Designing an IA

MrChips

Joined Oct 2, 2009
27,674
I have added an extra TL082 x100 gain stage. 2.5mV input gives 5V output as requested.
Note that R2 has been added to trim the overall gain. V1 and V2 are increased to 9V each to accommodate the higher output voltage.

U3A stage has gain of 1. You can increase this and reduce the gain in U4A.

TL082 IN-AMP x2000.jpg
 

Thread Starter

MMM**MMM

Joined May 30, 2022
45
hi MMM,
I have one of those, the elements are 1K each, use 5Vdc and it outputs 1mV/Volt.
So full scale 1kG gives 5mV.
They work well with the HX711 load cell amplifier and data output.

What are you driving with the INA amplifier output voltage.?
E
I have to design the amplification circuit, so I cannot use the HX711
 

Thread Starter

MMM**MMM

Joined May 30, 2022
45
hi MMM,
I have one of those, the elements are 1K each, use 5Vdc and it outputs 1mV/Volt.
So full scale 1kG gives 5mV.
They work well with the HX711 load cell amplifier and data output.

What are you driving with the INA amplifier output voltage.?
E
The full scale 1 KG gives 5.3 mV in my case instead of 5mV
 

ericgibbs

Joined Jan 29, 2010
16,761
Hi
Connect the R8 to 0v.
The V4 was used to check an offset, not used on your project.
>
What are you driving with the INA amplifier output voltage.?
E
@MMM**MMM
Connect your Load Cell as test voltage source

Note: V3 represents the voltage output of the load cell when it has a weight of1kG on it.
The Gain is set by R3 50 Ohms. to Gain= 400
 
Last edited:

WBahn

Joined Mar 31, 2012
27,866
Ok, I am working on a weight sensor, this sensor only gives a DC (using oscilloscope to view the signal, it was very close to a DC as long as the input is the same, when a change occurs in the input, the DC level will change but it is still DC). I just want to mention that there is no AC source, capacitors, or inductors, Thus, I think it is totally DC.
Again, many thanks for your contribution, this opens a new door for me to look for other ways :D
ANY change in a signal involves an AC component. In your case, any time the weight changes, the signal changes. The faster you want your system to respond to that change, the higher the frequency that the components of that signal change have.

How long after a change in the weight do you need the output signal of your amplifier to have stabilized at the new value? If it's a few seconds, then you only need a very low bandwidth of a few hertz. If you need it to respond in a few milliseconds, then you need several kilohertz of bandwidth.
 

ericgibbs

Joined Jan 29, 2010
16,761
The full scale 1 KG gives 5.3 mV in my case instead of 5mV
hi MM,
If you use a voltage source greater than 5V to power the Load Cell, it will give a Higher output than 5mV on full load of 1kG.
The nominal Full Load Spec is 1mV/V of Vexcitation
E
 

WBahn

Joined Mar 31, 2012
27,866
https://www.amazon.com/Uxcell-a14071900ux0078-Aluminum-Weighting-Sensor/dp/B00QC6M45A/ref=sr_1_4?crid=25F7VSTFIE9H4&keywords=load+cell+1kg&qid=1670492562&sprefix=load+cell+1k,aps,201&sr=8-4


The specs are written in the description of the product. I could not find its datasheet because I don't know the model of that weight sensor
What do you mean that you don't know the model of that weight sensor? It's right in the Amazon listing. It's a uxcell model YZC-131.

1670522790844.png
Go to the manufacturer's website and do a search for that model and you get.

https://www.uxcell.com/1kg-22lb-len...-cell-weighting-pressure-sensor-p-624187.html

1670522933089.png
 

Thread Starter

MMM**MMM

Joined May 30, 2022
45
hi MMM,
Please try this battery connection with your circuit +4.5 and -1.5V
Using your TL082 OPA's
V3 is a test 5mVolt voltage for the output of the load cell at 1kG load

Let us know what you see
E
R9,10,11,12 are the resistive elements of your load cell bridge

View attachment 282520
1670530817335.png

I think this is very similar to your circuit, still cannot get the output we need. What about the AD623AN and the multistage amplifier? I don't mind using both of them
 

Thread Starter

MMM**MMM

Joined May 30, 2022
45
Hi
Connect the R8 to 0v.
The V4 was used to check an offset, not used on your project.
>
What are you driving with the INA amplifier output voltage.?
E
@MMM**MMM
Connect your Load Cell as test voltage source

Note: V3 represents the voltage output of the load cell when it has a weight of1kG on it.
The Gain is set by R3 50 Ohms. to Gain= 400
What are you driving with the INA amplifier output voltage?
Briefly, a microcontroller.
More detailed answer:
This amplifier circuit will be one of many other circuits I will work on later on, then all of these will be cascaded, and the final output will be used as an input to a microcontroller. The microcontroller will be used to read the signal (by converting it to digital one), then the signal will pass on an error correction function, and at the end the microcontroller will display the output on a screen. One of the other circuits I will use is the comparator, just to make sure that the signal is within limit (from 0V to 5V). So, the INA amplifier is just a small part of a whole project.
 

Thread Starter

MMM**MMM

Joined May 30, 2022
45
ANY change in a signal involves an AC component. In your case, any time the weight changes, the signal changes. The faster you want your system to respond to that change, the higher the frequency that the components of that signal change have.

How long after a change in the weight do you need the output signal of your amplifier to have stabilized at the new value? If it's a few seconds, then you only need a very low bandwidth of a few hertz. If you need it to respond in a few milliseconds, then you need several kilohertz of bandwidth.
I am still confused, how does a weight sensor involve an AC component? The output is just due to the variations in resistances due to the deviation of the beam itself. So, there is no AC component, this is what I understand, please correct me if I am wrong.

"How long after a change in the weight do you need the output signal of your amplifier to have stabilized at the new value?" It doesn't matter, one second is fine, I don't need a very very high response.
 

MrChips

Joined Oct 2, 2009
27,674
There is common confusion in the use of the terms DC and AC.
In school we begin at knowing that DC means Direct Current and AC means Alternating Current.

As you progress in your electronics journey you come to learn that DC and AC have other usage.
In signal processing from a frequency perspective, all signals have DC and AC components.
DC refers to the component at 0Hz.
AC refers to all components that are not at 0Hz.

In every sensor signal, there is always DC and AC.
Even a DC power supply outputs DC voltage and AC voltage (ripple).
 

Audioguru again

Joined Oct 21, 2019
5,413
V5 should be the +4.5V and V9 should be the 1.5V supplies. You have them connected wrong!
No. A TL082 uses P-channel Jfets in its inputs and some of them Do Not Work if the input voltage is within 4V from the negative supply. It is recommended in Texas Instruments latest update to have a minimum supply that is +5V/-5V.
 

WBahn

Joined Mar 31, 2012
27,866
I am still confused, how does a weight sensor involve an AC component? The output is just due to the variations in resistances due to the deviation of the beam itself. So, there is no AC component, this is what I understand, please correct me if I am wrong.

"How long after a change in the weight do you need the output signal of your amplifier to have stabilized at the new value?" It doesn't matter, one second is fine, I don't need a very very high response.
Any signal can be represented as the superposition of a spectrum of sinusoidal waveforms of different amplitudes and frequencies from zero hertz to infinite hertz. The only way for there to be NO energy at ANY frequency other than 0 Hz (also known as "DC") is if the signal has existed since the beginning of time, has never changed, and will never change until the end of time. ANY signal that has ever or will ever undergo any change, has non-zero content at non-zero frequencies.

Look into Fourier series for more information.

If one second is fine, then you don't need much bandwidth (again, a few hertz), so you shouldn't be impeded too much by gain-bandwidth products.
 

ericgibbs

Joined Jan 29, 2010
16,761
V5 should be the +4.5V and V9 should be the 1.5V supplies. You have them connected wrong!
.
Hi Sarah,
You are correct, :) his drawing is Wrong in showing the power supply connections.

Others posting about PChannel FET's is not related to the TS's problem, the TS has the supplies reversed,
E
 
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