Use AD8295 for Loadcell Amplifier Board

Hello, I'm making a loadcell amplifier board. Input from -18mVDC to +18mVDC. Output +-2.5V. Input signal frequency from 0 to 14Hz. I have made a design as the attach. The total gain changes from about 140 to 50 according to real condition.

- First, I don't know why the output is very fluctuated and gaining is wrong.

- Second, I intend changing the low-pass filter to pre-amp + filter. The IA will became to post-amp.

Anyone help me what is the wrong in my design and how to solve it.



  • 0
    •  Analog Employees 
    on Jan 23, 2015 2:27 AM over 5 years ago

    Hi Ahncao,

    I'm sorry for the delay. The AD5259 is a 5.5V maximum device. Solid-state switches can't pass signals beyond their supplies. You can see from the data sheet that the Voltage Range at A, B, and W is 0 to Vdd.

    The AD8295 in-amp, on the other hand, has Superbeta NPN transistors at the inputs, meaning that the voltage at the Rg pin is equal to the voltage at the corresponding input minus about 0.5V. If your bridge is perfectly balanced, the common-mode voltage is 6V and your Rg pins each have about 5.5V. If this is beyond the AD5259 supply, fault current will flow and change the output of the AD8295.

    There are a few more common problems with a circuit like this as well. Have a look at this article for more info:

    Use a PGIA to avoid getting burned by switch parasitics

    If you could run your bridge on lower voltage, something like AD8556 or AD8557 programmable bridge sensor amplifiers might make a lot of sense.

    Otherwise, I believe the best solution is to take the minimum gain that you will need with the AD8295 instrumentation amplifier, and implement the PGA with the uncommitted op-amp, taking into account the voltage range and suggestions from MT-088 (specifically figure 28).

    I hope this helps.

    Best regards,


  • Thank you Scott,

    I intend use +-5V (total diff 10V) for loadcell. So the maximum common-mode voltage is about +-18mV. The voltage regulator is 7805 and 7905. What is the voltage at the Rg if input voltage is very low (mV) ? I haven't found any information about this in AD8295 at datasheet.

    I'm considering the AD8231. Can I use AD8231 as an pre-amp for the board? If the supply voltage is +-2.5V, the input voltage range is about +-2.2V ? I'm not sure this because I can't find the spec for dual-supply mode.



  • 0
    •  Analog Employees 
    on Feb 10, 2015 2:01 AM over 5 years ago

    Hello Anh,

    I am sorry once again for the delay.

    Assuming that the AD8295 in-amp inputs and outputs are within a working range, the Rg pin will be a small diode drop ~0.5V below the corresponding input. So with both inputs at a voltage near 0V, both Rg pins will be at about –0.5V. You can observe this with a DVM if you disconnect the digipot from the Rg pins. The internal feedback conditions for the in-amp ensure that this voltage drop is constant, which imposes your differential input voltage across Rg. See the AD8221 data sheet for a more detailed drawing of a similar instrumentation amplifier.

    The AD8231 input voltage range is relative to the supply, no matter whether it is a single supply or a dual supply. Looking at the +5V single supply table, you could rewrite the input range as -Vs + 0.05V to +Vs - 0.05V. ±2.5V is the same voltage span as +5V, so it will translate directly as -2.45V to +2.45V.

    To check all of the in-amp headroom limits for the AD8231, including input voltage range, output voltage range, and internal nodes, you can have a look at our in amp tool: Instrumentation Amplifier Tool - Common Mode vs. Output Voltage (Diamond Plot)

    I hope this helps.

    Best regards,