transistor output characteristic, question on load resistor

Discussion in 'Homework Help' started by bug13, May 23, 2012.

  1. bug13

    Thread Starter Well-Known Member

    Feb 13, 2012
    1,208
    38
    as seen in picture below, what's the purpose of R2, since Ib dictate Ic, why do we still need R2 anyway?

    If the print is too small to read,

    A better resolution pic can be seen here

    or

    I1 = constant 10uA, 20uA, 30uA
    R1 = 1kΩ
    R2 = 100Ω
    Q1 = 2N3055
    V2 = 0v to 2v step 0.0001v


    [​IMG]
     
  2. sbixby

    Active Member

    May 8, 2010
    57
    10
    I'm essentially a transistor tyro yet, but...

    The Ib current does control Ic, but only during active mode of the transistor. Once Ib pushes the transistor closer to saturation, there is very little resistance in the CE circuit, so we need something to keep the current down. Your simulation is only including Ib's within the active range of the transistor (I assume).

    Since the transient in this simulation is V2, going from 0 to 2 volts is forward biasing the CB, which allows the current to flow, hence the Ic growing from 0 to that which the Ib current stipulates the Ic current.

    Typically for transistors, you'd be looking at Ib to be transient, and Ic would then vary according to Ib, until the transistor is saturated, then Ic is almost completely dependent on the load resistance.

    I'm answering this because I struggled with the idea myself not terribly long ago.
     
    Last edited: May 23, 2012
  3. bug13

    Thread Starter Well-Known Member

    Feb 13, 2012
    1,208
    38
    Thanks sbixby, your explanation help me out a lot, now I can understand the purpose of the collector resistor!!

    So how do I know when a transistor operates in cut-off, saturation or active mode?

    Is it controlled by the Ib?

     
  4. WBahn

    Moderator

    Mar 31, 2012
    17,763
    4,800
    The purpose of R2 depends on the purpose of the circuit, which in most homework problems is not very well stated when all you are asked to do is analyze the circuit.

    Without R2, the only thing the circuit can due is pull a constant current from V2. What is the point of that? Perhaps there is one, but it's not obvious. With R2, you get a changing voltage at the collector of the transistor. Without it, the collector-emitter voltage is fixed at the value of V2 and, provided it is above Vcesat, you are unable to get the transistor to go into saturation.

    As for how to tell if the transistor is in saturation, that is very straightforward provided you aren't using too complex a model for the transistor. The basic model, which is sufficient the vast majority of the time, is that the transistor will be in the active region provided two things are true: (1) there is a forward-biased base-emitter junction and (2) the collector-emitter voltage is greater than the saturation voltage.

    In performing the analysis, you first determine if the base-emitter junction is forward biased. If it is, then you assume that the device is in the active region and that the collector current is beta times the base current (or you determine the collector voltage and/or current through analyzing the rest of the circuit under the assumption that the transistor is active). You then verify that the results of the active-assumption analysis are consistent with the device being in the active region by determining if the collector-emitter voltage does, in fact, exceed the saturation voltage. If it doesn't, then you know it isn't in the active region and you redo the analysis under a different assumption (either that it is in saturation, in which Vce will be equal to the saturation voltage or that the device is cut off).

    Another question that could be asked about this particular circuit is, what role is served by R1? If that value were changed (or if the resistor were removed entirely) what difference would the transistor experience?
     
  5. sbixby

    Active Member

    May 8, 2010
    57
    10
    Not to differ from WBahn's reply in any way, but from my own perspective of "still looking for clarity", I would answer this as:

    The simplest description of cutoff->active->saturation occurs when you can ignore the CE side of the transistor. That is, CE is sufficiently forward biased so that Ic is completely dependent on Ib. This is what we would look at when trying to understand the basic response of a transistor.

    In this case:

    Ib=0 means Ic=0. (Cutoff)

    Ib increasing means that Ic also increases at a ratio of the transistor's Beta. (Active)

    At some point, however, increasing Ib no longer has a proportional increase in Ic - that's the "knee" of our Ib/Ie curve - and the transistor goes into saturation mode. This isn't a hard change-over, there's a bit of curvature to the response, so "saturation" isn't an exact Ib value.

    Once in saturation, you can continue to increase Ib, but you will no longer see an increase in Ic. (Actually, you will, but that's leakage inherent in semiconductor devices and it's such a tiny increase that it's generally ignored when doing basic analysis and design.)

    The final part of this is that every transistor is different in it's response. You need to look at the transistor's data sheet (or lab test it) to have an estimate* of what values of Ib will put the transistor in cutoff, active, and saturation.

    (*I say estimate because transistors vary wildly, within and without their own family, so the datasheet is a guideline and designers strive to leave sufficient leeway within that guideline so that replacing a transistor with another one will not significantly affect the operation of the circuit.)

    (OK, I'm done pretending I have a clue, for this post. :) )
     
  6. bug13

    Thread Starter Well-Known Member

    Feb 13, 2012
    1,208
    38
    Hi sbixby, your explanation is actually very easy to understand, thanks :)

     
  7. MrChips

    Moderator

    Oct 2, 2009
    12,449
    3,363
    A simple way of looking at the purpose of R2 is R2 is the load resistor.
    Without R2 there is no load.
    With no load the circuit has no useful function.
     
  8. bug13

    Thread Starter Well-Known Member

    Feb 13, 2012
    1,208
    38
    As of point 2, I am not quite understand, do you get the saturation voltage from a datasheet? and I think the saturation voltage is different if the Ib is different right?

    as of the saturation voltage, do you mean the Vce(sat) here (bc547 datasheet, Page 3)

    but I don't seen to understand
     
  9. WBahn

    Moderator

    Mar 31, 2012
    17,763
    4,800
    When all is said and done, virtually everything is a function of virtually everything else. The key is to separate the dependencies that have a significant effect from those that don't. A good rule of thumb is that the Vcesat for a bipolar transistor is about a quarter of a volt. Yes, it can be less than a hundred millivolts under some circumstances and a bit over half a volt under others. But, in most situations, that has only a negligible effect.

    In the data sheet you reference, you can see that the saturation voltage is going to be about matter from those 90mV to 250mV at relatively low base currents but will only double or so even if the base current increases by an order of magnitude. So pick any value in that range, say 200mV, and perform the analysis. If the conclusion is that the transistor is in saturation with a base current of, say, 3mA, then figure that the Vce sat voltage under those conditions will be closer to the 250mV than to 90mV. If it turns out that the device is in saturation but the base current is 0.25mA, then figure that the Vce sat voltage is probably a bit lower than 90mV. In the end, it should make much of a difference. If it does, then you have bigger problems because, even at a fixed base current of 0.5mA, the Vce sat for the actual transistor in your circuit could have a Vce sat as high as 250mV and, though no minimum is given, could be somewhat lower than the typical 90mV. If your circuit is significantly affected by that possibility, then the circuit is a bad design. If it isn't significantly affected, then it shouldn't be significantly affected by the Vce sat turning out to be anything from 50mV to 500mV or somewhat higher.
     
  10. bug13

    Thread Starter Well-Known Member

    Feb 13, 2012
    1,208
    38
    funny enough, after I read here (another thread in this forum, P2-P3), I understand your explanation now.

    even I don't understand 98% of what they are talking about, one of my nerve in my brain must be connected at that point when I read it :eek:

    now I need to try to understand your last post on how to analyse it :) Thanks WBahn

     
  11. bug13

    Thread Starter Well-Known Member

    Feb 13, 2012
    1,208
    38
    thanks everyone's input, after you guys' explaination and some simulation on LTSpice, and some breadboarding, I understand BJT transistor a lot better now, appreciated!
     
Loading...