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The Art of Electronics: Solutions to Exercises?

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rstofer:

--- Quote from: masterquiper on April 06, 2017, 09:31:01 am ---
In this simpler case:
Using 5V as Input (from arduino), adding a resistor of 150 Ohms and a single LED of 2V and 20mA. The resistor will use 0.16amps and the LED 0.02? This looks quite inefficient, or am I wrong?

Really appreciate,
Masterquiper

--- End quote ---

5V source - 2V LED => 3V across the resistor.  20 mA forward current with 3V drop => 150 Ohm resistor and the current is the same in both the LED and resistor.  The resistor is dissipating E^2 / R watts => 9/150 or 0.06 Watts.

Every time you use a resistor, you are creating heat.  By definition, this is inefficient.  Unfortunately, this power loss is just a cost of doing electronics.  We work to minimize it but we can not totally avoid it.

iainwhite:

--- Quote from: Nusa on April 06, 2017, 11:54:29 am ---I can't correct your math, since you didn't show it. No idea where you got 0.16 amps.

--- End quote ---

I believe he used the E^2/R equation from a prior post i.e.  5x5v / 150 Ohms = 0.1666  Amps for the resistor.
Not saying that is correct - just my guess at the way it was worked out.

 

jensenr30:
I recently discovered this repository where people are writing solutions to exercises (in LaTeX!) for the 3rd edition of The Art of Electronics.

https://github.com/milesdai/TAoE3Solutions

The PDF has about 20 solutions so far.  I've added a few that should be merged into master soon.  Hopefully we can get more community involvement to speed up development of this collection of solutions.

PDF direct link: https://github.com/milesdai/TAoE3Solutions/blob/master/Chapter1.pdf


--- Quote from: Asmyldof on May 19, 2015, 10:16:23 pm ---I think I would be willing to contribute a little to some solutions and the like if the Wiki comes to life.

Keep me informed about any help I can provide and I will see what I can do.

--- End quote ---
Here is your opportunity!

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