# NTC thermistor series/parallel resistor calculation for MCP73833 linear Li-Ion charger

I am using MCP73833 which has the following datasheet https://ww1.microchip.com/downloads/en/DeviceDoc/22005a.pdf

Internal diagram as follows:

They says that "An internal 50 μA current source provides the bias for most common 10 kΩ negative-temperature coefficient thermistors (NTC). The MCP73833/4 compares the voltage at the THERM pin to factory set thersholds of 1.20V and 0.25V, typically"

Now the formulas shows that

For example, by utilizing a 10 kΩ at 25C NTC thermistor with a sensitivity index, β, of 3892, the charge temperature range can be set to 0C - 50C by placing a 1.54 kΩ resistor in series (RT1), and a 69.8 kΩ resistor in parallel (RT2) with the thermistor as depicted in Figure 6-1.

Now I just can't figure out how to properly calculate the resistors required series/parallel with NTC for desired window set, a more generic way based on RHOT and RCOLD (ie. for other range)

The best I came up with and I am stuck. Appreciate a little help.

1           1           1
----    =  ------ + --------
24-RT1      RT2       rcold

1           1           1
----    =  ------ + --------
5-RT1      RT2       rhot

• Sorry, I can't help but +1 for a well-written question. – Elliot Alderson Oct 15 '18 at 14:13

## 1 Answer

You're a good part of the way there. Simply subtract the two equations and you get

$$\R_{T1}^2 -29R_{T1}+m = 0\$$

where $$\ m = 24\cdot5 - \frac{19}{1/R_{HOT}-1/{R_{COLD}}}\$$ (units of k$$\\Omega^2\$$)

So $$\R_{T1}= \frac{29 \pm \sqrt{29^2 -4m}}{2}\$$ (standard solutions to a quadratic)

Pick the correct answer for RT1 (minus) and you can then subsitute RT1 back into either of the first two equations to get (say)

$$\R_{T2} = \frac{1}{\frac{1}{5K-R_{T1}}- \frac{1}{R_{HOT}}}\$$

(If you end with a negative answer for either resistor, you picked the wrong answer- assuming there is a solution for that thermistor).

• @Sphero, thank you! Very elegant. I should open my daughter's math book more often. – orfruit Oct 15 '18 at 15:41