AD8220

Question

Hello! 
 In a circuit I use AD8220 for measuring the sensor resistance between 100k ohm to 200k ohm. The problem is that I only get a good linearity from 100k to 110k at the output. Over 120k the output saturates. 
 How can I solve the circuit so a can get an linear output for 100k to 200k? 
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Arsol · Accepted Answer

Hi SMENA 
 Thank you for reaching out and sharing your inquiry. 
 To be able to recommend the correct recommendation, I would like to know the different design requirements of your application. 
 Isolating other components that may affect the AD8220 performance, Please cut the connection from Pin7 to (10uF cap and 10Kohm res) and confirm if this is correct.

What is your Sensor (s1) and intended applications? 
 Basing from your application circuit, AD8220 is using a single operation; +Vs =5V and -Vs =0V? 
 Vref (pin6) = 2.5V if variable resistor is 0ohms? If var res is 100K then, Vref = 2.62V?

Intended Gain =1?? 
 If S1 (Sens) = 100Kohms, what is the value for +In (Pin4), is this 2.5V? 
 If S1 (Sens) = 200Kohms, what is the value for +In (Pin4), is this 3.33? 
 
 Thanks and Best Regards,)

SMENA · Answer

Hello! 
 Thank you for helping me 
 I have disconnected from Pin 7. 
 The sensor is a gas sensor which changes the impedance depending on the concentration of the gas. For no gas it is 100k ohm for maximum gas concentration it is 200k ohm. 
 Yes I use single operation 5V. +Vs = 5V and -Vs = 0V 
 Vref (pin6) is used for offset trimming. 2.37V for 0 ohm and 2.62V for 100k. The offset shall be trimmed so between Earth ground and Vout (Pin7) shall be less then +/-1mV. 
 There is no resistor between Pin2 and Pin3. Gain = 1. 
 If S1(Sens) = 100k ohms +In (Pin4) is 2,503V and output Pin7 is 2,470V 
 If S1(Sens) = 200k ohms +In (Pin4) is 3,27V and output Pin7 is 2.632V 
 Why is the output so low for 200k? it should be at least 3,2V 
 What have I done wrong?

ADIApproved · Answer

Please let us know if you can provide these details being requested on your end. For the team to understand your inquiry further and be able recommend the suitable answers.

Arsol · Answer

Hi SMENA 
 Thank you for providing these design requirements for your intended gas-sensing applications. 
 I have additional questions on my end to confirm several items; please see below: 
 
 You mentioned that the applications will be used to measure the gas level/ concentration. Will it be for all gases, or is it only for specific types of gas? 
 For the sensor, is there a chance you can check what the typical tolerance (resistance equivalent) is? +/-5, +/-10%, etc.. 
 Can you please double-check the values on your -In (Pin1) for both conditions? If S1 =100K and S1=200K. 
 
 For the inquiries on your applications, 
 
 Vref (pin6) is used for offset trimming. 2.37V for 0 ohm and 2.62V for 100k. The offset shall be trimmed so between Earth ground and Vout (Pin7) shall be less then +/-1mV. 
 
 ArSol: The Vref should be half of 5V when the 100K is set to 0. Voltage divider = 5V * (1M/2M). There might be line parasitics that affect your Vref value and another reason is the resistor tolerances of the 1Mohm that you used. 
 
 If S1(Sens) = 100k ohms +In (Pin4) is 2,503V and output Pin7 is 2.470V and If S1(Sens) = 200k ohms +In (Pin4) is 3,27V and output Pin7 is 2.632V 
 
 ArSol: If S1 =100K. I&rsquo;m seeing 30mV offset on the output. Considering both inputs should be equal (2.5V), and your Vref set is 2.5V. Again, this might be the effect of the tolerances used in 1Mohm. 
 ArSol: If S1 =200K, the amplifier starts to saturate, at ambient temp, the maximum allowable Input voltage range on both pins (1, 4) is +Vs -2. In this case, it would be 3V only.

My recommendations 
 To satisfy the input voltage range (IVR) of AD8220, If you can modify the divider ratio on the -In (pin1) side; suggest using 1/3 ratio ( 1K||2K)

Better to use a precision resistor (find for 1% value) for 1M and resistors on the right-left side of the bridge. 
 If any chances there are precision varistor for your 100K, it will helps

I hope these suggestions will suit your requirements. 
 Regards

Arsol · Answer

Hi SMENA 
 Thank you for sharing the instruments that you are using. 
 I forgot to mention that we should keep the node voltage on the right side of the bridge (+IN; pin4) < (+Vs -2V) to prevent the amplifier from going into saturation. Hence, I propose two (2) conditions: varying the value of R1 and R3. I reattached your design applications schematics here and added several designators to ensure everything is clear. Please see below:

For clarity, I made a table correlating the expected vs the measured values. Please retake the measurements using the conditions below:

Condition 1

EXPECTED

MEASURED

R1 = 1K , R2 =1K , R3 = 200K , VR1 =0 Ohms

+IN

-IN

REF

OUT

+IN

-IN

REF

OUT

S1(min) = 100K

1.666667

2.5

1.666667

S2(max) = 200K

2.5

Condition 2

EXPECTED

MEASURED

R1 = 2K , R2 =1K , R3 = 200K , VR1 =0 Ohms

+IN

-IN

REF

OUT

+IN

-IN

REF

OUT

S1(min) = 100K

1.666667

2.5

S2(max) = 200K

2.5

1.666667

2.5

3.333333

I hope this will help you 
 
 Regards

ADIApproved · Answer

Hi SMENA 
 
 Our engineers are currently looking into this query. We will get back to you as soon we can. Thank you for your patience. 
 
 -L

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