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Thread: Directly driving a TIP120 Darlington NPN from 3.3V output

  1. #1
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    Directly driving a TIP120 Darlington NPN from 3.3V output

    I was wondering if someone could verify my electronic maths for directly driving a TIP120 Darlington NPN from 3.3V output, see datasheet diagram below.

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    I want to use it to switch a 9W LED powered from a 5 cell Li-Ion battery. If Watts = Volts x Amps then Amps = Watts / Volts = 9/21 = 0.43A . The TIP120 is rated at 5A so I am well within its limits, somewhat overkill, I am only using it because I already have some. As I see it if I directly connect the base to a 3.3V digital output I will get the following current draw from that output:

    R1 + R2 = 7000 + 70 = 7070 Ohms
    Teensy digital output voltage = 3.3V
    I = V / R therefore I = 3.3/7070 = 4.67 E-04 = 0.000467 = 0.467 mA

    So as I see it the TIP120 will draw 0.467mA from the Teensy digital output when the output is switched high.

    Is my maths correct or am I missing something?

    I want to drive this from a Teensy LC which from the datasheet has digital outputs rated at 5mA when driving at 3.3V so I am once again well within limits if my maths is correct.

    Anyone?
    Last edited by Experimentalist; 09-17-2019 at 11:36 AM.

  2. #2
    Senior Member+ Theremingenieur's Avatar
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    The resistors R1 and R2 are in parallel with the base-emitter junctions of both transistors which will also draw an exponentially growing current as soon as the individual junction voltage starts exceeding ~0.55V. So, your maths aren't wrong but they don't apply here. And your schematic doesn't tell how power supply, LEDs and the Transistor are wired up.
    Guessing that the LEDs will be in the collector circuit and the emitter will be connected to GND, the voltage at the base is hard limited by the two BE junctions to 1.1V. Driving from 3.3V means that there must be a current limiter resistor between the Teensy pin and the base which will "eat up" the 2.2V difference at 5mA. And 2.2V/0.005A makes 440R, so you would take the next R12 series value of 470R.

  3. #3
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    Quote Originally Posted by Theremingenieur View Post
    The resistors R1 and R2 are in parallel with the base-emitter junctions of both transistors which will also draw an exponentially growing current as soon as the individual junction voltage starts exceeding ~0.55V. So, your maths aren't wrong but they don't apply here. And your schematic doesn't tell how power supply, LEDs and the Transistor are wired up.
    Guessing that the LEDs will be in the collector circuit and the emitter will be connected to GND, the voltage at the base is hard limited by the two BE junctions to 1.1V. Driving from 3.3V means that there must be a current limiter resistor between the Teensy pin and the base which will "eat up" the 2.2V difference at 5mA. And 2.2V/0.005A makes 440R, so you would take the next R12 series value of 470R.
    That's great, thanks for your help . . . I knew I must be missing something. I had intended to wire it as you suggested so your answer is spot on, thanks again.

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