2/3

AN1761

2

CALCULATIONS OF THE ACCURACY

Based on the schematic used to program the voltage reference, we can write:

Where V stands for Cathode voltage,

Vref is for the voltage present on the ref pin and Iref is the input

reference pin current.

Iref is a very low current in the range of a few µA we will neglect if the term R1.Iref

is low enough regarding the other terms of the equation. Most of the time, this simplification can be done.

Thus, if we write the total equation of the partial derivative, we obtain:

with

So, the total equation gives

And if we try to make terms like dx / x appear, we obtain:

And by operating some simplifications, we finally have:

Then, if we consider small variations of all parameters around their values, we have the final expression

which can be used to calculate the relative error:

precision. Using the above formula, we can calculate the 5V output precision:

V

1

R1

R2

-------

+





Vref

•

R1

Iref

•

+

=

dV

Vref

∂

∂V

Vref

d

•

R1

∂

∂V

R1

d

•

R2

∂

∂V

R2

d

•

++

=

Vref

∂

∂V

1

R1

R2

-------

+

=

R1

∂

∂VVref

R2

-----------

=

R2

∂

∂VR1

–

Vref

•

R2

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

=

V

d

1

R1

R2

-------

+





Vref

d

•

Vref

R2

-----------

R1

d

•

R1

Vref

•

R2

2

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

R2

d

•

–

+

=

V

d

1

R1

R2

-------

+





Vref

dVref

Vref

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





•

R1

R2

-------





Vref

R1

d

R1

----------





•

R1

R2

-------





Vref

dR2

R2

----------





•

–

+

=

dV

V

-------

dVref

Vref

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

R1

R1

R2

+

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

dR1

R1

----------

dR2

R2

----------

–





•

+

=

∆V

V

-------

∆Vref

Vref

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

R1

R1

R2

+

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

∆R1

R1

-----------

∆R2

R2

-----------

+





•

+

=

∆V

V

-------

1

1

2

---2

2

+

()

•

+3%

±

==