Help with using a transistor instead of a relay please

Discussion in 'The Projects Forum' started by Markw996, Dec 20, 2009.

  1. Markw996

    Thread Starter New Member

    Dec 20, 2009
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    0
    Hi,

    I am currently trying to design a circuit which will hopefully in effect work the same as an automotive 'lights left on' warning.

    At present I'm using a relay with it's NC poles connected through supply power to a buzzer which sounds all the time. The relay is closed when the switched power is turned on, this removes supply power and hence silences the buzzer.

    This seems to be a very crude way of doing things, although in my application it actually does work perfectly and doesn't draw current all the time as you might expect because the supply is not always on.

    [​IMG]

    What I want to do is use transistors to switch the buzzer instead of a clunky relay.

    I have included a schematic of my transistor solution which doesn't work... can anyone suggest what I have done wrong, or suggest a better way of utilising a transistor to do what I want?

    Thanks,
    Mark.
     
  2. Wendy

    Moderator

    Mar 24, 2008
    20,765
    2,535
    Move the resistor between the Base Emitter to the Base Ground, and drop it's value.

    You may want to eliminate the Base to Signal, I need to think about it.

    Basically it is ground on the base that will turn this transistor on (current from base to ground to be exact). You're current schematic has no allowance for this.

    The +12VDC signal turns the transistor off, hard, but nothing is there to turn it on. There are other configurations that would work better, but again, I need to think about it. Before I get back I suspect someone will have a better idea.
     
  3. Markw996

    Thread Starter New Member

    Dec 20, 2009
    14
    0
    Thanks, that works great.

    I have mocked it up on a breadboard but I don't know how to calculate the resistor values of PNP transistors?

    I only know NPN transistors where I'm making sure they are saturated by providing approx 3 times base current...

    Is this the same for PNP's?

    [​IMG]

    The current draw of the buzzer is 20mA
    The transistor I'm using has a HFE of 60 and a typical VBE of 1v

    Thanks,
    Mark.
     
  4. SgtWookie

    Expert

    Jul 17, 2007
    22,182
    1,728
    The "rule of thumb" for transistor saturation is Ib=Ic/10

    So, if your desired Ic=20mA, then Ib=20mA/10=2mA.
     
  5. Markw996

    Thread Starter New Member

    Dec 20, 2009
    14
    0
    Based on the above formula then my resistor R102 will need to be 5.6kΩ.

    I read somewhere that my other resistor R103 should be 10 times the value of R102 so I have made it 56kΩ.

    [​IMG]

    When I put it on the breadboard it doesn't work....

    However if I increase R103 to around 130kΩ to 140kΩ then it works...

    Rather than just by trial and error, how do I actually calculate the correct value of the R103 resistor that goes from base to earth?

    Am I correct in thinking that I should just use the highest value resistor that I can lay my hands on!
    That way the base voltage will be as close as I can get it to the supply voltage?

    Thanks,
    Mark.
     
    Last edited: Dec 20, 2009
  6. Wendy

    Moderator

    Mar 24, 2008
    20,765
    2,535
    Since it is the main current turning on the reisstor, the 1/10 rule applies here. Measure the current of the load (the buzzer), and divide by 10. This is the value guaranteed to work, you can increase the resistor and it might work, but rules of thumb are not trial and error, quite the contrary as a matter of fact.
     
  7. Markw996

    Thread Starter New Member

    Dec 20, 2009
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    0
    Sorry if I wasn't clear, I understand what you are saying regarding R102 and I have calculated that a 5.6kΩ resistor will supply the 2mA required.

    I now need to calculate the value of R103.

    Using the Voltage Divider formula I have calculated that based on my supply of 12v and my base resistor of 5.6kΩ, an R103 value of 470kΩ will give me a base voltage of 11.86v on my transistor.
    This will be just a 0.14v base-emitter difference which will be low enough for the transistor to stay switched off.

    Could someone please correct me if I'm doing this wrong as it's been about 18 years since I last played around with electronic circuits back in my college days! :D
     
    Last edited: Dec 20, 2009
  8. ianwong

    New Member

    Jun 22, 2009
    1
    0
    It is R103 that turns on T101. Thus, your calculated value should be used in R103 instead of R102. Since you know hfe of your transistor, your actual needed base current is 20mA/60. However, your hfe usually is only the typical value. You'd better check what min. hfe is. R102, together with R103 is a potential divider to turn T101 off and should tune to set Vb above 12V-0.7V or higher.
     
  9. Markw996

    Thread Starter New Member

    Dec 20, 2009
    14
    0
    Thanks to everyone who replied.

    I need a bit of a refresher as my knowledge is obviously very rusty! :rolleyes:
     
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