convert output pin arduino to higher voltage and current

Discussion in 'Embedded Systems and Microcontrollers' started by mohammad2050, Aug 9, 2015.

  1. mohammad2050

    Thread Starter Member

    Nov 14, 2014
    56
    1
    hi,
    i want to convert output pin of arduino (5 vdc , 40 mA) to 24 vdc, 200 mA .i can not use transistor for saturation problem , please suggest appropriate circuit ?
     
  2. kubeek

    AAC Fanatic!

    Sep 20, 2005
    4,670
    804
    what saturation problem are you talking about?
     
  3. mohammad2050

    Thread Starter Member

    Nov 14, 2014
    56
    1
    because:
    Ib= (5-0.7)/2.15 = 2mA ,,, Ic = B*Ib=200 mA ,,, if load be 1k=> VCE= 24-200==SATURATION
    so
    VCE=0.2 => IC= 24-0.2/1=23.8 mA no 200mA
     
  4. MikeML

    AAC Fanatic!

    Oct 2, 2009
    5,450
    1,066
    So use a Darlington,

    D.gif

    or an NFet

    f.gif
     
    Last edited: Aug 9, 2015
    mohammad2050 likes this.
  5. mohammad2050

    Thread Starter Member

    Nov 14, 2014
    56
    1
    so i must calculate parameters for that , in your opinion how much load can support Darlington or NFet for 24 Vdc,200 mA?
     
  6. ScottWang

    Moderator

    Aug 23, 2012
    4,855
    767
    If the output of arduino is 5V/40mA and add a npn and pnp then you can get the current close to 4A.
     
  7. mohammad2050

    Thread Starter Member

    Nov 14, 2014
    56
    1
    but this is not possible because occur satturation before it reach to higher current .
     
  8. djsfantasi

    AAC Fanatic!

    Apr 11, 2010
    2,812
    834
    For my own curiosity, where did the value 2.15 come from when calculating Ib? From that point on, I am confused where the equations are coming from. Did you mean to divide by 1000 instead of 1? And Vce=24-200? Are you mixing units and precision?
     
  9. mohammad2050

    Thread Starter Member

    Nov 14, 2014
    56
    1
    excuse me ,
    Ib= (5v-0.7v)/2.15k = 2mA ,,, 5v is arduino output, vbe=0.7 ,Rb=2.15k ...because beta is 100 and i want to reach 200 mA
    Ic = B*Ib=200 mA ,,, if load be 1k=> VCE= 24v-200v==SATURATION
    so
    VCE=0.2v => IC= (24v-0.2v)/1k=23.8 mA no 200mA
     
    djsfantasi likes this.
  10. MikeML

    AAC Fanatic!

    Oct 2, 2009
    5,450
    1,066
    Go back and look at post #4

    However, here is the Arduino pin driving a 2n2222 directly. Note that I set the base drive to be 20mA (within the capability of the Arduino pin). This guarantees saturation of the 2N2222, minimizing its power dissipation when on.

    np.gif
     
    Last edited: Aug 9, 2015
    mohammad2050 likes this.
  11. kubeek

    AAC Fanatic!

    Sep 20, 2005
    4,670
    804
    Beta is not a constant and gets lower with higher current. Usually you want to use Ib=0.1 Ic in switch application. So just use 120 ohm base resistor and be done with it.
     
    mohammad2050 likes this.
  12. mohammad2050

    Thread Starter Member

    Nov 14, 2014
    56
    1
    thanks a lot MikeML
     
  13. mohammad2050

    Thread Starter Member

    Nov 14, 2014
    56
    1
    thanks a lot kubeek , you right, i was in wrong.
     
  14. ScottWang

    Moderator

    Aug 23, 2012
    4,855
    767
    When the bjt get into the saturation status, the hFE as 10 and then the current will be like this:
    Arduino (40mA) → NPN (400mA) → PNP (4A)
     
    mohammad2050 likes this.
  15. ScottWang

    Moderator

    Aug 23, 2012
    4,855
    767
    So if you want to get a 200 mA current from Arduino 5V/40mA then you can do as this:
    Arduino 5V/40mA → R_limit (2mA) → NPN (20mA) → PNP (200mA)

    NPN can be use as 2N3904, PNP can be use as 2SA684(1A) or similar.
     
    mohammad2050 likes this.
  16. mohammad2050

    Thread Starter Member

    Nov 14, 2014
    56
    1
    thanks ScottWang, in your opinion, how much maximum load without any problem in this case?
     
  17. ScottWang

    Moderator

    Aug 23, 2012
    4,855
    767
    The normally you will need to add some more current at least add about 20%, so the calculation will be as:
    Arduino 5V/40mA → R_limit (2.4mA) → NPN (24mA) → PNP (240mA)
    That will be enough for many applications.

    You have to test how is the real current for the load needs and not just look at infos from the datasheet.
     
    mohammad2050 likes this.
  18. mohammad2050

    Thread Starter Member

    Nov 14, 2014
    56
    1
    ok, thanks.
     
  19. ScottWang

    Moderator

    Aug 23, 2012
    4,855
    767
    What you need is just like the circuit shows in Fig-07, the values of resistors will be change according to the current.

    [​IMG]
     
    mohammad2050 likes this.
  20. mohammad2050

    Thread Starter Member

    Nov 14, 2014
    56
    1
    thanks a lot for circuits ScottWang,
     
Loading...