Hobby Series - Magic Smoke... Mystery Cache

This cache has been archived.

Hydroid: Due to the passing of one of my four legged partners, and advancing age of my remaining fur baby, and the constant threat of ticks - which I want NO part of, I'm limiting my caching activities to the cold weather months. It's not fair to ignore caches throughout the rest of the year, so any of my caches that require maintenence before then will be archived. THIS cache now falls into that category. To those that have logged a DNF, please feel free to log your find on the same day that you logged your DNF.

Regards, Hydroid and the mutt :-)

As this cache is being archived,, but some of the rest of the series - including the FINAL - are still active, the information you would have gleened had the cache been found is:

IR10 = 10.86 mA

More
Hidden : 3/20/2016
Difficulty:
3 out of 5
Terrain:
2.5 out of 5

Size: Size: micro (micro)

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Geocache Description:


There is nothing at the posted coordinates, other than the start of a trail,
but it does make a good place to park. The cache is within 1km.

As this cache is located on Nonquon Crown land, and hunting is permitted, it would be advisable to wear bright colours (Blaze Orange, perhaps) during hunting season. Also, you might wish to keep pets on a leash and refrain from suggesting that the 'better-half' wear that 'Antler Hat', just for fun...

This is the fourth part of a seven part series (6 caches plus a FINAL). Each part will build on the preceding part, so while not impossible to do out of sequence if you have the knowledge, it is suggested  that you do the caches in order:

1 - GC6DDCH - Resistors
2 - GC6DKK8 - Series & Parallel
3 - GC6DR23 - Ohm's Law
4 - GC6DR2X - Magic Smoke (This listing)
5 - GC6DR3G - Breadboards
6 - GC6EG5M - PCBs
7 - GC6DDDW - FINAL

Note: Each of parts 1 through 6 contain bonus information that's written on the top of the log. If you plan on doing the FINAL, please record this information as it will be required.

Magic Smoke:

As you're no doubt aware by now, there's a lot of math involved with electronics. For the most part, in the hobby anyway, you can usually get away without doing a lot of the calculations and just put something together. You call upon past successes - and failures! It will either work, or it won't. However, there are times that an error can result in a range of results from a slight burning smell, from overheated components (been there, done that), to spectacular obliteration of parts - done that too... Generally, there are two causes for the bad outcome. Either you hooked something up in reverse - most electronic parts are polarity sensitive - or, you pushed them beyond their limits to a minor, or major, overload. My high school electronics shop teacher had a saying which I still remember fondly to this day. He used to say that electronic parts were mysterious. They worked because they contained 'Magic Smoke' and if you abused the parts such that they emitted that magic smoke, they stopped working, forever...

One of the ways to easily destroy an electronic part, or anything really, is to overheat it. Parts usually have "spec" sheets associated with them that specify operating limits: Absolute Maximum ratings. One of the most important is power dissipation and is expressed in Watts (W) for higher powered parts, or milli-Watts (mW) for smaller components. Just as a light bulb of higher wattage runs hotter (and brighter), so do parts. But what is a Watt?

Power - Watts:

The Watt is named after Scottish engineer James Watt and is a way to express the rate of energy conversion with respect to time. Think about things you have around the house: Microwave - 1,000 Watts, Toaster - 1,500 Watts, Incandescent light bulbs - 40, 60 or 100 Watts (typically). The higher the wattage, the more power it consumes. It can express non-electrical work too. For example, a 80kg person climbing a 5m high ladder in 10 seconds is doing work at the rate of approx. 400 Watts. We are only going to deal with electrical aspects though: If 1 ampere (1 A) is flowing through a component which has 1 Volt (1 V) across it, then the rate at which work is being done is 1 Watt (1 W).

So, the formula for Power is:
P = V x I

where, P is Power measured in Watts (W). V and I were already covered in part 3, Ohm's Law.

Remembering from Ohm's Law that V = I x R and I = V / R, and doing some substitution into the Power formula above, yields two additional formulas that are useful:
P = I2 x R  and  P = V2 / R

An example:  A 220Ω resistor has a current of 75mA flowing through it. What power is it dissipating? Using P = I2 x R gives, P = 0.0752 x 220 = 1.24W. The standard wattages of small everyday resistors are 1/8, 1/4 or 1/2 Watt. In this case, we would be exceeding the largest of those (1/2 Watt or 0.5W) by more than double. Our resistor would heat up to the point of failure - likely in smoky fashion.

Another example: You buy a 2,000W heater at a garage sale and plug it in at home into a 120V outlet whose circuit is protected by a 15A breaker. Is it likely that the circuit breaker will trip? Using P = V x I, and re-arranging to isolate I gives I = P / V. Thus I = 2,000W / 120V = 16.66A - so yes, the circuit breaker is going to trip.


Retaining the Magic Smoke:

Let's do some simple design ensuring we design to retain the magic smoke...

Say you bought a used vehicle that came with aftermarket fog lights. One thing that has always irritated you is that there is no indication on the dash that the fog lights are on. You mention this to a friend - one of those friends that seem to have a bit of everything, but not much knowledge of what any of it does... He hands you a dusty blister package from Radio Shack (remember them?) labelled "5mm Green L.E.D." and suggests you mount it in your dash, fed from the fog light switch so it lights up when the fog lights are on... The only thing on the back is a diagram depicting an outline of the LED with the words "Anode" and "Cathode" and a line that reads, "Vf = 3.3V @ 50mA". So to see if it works, you temporarily connect it directly to your car battery using some lengths of wire to reach the terminals (as in the left hand side of the above diagram). Hmm, nothing, must be a dud... Remembering that most electronic components are polarity sensitive - which is to say the positive and negative matters, you reverse the wires... There is a momentary flash of bright light, followed by an impressive obliteration of the LED.. Oh yeah, and some smoke.. The magic smoke 

Hmm. You Google "LED on a 12V circuit" and find many, many diagrams. One thing they all seem to have in common is a series resistor inline with the LED (Like on the right hand side of the above diagram). You research some more and discover that when it comes to the term "Vf = X volts @ Y mA" it means Forward Voltage = X volts when the LED has Y mA flowing through it.. It's explained that to achieve the rated light output, you have to pass Y mA through the LED, which causes the voltage drop across the LED to be X volts...

You have an Aha moment, thinking back to Ohm's Law, and realize that you can calculate what resistor size is required. You know the battery is 12V, and at a current of 50mA the LED will have a voltage drop of 3.3V, so that means the resistor must drop 8.7V (12V - 3.3V). Using the Ohm's Law triangle, and covering the 'R' (what you need to find) leaves V / I. You know V = 8.7V and I = 50mA, so V/I = 8.7V / 0.050A = 174Ω.

Your buddy just happens to have another package that's labelled "174Ω - 1/8W" and another LED - Not likely, but just play along here . You connect the LED to the battery as before, but this time with the 174Ω resistor in series with the LED. Awesome, it lights up and is looking great and... what's that smell? Why is the resistor beginning to smoke... Oh no... You quickly disconnect the wires to save the day. It's overheating, but why?

When hooked up, what power is the resistor operating at? Well, from the Power formulas above, you realize that you know all the variables for the resistor (V, I and R), so you can use any of them to calculate power (P). Let's stick with the easiest, P = V x I. This gives, P = 8.7V x 50mA = 8.7V x 0.050A = 0.435W or 435mW. The package said the resistor was 1/8W. 1/8W = 0.125W or 125mW - so the resistor is operating at just under 3.5 times its rating (435/125=3.48). The resistor can't disipate heat fast enough and is thus overheating.

What to do? Your friend, this guy's amazing, just happens to have two more resistors. The packages read, "330Ω - 1/4W" and "370Ω - 1/4W". Would they work? Well, if the resistors were in parallel, their total resistance would be... Oh heck, let's draw it out to keep it straight:

OK, first thing. We know we needed a series resistor of 174Ω, so what is the total resistance of  the parallel R1-R2 pair? Well, we have two - and only two - resistors in parallel, so the special formula of RT = (R1 x R2) / (R1 + R2) will tell us: RT = (330 x 370) / (330 + 370) = 122,100 / 700 = 174.4Ω Close enough that we'll just say it's exactly 174Ω.. We know the voltage drop across the resistors is 8.7V for each because they're in parallel, so to calculate the power of each resistor, we can use P = V2 / R. So for R1, PR1 = VR12 / R1 = 8.72 / 330 = 75.69 / 330 = 0.229W or 229mW. For R2, PR2 = VR22 / R2 = 8.72 / 370 = 75.69 / 370 = 0.205W or 205mW. The packages said that both resistors were 1/4W which is 250mW (1/4 = 0.250W = 250mW), so each resistor is operating within its ratings and will be fine.

One last thing that hasn't been mentioned yet, but is true. Total power in a circuit is the SUM of all the individual powers. It doesn't matter if the parts are in series or parallel, power is just the SUM. So in our example above, the total power being consumed, or disipated, by the circuit is PLED + PR1 + PR2. We haven't mentioned the power of the LED, but since we know it's operating at 3.3V and 50mA, its power is: PLED = VLED x ILED = 3.3V x 50mA = 3.3V x 0.050 = 0.165W or 165mW. So the total power being supplied by the battery is 165mW for the LED, 229mW for R1 and 205mW for R2 for a total of 599mW. It should come as no surprise that if you use P = V x I and see that the battery is 12V and is supplying 50mA, and do the math, you get 600mW (12V x 50mA = 12V x 0.050A = 0.6W or 600mW). The missing 1mW is due to roundoff error when calculating the individual component powers...

The Cache:

The cache is located at:
N44 07.X  and  W078 59.Y

where:
X = VR4 rounded to the nearest hundredth of a volt and then divided by 10

Y = PR2 rounded to the nearest mW

In case you want to double check your answer before heading out, click here
Congratulations to MythicLionMan on his third FTF of this series

Notes:
- Please bring your own pen / pencil for the log.
- This trail, depending on the season, can be icy, deep in snow, or muddy - use caution.

Additional Hints (Decrypt)

Pnpur: - Ovttre guna n svyz pna. - Nobhg 10z bss gur genvy gb gur rnfg - Jung gur Sevraqyl Tvnag jbhyq fnl ng gur ortvaavat bs uvf fubj... - Hfr n arneol gbby, be snfuvba lbhe bja...

Decryption Key

A|B|C|D|E|F|G|H|I|J|K|L|M
-------------------------
N|O|P|Q|R|S|T|U|V|W|X|Y|Z

(letter above equals below, and vice versa)


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