The KCRC TechNet Puzzles For 2017

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The KCRC TechNet Puzzle for December 27, 2017


Which of the following terms describes the ratio of the minimum discernible signal a receiver can detect and the maximum signal that will not be distorted?

A)  sensitivity
B)  selectivity
C)  dynamic range
D)  signal to noise performance.


And now for the answer to our December 27th, 2017 TechNet Puzzle

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

The best approach for this question is the definition of the terms provided as possible answers:

Sensitivity is the lowest power level at which a receiver can still detect an RF signal and demodulate it into an intelligible signal, often expressed in Microvolts. The more sensitive your receiver is, the easier it is to make sense of those very weak signals.

Selectivity is the ratio of amplitudes, often expressed in decibels, that a receiver can reject signals close to, but not located exactly on the frequency that the receiver is tuned for. The better your receiver’s selectivity, the easier it is to understand a transmission, free of interference by other stations at nearby frequencies on the band.

Dynamic Range is the ratio of amplitudes, usually expressed in decibels, of the weakest signal that can be readable to the strongest signal that will not overdrive the receiver’s circuitry and cause distortion. How quiet whisper can your receiver reproduce, and how ell can it reproduce a yell!

(Sound familiar?)

Signal to Noise Performance is the ratio of the amplitude of the received signal to the background noise level, expressed in decibels. The better the Signal to noise ratio, the easier it is to make sense of those very weak signals.

The definition for Dynamic Range sounds exactly like the question, and that is why the answer to this puzzle is “C” dynamic range.

Sometimes you just have to remember what the words mean.


The KCRC TechNet Puzzle for December 13, 2017

And now for the answer to our December 13th, 2017 TechNet Puzzle

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

Well, this question asks you to recognize the difference between a low pass filter (“A”), a high pass filter (“B”), a notch (band reject) filter (“C”), and a low bandwidth (high Q) band pass filter (“D”). I find that if I don’t recognize the circuit, I take a look at where the inductors and capacitors are arranged. Which are in series, and which are in parallel?

Low bandwidth passband filters require tuned LC circuits either in series to pass the frequency or in parallel to reject the frequency. The circuit above is a bit too simple to offer a low bandwidth resonant frequency, so “C” and “D” are unlikely candidates. Low pass filters usually have, at their simplest, an inductor which blocks high-frequency current, in series and a shorting capacitor, in parallel, that shorts high-frequency current to ground. The above circuit is the exact opposite. It has a capacitor in series, passing higher frequency current through, while inductors, connected in parallel, short out lower frequency currents.

This circuit passes higher frequencies and shorts out lower frequencies – it is, by definition, a high pass filter, so answer “B” is the right one!

Logical, right?


The KCRC TechNet Puzzle for November 22, 2017

A balun transformer has two separate coils to provide a balanced output from an unbalanced input. The RF energy is transferred to the output due to the

A)  mutual inductance
B)  transformation ratio
C)  impedance matching
D)  symmetrical output.

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

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

Well, let’s take your choice from the bottom up!

Symmetrical output? That just means that a circuit has a symmetrical, balanced output, not how. It’s sorta like asking “why is the sky blue?” and being told that it’s because we call that color “blue”… Nope, not this one!

Impedance matching? Impedance matching might be important with you are going from an impedance other than 50 ohms. End fed antennas are often connected to their transmission line with a 9:1 transforming balun, but you can also transform an unbalanced input to a balanced output without changing its impedance whatsoever. So, no, it ain’t this one!

Transformation ratio? Well, what’s being transformed? Why, it’s the impedance, so just look at the previous explanation, which hasn’t changed since then. So, this one is just as wrong!

Mutual inductance? Well, there was probably a good reason that I answered these choices in reverse order! Mutual Inductance! RF currents in the primary winding induces a surrounding rapidly fluctuating magnetic field, this fluctuating magnetic field induces another RF current onto the secondary winding! Unbalanced RF current in -> balanced RF current out, thanks to the miracle of mutual inductance, so answer “A” is the right one this time!


The KCRC TechNet Puzzle for November 8, 2017

An RF power amplifier is rated at 100W output and a gain of 16dB. For full output power without distortion, the input should be…

A)  2.5W
B)  5W
C)  10W
D)  16W


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

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

This requires you to understand how decibels work, which you should if you have a license. Decibels are “tenths of a Bel”, a log base 10 of ratio of quantities. 10 dB is one Bel and one Bel is

1 Bel = Increase value by 101 = 10 fold increase
Decibel values to keep in the back of your head:

   3dB = 2 fold increase
  -3dB = 2 fold decrease
 10dB = 10 fold increase
-10dB = 10 fold decrease

The nice thing about exponential and logarithms are that adding these values is the same as multiplying their increase or decrease, so…

16dB = 10db + 3dB + 3dB = 10 x 2 x 2 = 40 fold increase

How many watts do you need to boost it 40 fold for an output of 100 watts, why 100/40 or 2.5 watts!

So, answer “A” is the right one!


The KCRC TechNet Puzzle for October 25, 2017

An HF dipole is fitted with a trap in each leg to permit matching at two different frequencies. The traps should be resonant at the…

A)  Higher of the two frequencies to cut off the outer section.
B)  Higher of the two frequencies to include the outer section.
C)  The lower of the two frequencies to cut off the outer section.
D)  The lower of the two frequencies to include the outer section.


And now for the answer to our October 25th, 2017 TechNet Puzzle

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

Well you can use a simple understanding of basic radio knowledge to discount two possible answers as technologically impossible. Dipoles are balanced antennas with two legs, each ¼ of a wavelength long, directed in opposite directions. As the resonant frequency gets higher, the dipole leg lengths most become shorter. This alone disallows answers “B” and “C”, since the filters would have the exact opposite effect, lengthening the dipole leg lengths for a higher frequency (“B”) or shortening the dipole leg lengths for a lower frequency (“C”).

Unfortunately, in order to decide whether the remaining possibilities, “A” or “D” are correct, you need to know what kind of tuned circuits are commonly used in trap dipoles. A trap dipole uses a parallel tuned trap to show a higher impedance above a given frequency (they act as low band pass filters) – the increased impedance acts as though the antenna length is shorter than it mechanically seems to be. That is consistent with answer “A”, which is the correct answer for this question!

Although many people own and enjoy their trap dipole antennas, keep in mind that they tend to be more expensive, more difficult to properly adjust, less efficient/more lossy, and subject to deteriorate over time. Your mileage might vary!



The KCRC TechNet Puzzle for October 11, 2017


A 12V transmitter is being tested on the bench and powered by a 12V battery. When on transmit into the station antenna on the roof of the property, it is noticed that the supply voltage at the battery has fallen to about 10V. This is most likely to be due to the…

A   Antenna being too close to the battery and causing RF interference.
B   The power cable between the battery and transmitter dropping too much voltage.
C   Internal resistance of the battery.
D   Increased losses in the transmitter.


And now for the answer to our October 11th, 2017 TechNet Puzzle

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

Well, Radio Frequency Interference (RFI) can do a lot of bad things, but it can’t lower the voltage of your battery (enough of it may be able to fry your battery, but then it would just die and show 0 Volts, not 10 Volts). The answer “A” is not good at all.

We are all told over and over again to keep our power cables short or at least make them very thick – this is because, although power cables have low resistances, they still have some resistance and the longer you make them and the thinner they are, the higher their total resistance and the greater the voltage drop from the source to the load. With this question, it is important to read it in its entirety – the radio and the power supply are both on his bench. If his radio was up on the roof and his power supply was on his bench this could have been a reasonable answer, or at least if the voltage drop were at the radio, rather than the battery’s output. With both the radio and power supply on his bench, he would have to be using a #30 gauge wire to have enough resistance to cause this 2 volt drop, and the wire would incinerate upon his first transmission, and the 2 volt drop would be at the radio’s power input. So “B” is wrong as well…

All electrical devices in the real world have efficiencies of less than 100%, so any transmitter in this question will have some losses, but those are losses INSIDE the transmitter, not going backwards into the battery’s output – so “D” is absurd…

And so, we are left with answer “C”! Although we usually think of batteries as purely energy sources, every device, be they capacitors, inductors, or anything else for that matter, possess an internal resistance. It is the same with our beloved battery – not all battery chemistries possess the same internal resistance, but they all possess an internal resistance. This is not very important when they are connected to a circuit that has a low current. Hopefully everyone remembers Ohm’s Law

E = IR
E/I = R

So, for a given voltage output of your battery, the lower the circuit’s current is, the higher its resistance is. Since a battery’s internal resistance is constant, in series, when the circuit’s resistance is MUCH greater than the battery’s internal resistance, the voltage drop produced by the battery’s internal resistance in negligible. When you key up your transmitter, though, you increase the current being drawn A LOT and lower that circuit’s resistance – at that point, depending upon the internal resistance of a given battery, that can produce a voltage drop that is noticeable and can be deleterious to your equipment. That seems to be the case in answer “C”, the correct answer.

How can you fix this problem? Switch to a battery with a lower internal resistance (usually marketing as a battery that can handle high current loads)!



During the previous KCRC TechNet your humble Net Control Operator noticed something he found curious. He operates the Net in duplex – he uses his Flex 6700 to transmit and receive BUT he also uses his IC-7100 to receive and record both sides of the QSOs as well as visualized the repeater’s output throughout the TechNet. Each radio has its own antenna, approximately 40 feet apart from each other. At the beginning of this TechNet he noticed that, although everyone else’s signal from the repeater was fine, his signal was VERY noisy. This caused him to request, a transmission check, and the responses indicated that his signal was great. Our intrepid Net Control Operator fiddled with the settings, but since the start of the TechNet was going to be any second and everyone else seemed to hear him fine, he decided not to try anything until after the TechNet. After the TechNet was over and an archive of his noisy transmissions was secured for posterity, he pondered this mystery and tested a theory and corrected the underlying problem. Any of you guys have any theories that might have fixed this anomaly? Points for creativity, even more points for the correct answer…


And now for the answer to our October 11th, 2017 TechNet Bonus Puzzle

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

This problem can be explained in one single word: Desensing! The first stage of a receiver is called the front end and is made up of passband filters and low noise amplifiers. The front end’s radio signal sensitivity is usually limited due to thermal noise – warm objects radiate random electromagnetic noise. Because radios are not purely linear devices, strong transmitted signals nearby can produce spurious signals, high above a receiver’s normal thermal noise floor, decreasing the sensitivity of the receiver – desensing the receiver. When my own transmitter was sending my input to the Repeater, my other receiver’s sensitivity plummeted and the Repeater’s output transmissions could barely be received by me. My transmission was fine, the Repeater’s reception and re-transmission was fine, and when I wasn’t desensing my IC-7100 with my Flex 6700’s transmission, my received signal was fine. When I pressed my press to talk button, it got way too noisy!

How to fix this? Well, you can either widen the distance between your receiving and transmitting antennas, or LOWER YOUR TRANSMITTING POWER. Sometimes the latter approach will feel – you may need to lower your transmitter so much, that the Repeater will only receive another very noisy signal, which it will rebroadcast just as noisily. In this case, the power could be lowered with a resulting greater improvement in signal quality.

In this case LOWERING the transmission power actually improved the signal! Who would have thunk it!


The KCRC TechNet Puzzle for September 27, 2017


Which statement below best describes the function of an antenna tuning unit (ATU)?

A    An ATU removes the reactive portion of the feed impedance.
B    An ATU adjusts the length of the antenna to bring it to resonance.
C    An ATU adjusts the input impedance so that Z=50Ω.
D    An ATU brings the input impedance to 50Ω resistive and non-reactive.


And now for the answer to our September 27th, 2017 TechNet Puzzle
(It is in “Invisotext” and will be visible if you highlight the area below!)

Well, it would help if you knew how an antenna tuner actually worked, which could be found at the end of the KCRC Minutes for April 2017:

In short, an antenna tuner is ultimately just an antenna system impedance matching circuit. Your transmitter is designed to pour all of its power out into a 50 ohm, purely resistive impedance – the farther away your antenna system’s load is from that optimal value, the less power is transferred from the power amplifier stage to where it would be useful in radiating a decent radio signal out of your antenna. Even worse, a mismatch will produce power reflections that might turn the important part of your final power transistors into sand.

So, at the very least you will need an impedance of 50 ohms, but does it matter how much of that impedance is pure resistance or reactance?

Unfortunately, it does matter – energy is only dissipated in pure resistances. When an antenna is made resonant it means that its inherent inductive and capacitive reactance gets zeroed out, leaving pure resistance, be it due to ground loss, insertion loss, or the very beneficial radiation resistance.

So, answer “D” is the best answer. Answer “C” is close, but “no cigar”! Answer “A” is partially correct as was answer “C” is, but if your final impedance, although purely resistive, isn’t near 50 ohms, you will still have problems with your power circuit’s energy transfer, as well as the power reflections that might blow out your power module!

Answer “B” is the wackiest choice – antenna tuners amount to a bunch of inductors and capacitors, arranged as impedance transformers. The length of your antenna will not grow, like Pinocchio’s nose, with a turn of your antenna tuner’s dial!?


The KCRC TechNet Puzzle for September 13, 2017

And now for the answer to our September 13th, 2017 TechNet Puzzle
(It is in “Invisotext” and will be visible if you highlight the area below!)


First off, RMS stands for “Root Mean Square”. It has many different meanings in different disciplines, but for cyclical alternating electrical current, like those seen in RF (radio frequency) generation, RMS is equal to the value of the direct current that would produce an equivalent amount of power. For Sine waves (sinusoidal waves), that are usually used in RF generation, this amounts to “peak amplitude” (labeled arrow “2” in the above illustration) divided by the square root of two, or 1.4142135623730950488016887242097… (you get the idea, it’s an irrational number). The only amplitude that is lower than the peak amplitude in this illustration is arrow “1”.

Arrow “3” is the Peak-to-Peak amplitude, and arrow “4” is the period of the wave = 1/f(Hz).

So, the correct answer is “A” Arrow “1”!


The KCRC TechNet Puzzle for August 23, 2017

And now for the answer to our August 23rd, 2017 TechNet Puzzle

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

Probably the first thing to consider is what the heck a “Q Factor” is? Yeah, sure, go look it up and you will see that it is a “Quality Factor” of an oscillator of some kind, but that isn’t how it started. An engineer by the name of K.S. Johnson over at Western Electric Co. just used “Q” because it was one of the few letters not overused in engineering terms. The “Quality” explanation came later on…

Q Factor” was first related to how damped a given oscillator was, but it has found many definitions that explore different aspects of the same inherent characteristic.

One of the many definitions for “Q Factor”, which has a unitless value, is that as it increases, an oscillator has greater amplitude oscillations over a narrower frequency range. Some circuits can have a “Q” in the single digits, but some crystal based oscillator circuits can have a “Q” as high as in the millions!

One definition for a circuit’s “Q” is the resonant frequency divided by the frequency range before the signal drops by 3 dB (in half). If you look at the above illustration you can “guesstimate” that at -3dB (a third of the way down in the first “grid line”, the “offset” intercept looks like more than 1,000 Hz in each direction. So, let’s say that the offset is 1.4kHz on one side and 1.4kHz on the other, and the frequency range is something like 2.8kHz and the resonant frequency is 10.7MHz or 10,700kHz, so the “Q” would be 10,700/2.8 which is 3,821. The closest number that is available from your “eyeballing” of the 3dB point is answer “C” 3200, which in this case is the correct answer!

Now go out there and calculate some Q Factors!



The KCRC TechNet Puzzle for August 9, 2017


Which schematic shows a way to make a non-polar capacitor from polar electrolytic capacitors?

Will the total capacitance be higher or lower than the individual capacitor’s capacitance?


And now for the answer to our August 9th, 2017 TechNet Puzzle

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

An extremely common polarized capacitor is the ubiquitous electrolytic capacitor – it has an anode (+) made from a metal that can form an oxidized insulating layer by the process of anodization (think of the process that gives many Apple products their colored aluminum finish). Any reverse DC current might destroy this anodized insulating layer and “poof” goes your capacitor!

“A” is the worst of all worlds – the diode will only conduct when your polarized capacitor is getting current in the opposite direction that it was designed for, when connected with the correct polarity for the capacitor, the diode will block current!

“B” is just two polarized capacitors in series – their total capacitance is reduced to 5 μF, and if you reverse the current you still are going to kill at least one of the capacitors and the other one would soon follow it and fail as well.

“D” is almost as bad as “A” – one of those two capacitors will be ruined the moment that the current travels through them in the wrong direction, and if you were lucky enough to have the capacitor fail as an open circuit, the remaining capacitor is only safe as long as the current does not change direction!

Well, by the process of elimination, it would seem that “C” was the correct answer, and it IS! This is called a bi-polar electrolytic capacitor, with their anodes connected in series. The total capacitance is lessened to 5μF, but at least you don’t have to worry about how to install it!




The KCRC TechNet Puzzle for July 26, 2017


The primary winding of a transformer has 80 turns and the secondary has 40 turns. If the input impedance is 250Ω, what is the impedance across the secondary terminals?

…….. A)   62.5Ω
…….. B)  125Ω
…….. C)  176Ω
…….. D)  500Ω


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

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

Transformers are AC devices. Effectively, they are magnetically coupled electromagnets. Alternating currents on the first, “primary winding” induce an always changing magnetic field, and that induced alternating magnetic field, in turn, induces another electric field onto the second “secondary winding”. How the output voltage and current on the secondary winding, compares to the voltage and current input on the primary winding, is determined by the ratio of the number of wire turns used in a given transformer.


EP = Input AC Voltage  
ES = Output AC Voltage 
NP = Number of turns on the primary winding
NS = Number of turns of the secondary winding


 If your primary and secondary windings have the same number of turns, you have a 1:1 ratio, the voltage and current induced in the secondary winding is identical to the voltage and current applied to the primary winding – you would have something used all the time – an isolation transformer, that isolates the electrical power from its source, usually for safety measures, or ground loop protection. What about the winding ratio of the question? 80 turns on the primary and 40 turns on the secondary give a 2:1 ratio – Two volts of AC in the input will get you 1 volt on the output, 20 volts on the input will get you 10 volts on the output…

But the question is not about a transformed “voltage” – they are asking for the transformed “impedance”?

We mentioned that your transformer can induce half the voltage it had in the primary winding – does that mean your transformer can make your energy disappear? Sorry! The Law of Conservation of Mass-Energy is not a suggestion, it’s THE LAW!

So, what happens to the current when you are halving the voltage?


IP = Input AC Current  
IS = Output AC Current 
NP = Number of turns on the primary winding
NS = Number of turns of the primary winding

So, the current is being doubled! No energy disappeared today!

Of course, this is quite theoretical, there are other factors afoot that waste energy, turning it into heat that escapes into the system during the transfer of energy. The devil is always in the details…

Everyone remembers Ohm’s Law:

E = IR or R=E/I

Well, Ohms Law for us big boys is actually


Impedance is equal to the Voltage divided by the Current.

So, the transformer cited in the question halves the input voltage while doubling the input’s current on its output winding:


ZP = Input Impedance  
ZS = Output Impedance 

So, halve the turn ration and you halve the output voltage, you double the output current and you transform the impedance from its value at the input to one fourth its original value on the output!

250Ω/4 = 62.5Ω

Answer “A” is the right answer!

(Or you can just remember that a transformer “transforms” the input impedance by the square of its winding ratio.)



The KCRC TechNet Puzzle for July 12, 2017

The antenna shown in the diagram is a

…….. A) Quad
…….. B) Delta loop
…….. C) Trap dipole
…….. D) Folded dipole



And now for the answer to our June 28th, 2017 TechNet Puzzle

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

Well, even if you are not well versed with each type of antenna, look at their names as an indicator of their likely shapes. A quad is a four-sided loop – it can simply be a single rectangular radiator, or much like a Yagi, it can be made of additional rectangular loops, of a slightly larger dimension behind the radiating element, working as a reflector and a series of smaller dimension rectangular loops in front of the radiator acting as directors, improving the antenna directionality!

A “delta loop” is named for the Greek letter delta “Δ”, so it’s a loop shaped like a triangle. The picture above is clearly not triangular!

A trap dipole is antenna with two equal lengths of wire, going in opposite directions, each interrupted by a tuned circuit “traps”. The trap operates as a low pass filter, by letting the RF pass through it at a lower frequency, for which the entire length of the dipole is resonant, but does not conduct the RF at a high enough frequency, effectively shortening the antenna to a resonant length at the higher frequency. In this way, a single length dipole can be made resonant for two or more different frequencies. The same principle can be used with numerous “traps” on the same dipole for even more resonant frequencies! The price that you pay is the power loss inefficiency introduced by the numerous “trap” filters. The picture above is not a length of two wires going off in opposite directions.

A folded dipole is just a simple pair of wires extending in the opposite direction from a central feed point, that have been folded back upon itself either once or more than once. The picture above does not appear to be any form of a dipole, neither a trap dipole, nor a folded dipole…

So, the answer would seem to be “A” a quad!


The KCRC TechNet Puzzle for June 28, 2017

If a 9-turn inductor has a measured value of 20μH and the turns are squeezed closer together the inductance might change to approximately

………A) 20mH
………B) 22μH
………C) 18μH
………D) 10μH


And now for the answer to our June 28th, 2017 TechNet Puzzle

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

There is an “easy” way to answer this question and a slightly harder one – let’s try the easy one first!

This method employs the amazing power of sheer logic, without any fundamental understanding of inductance! If you change the physical characteristic of an object there are only three possible outcomes – it will “increase” some property, it will “decrease” some property or it will leave that random property unchanged. Inductors are infamous for being affected by its surrounding environment – planning to adjust a variable slug inductor shielded in an IF “can”? Better use a nonmetallic screwdriver! Winding some wire with only air around it can make a modest inductor – wrap it around a ferrite doughnut and you have a MUCH more powerful inductor!

Your inductor is unlikely to be unchanged, so answer “A” is equally unlikely to be true. So, let’s guess that it changes its inductance for the worse – the trouble is that both answers “C” and “D” could be true, and there really isn’t that much information to tell you which value would be more likely! So, logically alone, you would be left with answer “B”  22μH, which is the correct answer!

Now for the less easy way:

Capacitors and inductors – two very important types of devices in electronics and especially so in the field of RF electronics. Capacitors store and release the energy of a stored electrical charge into a voltage potential and the flow of current, and inductors store and release the energy of a stored magnetic field into a voltage potential and the flow of current! Inductance causes current flow to “lag” voltage potential, and capacitance causes voltage potential to “lag” current flow! One cancels the other.

Capacitance values are relatively simply determined by the surface area and geometry of the opposing conductive plates, the distance between them and the “dielectric constant” of the material separating them. The equation describing the inductance of an inductor is far less clear cut:

L = (1/ℓ)(μ0KN2A)


L = inductance in Henries (H)
μ0 = permeability of free space = 4  × 10−7 H/m
K = Nagaoka coefficient
N = number of turns
A = area of cross-section of the coil in square meters (m2)
= length of coil in meters (m)

(ℓ is the important variable in your question)

Sufficiently “clear as mud”?

Suffice it to say, there are four factors that principally determine the inductance of a given inductor:

The number of turns of the coils – the more coils, the more inductance.

The coil area – the greater the coil area’s cross-section (the bigger the coils), the more the inductance.

The core material – the greater the magnetic permeability of the core that the coil is wrapped around, the greater the resultant inductance.

The coil length – the greater the coil length, the lesser the resulting inductance (with a longer path for the magnetic field flux to travel there is more opposition to the formation of magnetic flux, for any given amount of field force). Stretch a given coil out until it is a straight piece of wire and the inductance shrinks to its smallest amount (never actually zero – all wire has some inherent inductance AND capacitance!).

So, bunch up a given coil, and its resultant inductance increases…

Giving you answer “B”  22μH!




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 immediately) – the superheterodyne radio receiver circuit!

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

………. A) 1898
………. B) 1918
………. C) 1929
………. D) 1942

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”.

3) Who invented the superheterodyne radio?

………. A) Edwin Armstrong
………. B) Lee DeForest
………. C) Guglielmo Marconi
………. D) Hedy Lamar


What event, framed the end of life of the superheterodyne inventor?

………. A) A Nobel Prize
………. B) An Oscar Lifetime Achievement Award
………. C) An episode of “This Is Your Life”
………. D) Took a “gainer” out of an open window in their 13th floor apartment, plummeting to their death on a nearby roof.


And now for the answer to our June 14th, 2017 TechNet Puzzle
(It is in “Invisotext” and will be visible if you highlight the area below!)


So, here are the answers:

1) “B” – Yup, that brand new shiny rig is, at its heart, operating using technology patented almost a century ago! Not exactly bleeding edge technology there…

2) “C” – They tagged a “supersonic” on heterodyne technology, supposedly because the intermediate frequency used in this technology, was above the range of human hearing? Personally, it sounds more like marketing BS, but that was their reason and they’re sticking with it.

3) “A” –  Yes, Edwin Armstrong, one of the most brilliant engineers of the early 20th Century. He developed all kinds of things that you may have heard of, like the super-regenerative circuit, and FM radio. He was involved in a legal patent case, during which he testified on how DeForest’s Audion tube actually worked (Le DeForest didn’t seem to have a clue at the time!). Marconi’s research took place for the most part before the turn of the 20th Century and was more involved in promoting the technology rather than developing it further when Armstrong was active. Hedy Lamar, in addition to being a successful actress, was a smart lady and was very interested in military weapon technology and came up with the principle on which spread spectrum radio technology is based.

Extra Credit Question: “D” – Yup, life ain’t fair. Edwin Armstrong was a brilliant engineer, but an awful patent lawyer. A parade of larcenous individuals cheated him out of most of the royalties from his designs. In the throes of financial ruin and despair, he sought solace his 13th Floor window and the nature of gravity and ended his own problems, the hard way. His estate was able to arrange for a more equitable sharing of royalties that he deserved, but that didn’t really help him much after he killed himself.

Remember George Santayana “Those who do not learn from history are doomed to repeat it”! So, if you find yourself in a similar situation, spend your time looking for a good patent lawyer, not a window on the top floor!



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


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:

Evoltage drop = (R1/(R1+R2)) x Esource

But isn’t it nicer knowing why?



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.!



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…



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!



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).

One picture is worth a thousand words, so here’s a link to a very nice animated GIF to show this in action!

So, the answer is B, drawing 2!

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.)


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