3

AVR401

0953C–AVR–02/03

The calibration cycle is executed by holding the PB7-pin high during Power-up. The cal-

ibration voltage is then applied, and the PB7-pin is set low. This starts calibration, and

once performed, the value of the reference voltage is stored in EEPROM. During normal

operation, the reference value is read from EEPROM, and the input voltage is calculated

using Equation 1.

Configuration

Example

As the resulting output is to be eight bits, the timer should be of at least nine bits to

maintain the resolution. The components should be chosen so that the nominal time

charging the capacitor up to V

component values and temperature changes are allowed, without causing the charging

time to be longer than the maximum timer period, or too short, giving lower resolution.

To achieve sufficient accuracy, a prescaler factor of eight or higher should be used. The

AT90S1200 Timer/Counter0 is of only eight bits, so the ninth bit must be handled in soft-

ware. The following example illustrates how the component values can be found.

First, decide which crystal frequency to operate at. With a 4 MHz crystal, the clock

period is 250 ns. By setting the prescaler to CK/8, the Timer is incremented every 2 µs.

The maximum timer period with nine bits is 512 x 2 µs = 1,024 µs. From this, we set

2x T

The charging of a capacitor whit a constant current is described by the equation:

Equation 4

We can find the required current when the capacitor size, the time and the voltage differ-

ence is known:

Equation 5

The capacitor will be charged up to V

tor must supply a current of 2.15 mA. The R

h

sistor ideal for use, since any h

sure the full range in h

tions. The resulting base current is 11.4

µA.

The transistor is turned on by applying a “0” on the corresponding pin. At this current

values, the transistor base-emitter voltage is about ÷0.1 V. The base resistor is found to

be:

Equation 6

be much larger than these two, so that the input voltage will not influence with the refer-

ence voltage. 100 k

Ω for R

∆V

I

C

----

∆t

×

=

I

∆VC

×

∆t

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

=

R

B

V

CC

V

BE

+

I

B

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

4.9V

11.4

µA

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

430k

Ω

==

=