How does an instrumentation op-amp(AD620) work as a comparator?

Thread Starter

Isaac Po

Joined Nov 27, 2022
26
Hi, I am currently trying to build a fire alarm using LM35 and AD620 IC.


1669781185934.png


1669781216208.png

How does the LED1& LED2 lights up separately when the AD620 IC's input voltage(+) is higher than my input voltage(-) as well as when the input voltage(+) is lower than my input voltage(-)?

--------------------------------------------------------------------------------------------------------------------------------------------------------------


1669783038060.png


1669783068708.png

Additionally may I know why the output voltage differs greatly from each other with one being 12.5V while the other one is -8.59V?
When i calculate using Ohm's law & voltage divider rule, it doesnt come out as I calculated.
 

Thread Starter

Isaac Po

Joined Nov 27, 2022
26
I mistakenly use the wrong op-amp for the measurements part, this is the one for creating fire alarm using LM35 & AD620.

1669783649075.png


1669783685204.png


May I know why the output voltage differs greatly from each other with one being 9.53V while the other one is -5.13V?
When i calculate using Ohm's law & voltage divider rule, it doesnt come out as I calculated.
 

Thread Starter

Isaac Po

Joined Nov 27, 2022
26
I mistakenly use the wrong op-amp for the measurements part, this is the one for creating fire alarm using LM35 & AD620.

View attachment 281816


View attachment 281817


May I know why the output voltage differs greatly from each other with one being 9.53V while the other one is -5.13V?
When i calculate using Ohm's law & voltage divider rule, it doesnt come out as I calculated.
 

crutschow

Joined Mar 14, 2008
34,285
May I know why the output voltage differs greatly from each other with one being 9.53V while the other one is -5.13V?
Likely because the AD620 was designed as a linear amplifier driving a high impedance (2-10kΩ) load, and was not designed to be used as a comparator driving a low-impedance load, so has different saturated output voltages for different plus and minus output loads (which your circuit has).
Monitor the output current of the amp to see what it is for both states.
(You can put a 0V voltage source with in series with the output to see what it is in the simulation.)

Below is from the data sheet showing the output voltage versus load resistance.

1669789767594.png
 

LowQCab

Joined Nov 6, 2012
4,026
You don't need a very expensive Instrumentation-Op-Amp to
do the work of a cheap, generic Comparitor.

Use something like a TLC3702IP ( DigiKey 296-10300-5-ND ), instead.
It can put out ~20mA for your LEDs with no problems.
.
.
.
 

Ya’akov

Joined Jan 27, 2019
9,070
Because the TS identifies as "Student" and the nature of the question suggests assigned work, this post is being moved to the Homework Help forum and falls under the rules of that forum.

Please help with resolving questions about the TS' own work but not about the problem itself. Thanks!
 
Top