Transistors (Schottky)/Voltage is confusing me.

Discussion in 'General Electronics Chat' started by Mercfh, Jan 15, 2013.

  1. Mercfh

    Thread Starter New Member

    Apr 24, 2012
    11
    0
    So I understand the "basics" of how transistors work but when seeing a Schottky transistor I got confused.

    I get that a Current Applied to the base controls the flow from Collector to Emitter.....ok makes sense

    So I asked a question on electronics.stackexchange. (see below)

    http://electronics.stackexchange.com/questions/55073/schottky-transistor-not-sure-i-understand-it


    Basically someone said "This "steals" current from the base, preventing the transistor turning on more and the collector reaching it's saturation voltage." Ok that makes sense....but then I went on to read about Saturation of a Transistor.....and that the Emitter is actually going lower and lower in voltage as the Transistor Saturates........This makes no sense to me?


    I understand that it has a forward voltage of .2......but apparently when it's "Saturated" or "On" it's voltage slowly lowers down to that .2v and Schottky prevents it from Lowering it down this much correct?

    The only way I can visualize why (and I may be wrong) but using the "water" analogy....when the Transistor/Valve is "Off" water is building up pressure at the valve opening (collector) when it turns on the "Pressure/Voltage" lowers since current is finally going through......(does this make sense?)
     
  2. SPQR

    Member

    Nov 4, 2011
    379
    48
    Hi,
    It's been about seven hours since you posted and no responses.
    So I'd like to take a Schottky at this:), and see how I do.
    I'm going through a "transistor phase" and trying to learn more about them.

    The key to having a transistor switch "quickly" between states is to keep it out of saturation in the active region of its operation.

    The transistor will start to turn on when the Vbe reaches a certain point, and then Vce voltage will start to drop.
    It will continue to drop until Vce is about 0.2V, and when it reaches that point, the transistor is considered "saturated".

    So the question is, "how can you keep the Vce from dropping to less than 0.2V?".

    One answer would be to feed more current into the Vce junction - this additional current will keep the Vce higher at any given Ibe.
    By injecting more current into the CE system, you keep the Vce above the 0.2V.

    One way to do this is by adding "feed forward" current via a Schottky diode, thus at any given Ibe, there will be a higher Vce, and that pesky 0.2V will never be reached.

    The transistor stays in the "active" region, and voila! you have a faster transistor.

    Now, let's wait for an expert or two to come by and see if I got this right.
     
Loading...