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Basic Diode & Transisitor Question about Turn on Voltage

Discussion in 'Electronic Basics' started by TVisitor, Jul 6, 2007.

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  1. TVisitor

    TVisitor Guest

    Let's say I have a very simple circuit.

    +5V DC in series with a 100ohm resistor & diode (anode on the
    resistor, cathode to the - terminal of supply.
    Diode forward voltage for turn on is 0.7V. I understand this
    circuit. The cathode is already grounded, so it's very easy to see
    that all you need is 0.7V on the anode of the diode to get it to turn

    Now, let's say you have the above circuit, but TWO diodes in series
    with each other. I know that you need 1.4V measured from Anode #1 to
    ground to get the diodes to turn on and complete the circuit.

    However, my conceptual problem is this: Each diode is essentially an
    open switch. The first diode needs 0.7V across it to turn on. How do
    you reference the voltage of cathode #1 to ground? To me, it
    essentially looks like it's floating, so how do you ever get 0.7V
    across the diode?

    I know this is simple, but it's got me baffled.
  2. Your idea of how a diode works is a bit oversimplified. The
    concept that they suddenly turn on when the voltage across
    them reaches .7 volts is just not quite right. They conduct
    some current at all forward voltage, just not an amount that
    is proportional to the voltage. Take a look at this data
    sheet for a diode:
    On page 2 there is a graph called Forward Characteristics.
    This is a diode rated for 1 ampere of forward current, and
    the forward voltage drop that corresponds to the 1 amp point
    on this curve is about 0.95 volts (not 0.7 volts, in this
    case). But at 0.8 volts, it will pass 0.2 amps, and at 0.6
    volts it will pass 0.01 amps. The curve just keeps going
    all the way to zero volts. Diodes are not switches, but
    very nonlinear resistors.
  3. Hi,
    A diode is not an ON/OFF device. It is ON all the way up to 0.7V
    and beyond. It is just that at the so-called 'turn-on point', it is
    more 'ON' than at 0.5 volts. In 'Math-Speak' the resistance of the
    diode is said to be non-linear. You can show this by increasing the
    value of the series resistance to a high value until the voltage
    across the diode drops and where you will find that current is still
    flowing; plot the values then and look at the curve.

    BTW, the actual diode voltage also depends on the current through
    it and the area of the die among other things. Think about the voltage
    across a bunch of them in parallel at the same total current as a
    single one. It will be slightly lower.

    Cheers - Joe
  4. default

    default Guest

    You have to stop thinking about the diode as just a component to be
    used as a one way valve - you have to understand the atomic theory.

    Both P and N material conduct electricity - not well, but they do

    Ordinarily the silicon crystal is a good insulator. The manufacturer
    adds controlled amounts of contaminants to unbalance the insulating
    properties and this causes it to have a either a surplus of electrons
    (N type material) or a deficit of electrons (P type material) - these
    "carriers" allow the silicon to pass current.

    No big deal so far.

    Now, when a junction is formed with N/P types of material, an
    interesting phenomena takes place for a very thin plane right at the
    junction. Some of the "majority carriers" of the N material are
    attracted to the majority carriers of the P type material and they
    swap places and form a sort of curious insulator because once again
    the atoms are in a balanced state.

    This area where they swap, is called the "barrier region." It is an
    insulator, but not a very strong one. For the diode to conduct you
    merely have to overcome the attraction that is forming the insulator
    (barrier region) with enough VOLTAGE to get the electrons to move
    across the barrier.

    The barrier can be thought of as a small battery with a .6.7 volt

    Put two diodes in series, and they both respond to the same VOLTAGE,
    each will begin conducting CURRENT when both barriers are overcome.

    And diodes do conduct electricity even before the barrier voltage is
    satisfied - a voltage of just a few millivolts will cause some
    conduction but is just a leakage amount. It isn't an on - off thing
    or not perfectly on or off - diodes are non linear in the amount they
    conduct with potential difference.

    If you look at the curves for diodes, a silicon diode will show a
    slight current that will increase as voltage increases until the
    barrier voltage is overcome then the conduction takes off very sharply
    as voltage increases further. Same thing happens with a reverse
    biased diode current increases slowly until the diode breaks down (and
    as long as the current is limited - breakdown doesn't destroy the

    When a diode is reversed biased the depletion region grows and becomes
    thicker - but there's still some small leakage current.

    Any book on basic electronics could probably explain it better than I
    just did . . .
  5. David Harmon

    David Harmon Guest

    On Fri, 06 Jul 2007 08:03:27 -0700 in sci.electronics.basics, TVisitor
    It's not all that completely off. That's a simplified model.

    Slightly better: the diode is almost all off with slight leakage,
    which is enough to equalize the voltage on both diodes (same leakage
    current through both.)

    For a better model, look at the actual graph in the datasheet for the
    actual diode you are using. For example Figure 3. from
  6. TVisitor

    TVisitor Guest

    Actually I think everyone did a good job of explaining my
    "fundamental" problem - thinking of it as a simple on-off switch.
    With it always conducting some current through it, it now makes more
    sense; there is some drop across it, and my head is a lot clearer on
    this now. It also helps as I was having the same fundamental problem
    with transistors.

    (I'll go back and consult my texts for some finer reading, this is the
    problem when you did electronics a long time ago, but since then have
    only been a software guy!)

    Thanks you gents!
  7. I might as well add something else that may help a little. The
    junction voltage barrier is proportional to a logarithm of current
    through it plus an amount related to the 'saturation current' (which
    is just where an extrapolated line intersects the y-axis.) So the
    voltage across goes something like 60mV/10X-change-in-current or some
    not so distant multiplier of that. (10X is a 'decade' change.) That's
    often close for small signal transistors, for example. Diodes are a
    little different from the base-emitter junction of a transistor, in
    that there is more commonly something closer to double that rate --
    maybe 100mV/decade or more. There is a number of factors leading to
    the difference, but these are often combined into a single number
    called the 'emission coefficient.' For BJTs, it's often near 1. For
    diodes, often closer to 2, though it can vary to maybe 3-ish.

    Anyway, without necessarily knowing the exact figure of change, it
    gives you a rough handle to just know that there is something on the
    order of a 100mV change in voltage across a diode for each 10-fold
    change in current through it. If you know that there is, for example,
    0.6V at 1mA, then you can guess that there will be about 0.5V at
    100uA, 0.4V at 10uA, 0.3V at 1uA, and 0.7V at 10mA. Something like
    that. If you continue that process into lower currents for the
    example diode I'm talking about, you will see 0V at 1nA. That would
    be the 'saturation current' of the diode, the concept I mentioned
    earlier as the y-axis intercept.

    For BJTs, that will probably be closer to around 60mV/10-fold, but
    that gets the idea across. Also, they will usually have a somewhat
    higher voltage at 10uA base current, maybe 0.6V or so. So with
    smaller steps (60mV) for each decade down, you can see that it would
    take 10 decades of current to get to 0V, which implies a smaller
    'saturation current' of maybe 10^-15 (1fA) or so.

    You usually don't worry so much about all this. You just use diodes
    and can roughly guess their voltage drops for some ballpark current.
    100mA would be about 0.8V, as a guess. And that usually is good
    enough. If you are off by a factor of 2 in your guess about the
    current, it won't make all that much difference in the voltage you
    guessed at.

    There is another confounding factor at high currents, which is a kind
    of current crowding. When you start pushing a lot of current through
    a small signal diode, maybe more than 100mA or so (maybe a factor of
    100-1000 less than this for small signal BJT base currents), you start
    to see the voltage rise more rapidly than you'd expect. So if you
    were talking about an amp or more, I'd guess that the rate of voltage
    change would be significantly higher than 100mV/decade for a small
    signal diode. More current-capable diodes will be uniform for much
    higher currents, with current crowding taking place at much higher
    current densities.

    So, don't crowd your current (pick a diode or transistor designed for
    the currents you are working with) and the rules work pretty good.

    I'm a hobbyist, by the way, and have _NO_ formal training at all. If
    a professional corrects me on the above, listen to them.

  8. Don Bowey

    Don Bowey Guest

    On the other hand........ I don't believe I have seen a case of anyone
    correcting Jon.
  9. neon


    Oct 21, 2006
    diodes do not turn on ever but if ever see the charateristic of one you will understand that they exibit a hi impedance by boltzman constant equation to a point where the inpedance remain very much the same for ever incresing current making it a .7v germanium diodes are .2v and tunnel diodes they will oscillate with the proper bias. the .7 is an assumption not a rule the voltage can change with current and temperature because you are changing the qiuescent point.
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