Calculating voltage divider bias, using a thermistor for transistor circuit

Discussion in 'General Electronics Chat' started by Cretin, May 4, 2013.

  1. Cretin

    Thread Starter Member

    Dec 13, 2012
    hey all, i have been working on this problem for days now...and it should not be this hard but i can not get it to work.

    I need to design a simple circuit so that when a thermistor is heated with a lighter, its resistance drops and creates a bias voltage of at least .7V volts through an npn transistor, and then this transistor lights an LED.

    Could someone please explain this circuit, how it works, and perhaps even draw a simple schematic? I can only thank you so so much.

    I can't even get the right resistor value, but here are the values you have

    Resistance of thermistor when at room temperature: 4500kohms,
    Resistance of thermistor when heated up: 3500 kohms.

    Resistance of collector: 1000 ohms

    VCC: 5 volts

    I need to find R2, please help lead me through a voltage divider calculation and please help me draw out the schematic. I've spent way too much time on this
  2. Dodgydave

    Distinguished Member

    Jun 22, 2012
    Is this an homework project?
  3. Cretin

    Thread Starter Member

    Dec 13, 2012
    No Dave, it's part of a side project im working on
  4. Jony130

    AAC Fanatic!

    Feb 17, 2009

    4500Komh = 4.5Mohm ??

    Or maybe 4500ohm = 4.5Kohm ?
  5. Cretin

    Thread Starter Member

    Dec 13, 2012
    sorry jony, you are correct, 4500ohms = 4.5khm not apologies
  6. Jony130

    AAC Fanatic!

    Feb 17, 2009
    First what you need to understand is that BJT transistor don't have "trigger" voltage which turn-ON the transistor instantly .
    In reality BJT start to conduct the current from about 0.5V voltages present between base and emitter (Vbe). And BJT is "Full-ON" when Vbe > 0.6V.

    But I would encourage you to try to build this circuit on the breadboard.


    Where RT is yours NTC thermistor.
    And we can find R1 value quite easily. We simply can use a voltage divider equation.
    You want for 4.5K BJT to be OFF. So we have

    0.5V = 5V * (R1)/(R1 + 4.5K)

    0.5 = (5 * R1)/(R1 + 4.5K)

    0.5*(R1 + 4.5K) = 5R1

    5R1 = 0.5*(R1 + 4.5K)

    5R1 = 0.5R1 + 2.25K

    4.5R1 = 2.25K

    R1 = 2.25K/4.5 = 500Ω

    So we need to use a 510Ω resistor.

    But in real life we a force to use a potentiometer instead of a R1 resistor.
    Why we need a POT? Well the answer is simple.
    Even if you just have a bucket full of transistors of the same type (2N2222). Each of them will have a slightly different turn-ON voltage.
    So we need to use a 4.7K POT and then on the on the bench trim the POT.
    Until we get the desired result.
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