NA105 Precision for Voltage Addition/ Subtraction


Thread Starter


I'm doing a simple project which will do voltage calculations(A+D)-(B+C)/(A+B+C+D).

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. This is problematic because the precision of the INA105 itself is not 100%. In this way, some noise may occur.

I'll add a VR in the feedback loop to fine-tune, I guess. But I may be wrong. Any suggestions to help ease the design process would be much appreciated.

Cheers mates!
I don't really understand your terminology.

Is Vcc +5 or +10? What is Vdd?
And is it -5 or -10?

Let me know and I might be able to help.
I have been thinking about this one.

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.

This part is not addition. It is subtraction.

From your post, I get the impression that you set A=-5, B=-5, and
measured the result of A+B at -10.

I don't actually think that is inaccurate now. I think it's correct.
Hmm. Sorry for the confusion.

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

Sorry. The A+B i got is actually -9V.

A variable Resistor.
You probably meant "Vcc" and "Vee." I'd stick w/ +/-15 Volts.

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

The chip itself should be capable of much better results.

you could always start a new thread and post your full schematic.
Just updating to say this worked out with your help. It wasn't easy, per say, but I definitely used your critique.

Thanks, mate!
Thread starter Similar threads Forum Replies Date
A Power Conversion 2