I would like to ask for help regarding a certain phenomenon related to amplification by the Analog Devices AD620 instrumentation amplifier (device home page, device datasheet).

schematic diagram

Vcc is 5V.

The AD620 receives a differential voltage from the KMZ10B magnetoresistive sensor, which has a Wheatstone bridge construction.


I wanted to amplify voltage from KMZ10B (which is about several tens of millivolts). In my circuit this voltage depends on direction of the magnetic induction of the neodymium magnet. A direct voltage results gives such a characteristic:

enter image description here

On the above graph:

  • X-axis = Neodymium magnet angle (degrees)
  • Y-axis = voltage (V)

Then I decided to implement the voltage reading from AD620.


I used 4.7 kOhm resistor between pin 1 and 8. Datasheet of AD620 says that this resistance (called Rg) must be calculated like this:

Rg = 49.4 kOhm/(G-1)

G - gain

I used 4.7 kOhm resistor, so gain is 11.51. Reference voltage is about 2.5 V (from second op-amp: MCP6022) and voltage divider (half of 5V from Vcc)

I created a chart of this circuit (voltage measured between output of AD620 and GND) and it looks like below:

enter image description here

Now I decided to compare signals before and after amplifying.

I took the results in Excel made before using AD620, I multiplied them with Gain (11.51) and added a 2.5 volt reference (below).


It was supposed to show if the amplifier really strengthens as the calculations show. And look, I compared both charts:

enter image description here


  • Blue chart is a real, amplified signal
  • Orange chart is a calculated signal.


It looks like something's going wrong and in my opinion it depends on in-amp AD620. The signal rise up quickly and falls slowly. What can it be caused by?


1 Answer 1


In your diagram, you have placed your 47k resistor across the outputs of the Wheatstone bridge (+Vo and -Vo). This has loaded up the bridge, resulting in measurement error.

The whole reason we use instrumentation amplifiers like the AD620 is to provide a high input impedance and avoid loading effects.

Remove the 47k resistor from the +/- inputs of the AD620, and place it across the RG pins (pins 1 and 8 as per the datasheet).

Let us know if you're still having trouble afterwards.

  • \$\begingroup\$ Oh, I'm really sorry, my fault, it wasn't 47k - it's only 4,7k (missed comma). Let me correct this on schematics. Calculating (49400/4700 + 1) gave me 11,51 of gain. But about removing this Rg at all - are you sure? :( I need gain, because in next step this signal is measured in ADC in my ATmega8. Signal without amplifying could be too hard to measure on Atmega and that's the reason. \$\endgroup\$
    – Proboszcz
    Commented Sep 26, 2018 at 23:20
  • \$\begingroup\$ You aren't totally removing Rg, you are placing it across the dedicated Rg pins (pins 1 and 8). Right now it's across the input, which is incorrect. Placing it across the Rg pins will give you your gain and your necessary high impedance input. \$\endgroup\$ Commented Sep 26, 2018 at 23:27

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