# current of forward-biased pn junction

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

$I = A * (J_p + J_n)$

where $A$ is the surface of the pn junction, $J_p$ is the current density generated by the diffusion of the free holes into the n-type region and $J_n$ 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 $J_p$ 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 $J_p$ is a declining exponential function of $x$, so how can it be constant?

Or is it simply the sum of $J_p$ and $J_n$ 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 ($x$ = $x_n$) 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 $J_p = q * \frac{D_p}{L_p} * p_n_0 * (exp(V/V_T) -1)$

A similar analysis can be performed for the electrons injected across the junction into the p region resulting in the electron-current component Jn, $J_n = q * \frac{D_n}{L_n} * n_p_0 * (exp(V/V_T) - 1)$

where $L_n$ is the diffusion length of electrons in the p region. Since $J_p$ and $J_n$ are in the same direction, they can be added and multiplied by the junction cross-sectional area A to obtain the total current $I$.

• This isn't an area where I'm expert. And furthermore, I haven't a copy of Sedra & Smith to refer to, either. (You might consider using google books to get access to a fair-use section for us to read.) But the total current, I, is constant across the diode and must be the sum of the minority hole current and the majority electron current, each of which do vary as a function of $x$. The sum is constant throughout. But each, individually, does vary with $x$. So I don't see a problem there. – jonk Dec 27 '16 at 23:44
• Let me know if that makes sense to you, or not. I can provide curves to illustrate better what I mean. But I'd have to turn it into an answer to achieve that. I think what I wrote should be enough to get the point across, though. Electrons are injected in the N material. Holes are injected in the P material and the ratio of holes to electrons varies with $x$, from one end to the other. – jonk Dec 28 '16 at 7:55
• I've found the relevant snippet from the book's english version and added it and it does seem clearer than the translated version I've been using. Apparently, since the total current is constant across the diode, we only need to calculate its value at one specific point of the x axis. – Red Herring Dec 28 '16 at 16:34
• But how does adding the minority hole current at the edge of the n-type region and the minority electron current at the edge of the p-type region accomplish that? Does no recombination take place in the depletion region or is it simply that the depletion region has diminished so much that we can safely assume that the borders of the two regions virtually overlap? – Red Herring Dec 28 '16 at 17:16
• (I don't know if you took note of it, but the usual case has high doping for P and low doping for N, I think.) In any case, as the holes approach the junction some of them will recombine with the electrons, which are being injected into the P side from the N side. So the drift hole current decreases slightly towards the junction. What remains crosses the junction into N and becomes diffusion hole current which never makes it all the way to the end of the N material. The reverse discussion describes the electron drift current coming in from the N side. It's a bipolar current. – jonk Dec 28 '16 at 19:36