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Herky 061 - Amp it up Mystery Cache

A cache by L7! Message this owner
Hidden : 1/30/2016
Difficulty:
4.5 out of 5
Terrain:
1.5 out of 5

Size: Size:   micro (micro)

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Geocache Description:

THE CACHE IS NOT AT THE GIVEN COORDINATES.

You will need to solve the puzzle to find the actual coordinates.


Welcome to the University of Iowa Hawkeye GeoArt caches!

 

This is a collection of 125 puzzle caches of varying degree of difficulty. Some of the puzzles might be more challenging and some are easy enough to be solved by anyone from the other two state universities (Just kidding, Iowa has 3 fantastic state universities). ;-) Hopefully everyone will find something that they enjoy. Geocheckers will be provided on some, but not all, of the caches. All caches are within 4 miles of the given coordinates. Permission was obtained for deviation from the suggested 2-mile radius due to permit restrictions and housing density in the area surrounding the geoart.

Good luck and have fun!


Simple Transistor Amplifier


Question for “A
What type of transistor is this?
- PNP: A = 0
- NPN: A = 9
- Ceramic: A = 3
- Sputter: A = 6

Question for “B
What type of transistor amplifier configuration is this?
- Common Emitter: B = 3
- Common Base: B = 5
- Common Collector: B = 7
- Base Follower: B = 9

Question for “C
The input is applied to the?
- Collector: C = 1
- Emitter: C = 4
- Gate: C = 6
- Base: C = 3

Question for “D
Notice that the input is 180° inverted to the output. Why is this?
- +Vcc is a negative voltage causing an avalanche effect: D = 0
- A sputter transistor will always invert the output as long as +Vcc is positive: D = 1
- As the input goes high the transistor conducts more and drops less voltage, as the input goes low the transistor conducts less and drops more voltage: D = 3
- When a Common Collector transistor amplifier has a polarized capacitor on the emitter the transistor will turn on and off due to the capacitor discharging through the emitter: D = 9

Question for “E
What is the purpose of “RL”?
- It serves as the gate keeper for the entire universe: E = 0
- It allows the input voltage to remain constant: E = 7
- It allows the voltage between the gate and the source to remain consistent: E = 5
- It drops the voltage that the transistor (and RE) doesn't drop: E = 3

Question for “F
If the transistor was saturated (fully conducting) the output would be?
- Very low: F = 6
- Very high: F = 8
- Unstable: F = 0
- Greater than +Vcc: F = 7

The Cache is at:
N 41° 39.ABC W 091° 35.DEF

Check your solution

Additional Hints (Decrypt)

Guvf chmmyr yvirf hc gb vgf qvssvphygl engvat, ohg tbbtyr vf lbhe sevraq. Fgneg jvgu gur jbeq "nzcyvsvre" naq gel cnfgvat gur dhrfgvbaf vagb gur frnepu one. Jung qb lbh svaq jura lbh qb guvf jvgu dhrfgvba Q? Ubj zvtug guvf yrnq lbh gb bgure nafjref?

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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