## The KCRC TechNet Puzzle for June 14, 2017

And now for something completely different – a little amateur radio trivia about a revolutionary technological invention (try not to “Google” it – the superheterodyne radio receiver circuit!

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1) When was the superheterodyne (superhet) radio first invented?

………. A) 1898

………. B) 1918

………. C) 1929

………. D) 1942

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2) What does the “super” in superheterodyne stand for?

………. A) Nothing, it’s just a marketing gimmick.

………. B) It stands for “super-duper”!

………. C) It stands for “supersonic”.

………. D) It stands for “super accurate”.

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3) Who invented the superheterodyne radio?

………. A) Edwin Armstrong

………. B) Lee DeForest

………. C) Guglielmo Marconi

………. D) Hedy Lamar

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## The KCRC TechNet Puzzle for May 24, 2017

.And now for the answer to our May 24th, 2017 TechNet Puzzle

(It is in “Invisotext” and will be visible if you highlight the area below!)

Well, this one is kind of easy to guess – it is a voltage divider. The voltage drops across each resistor, and adds up to the total voltage of the source – 20 volts. It would be great if it multiplied voltages, but that would defy a couple of laws of physics, so answer “D”, 82 volts is definitely out!

If it divides voltage it couldn’t be answer “C” 20 volts – that’s the entire source’s voltage potential!

Most people with an idea of how a voltage divider works should realize that the only way to evenly divide the voltage drop equally would be to have both resistors be of the same value, which is NOT the case here, so answer “B” 10 volts can’t be right.

So, by the simple process of elimination the answer is “A” 2 volts!

But wouldn’t it be nice to know why the answer is “A”?

Consider good olde “Ohm’s Law”:

E = IR

Or

I = E/R

Simplify the circuit by adding both resistors, as you can quite easily do when they are in series, and the total current flowing through the circuit is:

20 volts/ 92,000 ohms = 217 microamps

Now, by using ohms law, E= IR you can figure out the voltage potential drop occurring across each resistor – the 82K ohm resistor would be:

E = 217 microamp x 82,000 ohms = 17.8 volts

And the 10K ohm resistor would be:

E = 217microamps x 10,000 ohms = 2.2 volts

If you want a formula, the voltage drop across R1 with two resistors in series, R1 and R2 could be written as:

E_{voltage drop} = (R1/(R1+R2)) x E_{source}

But isn’t it nicer knowing why?

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## The KCRC TechNet Puzzle for May 10, 2017

And now for the answer to our May 10th, 2017 TechNet Puzzle

(It is in “Invisotext” and will be visible if you highlight the area below!)

Sorry, but you just have to figure out what the darn device is. It looks a lot like a *strip line* connected to coaxial cables externally with a pair of *strip lines* internally inductively coupled to the main *strip line*. Why two couplers? Well, if you take a look at the way the diodes are connected, they appear to be *directional * power couplers, one detecting *forward *power, the other, reverse power. So, what would switch “S” do? It switches the forward power coupler to to the power meter in one position, and the reverse power coupler in the other position!

Just like choice “D” Forward-Reverse.!

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## The KCRC TechNet Puzzle for April 26, 2017

And now for the answer to our April 26th, 2017 TechNet Puzzle

(It is in “Invisotext” and will be visible if you highlight the area below!)

Well, the illustrated schematic is a simple RC circuit. The instant that the switch closes the circuit, current flows into the charging capacitor slowed down by the resistor in series with it – over time as the capacitor develops a greater and greater charge the current flow declines until at the point at which the capacitor is fully charged when the current ceases to flow at all.

At “zero time” the capacitor acts as if it were a short circuit and the voltage potential between both sides of it are zero – it is only after a charge begins to develop in the capacitor, and current flow is diminished by the charge repulsion of the partially charged capacitor, that a voltage differential, voltage drop, rises and finally approaches a constant value, giving you a curve that looks like this:

Does that look anything like the curve in the question?

Nope, so it ain’t “A”.

Talk about coincidence – that’s pretty much the same curve for the charge developing over time on the capacitor…

So, it ain’t “B” either!

Well, what about it being a curve showing the time constant, like answer “C”? The problem is that an RC circuit’s time constant isn’t a curve it’s a…

Constant!

τ=RC

Or, in the wise words of Wikipedia:

“Physically, the time constant represents the elapsed time required for the system response to decay to zero if the system had continued to decay at the initial rate, because of the progressive change in the rate of decay the response will have actually decreased in value to 1/e ≈ 36.8 % in this time (say from a step decrease). In an increasing system, the time constant is the time for the system’s step response to reach 1-1/e ≈ 63.2% of its final (asymptotic) value (say from a step increase).”

(Try to say that three times fast!)

So, if the graph looked like this:

It might be choice “C”…

But, since that isn’t the same illustration in the question…

It ain’t “C” either!

So, by the process of elimination (or knowing what the voltage potential drop between a resistor in an RC circuit actually looks like) you have the answer, “D”! Initially when the circuit is closed the current flows as if the capacitor was a simple piece of conducting wire – under that circumstance the voltage drop between the two leads of a single resistor in a simple resistor circuit would be the entire voltage potential applied to the circuit, BUT as the capacitor develops its charge, the current slows down with the capacitor acting as a component whose resistance to the flow of current, increases over time. Eventually the capacitor acts as an open circuit not allowing any current to flow – at that point the voltage potential difference between both sides of a resistor, its voltage drop, when NOT connected to a circuit with an actual current, is zero.

So, “D” is the answer, and if ya think about it for just a little while, it makes a lot of sense!

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## The KCRC TechNet Puzzle for April 12, 2017

And now for the answer to our April 12th, 2017 TechNet Puzzle

(It is in “Invisotext” and will be visible if you highlight the area below!)

Well, there are four basic types of filters:

Curve 1 isn’t really a filter, it is a straight wire connection that allows the entire frequency spectrum to be passed unaffected. Curve 2 is a notch filter – it allows DC to low frequencies as well as high frequencies be passed and only blocks a small selection of frequencies in the middle range. Curve 4 is a low pass filter, blocking higher frequency signals. If you look at the circuit shown and just think of how it would deal with the ultimate low frequency 0 Hz – DC, you can see that a DC current would short itself out by the shorting inductor – this would rule out curves 1, 2, and 4 and only leave a band pass filter curve 3! Indeed, the schematic is for a band pass filter, but even if you did not recognize that, you could still work out how it would behave logically!

So the answer is “C” curve 3!

## The KCRC TechNet Puzzle for March 22, 2017

And now for the answer to our March 22nd, 2017 TechNet Puzzle

(It is in “Invisotext” and will be visible if you highlight the area below!)

At its resonant frequency a quarter wave vertical acts just like any other serial resonant LC circuit – this means that it’s voltage peaks at its tip (this is why it is more risky to get an RF burn at the tips of a resonant antenna), and its current peaks at the site of the antenna feed (this is why the position of the antenna feed is more important than any other portion of the antenna and why you see A LOT more inverted “V” dipoles than you do noninverted “V” dipoles (the radiating centers are higher and less likely to suffer from ground wave absorption).

So, the answer is B, drawing 2!

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## The KCRC TechNet Puzzle for March 8, 2017

And now for the answer to our March 8th, 2017 TechNet Puzzle

(It is in “Invisotext” and will be visible if you highlight the area below!)

Ultimately, this question is asking for the length of a quarter-wave coaxial stub for 145 MHz. You can calculate wavelength as:

Wavelength = The speed of light in a vacuum / the frequency of the wave

Wavelength = 299,792,458 meters per second / 145,000,000 cycles per second

Wavelength = 2.06 Meters

1/4 wavelength = 1/4 x 2.06 Meters = 0.50 meters = 50 cm

(Of course the simplest way to approximate is:)

Wavelength = 300/Frequency in MHz = 300/145 = 2.06 Meters

1/4 Wavelength = 1/4 x 2.06 Meters = 50 cm

So…. answer “B” is, right??????

WRONG! Notice that you have been using the speed of light IN A VACUUM. Coaxial cable is nothing like a vacuum, it has a high dielectric material that causes the transmission of electromagnetic waves to slow down…

That’s what ** velocity factor** means – the percentage of the speed in a medium compared to its speed in a vacuum.

In the case of a solid polyethylene dielectric coaxial cable its * velocity factor* is 66 percent (in the case of something like LMR400, which uses a closed cell foam it is 85 percent).

As you slow down a wave’s velocity at the same frequency, you shorten its wavelength. In this case you shorten it to 66% of your previous calculation:

50 cm x 66% = 33 cm

So, the answer is “A”, not “B” and you are now more aware of how *velocity** factor *can screw up a calculation!

## The KCRC TechNet Puzzle for February 22, 2017

And now for the answer to our February 22, 2017 TechNet Puzzle.

(It is in “Invisotext” and will be visible if you highlight the area below!)

Simple diodes may supply a voltage drop when forward biased, but they do not regulate the voltage output. Zener diodes are designed for a specific reverse bias voltage breakdown threshold – if you reverse bias the Zener diode across the output power leads, when the voltage supply is a higher than the reverse bias breakdown voltage, the Zener diode conducts, until the voltage output is just below the breakdown voltage. It’s not efficient, but it works. Zener diodes are somewhat affected by temperature variations and current drain. Regulator IC are probably the best method to regulate power, and relatively inexpensive. The only schematic showing a Zener diode reverse biassed is “circuit 2”, which is answer “B”!

## The KCRC TechNet Puzzle for February 8, 2017

The circuit shown is fed from an AC source and two of the three voltmeters show the readings obtained. What should the third meter read?

A 4V

B 5V

C 6V

D 7V

E 8V

And now for the answer to our February 8, 2017 TechNet Puzzle.

(It is in “Invisotext” and will be visible if you highlight the area below!)

Everyone here knows Ohm’s Law:

E = I R

Well, you can expand that a bit for alternating currents:

E = I Z

Where “Z” is, of course, the complex impedance, rather than the pure resistance of a circuit.

In a series circuit the current going through all components is identical, so you can consider it to be a constant “c”:

E = cZ

In other words, voltage is PROPORTIONAL to impedance.

Therefore, if you sum the AC voltage drops as you would their impedances, you can discover the AC voltage of the power source powering this circuit.

Whether you wish to solve it as a rudimentary bit of vector math or a simple Pythagorean Equation, is up to you. The equation for the sum of the voltage drop across a resistor, because of resistance (3V) and an inductor, because of inductive reactance (4V) is:

sqrt(3^2 + 4^2) = sqrt(9 + 16) = sqrt(25) = 5

So, the answer IS 5 Volts (choice “B”). If both components were resistors (or inductors for that matter) the answer would have been 7 Volts, or choice “D”

THE END (and the resistor and the inductor lived happily ever after.)