0
\$\begingroup\$

Premium motherboards (Asus Crosshair Hero VII) have T_in 2pin header for 10k thermistor temperature monitoring. Probing around reveals internal voltage divider with ~12KOhm rTop resistor (as per attached picture).

Goals for sensor read @ T_in: 25C (aka 10K) when GPIO is off or disconnected. 60C+ (aka ~2K) when GPIO is high.

Rough calculations, voltage drop T_in-1 to T_in-2 be ~1V for ~2k rBottom equivalent resistance.

I thought I could use NPN transistor as a simple switch, plus some resistors to act like a programmable parallel resistor network (10k at low, 2k at high). Really its much more complicated.

I don't have exact datasheet, only markings on transistor C945, scavenged parts bin.

I have a background in software engineering but its been a while for transistor logic. I built attached circuit and IT WORKS! 1.08V vCollector to Gnd. Asus reports 68C at GPIO high, and 25C at GPIO low. 10Kohm resistors were chosen mostly because they were convenient in my tickle trunk.

Questions:

  1. I fudged around with the resistors until I got what I wanted. How would I go about calculating the values?
  2. What if I wanted 60C(~2500ohm) instead of 68C ?
  3. Am I over complicating this / easier circuit to build ?

enter image description here

\$\endgroup\$
1
\$\begingroup\$

schematic

simulate this circuit – Schematic created using CircuitLab

If desired thermistor resistance is Rt then R3 = RtR4/(Rt-R4). For Rt = 2K -> R3 = 2.5k.

A small error will be due Q1 Vce_sat of 2mV for R1 = 1k or 3mV for R1 = 10k

\$\endgroup\$
  • \$\begingroup\$ it works more predictably then my previous circuit but the formula does not work, or the board has a large offset. R3=3.3k=67C, 4.7K=59C. How did you come to that formula? \$\endgroup\$ – kevinf May 16 '18 at 4:08
  • \$\begingroup\$ 67C (1942 ohm thermistor), 59C (2578 ohm thermistor) \$\endgroup\$ – kevinf May 16 '18 at 4:18
  • 1
    \$\begingroup\$ @kevinf R4||R3 =Rt , R3R4/R3+R4 = Rt , R3(R4-Rt) =RtR4. I neglected emitter collector saturation voltage of few mV . A MOSFET Q1 would be much better instead a bjt but you said you used existent parts. \$\endgroup\$ – Dorian May 16 '18 at 5:29
1
\$\begingroup\$

In cases like these, perhaps experimentally gathered results are more useful then a formula. vR3=vR4=T_in. vEmitter=0V. vBase=0.75V consistently. vCollector=0.008V

With the circuit provided by Dorian, replacing R3 with the following resistors:

 R3       = T_in   = T_in   = Thermistor Equivalent
 1.0 kOhm = 103 °C = 0.51 V = ~622 Ohm
 2.2 kOhm = 78  °C = 0.88 V = ~1341 Ohm
 3.3 kOhm = 67  °C = 1.10 V = ~1942 Ohm
 4.7 kOhm = 59  °C = 1.28 V = ~2578 Ohm
 10  kOhm = 45  °C = 1.65 V = ~4367 Ohm
 47  kOhm = 31  °C = 2.03 V = ~7721 Ohm
 100 kOhm = 25  °C = 2.16 V = ~10000 Ohm

Parallel resistance formula between R3 and R4 increases in error from 2 °C at 100kOhm to 13 °C at 1kOhm due to the increase in current draw through R5 to collector.

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.