QUESTIONS & ANSWERS
at 35 degrees C when the resistance of the external
thermistor is 6,535 ohms.
I calculate the switching voltage at 26 degrees just to
be sure that switching will take place at 25 degrees. That
voltage is 18V/(109.572K)* 9.572K = 1.572V. The low
switching point is 18V/(106.535K)* 6.535 = 1.104V.
Plugging these numbers into an equation I developed
some years ago (see Figure 6) results in the values in
Figure 5. The program I used is obsolete, but you could
do the same thing with a spreadsheet.
The positive feedback in the IC2B circuit is just to
prevent it from switching on noise, so pulling the input
down 10 mV will be sufficient. When the output is high,
I don’t want excess current through IC3, so that is the
reason for D3. I want the charger to become active again
when the battery voltage gets down to 12 volts ( 12.6V at
R8), so the input voltage can’t go lower than
12.6/(110K+ 39.2K)* 39.2K = 3.31V.
At that voltage, the drop across R10 is 18-( 3. 31+. 5) =
14.19V and the current is 14.19V/75K = .1892 mA. The
same current flows through R11 and the IC3 current is
zero, so R11 = 3.31/.1892 = 17.5K. A standard higher
value is 18.2K which will insure that the voltage is
maintained to at least 12 volts. NV
Program in BASIC
64-bit Floating Point
Sin, Cos, Log and more
30k RAM, 64k Flash
3.3V to 5V
Bi-Directional Level Shifter
with code “NUTS”
February 2013 27