Need help understanding this simple transistor sensor circuit

Discussion in 'The Projects Forum' started by Cretin, Apr 29, 2013.

  1. Cretin

    Thread Starter Member

    Dec 13, 2012
    Hey all, take a look at the circuit below.


    Keep in mind that this circuit has been modified, and instead of R1 there is a reed switch, and Vout goes to a LED

    I have a few questions

    1: Is my interpretation of the theory of this circuit correct?

    If the contacts of the magnetic reedswitch are closed, then current flows through the reed switch and sends enough current to the base to "open it" and allow current to flow from the emitter to the collector. current then passes into Vout as a result.

    2: If this is correct, then how come current doesn't pass through Vout all the time, why is this the case only when the base has a small amount of current flowing into it? I know the transistor operates like a switch, but I'm having a hard time interpreting its function as i apply it to other circuits.

    3: How do i determine the resistance values for the base?
    Last edited by a moderator: Apr 29, 2013
  2. ScottWang


    Aug 23, 2012
    I can't see the circuit, you may attach it on this site.

    Using the function of Additional Options/Management attachments.
  3. wayneh


    Sep 9, 2010
    A normal NPN transistor will conduct from collector to emitter whenever its base voltage is about 0.65V or more above the emitter voltage. The current required to maintain that base voltage is - rule of thumb - 10% of the collector-emitter current. That's to achieve clean switching. In less saturated applications (e.g.. linear audio amplifier), the current gain might be 100 or more.

    Anyway, you want to be sure the transistor is protected so that C-E current doesn't exceed its rating, and choose a resistor for the base current to limit it to ~10% of that. There's no reason to allow more base current than that.

    Of course you need to know the voltages involved, so you can calculate the needed resistor values. The B-E voltage drop will always be ~0.7V when conducting, and the C-E voltage drop is usually less than that - you can use 0V for approximations.
    Last edited: Apr 29, 2013
  4. WBahn


    Mar 31, 2012
    Typo: You meant to say more than about 0.65V above the emitter voltage.
  5. wayneh


    Sep 9, 2010
    Dang it, I always get those backwards. Of course you are correct and I've edited the post above to correct it.