Kirchhoff Current Loop in Resistor Diode Ladder Network

This looks like a homework question to me. From the CR4 FAQ #14:

• Do your own homework. CR4 is not a homework cheat site. While some here might relish the opportunity to sharpen up old rusty skills by working the homework problem, consider the following and consider it well. If you cheat on your homework by using someone else's answers, you are only cheating yourself, because the purpose of any homework or other academic assignments is to help you learn - by practice, repetition, and self-discovery.

Here are two big hints. You should already be familiar with the Schokley diode equation for your class. Google is your friend.

How would you solve this were the diodes resistors instead?

Good point. I believe it is actually easier to solve this problem with identical diodes than all resistors but the fundamentals are still the same.

Everything to the right of the first resistor and diode is just a repeat of what you are trying to compute. This reduces the problem to a simple quadratic equation.

The solution depends on the accuracy needed as usual. If ideal diodes are considered the following will apply:

Looking at the circuit from the left, when the top terminal is positive to the bottom terminal the current may be calculated as (Vsupply - Vdiode)/R(first). (Vdiode is usually 0.6V for Si diodes). All other diodes will not conduct as they can only get a portion of the break down voltage of the first diode, which will not be adequate to break down any other diode. In other words only the first loop is conductive - all other loops are open circuit.

When the polarity is reversed all (ideal) diodes are open circuit, so the current will be zero. In other words all loops are open circuit.

If practical diodes are considered, one must reconcile Vf versus If characteristics of each diode, the band of variations, etc, which may result in the problem being too complex to solve mathematically, demanding simulations and what-if scenarios.

Hope this helps.

assuming an exponential (approx. Ebers-Moll) diode model, the voltages on the diodes will be close to k * log(log(log(log....Vf))), Vf>:0.

If the resistors and diodes are identical then: Rt = R/N

Otherwise: 1/Rt = 1/R1 + 1/R2 +1/Rx

No. You never get a linear network when half of the parts are non-linear parts. Your equations imply a linear network.