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
(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!
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
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
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.