Connect with us

semiconductors & ohm's law

Discussion in 'Electronic Basics' started by Mofman, Jan 25, 2004.

Scroll to continue with content
  1. Mofman

    Mofman Guest

    Hi All,

    I am an electronics newbie trying to understand how to calculate power
    dissipation/voltage drops/current for simple circuits involving
    semiconductor components.

    Calculating these values is easy if only "passive components" are involved.
    However, as soon as you introduce diodes or transistors, I start having
    trouble and I have reached the limits of my research (although not much). I
    have the following questions:

    How do you calculate the voltage drop across a diode?
    How do you calculate the base-emitter voltage drop for a transistor? What
    about the collector-emitter voltage drop?


  2. Don Pearce

    Don Pearce Guest

    The answer is pretty much you don't. You have the options of looking
    them up in a data sheet if it is comprehensive enough, or measuring
    some real devices.

    Spice models for these parameters are generally pretty good, so you
    could use these to find the answers.


  3. There is a formula, V = (kT/q) ln(I/Is), which links the voltage V with
    the current I for an ideal junction with no series resistances. To use
    it, you need to know Is for the particular device. k = Boltzmann's
    constant, T = absolute temperature, q = charge on the electron, Is = a
    characteristic of the device, hardly ever given in the data sheet.
    Instead, you get curves of typical, and, if you are very lucky, maximum
    and minimum forward voltage against forward current, maybe at two

    But for normal silicon diodes at low currents (up to a few tens of
    milliamps), the value of V is between 600 and 700 mV approximately. For
    Schottky diodes, it is between 300 and 450 mV, AIUI. For germanium
    diodes, about 200 mV.

    If the diode is reverse-biased, the voltage depends on the rest of the
    Use the above formula, I being the base current. But you may need to add
    in I(rbb' + ßree') to allow for series resistances. The point about the
    voltages being rather generic to the *material*, not the device,
    That depends on the operating conditions, and is likely to be something
    you optimise when designing the circuit.
  4. Olaf

    Olaf Guest

    take a look at , the intermediate
    section contains answers and explanations for both diodes and 'normal'
    transistors (junction transistor) on newbie level. Usually you don't
    calculate the voltage drop. Take a default value (see the link) or look in
    the datasheet.

    bye, Olaf
  5. One of the things about newbies is that you never know whether they
    happen to have a first-class degree in maths, and really WANT the
    equations. (;-)
  6. Olaf

    Olaf Guest

    You're right, a newbie to electronics might be a calculus whizard. So I
    didn't answer OP's question and you did. ;-) Maybe M will let us know what
    he was looking for after all...

    bye, Olaf
  7. Leon Heller

    Leon Heller Guest

    It's about 0.7V for an ordinary Si diode. The data sheet should give it
    for various currents. Temperature comes into it as well.
    As above.

    You usually calculate this as part of the design process, when biasing
    the device.

  8. So does everyone else, in certain cases. The nonlinear math can be a
    bear. This is one of the reasons circuit simulators were created.
    There is a logarithmic function that describes the typical current to
    voltage relationship, but most diodes drop about 6. to 1 volt in their
    normal operating range. To see how a particular diode will act, it is
    a good idea to check the data sheet. This usually contains a graph of
    voltage drop versus current with three curves for three different
    temperatures, room temperature, very cold and very hot, to give you an
    idea of the range.
    The base to emitter voltage is a diode junction, so the generalities
    above apply. Again, the data sheet often has a lot of detail on this,
    including how it changes when the transistor saturates.

    If you want to wade in to the math, here is a paper that compared two
    models of the voltage to current relationship. Don't be put off by
    the greek letters or the fact that they use U for voltage.

    The collector to emitter voltage is totally dependent on the base bias
    current and the collector load, as well as the current gain and other
    details of the particular transistor, so there can be no simple rule
    to guess this voltage. You will have to learn a lot more about how
    transistors actually work to make good estimates of this.

    But the one rule that always works is that the power going into a
    device in any given instant is the product of the current through it
    times the voltage across it. This includes the case where the current
    is going opposite the way the voltage is trying to push it, except,
    there the power is negative (applies to batteries and other energy
    storage components).
Ask a Question
Want to reply to this thread or ask your own question?
You'll need to choose a username for the site, which only take a couple of moments (here). After that, you can post your question and our members will help you out.
Electronics Point Logo
Continue to site
Quote of the day