Simple NPN Switch

Discussion in 'General Electronics Chat' started by philipm, Dec 30, 2013.

  1. philipm

    Thread Starter Member

    Jun 27, 2012
    47
    3
    I'd like to use an NPN to switch a small load, either saturated or off.

    In the attached circuit, +VBAT varies between 3V and 4.2V and the base logic operates at 2.8V.

    I'm planning to use a BC817-40 (datasheet here).

    My calculations are:
    Ic = 4.2 / 10000 = 0.42mA
    Ib = 2.8 / 4700 = 0.50 mA
    I know it's not usual to have Ib>Ic, but with Hfe of approx 400, it'll be saturated!

    If I'm reading the datasheet correctly, this condition should give:
    Vce(sat) of 0.01v
    Vbe(sat) of 0.5v

    Does that all seem correct? Anything to watch out for with this circuit?
     
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  2. crutschow

    Expert

    Mar 14, 2008
    13,053
    3,244
    It will work, but normally you don't need a base current higher that 1/10th of the collector current for full saturation. Under those conditions the base-emitter voltage would be closer to about 0.6V. For that R2 = (2.8v - 0.6V) / .0042ma = 520k ohms.
     
  3. MaxHeadRoom

    Expert

    Jul 18, 2013
    10,566
    2,379
    2N7000 'Fetlington'
    Max.
     
  4. philipm

    Thread Starter Member

    Jun 27, 2012
    47
    3
    Many thanks. I have spare resistors in an array and another BC817 on the board, hence keeping the design simple with fewer lines on the BOM. =)
     
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