Instrumentation amplifier problem

Thread Starter

BramLabs

Joined Nov 21, 2013
98
Hi all,

I have a project back then 1 month a go. Using PT-100 as a temperature sensor and read it using ADC of my microcontroller. And, i just want to 'experiment' using an instrumentation amplifier as an amplifier with gain needed was 61.7488 times. And i've calculated that i need a resistor as you can see on the picture below.
tes instrumentation amplifier.PNG
I've calculated also the 'self-heating' problem, so i used 6k6 ohm resistor in the input. And also i've calculated the thevenin of the wheatstone bridge voltage and impedance.

What i want to ask is that, the problem is within the amplifier. The voltage divider output as the result of wheatstone bridge looked so nice in reality. But when i want to amplify it with instrumentation amplifier (i made an instrumentation amplifier using LM358, so i made with 2 LM358. I didn't use a single chip of instrumentation amplifier like INA122 or any else), it goes wrong with the result. The result is like nothing has been amplify by the instrumentation amplifier.

From my opinion of view, is that, the instrumentation amplifier, can't amplify gain to 61.7488 times. And if we want to use an instrumentation amplifier, we need maybe a cascade configuration, like maybe 5 times * 4 times * 3 times ( the 3 times gain using instrumentation amplifier or at the 5 times, the remaining amplifier just used a non-inverting amplifier or inverting amplifier ).
That was the first of my thought.
My second thought is that, we need a single instrumentation amplifier that is made into a single chip like INA122. So it could give us a gain up to 61.7488 times ( no need a cascade configuration ).
Am i right for my opinions ?

Thank you before for your help, i really appriciate that.

(At the end, because i failed to use an instrumentation amplifier, i made the reading of PT-100 using another easy configuration and it works as you can see below)
test2.PNG
 

Attachments

Picbuster

Joined Dec 2, 2013
990
Well this a problem of all floating.
using a opamp with + and - voltage it means + and minus from gnd(common).
Or when a single pwr chip is used minus gnd(common) as stated by bordoynov
current schematic is incorrect.
 

Thread Starter

BramLabs

Joined Nov 21, 2013
98
Why do you have in the first scheme floating voltage source 12V? You need to connect minus of source to ground.

Well this a problem of all floating.
using a opamp with + and - voltage it means + and minus from gnd(common).
Or when a single pwr chip is used minus gnd(common) as stated by bordoynov
current schematic is incorrect.
Thank you sir both of you for the answer. I forgot to connect it to the ground. That was the result i got back then and it was wrong.
Yeah, i tried it again with my simulation and got 0.283 v.

I already correct it with the right circuittes instrumentation amplifier.PNG

and

test2.PNG

Then i want to ask again sir, why did i get 0.283 v as the result ? I guess i calculate the operational amplifier correctly. As i looked it from wikipedia ( https://en.wikipedia.org/wiki/Instrumentation_amplifier ), i got the gain is test3.PNG

I ady set the R3 and R2 is the same, so the gain is (1+2R1/Rgain) as the configuration below :test4.PNG
Why my gain is wrong ? Because with my configuration, i should get gain 61.8 times (because my R1 = 152k and my Rgain = 5k, it means that my gain would be (1+2*152/5) or it's 61.8 times).

Thank you sir both of you for the help.
 

Attachments

Jony130

Joined Feb 17, 2009
5,127
Your amp doesn't work because you don't understand how this circuit should work. For this circuit you need to use a symmetrical power supply (split power supply) not a single one.
 

Thread Starter

BramLabs

Joined Nov 21, 2013
98
Your amp doesn't work because you don't understand how this circuit should work. For this circuit you need to use a symmetrical power supply (split power supply) not a single one.
Thank you sir for your reply.

I use LM358 which is contain 2 op-amp inside and has only 1 port of Vcc and 1 port of Gnd. And at the end of instrumentation amplifier which is differential amplifier ( i set the gain is 1 for high CMRR ), i use also another LM358.
My question is, you mean, i should go like this ?

test5.PNG

Thank you before.
 

Thread Starter

BramLabs

Joined Nov 21, 2013
98
Wow sir, thank you very much.
I thought that dual supply only works with LM741. I think i made a mistake.
Sir, i want to ask, LM358 pin 4 is for Gnd. And from the circuit, it's connected to the minus of power supply. Why should it connect to the minus of power supply ? Why it doesn't connect to the Gnd instead ?
I'm really poor at the Dual supply or single supply or something like that T___T.
And if i'm not wrong, if i connect the Gnd pin of LM358 to the minus of power supply, am i getting a voltage from Gnd to -12 V sir ?

Oh yeah sir, how about my first circuit ? Is that correct ? (Because i use it and it works good with a little noise going in with my ADC read)
test2.PNG

I really appreciate for your help sir !
 
Last edited:

Thread Starter

BramLabs

Joined Nov 21, 2013
98
Oh yeah sir, as an additional, am i getting a voltage from power supply 24 volt for my LM358 (12 volt to the Vcc pin of the LM358, and Gnd pin of LM358 to the -12 volt of dual supply) from using your scheme ?
 

bertus

Joined Apr 5, 2008
20,230
Hello,

You are correct.
Jony's circuit uses 2X12 Volts so the total voltage on the LM358 is 24 Volts, centre tapped.
This is no problem as the chip may work upto 2X16 Volts:

LM358_features.png

See the datasheet for more details.

Bertus
 

Attachments

Jony130

Joined Feb 17, 2009
5,127
All op amp needs dual supply. But they can also be used with single supply , but in this case you need a "virtual ground".
Because every real life "active device" to work properly as an amplifier supplied from a single source need proper bias circuit.
When you use BJT as a CE amplifier, you use a voltage divider to bias the active device somewhere in the "linear region".
In case of single supply op amp you have to do the same think. You need to bias the op amp somewhere in the middle of his "linear region".
 

Thread Starter

BramLabs

Joined Nov 21, 2013
98
Hello,

You are correct.
Jony's circuit uses 2X12 Volts so the total voltage on the LM358 is 24 Volts, centre tapped.
This is no problem as the chip may work upto 2X16 Volts:

View attachment 96868

See the datasheet for more details.

Bertus
All op amp needs dual supply. But they can also be used with single supply , but in this case you need a "virtual ground".
Because every real life "active device" to work properly as an amplifier supplied from a single source need proper bias circuit.
When you use BJT as a CE amplifier, you use a voltage divider to bias the active device somewhere in the "linear region".
In case of single supply op amp you have to do the same think. You need to bias the op amp somewhere in the middle of his "linear region".
Thank you sir for your help ^^

I'm really sorry if my question is really silly.
Oh yeah, i want to ask again sir. ahahaaha....
You said that i need a 'virtual ground' to biased the op-amp to work in the 'linear region'. When you mentioned the 'virtual ground', do you refer to the instrumentation analytical process ( i mean like, the instrumentation work with the superposition theorem, and it works with inverting and non-inverting amplifier mode at the top of the op-amp [node Va] and then it works also at the bottom of op-amp [node Vb] and then it goes to the differential amplifier ) ?
test4_1.png

So, what i understand is that, if i want to use an active device amplifier using op-amp, when i needed a virtual ground, i should use a dual supply ( which mean that if i only use an inverting amplifier in a basic configuration, i should use a dual supply like the scheme bellow. Node Vx is the virtual ground ) ?
test6_1.png
And in the case of Non-inverting amplifier, i didn't have to use a dual supply ?
test6_3.png
Because there's no virtual ground over the Vx ( in the input minus of the op-amp ) ?

Thank you sir for both of you.
Both of you and the others really help me ^^
Thanks a lot
 

Jony130

Joined Feb 17, 2009
5,127
Node Vx is the virtual ground ) ?
View attachment 96889
Yes, the node Vx is at virtual ground as long as opamp work in the linear region.
But notice that for a single supply this circuit work only if Vout is positive and Vin is negative.
Which is not the case with single supply.
And in the case of Non-inverting amplifier, i didn't have to use a dual supply ?
View attachment 96892
In this circuit we have a "virtual short" Vin = Vx, and again this is true as long as the op amp work in the linear region. But again this circuit can only amplifier the positive voltage.

And I was talking about different type of a "virtual ground ".
Try this links
http://www.swarthmore.edu/NatSci/echeeve1/Ref/SingleSupply/SingleSupply.html
 

Thread Starter

BramLabs

Joined Nov 21, 2013
98
Yes, the node Vx is at virtual ground as long as opamp work in the linear region.
But notice that for a single supply this circuit work only if Vout is positive and Vin is negative.
Which is not the case with single supply.

In this circuit we have a "virtual short" Vin = Vx, and again this is true as long as the op amp work in the linear region. But again this circuit can only amplifier the positive voltage.

And I was talking about different type of a "virtual ground ".
Try this links
http://www.swarthmore.edu/NatSci/echeeve1/Ref/SingleSupply/SingleSupply.html
Woah, your video really help me to understand the split supply / dual supply and single supply.
Now i know, that if we want to use a single supply, we should bias it to the 'virtual ground' or the linear region, so it could have a better swing ( up and down, and not being clamped at the top signal or the bottom signal, or in easy way is that like if we want to bias a BJT using as CE amplifier, we just need to make sure that the signal operate at Vcc/2, so it could get nice swing up and down )
And if we use a dual supply, it gives us more easier way to bias the op-amp to operate at the linear region ( no need to bias literally, but the power supply gives us more space for up and down signal and not being clamped at the top signal or the bottom signal ).
So, my instrumentation amplifier was wrong because it's like having a clamped input signal and amplify it and get a low signal output because of the clamped signal at the input signal and i didn't bias it to the proper region that gives me 61.7488 times amplify.
Am i right ?


ahahaha....

Big thanks for you sir !
Thank you a lot for your lecture time >_<v

I only know how to calculate in the theoretical an design the op-amp, but don't know how to implement it in the reality because of the lack of understanding 'biased region for the op-amp' and supply things (split supply/dual supply and single supply).
Woah, i really love op-amp and power electronics !
 
Top