Zener TVS

Discussion in 'The Projects Forum' started by Schniz2, Sep 10, 2011.

  1. Schniz2

    Thread Starter New Member

    Apr 15, 2010
    22
    0
    I need to place a ~5V limit on the output of a current sensor which could potentially spike to 8.3V for a short period of time (~18msec) (manufacturer part number ACS712ELCTR-05B-T).

    As far as i understand, i just connect the cathode to the positive voltage and the anode to ground...

    I need help selecting the zener diode, i was looking at the following "1N5337BG" (characteristics in attached image zenerdata).
    It says the Zener impedance Z_ZT @ reverse current I_ZT is 2.Ohms. Is this the correct reasoning?>>> if there was a 8.3V voltage spike, and the current was diverted through the zener it will see an impedance of 2.Ohms (or will it see a lot less than this?)... , so the zener will need to dissipate P=V^2/R = (8.3^2)/2 = 34.4Watts? It says the maximum steady state power dissipation is 5W and the surge current rating is 15.3A (current would be 8.3/2 = 4.15A)

    The main thing i am worried about is what impedance the current will see at voltages greater than the zener voltage??
     
  2. vrainom

    Member

    Sep 8, 2011
    109
    19
    The surge current rating is for a non repetitive short pulse (8ms) so you should not rely on that for normal duty. The safest would be to put a limiting resistor from the sensor to the zener as the output to mantain it within its electrical limits.

    For a maximum input of 8.3v a resistance as low as 3.9ohms would do: (Vin max - Vz min) / Izm = R (8.3 - 4.7 ) / .930 =~3.9 ohm

    However if you only need to read the sensor's voltage drop as a signal this is way overkill.
     
    Last edited: Sep 11, 2011
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  3. Schniz2

    Thread Starter New Member

    Apr 15, 2010
    22
    0
    Is this what you mean? - see attachment (note there will be a ground reference at the bottom.. i forgot to put it in)
     
  4. bertus

    Administrator

    Apr 5, 2008
    15,648
    2,347
    Hello,

    You still will have the full voltage on the measurement part.

    Here is a "corrected" corcuit:

    [​IMG]

    Now the voltage on the measurement part will not go higher as the zener voltage.
    The 3.9 Ohms resistor may be larger, as the microcontroller will not take much current.

    Bertus
     
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