Hi,
New to this forum, but the article says to notify you of any errors we catch. So here I am...

In the Zener Diodes article, mid way through there is a discussion of adding a load to the circuit and calculating the current, etc. More importantly, finding out if the zener is even turned on looking at the R-dropping resistor.
At the end of the calculations box figuring out (using Ohm's law) what the minimum R-dropping resistor should be is a paragraph described below:
"Thus, if the load resistance is exactly 38.889 kΩ, there will be 12.6 volts across it, diode or no diode. Any load resistance smaller than 38.889 kΩ will result in a load voltage less than 12.6 volts, diode or no diode. With the diode in place, the load voltage will be regulated to a maximum of 12.6 volts for any load resistance greater than 38.889 kΩ."

okay, a couple of words are reversed - Any load resistance 'smaller' - should be 'greater'. (as the example of 100K proves). Like wise - the last line ....for any load resistance 'greater' - should be 'smaller' to keep the 12.6 volts.

Hope this helps.

 

Part of the problem is you are taking the paragraph out of context. An earlier set of paragraphs set up the problem as follows.

Now consider our “power-saving” regulator circuit with the 100 kΩ dropping resistor, delivering power to the same 500 Ω load. What it is supposed to do is maintain 12.6 volts across the load, just like the last circuit. However, as we will see, it cannot accomplish this task. (Figure below)

Zener non-regulator with 100 KΩ series resistor with 500 Ω load.>
With the larger value of dropping resistor in place, there will only be about 224 mV of voltage across the 500 Ω load, far less than the expected value of 12.6 volts! Why is this? If we actually had 12.6 volts across the load, it would draw 25.2 mA of current, as before. This load current would have to go through the series dropping resistor as it did before, but with a new (much larger!) dropping resistor in place, the voltage dropped across that resistor with 25.2 mA of current going through it would be 2,520 volts! Since we obviously don't have that much voltage supplied by the battery, this cannot happen.

The situation is easier to comprehend if we temporarily remove the zener diode from the circuit and analyze the behavior of the two resistors alone in Figure below.

Non-regulator with Zener removed.
Both the 100 kΩ dropping resistor and the 500 Ω load resistance are in series with each other, giving a total circuit resistance of 100.5 kΩ. With a total voltage of 45 volts and a total resistance of 100.5 kΩ, Ohm's Law (I=E/R) tells us that the current will be 447.76 µA. Figuring voltage drops across both resistors (E=IR), we arrive at 44.776 volts and 224 mV, respectively. If we were to re-install the zener diode at this point, it would “see” 224 mV across it as well, being in parallel with the load resistance. This is far below the zener breakdown voltage of the diode and so it will not “break down” and conduct current. For that matter, at this low voltage the diode wouldn't conduct even if it were forward-biased! Thus, the diode ceases to regulate voltage. At least 12.6 volts must be dropped across to “activate” it.

After several problem worksheets to demonstrate the principle it follows with your problem paragraph.

Thus, if the load resistance is exactly 38.889 kΩ, there will be 12.6 volts across it, diode or no diode. Any load resistance smaller than 38.889 kΩ will result in a load voltage less than 12.6 volts, diode or no diode. With the diode in place, the load voltage will be regulated to a maximum of 12.6 volts for any load resistance greater than 38.889 kΩ.



With the original value of 1 kΩ for the dropping resistor, our regulator circuit was able to adequately regulate voltage even for a load resistance as low as 500 Ω. What we see is a tradeoff between power dissipation and acceptable load resistance. The higher-value dropping resistor gave us less power dissipation, at the expense of raising the acceptable minimum load resistance value. If we wish to regulate voltage for low-value load resistances, the circuit must be prepared to handle higher power dissipation.

As you can see, we are talking two different cases, two different circuits. The author was discussing what the minimum resistance could be (maximum load).

A smaller resistance will result in less voltage across the load. This is a true statement. The loading (not resistance, but amps) would be greater.