The precision of INA105 for voltage addition and voltage sub

I'm doing a simple project which will do voltage
calculations(A+D)-(B+C)/(A+B+C+D) which some of you have known through
my previous post.

I'm using INA105 for the addition part. With Vcc +5 (+10) and Vdd -5
(-10), i obtained very poor results. For example, A=-5, B=-5, A+B=-10
which is very inaccurate. Based on your experience, are there any
problems with the precisions of the INA105 itself which is not 100%
unity gain or some noise may be incurred?

I'll add a VR in the feedback loop to fine tune. I may be wrong. Do you
have any suggestions/opinions which may help to quicken the design
process?

Thanks in advance.

Regards,
Albert

 

[Mac]

On Sun, 16 Jan 2005 18:21:40 -0800, albertleng wrote:

Mac wrote:
Thanks Mac for your help. I'm really sorry for the confusion and
ambiguity.
On Sun, 16 Jan 2005 08:08:05 -0800, albertleng wrote:

I'm doing a simple project which will do voltage
calculations(A+D)-(B+C)/(A+B+C+D) which some of you have known
through
my previous post.

I'm using INA105 for the addition part. With Vcc +5 (+10) and Vdd
-5
(-10), i obtained very poor results. For example, A=-5, B=-5,
A+B=-10
which is very inaccurate. Based on your experience, are there any
problems with the precisions of the INA105 itself which is not 100%
unity gain or some noise may be incurred?


I don't understand your terminology. Is Vcc +5 or +10? What is Vdd?
And is it -5 or -10?

I tested with 1) Vcc =+5, Vdd=-5. 2) Vcc=+10V, Vdd=-10V and
3)Vcc=+15V, Vdd=-15V

I think you mean "Vcc" and "Vee." I suggest you stick with +/-15 Volts.

"Vdd" is usually used in place of "Vcc" in MOS parts. For bipolar process
parts, you would use "Vcc" for the positive power supply, and "Vee" for
the negative supply if there is one. In MOS parts, the negative supply
would be Vss.

From the datasheet for the part (which is a Texas Instruments part
now
that TI bought Burr-Brown) V+ (Vcc) should be +15 Volts if you want
to
input and output signals up to 10V. And V- (Vee) should be -15. Also,
the
part doesn't perform addition. It performs subtraction.


The IC is designed for subtraction but it can be modified to do
addition also. You can refer to the datasheet. There's one diagram
about addition and etc.

I guess you are referring to figure 13?

From your post I get the impression that you set A=-5, B=-5, and
measured the result of A+B at -10. Why do you call that inaccurate?
That sounds correct to me.
Sorry. The A+B i got is actually -9V.

Oh, OK. The chip itself should be capable of much better results than
that. It might be a good idea to start a new thread, and post a full
schematic showing your whole circuit (including resistor tolerances) and
detailing the results you got. I don't think anyone is watching this
thread anymore. If you can put the schematic on a web page you control
somewhere, that would work. If not, use savefile.com

http://www.savefile.com/filehosting/

Either way, post a link to the schematic.

I'll add a VR in the feedback loop to fine tune. I may be wrong. Do
you
have any suggestions/opinions which may help to quicken the design
process?


What do you mean by VR?

Variable resistor.

You should be able to get very close using 1% (or better) tolerance
resistors without resorting to a potentiometer (variable resistor).

Thanks in advance.

Regards,
Albert

--Mac

--Mac

 

Mac wrote:
Thanks Mac for your help. I'm really sorry for the confusion and
ambiguity.

On Sun, 16 Jan 2005 08:08:05 -0800, albertleng wrote:

I'm doing a simple project which will do voltage
calculations(A+D)-(B+C)/(A+B+C+D) which some of you have known
through
my previous post.

I'm using INA105 for the addition part. With Vcc +5 (+10) and Vdd
-5
(-10), i obtained very poor results. For example, A=-5, B=-5,
A+B=-10
which is very inaccurate. Based on your experience, are there any
problems with the precisions of the INA105 itself which is not 100%
unity gain or some noise may be incurred?


I don't understand your terminology. Is Vcc +5 or +10? What is Vdd?
And is it -5 or -10?

I tested with 1) Vcc =+5, Vdd=-5. 2) Vcc=+10V, Vdd=-10V and

3)Vcc=+15V, Vdd=-15V

From the datasheet for the part (which is a Texas Instruments part
now
that TI bought Burr-Brown) V+ (Vcc) should be +15 Volts if you want
to
input and output signals up to 10V. And V- (Vee) should be -15. Also,
the
part doesn't perform addition. It performs subtraction.

The IC is designed for subtraction but it can be modified to do
addition also. You can refer to the datasheet. There's one diagram
about addition and etc.

From your post I get the impression that you set A=-5, B=-5, and
measured the result of A+B at -10. Why do you call that inaccurate?
That sounds correct to me.
Sorry. The A+B i got is actually -9V.

I'll add a VR in the feedback loop to fine tune. I may be wrong. Do
you
have any suggestions/opinions which may help to quicken the design
process?


What do you mean by VR?

Variable resistor.

Thanks in advance.

Regards,
Albert

--Mac