All About Circuits Forum E-book Correction Zener Diodes section - error after ohms law section
 Register Blogs FAQ Members List Today's Posts Search Today's Posts Mark Forums Read

 Notices Welcome to the All About Circuits forums.Our forum is a place where thousands of students, hobbyists and professionals from around the world share knowledge and ideas. You are currently viewing our boards as a guest which gives you limited access to view most discussions and access our other features. By joining our free community you will have access to post topics, communicate privately with other members (PM), respond to polls, upload content and access many other special features. Registration is fast, simple and absolutely free so please, join our community today! If you have any problems with the registration process or your account login, please contact contact us.

 Feedback and Suggestions Forum for providing feedback and suggestions about All About Circuits, including corrections to the e-book. This forum is not for getting help with technical questions.

#1
04-25-2012, 03:32 PM
 Recrex New Member Join Date: Apr 2012 Location: Indiana Posts: 1
Zener Diodes section - error after ohms law section

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.
#2
04-25-2012, 08:06 PM
 Bill_Marsden Super Moderator Join Date: Mar 2008 Location: Dallas, TX (GMT-5 w/ DST) Posts: 19,042 Blog Entries: 5

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

Quote:
 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 ľA. 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.

Quote:
 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.
__________________
..
"Good enough is enemy of the best." An old engineering saying, Author unknown.

General info:
If you have a question, please start a thread/topic. I do not provide gratis assistance via PM nor E-mail, as that would violate the intent of this Board, which is sharing knowledge ... and deprives you of other knowledgeable input. Thanks for the verbage Wookie.

 Related Site Pages Section Title Worksheet Performance-based assessments for semiconductor circuit competencies Worksheet Regulated power sources Worksheet Zener diodes Worksheet Th'evenin's, Norton's, and Maximum Power Transfer theorems Worksheet Ohm's Law Textbook Transistor as a switch : Discrete Semiconductor Circuits Textbook Voltage regulator : Discrete Semiconductor Circuits Textbook Zener diodes : Diodes And Rectifiers Textbook Voltmeter design : Dc Metering Circuits Textbook Simple series circuits : Series And Parallel Circuits

 Similar Threads Thread Thread Starter Forum Replies Last Post Ratch Physics 3 05-17-2008 03:41 AM Joey_pr General Electronics Chat 4 11-06-2007 07:59 AM Ubiyca General Electronics Chat 1 09-18-2007 10:54 PM Chernokril Feedback and Suggestions 2 03-14-2007 11:02 PM gemawannabe Homework Help 11 09-26-2006 09:52 PM

 Thread Tools Display Modes Linear Mode

 Posting Rules You may not post new threads You may not post replies You may not post attachments You may not edit your posts BB code is On Smilies are On [IMG] code is On HTML code is Off Forum Rules
 Forum Jump User Control Panel Private Messages Subscriptions Who's Online Search Forums Forums Home Electronics Forums     General Electronics Chat     The Projects Forum     Homework Help     Electronics Resources Software, Microcomputing, and Communications Forums     Programmer's Corner     Embedded Systems and Microcontrollers     Computing and Networks     Radio and Communications Circuits and Projects     The Completed Projects Collection Abstract Forums     Math     Physics     General Science All About Circuits Commmunity Forums     Off-Topic     The Flea Market     Feedback and Suggestions

All times are GMT. The time now is 02:35 PM.