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Using a bipolar transistor to turn a load on and off

For situations when your microcontroller (or whatever) can't do it on its own.

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  1. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

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    (*steve*) submitted a new resource:

    Using a bipolar transistor to turn a load on and off - For situations when your microcontroller (or whatever) can't do it on its own.

    Read more about this resource...
     
  2. KrisBlueNZ

    KrisBlueNZ Sadly passed away in 2015

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    That's excellent Steve. I have a few suggestions:

    Suggest 2N3904 as well as BC548 - Americans seem to not know, or dislike, the BC series.

    Mention relay coils and solenoids, as well as motors, as types of inductive loads.

    Suggest removing the + and - markings in the diode picture - they don't really tell you anything useful and they can lead to confusion, and they're backwards in this case anyway!

    In section 4.2, "What's that 0.8?", your comment under the transfer characteristic graph that "BBE (typo - should be VBE) doesn't hit 0.8V until 100 mA base current" is wrong - it's the collector current on the graph, not the base current. So 0.8 is not such a crazy number to use.

    I think you should mention that you need to ensure that the driving source can comfortably supply the required base current. This mightn't be the case with low power devices or when the transistor is switching high-ish currents (you mentioned the 2N2222 switching up to 500 mA), especially if the transistor's current gain is not high. And to check the droop in output voltage under load.

    Also you mentioned earlier that the driving voltage needs to be fairly high - I suggested 3.3V or more as being plenty in my resource - to ensure plenty of voltage across the base resistor.

    You might perhaps want to mention that for quick turn-off, a base-emitter resistor helps.

    That's an excellent resource Steve. It covers a much wider scope than mine. The subjects overlap but I don't think they would fit together very comfortably because of the different approaches.

    If you think otherwise, please feel free to merge them together. Otherwise I guess they can remain as separate resources.
     
  3. Arouse1973

    Arouse1973 Adam

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    Dec 18, 2013
    Nice work Steve.
    Ok I am going to probably embarrass my self now. The magic number 3 is that multiplying the current? So the transistor could draw 900mA? With a gain of 100 isn't it 466.7R and not 4667R. Just can't work out how you did it.
    Cheers
    Adam
     
  4. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

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    Perhaps I did it incorrectly. I'll check.
     
  5. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

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    You should be embarrassed! :D I clearly multiplied 4.2 by 100 and got 4200 instead of 420.

    Seriously, thanks for pointing that out. I am going to hang my head in shame. I remember thinking at the time that the resistor value seemed quite large, but I (stupidly) didn't check my calculations.

    I appreciate that you've done the sanity check.
     
  6. Arouse1973

    Arouse1973 Adam

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    No Probs, easy to make mistake. I do it all the time as you know. I'll have a look at the rest of the document in more detail. I don't expect there will be anymore :)
    Adam
     
  7. Anon_LG

    Anon_LG

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    Does having the load from the collector have any advantage of having the load following from the emitter, if so what is the advantage and why does it occur?
     
  8. Arouse1973

    Arouse1973 Adam

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    When the transistor is used as an emitter follower (remove base resistor) and the base voltage is the same as the collector voltage which is quite common for current amplification, ok you loose a diode drop of voltage which is usually acceptable. This configuration develops more power in the transistor, this is because it will have a higher voltage across it compared to the common emitter configuration that Steve shows. Steve circuit also allows you to switch current from a higher collector voltage like 5V switching a 12V load.
    Adam
     
  9. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

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    A load in the emitter is rarely used for switching because it is harder to saturate the transistor.

    If the load has to be connected to ground then you might use a PNP transistor in common emitter configuration. In this case the emitter is connected to the load's positive supply rail.

    Of course there are always exceptions and circumstances where you need to do something different.
     
  10. Arouse1973

    Arouse1973 Adam

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    Yes indeed, you would need to increase the base voltage to saturate it, which could be quite considerable for high load currents.
    Adam
     
  11. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

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    Mathematically, in a common emitter configuration VB needs to be around VBE for any current IC to allow Vload to get within VCE(sat) of VCC. This means a resistor between VCC and the base of the transistor can be chosen to allow the transistor to be saturated for a given IC.

    In a common collector (emitter follower) circuit, VBE needs to be around VBE + Vload to allow Vload to get within VCE(sat) of VCC. Even with a simplification that VBE < VCE(sat), the voltage at the base will get very close to VCC. This makes it hard to select a resistor VB (which is arguably not required), but does require that the driving signal be at or as close as possible to VCC, and able to source current. This alone is difficult. However, when we realise that VBE > VCE(sat) in saturation, VB now has to exceed VCC, a task not possible unless the driving circuit is powered from a voltage higher than VCC. Once we do that, we then need a base resistor to limit current.

    If VCC is the highest voltage available then using an NPN or a N Channel mosfet (anything "N") as a high side switch is not possible (or not efficient).

    This is why you commonly see "N" devices (NPN transistors and N channel mosfets) as low side switches, and "P" type devices (PNP atransistors and P channel mosfets) as high side switches.

    Because "N" devices (NPN, N-CH mosfets and IGBT's) are "better" than the "P" devices (all to do with mobility of electrons) then there are special drivers to allow N channel mosfets and IGBT's (which are only available as NPN/N-CH equivalent) to be used as a high side switch. In these drivers there is typically a switched capacitor or other type of voltage multiplier to generate a voltage rail above VCC (VDD for a mosfet) that can be used to drive the gate. Similar devices are not available for bipolar transistors because of the higher current required for the base electrode.
     
  12. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

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    I just found your comment in the original thread that I've copied here because I may not have addressed them yet.
     
  13. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

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    Updated to address Kris' concerns.
     
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