Example questions from the General class question pool

## What is the peak-inverse-voltage across the rectifier in a half-wave power supply? [G7A04]

A. One-half the normal peak output voltage of the power supply
B. One-half the normal output voltage of the power supply
C. Equal to the normal output voltage of the power supply
D. Two times the normal peak output voltage of the power supply

To answer this we must know what is going on in a half wave power supply.

A half-wave rectifier takes, as you might suspect, only half of a given alternating current (AC) input voltage, and turns that into (alright, well, sort of) direct current (DC.)  The above image shows what that basically looks like.  In this instance, it takes the positive voltage and discards the negative voltage.

The peak inverse voltage is the maximum input AC voltage that the device can withstand before it is operating outside of specifications, possibly even to the point of damage.  In a half-wave power supply, we are discarding one half of the wave, so it must be able to take a peak-to-peak input voltage of at least twice its output.

The answer then, is D. Two times the normal peak output voltage of the power supply.

## What information is traditionally contained in a station log? [G2D09]

A. Date and time of contact
B. Band and/or frequency of the contact
C. Call sign of station contacted and the signal report given
D. All of these choices are correct

When keeping logs for your station, you need certain pieces of information for the contact, or QSO to be considered valid.  If you don’t have the correct information in your log, when you go to confirm this contact with the other party, the information may not match, and the QSO might not be confirmed.  In a contest this can cost you points, or you might miss out on an award, or even something as simple as a QSL card.

At a bare minimum, you need the date and time of the contact, in UTC or “zulu” time.  This is also known as Greenwich Mean Time (GMT) and is basically the “world reference” for time.  You need the band, preferably the exact frequency (from which you can infer what band it was.)  You also need the call sign of the operator you contacted, and an additional piece of information that you (and they) can confirm as having been received (and sent.)  In contest settings this is usually a signal report.  Often times it will include other information relevant to the contest, like a location or sequence number.

Therefore, the answer is D. All of these choices are correct.  Now obviously you can record as much information about a QSO as you’d like, including recording the whole thing(!) But at a minimum you need the information outlined above.

## Which of the following is an advantage of using a Schottky diode in an RF switching circuit rather than a standard silicon diode? [G6A06]

A. Lower capacitance
B. Lower inductance
C. Longer switching times
D. Higher breakdown voltage

I know what you’re thinking.  Where did this bit of word soup come from, right?

First off, we need to know the difference between a Schottky diode and a standard silicon diode.  To start, they have a much lower breakdown voltage than a regular diode, which makes them generally better in high speed applications.

Since they aren’t capable of storing as much electric potential, and therefore electric charge, (which is the definition of a capacitor,) we can deduce that it would also have a A. Lower capacitance than a standard silicon diode.  This could make it quite useful in high frequency RF applications.

## Which of the following connectors would be a good choice for a serial data port? [G6B12]

A. PL-259
B. Type N
C. Type SMA
D. DE-9

The answer is D. DE-9.  Now, wait a minute! I thought those were DB-9 connectors! Well, technically no.  The “B” refers to the size of the shell, so the older style 25-pin DB-25 connectors could be called that.  The DB designator just followed along, because they’re used so often for serial ports also.  The connector in the above image is an example of the DE-9 connector.

The remaining three connectors in the question, PL-259, N-Type, and SMA are all center-pin/outer-shield 2 conductor style connectors, specifically made for coaxial cables, like your radio-antenna feed line.  A serial port requires at minimum four conductors, two for transmit, and two for receive.

## What does the Q signal “QRV” mean? [G2C11]

A. You are sending too fast
B. There is interference on the frequency
C. I am quitting for the day
D. I am ready to receive messages

Q-codes are one of those things you just have to memorize. There’s no good way around it.  I keep a list posted near my station, because there’s no way to remember all of them.  There are several that are used often, and this isn’t really one of them. Unless you’re really into CW and digital modes, you probably wouldn’t even think to use it at all.

The answer is D. I am ready to receive messages.  You can also add a “?” to the message, to turn it into the question: “Are you ready to receive messages?” That might actually get used more….

I’ve put a list of Q-codes in the Resources section of the website for your reference.

## At what height above ground is an NVIS antenna typically installed? [G9D03]

A. As close to 1/2 wavelength as possible
B. As close to one wavelength as possible
C. Height is not critical as long as it is significantly more than 1/2 wavelength
D. Between 1/10 and 1/4 wavelength

First off, what the heck is an NVIS antenna?

NVIS is an acronym for Near Vertical Incident Skywave.  The military uses HF antennas of this design in order to facilitate short-range communications.  As the name suggests, the idea is to send the signal almost nearly straight up to the sky, where it will be reflected down again.

The answer is D. Between 1/10 and 1/4 wavelength. At HF frequencies, this is pretty low. Low enough to make it easily deployable.  There’s a lot of math and physics involved, and rather than really try and dive into it without a lifejacket, lets just say that the other answers are incorrect.  The reason is actually pretty obvious when you think about the wavelengths involved.  To mount a horizontal dipole, which is how this technique is normally used, at such distances above ground is often impractical, if not impossible.

## Which of the following is an effect of overmodulation? [G8A08]

A. Insufficient audio
B. Insufficient bandwidth
C. Frequency drift
D. Excessive bandwidth

The answer is D. Excessive bandwidth.  Now, for the purposes of getting this question correct on the exam, I’ll just say this: Just remember the connection between overmodulation and excessive bandwidth.

The real reason this occurs, is technically fairly complex, and rather than try and rewrite and research it on my own, I’ll just point you at this informative piece at stackexchange: http://electronics.stackexchange.com/questions/107128/what-is-overmodulation

They have some animations that clearly show what is going on there, showing the development of the spurious emissions that develop in an AM radio signal when overmodulated.

The above image shows a normally modulated FM signal.  If I locate a decent example or demo of how this also applies in the FM realm, I’ll add to this entry.

## Which of the following frequencies is within the General Class portion of the 80-meter band? [G1A08, 97.301(d)]

A. 1855 kHz
B. 2560 kHz
C. 3560 kHz
D. 3650 kHz

Well, we can eliminate answers A. 1855 kHz and B. 2560 kHz as neither one of those is in the 80-meter band, which is at 3500 kHz to 4000 kHz.  The answer is C. 3560 kHz.  This lies in the CW/data range of the 80m band.  The “phone” or SSB (single sideband) voice portion of the 80m General class portion doesn’t start until 3800 kHz.

This is defined in FCC Part 97.301(d)

Unfortunately, there isn’t a good way to remember the band privileges, except to memorize them.  Sorry!  (If you have a good way to do this, please let me know!)

## What is the turns ratio of a transformer used to match an audio amplifier having 600 ohm output impedance to a speaker having 4 ohm impedance? [G5C07]

A. 12.2 to 1
B. 24.4 to 1
C. 150 to 1
D. 300 to 1

This is a simple ratio type equation, almost.  The equation for solving this problem looks like this:

$\frac{N_{p}}{N_{s}} = \sqrt{\frac{Z_{p}}{Z_{s}}}$

Where N is the number of turns, p and s are primary and secondary respectively, and Z is impedance.  If we substitute our numbers, we get

$\frac{N_{p}}{N_{s}} = \sqrt{\frac{600}{4}}$

$\frac{N_{p}}{N_{s}} = \sqrt{150}$

$\frac{N_{p}}{N_{s}} = 12.247....$

So our answer becomes: A. 12.2 to 1.  Note that if we had flipped the values in the equation, we would have wound up with exactly the same value, only inverted, that is, 1 to 12.2.

(image: https://commons.wikimedia.org/wiki/File:Line_level_audio_transformers.jpg)

## Why is high input impedance desirable for a voltmeter? [G4B05]

A. It improves the frequency response
B. It decreases battery consumption in the meter
C. It improves the resolution of the readings
D. It decreases the loading on circuits being measured

Remember that impedance is the property of a circuit that prevents current from moving through a circuit.  When measuring a circuit, we do not want to alter the operation of the circuit.  In practice this is impossible, there will always be a certain amount of change when we measure something.

However, by using a voltmeter with a high input impedance, we prevent the current in the circuit from splitting off into the meter.  “Path of least resistance” and all that, or in our case, impedance.  If we were to attach a low-impedance meter to the circuit, it would “load” or pull power away from the circuit.

The answer then, is D. It decreases the loading on circuits being measured.