guide.pdf
http://contactsystems.co.nz/files/Loadcell%20Guide.pdf
fwd03, that is a good pdf, thanks. So in your previous post are you saying if I use a 1k POT and a 5V supply I would get 3.8V output?
Here are the specs for my load cell: Rated Output: 1.0946mV/V; Lineraity: =<0.005%F.S; Hysteresis: =<0.005%F.S; Repeatability: =<0.005%F.S; Zero Balance: ±2%F.S; Temp Effect on Zero: =<0.005%F.S. 10°C;Temp Effect on Zero Output: =<0.005%F.S. 10°C; Creep: =<0.005%F.S./30minu; Crner Error: ±1uV; Input Impedance: 410±10?; Output Impedance: 350±3? ; Insulation: >5000M?/50V DC;
So is that 1.09mV output max or does that depend on the input voltage too?
Sorry for the questions, but as I said, I am over my head on this and trying to figure it all out.
Here are the specs for my load cell: Rated Output: 1.0946mV/V; Lineraity: =<0.005%F.S; Hysteresis: =<0.005%F.S; Repeatability: =<0.005%F.S; Zero Balance: ±2%F.S; Temp Effect on Zero: =<0.005%F.S. 10°C;Temp Effect on Zero Output: =<0.005%F.S. 10°C; Creep: =<0.005%F.S./30minu; Crner Error: ±1uV; Input Impedance: 410±10?; Output Impedance: 350±3? ; Insulation: >5000M?/50V DC;
So is that 1.09mV output max or does that depend on the input voltage too?
Sorry for the questions, but as I said, I am over my head on this and trying to figure it all out.
Actually, if you want to have accurate reading every time, the Rg should be a metal resistor, not pot. Pot is too easy to change its value.
Your load cell rated outout is 1.0946mv/V. That means if you use the 10V ref from that chip, wiring like Fig1 in page 10 of that TI datasheet, you will get max 10.946mV output with 5KG load. -- because that Fig 1 using 10V ref voltage. On page 11 of the INA125 datasheet, it stated that "Positive supply voltage must be 1.25V above the desired reference voltage." That means if you want to get 10V ref, your support voltage must be 11.25V and above.
Higher supply voltage will allow your load cell having higher output. If you power INA125 with 5V, The max Vref is 2.5V from that chip. Then your load cell output max will be 2.5V x 1.0946mV/V = 2.7365V. Your INA125 max output is 3.8V when its power supply is 5V. If you power the chip with 10V, you will get max output 7.6V. I think you want to use as high supply voltage as possible. For example, use 9V (or 12V whatever you are using) power that supplying the WebControl board.
Your max output voltage divided by the max load cell output, that is the gain you need to set. Say you use 9V power supply, and you will probably get about 6V or so max output from INA125. When using 9V, your Vref will be set to 5V, your load cell max output will be 5x 1.0946mv = 5.473mV. Then 6 / 0.005473 will be the gain you need to set. That is something like 1096. From the chart on the Fig 1, that Rg is less than 60 Ohm. On that same Fig 1 of the datasheet, you see there is a fomular:
G = 4 + 60K / Rg; from that, Rg = 60k/(G-4) = 60k / (1096 -4) = 54.95 Ohm.
Try to find a fixed value good quality metal film or wire resistor, that will help your setup accurate over the time.
Then, after all this working, you can put a known weight and read from WebControl analog input to see what is your known weight reading. It is better to have some kind of calibrate weight to find out their reading. Supposedly they are linear. With two known weights, you can figure out what is the curve looks like, then do a calculation in PLC code to determine the display value on VAR.
Your load cell rated outout is 1.0946mv/V. That means if you use the 10V ref from that chip, wiring like Fig1 in page 10 of that TI datasheet, you will get max 10.946mV output with 5KG load. -- because that Fig 1 using 10V ref voltage. On page 11 of the INA125 datasheet, it stated that "Positive supply voltage must be 1.25V above the desired reference voltage." That means if you want to get 10V ref, your support voltage must be 11.25V and above.
Higher supply voltage will allow your load cell having higher output. If you power INA125 with 5V, The max Vref is 2.5V from that chip. Then your load cell output max will be 2.5V x 1.0946mV/V = 2.7365V. Your INA125 max output is 3.8V when its power supply is 5V. If you power the chip with 10V, you will get max output 7.6V. I think you want to use as high supply voltage as possible. For example, use 9V (or 12V whatever you are using) power that supplying the WebControl board.
Your max output voltage divided by the max load cell output, that is the gain you need to set. Say you use 9V power supply, and you will probably get about 6V or so max output from INA125. When using 9V, your Vref will be set to 5V, your load cell max output will be 5x 1.0946mv = 5.473mV. Then 6 / 0.005473 will be the gain you need to set. That is something like 1096. From the chart on the Fig 1, that Rg is less than 60 Ohm. On that same Fig 1 of the datasheet, you see there is a fomular:
G = 4 + 60K / Rg; from that, Rg = 60k/(G-4) = 60k / (1096 -4) = 54.95 Ohm.
Try to find a fixed value good quality metal film or wire resistor, that will help your setup accurate over the time.
Then, after all this working, you can put a known weight and read from WebControl analog input to see what is your known weight reading. It is better to have some kind of calibrate weight to find out their reading. Supposedly they are linear. With two known weights, you can figure out what is the curve looks like, then do a calculation in PLC code to determine the display value on VAR.
@fwd03 - thanks again, you're a huge help. My INA125s should come this week so I can play with it then. My load cell can take an excite V of 5-10V, so I think I should try to use 10V to get the most variation in the range. So If I use a 12V supply to the INA125, I could use the 10V ref voltage to feed the load cell excite, or is Vref only to supply the INA125 chip?
So with 12 V supply to INA125 I should get 10Vmax from the chip. Also 10V load cell max it would be 10x 1.0946mv = 10.946mV. So gain would be 10/.010946 = 914
Rg = 60k/(914-4) = 65.93 Ohm?
Is that right? It just seems odd to me as I see other people set these up with the INA125 and they are using 1K Ohm for Rg.
So with 12 V supply to INA125 I should get 10Vmax from the chip. Also 10V load cell max it would be 10x 1.0946mv = 10.946mV. So gain would be 10/.010946 = 914
Rg = 60k/(914-4) = 65.93 Ohm?
Is that right? It just seems odd to me as I see other people set these up with the INA125 and they are using 1K Ohm for Rg.
INA125 has two parts in one chip. One part is to provide a stable reference voltage to power the load cell. That part output is 2.5V, 5V, or 10V. Your power to the chip will limit which output to select. Of course, there is another consideration. If I remember correctly the datasheet stated, the Vref output max current is 5mA. If your load cell is 410Ohms, there maybe better off to have additional circuit providing more current for the load cell, please look the Fig4 on page 12 of the datasheet.
The second part of INA125 chip is operational amplifier side that has no difference from LM2902, I think. Because no amplifier can output to the full supply voltage linearly, this INA125 is no exception. The datasheet stated somewhere when supply is 5V, its amplifier out[ut is max at 3.8V. You can proprotinally calculate the max output voltage based on your supply voltage. I think for WebControl max A/D input is 10V, having your most desired weight in the range about 2/3 of max A/D input will be good enough. Of course, if you can make 5KG weight output exactly 10V from INA125 that will be easier for calculation.
Since we already know 5V supply got 3.8V, when supply is 12V, we will get: 3.8V/5V = XV/12V, X = 9.12V. Then you calculate the gain and Rg. Your calculation looks good.
When Rg = 1K, your gain is less than 100. You will not read the max possible range analog to A/D. To get best accuracy, you want to use max Vref and max Vo to A/D.
The second part of INA125 chip is operational amplifier side that has no difference from LM2902, I think. Because no amplifier can output to the full supply voltage linearly, this INA125 is no exception. The datasheet stated somewhere when supply is 5V, its amplifier out[ut is max at 3.8V. You can proprotinally calculate the max output voltage based on your supply voltage. I think for WebControl max A/D input is 10V, having your most desired weight in the range about 2/3 of max A/D input will be good enough. Of course, if you can make 5KG weight output exactly 10V from INA125 that will be easier for calculation.
Since we already know 5V supply got 3.8V, when supply is 12V, we will get: 3.8V/5V = XV/12V, X = 9.12V. Then you calculate the gain and Rg. Your calculation looks good.
When Rg = 1K, your gain is less than 100. You will not read the max possible range analog to A/D. To get best accuracy, you want to use max Vref and max Vo to A/D.
The diagram you showed above looks more like an instrumentation amplifier than an opamp. (Yes, you can create an IA from Opamps, but they will usually have worst performance than a properly made one with precision on-chip matched components)
There are rail-to-rail opamps.
The ADC in the WC is actually 3.3V FSD. You can remove or replace the input resistors on the WC board to gain direct access to the 3.3V input, or make it 0-5V if you prefer. Any ADC is going to give you better results if you can arrange your inputs to have the maximum span. This includes offset zero if applicable. (Eg, if your output is going from say 3V to 6V (3V span), you would be far better off with a level-shift (make it 0-3V) and a gain of 3.3 (to make it 0-10V).
Also beware that in my own WC boards, I observed SEVERE crushing of the A/D input once you got to about 9V. This was because of the protection circuitry implemented by the CAI boys. I don't know if they've improved it in later revisions. I hope so
CAI_Support
Senior Member
Ross,
ADC input directly on chip is limited to the Vcc, which is 3.3V. The protection circuit is clamping about 3.2V after resistors, which is made of D8 and R31. I don;t know which revision boards do you have. For HW rev 2.2.2 board, the reading should be pretty linear. If you have older 2.0.2 rev boards, which was sold over 2 years ago, the R31 is too big, you could simply took out the D8 and R31, tied the middle point to Vcc(3.3V regulator) to get full 10V with protection.
ADC input directly on chip is limited to the Vcc, which is 3.3V. The protection circuit is clamping about 3.2V after resistors, which is made of D8 and R31. I don;t know which revision boards do you have. For HW rev 2.2.2 board, the reading should be pretty linear. If you have older 2.0.2 rev boards, which was sold over 2 years ago, the R31 is too big, you could simply took out the D8 and R31, tied the middle point to Vcc(3.3V regulator) to get full 10V with protection.
So in the PLC manual it says "Three analog inputs are available, each have an input range of 0 to 10Vdc" are you saying this isn't true? I was trying to get my output from the amp to be around 8-9V so I could use most of the 0-10V range, or are you saying that without changes the range is really only from 0-3.3V not 0-10V?
