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AD620 Instrumentation Amplifier Usage

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corvese210

New Member
Good Afternoon,

I am using a AD620 instrumentation amplifier for the purpose of subtracting two voltages and scaling it by 20. I am having problems with the functionality of this IC.

The data sheet can be found here:
http://www.electro-tech-online.com/custompdfs/2010/04/AD620.pdf

My circuit has a 2.6K resistor between pins 1 and 8 (according to pg 16 of the data sheet this will give me a gain of 20), I have a voltage source input to the inverting input at Pin 2 set at 1V. I have another voltage source at the non-inverting input Pin 3 set to 2V. I have a +/-15V power supply connected to the supply pins at 7 and 4, respectively. The grounds of my two voltage inputs are connected to the ref pin 5. I am measuring the output on a multimeter with probes on pins 6 (output pin) and pin 5(ref pin).

My multimeter is giving an ouput of 2.2V when I am assuming it should be 20V. (2V minus 1V with a gain of 20).

Has anyone used this IC before or looked at this data sheet to see what I am doing wrong?

Any help would be greatly appreciated.
 

ericgibbs

Well-Known Member
Most Helpful Member
Good Afternoon,

I am using a AD620 instrumentation amplifier for the purpose of subtracting two voltages and scaling it by 20. I am having problems with the functionality of this IC.

The data sheet can be found here:
http://www.electro-tech-online.com/custompdfs/2010/04/AD620-1.pdf

My circuit has a 2.6K resistor between pins 1 and 8 (according to pg 16 of the data sheet this will give me a gain of 20), I have a voltage source input to the inverting input at Pin 2 set at 1V. I have another voltage source at the non-inverting input Pin 3 set to 2V. I have a +/-15V power supply connected to the supply pins at 7 and 4, respectively. The grounds of my two voltage inputs are connected to the ref pin 5. I am measuring the output on a multimeter with probes on pins 6 (output pin) and pin 5(ref pin).

My multimeter is giving an ouput of 2.2V when I am assuming it should be 20V. (2V minus 1V with a gain of 20).

Has anyone used this IC before or looked at this data sheet to see what I am doing wrong?

Any help would be greatly appreciated.
hi,
The gain of 20 is from respect to 0V, so the +/15V supply will only allow an output swing of about +/-13V.
The AD620 in your app is saturated.
Do you follow.?
 

ericgibbs

Well-Known Member
Most Helpful Member
Not exactly....is it possible to make this work how I described?
ok,
Set the gain for 10 [ as a test] the output should show +10v.

Reverse the AD620 inputs and now the output will be -10v.
 

corvese210

New Member
ok,
Set the gain for 10 [ as a test] the output should show +10v.

Reverse the AD620 inputs and now the output will be -10v.
I substitued a resistance of 5.5K into the circuit between pins 1 and 8. According to the data sheet this should give me a gain of 10. I am still inputting 2V and 1V into the circuit. I am reading an output of 2.1V now.

When I remove the resistors between 1 and 8 I am getting a gain of 1. This is correct operation according to the datasheet (resistance = infinity, gain = 1). It is working properly until I try to change the gain with resistors.
 

ericgibbs

Well-Known Member
Most Helpful Member
I substitued a resistance of 5.5K into the circuit between pins 1 and 8. According to the data sheet this should give me a gain of 10. I am still inputting 2V and 1V into the circuit. I am reading an output of 2.1V now.

When I remove the resistors between 1 and 8 I am getting a gain of 1. This is correct operation according to the datasheet (resistance = infinity, gain = 1). It is working properly until I try to change the gain with resistors.
hi,
The REF pin should be connected to 0V in your app.

EDIT:
I now see in your first post that the Ref is at 0V or is it, I cannot be sure I follow your description.

Do you have a circuit diagram you could post.?
 
Last edited:

ericgibbs

Well-Known Member
Most Helpful Member
How can I do this without using another power supply?

Right now I have the ground of the input power supplys connected to the REF pin.
hi,
This is why a circuit diagram can help to avoid confusion, I use IA quite often, so I can help.

The signal inputs of the AD620, how do you have them connected.
There must be a 0V path for the input signals.

The Ref pin is only there to provide an OFFSET on the output voltage, you should connect to 0V.

Do you have a diagram to post.?:)
 
Last edited:

corvese210

New Member
hi,
This is why a circuit diagram can help to avoid confusion, I use IA quite often, so I can help.

The signal inputs of the AD620, how do you have them connected.
There must be a 0V path for the input signals.

The Ref pin is only there to provide an OFFSET on the output voltage, you should connect to 0V.

Do you have a diagram to post.?:)
Thanks for your help...here is my circuit diagram:
 

Attachments

crutschow

Well-Known Member
Most Helpful Member
Where is the common of your power supply connected? It needs to go to circuit common.
 
Last edited:

ericgibbs

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hi,
You must have the 0V to the +/-15V supplies.
Look at this edit.

Ad620...jpg
 

corvese210

New Member
Hi...like i mentioned this circuit works perfectly now but I have another question...

The final piece of my design is to limit this output of the IA between 0.5V and 14V...I had another post on this but this question relates to the IA.

I noticed that running the IA at +/-15V that the output maximum was being limited at around 14.1V already...is it possible to use power the rails of this IA with 15V and 0.5V just so the output would not swing above or below these vales?

Also, could this IA be powered using a single supply (15V and 0V)?
Thanks for any help you can give!
 

crutschow

Well-Known Member
Most Helpful Member
I noticed that running the IA at +/-15V that the output maximum was being limited at around 14.1V already...is it possible to use power the rails of this IA with 15V and 0.5V just so the output would not swing above or below these vales?

Also, could this IA be powered using a single supply (15V and 0V)?
Thanks for any help you can give!
Yes, lowering the supply voltage will limit the output voltage correspondingly. But you must follow in the input voltage limits.

As is frequently mentioned for this question, you must look at the common-mode range of the amplifier to determine how close to the supply rails it will operate. If you look at the "Input Voltage Range" on Pg. 3 of the AD620 data sheet, it states the limits. It varies from +1.9 to +2.1V above the minus supply voltage and -1.2 to -1.4V below the positive supply voltage. Thus if you powered if with 15V and 0V, then the voltages should be kept between about +2.1V to +13.6V for proper amp operation.
 

ericgibbs

Well-Known Member
Most Helpful Member
Hi...like i mentioned this circuit works perfectly now but I have another question...

The final piece of my design is to limit this output of the IA between 0.5V and 14V...I had another post on this but this question relates to the IA.

I noticed that running the IA at +/-15V that the output maximum was being limited at around 14.1V already...is it possible to use power the rails of this IA with 15V and 0.5V just so the output would not swing above or below these vales?

Also, could this IA be powered using a single supply (15V and 0V)?
Thanks for any help you can give!
hi,
If you want to operate the AD620 from a single supply, look at page 12 of the datasheet.
It does not work well with a single supply.
You can move the 'centre line' of the AD620 swing by using the REF pin.


EDIT:

look at the AD623 for single supply operation
 
Last edited:

ericgibbs

Well-Known Member
Most Helpful Member
The final piece of my design is to limit this output of the IA between 0.5V and 14V...I had another post on this but this question relates to the IA.
hi,
This simple example shows the result of using the REF pin.
If you limited the output swing of the existing circuit to say +/-7.5V, then applied +7.5V to the REF pin, the output swing would become 0V to +14.5V

AD620a..gif
 
problem in ad 620

i have followed the discussion on this forum about the working of the ad 620..
my specs are
Rg=5.48 kΩ
v(pin2)=3V
v(pin3)=5v
Vs=±15V

ref=GND;

i should get the o/p as 20V approx...but the multimeter is giving 10V...can you please guide me as to what the problem can be??
 

crutschow

Well-Known Member
Most Helpful Member
See post #2.
 
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