Saturday, March 14, 2015

@MAKE Electronics Experiment 20: Keypad Security System (Epilogue--May be the Book's Fault!)

As I noted that 555 Output Pin 3 voltage will be lower than the Vcc on Power Pin 8 by up to 1.7V.  That probably should have been the first thing I checked, but...  Anyway, I was puzzled by this--how can we flip a 5V relay if we design a 5V input into a chip that cannot put out 5V.  I went the book page on the O'Reilly website and got this response:

Your Errata Submission for Make: Electronics


11:53 PM (10 hours ago)
to meplattland
Hi Virgil Machine,

Thank you for submitting errata! The author of Make: Electronics, Charles Platt, has written you the following response.


I think you're right but I am traveling right now and do not have a copy of the book. I will try to address this soon.


We appreciate the time you took to write.

Kind regards,
O'Reilly Customer Service
O'Reilly Media, Inc.


Your Errata Submission:

Type: Serious technical mistake
Page: 200
Location: Questions, 1st paragraph
Many have had trouble getting the relay to activate due to insufficient voltage. This paragraph says that the reason for using a 555 was to deliver enough voltage. However, according to Charles' Encyclopedia, Vol 2, the voltage on the output pin will be up to 1.7V less than the input.  Since input is 5V, and that's what the relay needs, how can that be? In my case, I'm giving 4.86V to pin 8 and getting 3..77 volts on pin 3.  Is this an error or am I missing something?

So, this may be a dead end.  I have to figure out how to increase the person reports having success with a transistor, but I can't get that to work--I may not be understanding his wiring directions. More learning to do.