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I've worked through Forrest Mims' Getting Started in Electronics, but cannot figure out why the below circuit (on page 92 in the fourth edition) claims that LED1 and LED2 alternate in being lit depending on the state of input IN1. From everything I know, the path via LED2 will always be at least equally attractive (i.e. of a lower or equal impedance) to the 9V current in the circuit compared with the path via Q1, so the circuit will alternate between only LED2 lit and both LED1 & LED2 lit, IMHO. I built the circuit (slightly varying the values of the resistors and using a 6V power source) which indeed had LED1 & LED2 lit at the same time when IN1 is high. Am I wrong in my understanding, or is Forrest's diagram incorrect, and if so, how to correct the circuit so it works correctly (apart from inserting a separate PNP transistor). Note that I simulated the IN1 input using a push button and a constant 5V power source.

schematic

simulate this circuit – Schematic created using CircuitLab

[UPDATE] Just for the fun of it - here is Forrest Mims' response to my notification on this issue in his book:

Dear Philip,

Thanks very much for reporting the omission you described. You are correct. All of the 100 circuits in this book were built and tested 4 times each--but not this one.

I intended to print “RED” over LEDs 1 and 3 and “GREEN” over LEDs 2 and 4 and neglected to do so.

This is copied to the publisher. For the next printing I will add the missing colors over the 4 LEDs.

Thanks again for your kind note. I believe yours is only the second omission/error found so far in this book.

Best regards,

Forrest

Forrest M. Mims III

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    \$\begingroup\$ Well, you are correct, given these parameters LED2 should always be lit. There has definitely been an error, on the original diagram, the one here, or in the text. \$\endgroup\$ – Sean Boddy Apr 4 '15 at 2:57
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    \$\begingroup\$ Is maybe LED1 a red LED, and LED2 a green LED or a blue LED? \$\endgroup\$ – davidcary Apr 4 '15 at 13:11
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The only way this works is if the forward voltage on LED2 is higher than that of LED1. For example, LED2 might be a green or blue LED with a forward voltage of about 3.2V. LED1 could easily be a red LED with a forward voltage of about 1.7V. When Q1 turns ON, the forward voltage of LED1 pulls the collector voltage of Q1 below the turn-on voltage of LED2 and it goes out.

Basically, Q1 diverts all available current into LED1, thus turning LED2 off.

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    \$\begingroup\$ YES! After making LED2 a green LED and LED1 a red LED, the circuit indeed works as Forrest describes. He doesn't mention these LED characteristics in his text though. Thanks for the hint @Dwayne Reid \$\endgroup\$ – Phil B. Apr 4 '15 at 13:25
  • \$\begingroup\$ Now for a very basic, fundamental question: Why is the voltage at the collector of Q1 1.7V (so R2 covers the remaining 7.3V) and not 3.2V when Q1 is on? Is it because the impedance of LED2 is higher (hence the 3.2V forward voltage) and therefore the Q1 route has lower impedance and receives the current? \$\endgroup\$ – Phil B. Apr 4 '15 at 14:53
  • \$\begingroup\$ To understand this, you need to take a look at the voltage / current characteristics for your LED. In a nutshell, a LED presents a fairly-high impedance below its threshold (forward) voltage and a very low impedance as soon as the voltage exceeds that voltage. Sort of like a Zener diode. \$\endgroup\$ – Dwayne Reid Apr 6 '15 at 13:31
  • \$\begingroup\$ thanks for the response - I'm going to start a separate question for this, as what I am trying to understand is a bit more complex than just how 1 LED works. \$\endgroup\$ – Phil B. Apr 6 '15 at 16:18

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