hi,
I use the MCP6002 series, they operate from a 5V supply and are rail to rail output.
The 6002 is a dual but there is also the MCP6001 single OPA, not too expensive.

E

The whole that instrumental amplifier thing is very crappy about resistor nominal values. If You can with an ease warranty that 1,0 kOhm ir no less than 0,999 and no more than 1,001 then all is OK, but if not - then You shall see all possible wonders from horror-tales. Therefore my advice is take the instrumental amplifier tablet from already ready-made, where everything except the gain resistor is inside, is laser-trimmed and work with a warranty.

 

The 741 OPA is not the best choice for a Differential amplifier.

Use at least 0.1% tolerance resistors throughout the circuit.

As a general rule I use a 'true' Instrumentation amplifier for low level signals, the type shown post #15 by @OBW0549 are OK.
I also use the AD and INA series.

Are the resistor tolerances at least 0.1%.? 0.01% ideally should be used.

I would advise that you do not proceed with the 741 and 1% resistors, you will never get the performance you require.

If you can only build your own INA using LM358's , I would suggest a dual power supply, say +/-5V and use at least 0.1% resistors in the circuit, 0.01% are more typical.

hi Janis,
Why have you posted to me, I am not the Thread Starter.?
The above quotes are from my posts to the TS, as you can see I have been advising him of the problems with regarding trying to build his own INA, but he cannot get an actual INA in his area.

Eric

 

hi Janis,
Why have you posted to me, I am not the Thread Starter.?
The above quotes are from my posts to the TS, as you can see I have been advising him of the problems with regarding trying to build his own INA, but he cannot get an actual INA in his area.

Eric

yes.. this my situation.. i have to build my circuit with my availability...

 

The original circuit was described as a "half bridge" strain gauge, which presumably means that the two resistors making one side of the bridge are fixed in value and the other side of the bridge has the strain gauge and another fixed resistor.

Precise ratio matching of resistors in an instrumentation amplifier is required when the common mode voltage varies, either because the desired signal is "riding" on a varying voltage or there is noise equally coupled to both inputs of the amplifier. In this case, ignoring noise, the common mode voltage is fixed. The excitation for the bridge is well regulated - or it should be, since the output will be directly proportional to the excitation voltage. Mismatch of amplifier resistors therefore causes simple fixed DC offset error and/or simple fixed gain error. The amplifier resistor tolerance is no more important to the circuit performance than the tolerance of the fixed resistors in the bridge and the tolerance of the bridge excitation voltage. Common mode noise rejection is compromised by imperfect matching of the resistors.

 

yes.. this my situation.. i have to build my circuit with my availability...

If you are in India, AD620 instrumentation amplifier is freely available, though you need dual power supplies.

 

If you are in India, AD620 instrumentation amplifier is freely available, though you need dual power supplies.

hi ramaD,

okay but i looking single power supply and also i should use 5V or 24 V as a power supply in my circuit.

 

The problems from the beginning were not all with the instrumentation amplifier but for the parts of the circuit not shown. A power source must not be other than pure DC, and in addition the ground connection at the center of the supply voltages must be connected to the instrument amplifier common. Besides that, the bridge and half bridge supply must be just as good. I did not read all of the intermediate posts, but adequate supply stability is vital and also adequate shielding of all the connections is vital.
But probably the whole problem has somehow been resolved, I hope.

 

hi r,
I have run your circuit in LTSpice simulation, it works as expected.
I have set the Gain of the bridge amp to 1000, so 5mV from the bridge gives a Vout of 5V.
Inputting 0v to 5V to the V2I converter circuit gives 0mA thru 20mA.
Check thru the attached circuit.
E

dear eric,
i test the below circuit and get the output on testboard(here is share the circuit and ouput values)
now i assemble it on pcb board but in my circuit i did not get the proper output pin at U2 (IC) eventhough the ic get proper input U2.. how could i rectify the problem on pcb board.. ( all connection looks good and IC is on working condition)

sorry for the delay.. i was waiting for the PCB board

 

hi rajit,
I have downloaded your images, I will look them over.

E

EDIT:
The balance condition, showing 0.5mV imbalance on the bridge cannot give an amplifier Vout of 1.3V.
That is a gain of 2600!
Can you confirm the results for the amplifier.
With the bridge resistors still connected, short across Va Vb and remeasure Vout.

 

hi rajit,
I have downloaded your images, I will look them over.

E

EDIT:
The balance condition, showing 0.5mV imbalance on the bridge cannot give an amplifier Vout of 1.3V.
That is a gain of 2600!
Can you confirm the results for the amplifier.
With the bridge resistors still connected, short across Va Vb and remeasure Vout.

Eric,

i use 2.2 kohm for R6 instead of 5.3 K so thats why i get the 1.3V

 

My circuit condition is when the circuit is balanced i should get 4 mA and when i R1 is 330.7(maximum load) i should get maximum output 20mA (approx.)

 

hi,
If the bridge is in balance, ie: 0V across va,vb, then the Vout will be 0V. So that will make Iout =0mA.
You need to have Vout = 1V for the Iout =4mA.
One way is to inject a small current into the bridge, so at resistive balance, Vout =1V, do you know how to do that.?
E

 

hi rajit,
I have downloaded your images, I will look them over.

E

EDIT:
The balance condition, showing 0.5mV imbalance on the bridge cannot give an amplifier Vout of 1.3V.
That is a gain of 2600!
Can you confirm the results for the amplifier.
With the bridge resistors still connected, short across Va Vb and remeasure Vout.

eric
this circuit gives output on test board.. so i trial it on PCB..but i cannot able to get same output on pcb
in circuit U2 is not work... i get proper output at U1 so its affect on U3 non inverting terminal input value

My PCB board reading is

Va = 2.484 ; Vb= 2.482
U1 output is 4.90V and U2 output is 80.9mV so my U3 output opamp is gets affected U3 = 122mV

help me to handle this situation..

 

hi,
If the bridge is in balance, ie: 0V across va,vb, then the Vout will be 0V. So that will make Iout =0mA.
You need to have Vout = 1V for the Iout =4mA.
One way is to inject a small current into the bridge, so at resistive balance, Vout =1V, do you know how to do that.?
E

no eric,
can you explain it..

 

OK,
Couple of questions.
Is R1 resistor on the bridge the only resistor that changes value under load.? ie: is it a quarter bridge or full bridge?
Is it a home made bridge or a commercial bridge.?
If commercial, do you have a datasheet./
E

 

OK,
Couple of questions.
Is R1 resistor on the bridge the only resistor that changes value under load.? ie: is it a quarter bridge or full bridge?
Is it a home made bridge or a commercial bridge.?
If commercial, do you have a datasheet./
E

it is a quarter bridge. R1 resistor only changes the value
it is a home made bridge.. i assemble it use an 330 ohm resistor and potentiometer

 

OK,
I will work out a circuit to enable you to add the Vout =+1V when the R1 =330.
Give me a little while and I will post.
Is it like this image,?
E

 

o

OK,
I will work out a circuit to enable you to add the Vout =+1V when the R1 =330.
Give me a little while and I will post.
Is it like this image,?
E

okay eric

 

but i have an doubt...
in practically i get some mv at balanced condition.. in this state can i use the circuit?

OK,
I will work out a circuit to enable you to add the Vout =+1V when the R1 =330.
Give me a little while and I will post.
Is it like this image,?
E

but i have an doubt...
in practically i get some mv at balanced condition... in this state can i use the circuit?
you replace the R3 and R4 using variable resistance

 

hi,
Use this procedure,
Change R6 to 3k5 , this will increase the Gain.
1. Set R1 to 330R, increase the value of R3 to ~330.18R. [as you can see it is very small change in R3]
Vout should now read ~1V [ this is the base voltage which gives an Iout of 4mA]

2. Set R1 to 330.7R, Vout should read ~ 5V, so Iout should read =20mA

You may have to repeat from Step 1 around the loop until you get Vout =1V and 5V

Note: You could reduce R4 by a very small value instead of increasing R3 [ it may be easier, by adding a high value resistor]

E

EDIT:
This LTS sim shows the Vout and Iout for R1= 330R [ 1v, 4mA] and 330.7R, [5v,20mA]
Note R3 value.