I've been studying pn junctions from the 5th edition of Sedra & Smith's Microelectronic Circuits and I've hit a roadblock regarding how the junction works in forward bias and, more specifically, how its i-v equation is calculated.
From what I've understood so far, in forward bias, free electrons flow into the n-type region and cover up the bound positive charges of the depletion region, while free holes flow into the p-type region and cover up the bound negative charges of the depletion region, thereby reducing the width of the depletion region and reducing the electric field.
This allows the diffusion of free holes from the p-type region into the n-type region, where they become (excess) minority carriers. The concentration of excess minority carriers (free holes) is maximum at the edge of the n-type region, but reduces exponentially the further we go from the edge depletion region, as the free holes recombine with the free electrons of the n-type region, until it becomes 0 at x >= diffusion length.
In equilibrium, the external circuit provides a steady flow of free electrons to replenish the ones that are "lost" when they recombine with the incoming free holes, and the distribution of the concentration of excess minority carriers (free holes) across the n-type region does not change with time, with the excess minority carrier concentration being a declining exponential function of x as described above. (And the same happens with the free electrons diffusing into the p-type region.)
What I can't understand is how does the total current (which would be
where is the surface of the pn junction,
is the current density generated by the diffusion of the free holes into the n-type region and
is the current density generated by the diffusion of the free electrons into the p-type region.)
flowing through the diode remain constant across the length of the pn junction? The S&S book claims that somehow remains constant and equal to its maximum value (aka its value at the edge of the n-type region) across the entire n-type region because of equilibrium. But
is a declining exponential function of
, so how can it be constant?
Or is it simply the sum of and
that remains constant across the length of the pn junction?
From the book:
Observe that Jp is largest at the edge of the depletion region (
=
) and decays exponentially
with distance. The decay is, of course, due to the recombination with the majority electrons.
In the steady state, the majority carriers will have to be replenished, and thus electrons
will be supplied from the external circuit to the n region at a rate that will keep the current
constant at the value it has at x = x„. Thus the current density due to hole injection is given by
A similar analysis can be performed for the electrons injected across the junction into the p region resulting in the electron-current component Jn,
where
is the diffusion length of electrons in the p region. Since
and
are in the same
direction, they can be added and multiplied by the junction cross-sectional area A to obtain the total current.
Best Answer
What you need to understand, is that although Jp and Jn do decay exponentially with distance, at any given forward bias voltage, "electrons will be supplied... to the n region at a rate that will keep the current constant, at the value it has at X = Xn".
Another way to see this, is by analyzing the equation for Jn (or Jp) and noticing that for a particular diode at a given voltage (V), all other variables have a fixed value.
Since Jn (or Jp) ends up with a fixed value, the total current [I] (at a given voltage [V]), is constant!