Integrating zener diode into my circuit

Discussion in 'The Projects Forum' started by Fobio Design, May 18, 2016.

  1. Fobio Design

    Thread Starter New Member

    May 11, 2016
    17
    0
    Hello-

    Hope you are all having a wonderful day so far.

    I am looking to use a zener diode in my circuit (5.1V & 5W). My supply voltage is a 9V 600mAH battery.

    My question is what type of resistor and NPN transistor I would need to use in my circuit before the zener diode?

    Thank you!
     
  2. ScottWang

    Moderator

    Aug 23, 2012
    4,858
    768
    Is this a homework or personal study?
    Your circuit (5.1V & 5W), so I = W/E = 5W/5.1V = 0.98A
    If the above is true then the current of your 9V 600mAH battery will be not enough for the circuit.

    Are you trying to make a 5V/5W power supply from 9V battery (or any 9V power supply) using the npn R+zener diode+n bjt?

    What is your circuit (R+zener diode+bjt, no matter will it work or not)?
     
  3. Fobio Design

    Thread Starter New Member

    May 11, 2016
    17
    0
    Thanks for your reply. It is for a personal project. I am trying to implement a voltage regulator into my circuit. I have attached a picture of what I want the circuit to look like. I am just confused on what resistor and transistor I would need to use with the battery supply voltage I have to create 5V and 6W.
     
  4. crutschow

    Expert

    Mar 14, 2008
    13,056
    3,245
    You won't get anywhere near 6W from a 9V battery, at least not for very long.
     
  5. Veracohr

    Well-Known Member

    Jan 3, 2011
    552
    76
    I assumed 5W is just the rating of the zener the OP has, not the target capacity of the supply.

    You'd probably be better off using an IC regulator like 7805.
     
  6. ScottWang

    Moderator

    Aug 23, 2012
    4,858
    768
    I thought the circuit would be like that, because that is the basic circuit.
    What is your application of the Load?

    The Vout = Vz + Vbe
    Ic = Ib*10
    Iout = 6W/5V = 1.2A
    Ie = Iout
    = Ic+Ib
    = Ib*10+Ib
    = Ib*11

    Ib = Ie - Ic
    = 1.2A/11
    = 0.109 A
    = 109 mA

    If you using the zener diode + bjt circuit, when the load does not draw the current then the zener will be drawing too much current.

    If you using two stages as 2sc1384 + mje3055 + 6.8V zener diode and then the Vout will be as
    Vout = 6.8V-0.7V-0.8V = 5.3V

    Mje3055_Ic and Ib in sat.gif

    Another choice is using LM2576-5V_12V, 3A

    Whatever which one you choose, the most important thing is the input current should be as
    Ii = 1.2 A*(1+20%) = 1.44 A is better.
     
    Fobio Design likes this.
  7. Fobio Design

    Thread Starter New Member

    May 11, 2016
    17
    0
    Sorry for all the confusion. I have attached a new file of the setup I want. I am using a 5.1V 5W zener, I am just confused on what type of resistor and transistor I should be using.
     
    • kk.png
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  8. ScottWang

    Moderator

    Aug 23, 2012
    4,858
    768
    Iout = 5W/5V = 1A
    Ib = 1A/11 = 90.9 mA
    Vout = Vz -0.8V = 5.6V = 4.8V
    I_R1 = Ib + Izd = 90.9 mA + 10 mA = 100.9 mA
    V_R1 = 9V - 5.6V = 3.4V
    R1 = 3.4 V/100.9 mA = 33.7 Ω = 33 Ω

    As IN4734, it has Imax = 162 mA, and we will use it less then I = 162 mA(1/3) = 54 mA, and the Ib as 90.1 mA, so it has risk that you don't know when the zener will be burned, because too hot.

    Please reread my post on #6 to know more about the details and calculation.
     
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