Tuesday, September 29, 2009

Just one week left to go - and the question numbers have changed!!

Countdown timer: 1 week 0 days to go

Can you believe how fast time has gone around? And here I am, with just a week to go, an anxiety building, thinking about the subjects I still have to cram in !

Just a quick note on something I copped onto today. In Section A the number of questions on various subjects has CHANGED.

There are no longer 12 questions on theory and circuits etc, but just 10.
There will be 6 questions on propagation instead of 4.
There will be 7 questions on antennas and feeders instead of 4.

The above has been confirmed to me tonight by the IRTS. The change was obvious between sample paper revision 3 and revision 4 but to be honest I only saw it today!

Some people might consider this a boost. I certainly know some people who are anxious about the first part of Section A. It seems that the IRTS has rearranged section A slightly to focus on more practical areas and less on the theoretical.

My advice right now is this:

Make sure you can rhyme off all the formulae needed for Section A.

Ohm's Law and its variants - V=IR, I=V/R, R=V/I.
Power and its variants
Inductive reactance
Capacitive reactance
Resonance R = one over two pi square root of LC!

Here is the impedance formula:

How to calculate resistance, impedance and capacitance. Remember resistance and inductance in series = R1+R2, L1+L2 etc, and in parallel is R1xR2/R1+R2; and all this is the opposite with capacitors - i.e. CAPAcitors Capacitors Add in PArallel. So parallel = C1+C2, series = C1xC2/C1+C2.

It's handy to be able to identify various diagrams. These might include the filters (low-pass, highpass, band stop etc) and a balun and a transformer and an oscillator etc but don't stress out on this. Getting one question wrong won't make you fail.

Propagation = 6 questions, so use QADV and answer 30 questions at a time. You'll see a pattern emerging and eventually you won't be able to get anything wrong!

7 questions on antennas and feeders. Again, keep answering questions in QADV.

The other sections are the same as they were in Sample Paper revision 3. 15 questions in Section B covering the operating rules and procedures and the licensing conditions. Remember this is based on the new Radio Amateurs legislation introduced in June and not the previous Radio Experimenters legislation. Read document 09/45 on Comreg.ie to catch up if you haven't done so already. Remember at least one, but probably two, questions in Section B will relate to callsigns. See earlier posting for list of European prefixes. History would show it is European countries which are asked about and not anything obscure.

In Section C, electromagnetic compatibility and transmitter interference are now in one section of 7 questions. Again, hit QADV and answer a couple of hundred questions and you'll see a pattern emerging. QADV works. When you see the same questions coming up over and over and over again you eventually get them right!! I have done 1,400 questions in the past fortnight and got 83% of them right.

For safety (3 questions), one piece of advice is to learn the appropriate values for plug tops, thus:

3A = 550 watts
5A = 1100 watts
13A = 2850 watts

This is one area that I knew nothing about and hence got a question in Sample Paper rev.4 wrong.

Best of luck guys. If I don't update between now and next week, it's because I've got my head in the books!! See you there . . .

Tuesday, September 22, 2009

Diodes and transistors - the final countdown

Countdown timer: 2 weeks 0 days to go

Right. Just a fortnight to go. Last night was spent in the company of Brian EI7GVB to do some study on diodes and transistors. Now I'll be honest in saying that I have been daunted about approaching this subject. I think I've mentioned tackling it half a dozen times on here but never actually touched the subject. I did learn some things last night but the major thing I learned is that there is a certain amount of information to be taken in and it will require concentration to do so.

However, if you're at the stage I'm at and you've covered about 80% of the material you might be of the opinion that you could afford to skip over it. After all, you might only get one or two questions on this in Section A.

Sucker for punishment that I am, there's no option for me but to tackle it and see what I can learn!! More later . . .

Monday, September 21, 2009

Class Act! - A little bit on amplifiers

Countdown timer: 2 weeks 1 day to go
Right, there are four main classes of amplifiers. They are:

Class A
Class AB
Class B
Class C

The sample paper, and indeed the June 2009 paper, asked which class of amplifier is the most efficient, and the answer would be Class C. Class C amplifiers produce harmonics but are only suitable for FM and CW transmissions.

Here's what you need to know about the efficiencies of the various classes.

Class A would demonstrate an efficiency of around 25%.
Class B would demonstrate an efficiency of around 50%.
Class C would demonstrate an efficiency of between 60% and 90%.

In Class A, the input voltage signal and output voltage are 180 degrees out of phase.
Class B requires less bias, and current flows in the output circuit for alternate half-cycles only. As I said above, a practical operating efficiency is around 50%.

According to JOhn Bowyer's Towards the RAE, Class C amplifiers are often used in oscillators and tuned r.f. amplifiers. Expected efficiency is 65%-90%. A class C amplifier is rich in harmonics and can be used as a frequency multiplier.

The total gain of an amplifier in B is the sum of the individual gains, expressed in decibels (dB), of each stage.

Output power is often expressed in dBW, which is the power relative to 1W. More on that later.

You might get a question about the output power of an amplifier. For instance, try this:

A carefully designed class C amplifier is taking an input current of 1A at 24V. The expected output would be:
a 24 W
b 16 W
c 10 W
d 5 W

Now we know class C is 65-90% efficient. Now we need to calculate the power. Here's the formula: P = V*I, and its variants: P = I(2)*R, P = V(2)/R, R = V(2)/P, where the (2) means squared.

Here we know the voltage (V) - 24V - and the current (I) - 1A. So 24*1 = 24. We can see that answer a, 24W, is not possible because that would require 100% efficiency. Answer b is the correct answer, being 66% or thereabouts of 24W.

Sunday, September 20, 2009

A new method to remember the small units !

Countdown Timer: 2 weeks 2 days to go

OK, I was sleeping on this last night. I became a wee bit agitated, nay worried, in recent days about how I would remember the small units. You know, those tiny little parts which are measured in milli and micro and nano and pica? Yes indeed. Those. Well, I came up with a ridiculously simple and slightly bizarre way of remembering the order of them.

You see, when it comes to test day, you might be asked to calculate the total capacitance based on a 20 µF capacitor in series with a 100 nF capacitor. Now looking at the two sample papers, it seems that the trend is to stick to one measurement, ie in sample paper rev.3 you were asked what was the overall capacitance of a circuit with three capacitors all measured in nF. However, it is well to know this stuff just in case.

So here is my silly method. Just think of the following statement:

Man United No Players
Man United's New Players
Man United No Problem

(If you're not a Man U fan perhaps you could make something up yourself. Suggestions welcome!)

Man = m = milli = 10 to the power of minus 3 = .001
United = µ = micro = 10 to the power of minus 6 = .000001
No = n = nano = 10 to the power of minus 9 = .000000001
Players = p = pico = 10 to the power of minus 12 = .000000000001

OK, admittedly the symbol for micro is not a u, but rather that funny looking µ . However, you get the point!

Thursday, September 17, 2009

Some questions on measurements

Countdown Timer: 2 weeks 5 days to go

Right, onto one of those subjects I have not covered in much detail - measurements.

Which one of the following is an essential characteristic of a diode for use in a probe to measure RF?

a) A high current capacity
b) A low internal capacitance
c) A screened outer coating
d) A high peak inverse voltage (PIV)

The correct answer is b) A low internal capacitance.

Which of the following instruments can be used to detect the presence of harmonics?

a) Analogue multimeter
b) Oscilloscope
c) Absorption wavemeter
d) Digital counter

The correct answer is c) Absorption wavemeter.

What name is given to a resistor wired in series with a milliammeter to measure voltage?

a) Shunt resistor
b) Parallel resistor
c) Limiter resistor
d) Multiplier resistor

The answer is d) Mutiplier resistor.

A heterodyne wavemeter consists of an absorption wavemeter and a

a) frequency counter
b) SWR meter
c) digital readout
d) crystal oscillator

The answer is d) crystal oscillator.

What sort of voltage is applied to the horizontal deflection plates of an oscilloscope to produce the timebase?

a) Ramp
b) Square-wave
c) Triangular
d) Sine-wave

The correct answer is a) ramp.

When using an absorption frequency meter, a DC power source is connected

a) to the circuit being tested but NOT to the absorption frequency meter
b) to BOTH the circuit being tested AND to the absorption frequency meter
c) NEITHER to the circuit being tested NOR to the absorption frequency meter
d) to the absorption frequency meter but NOT to the circuit being tested

The correct answer is a) to the circuit being tested but NOT to the absorption frequency meter.

A resistor wired across the terminals of a milliammeter to enable it to measure a higher current is known as a

a) expanding resistor
b) current shunt
c) bridge
d) series resistor

The correct answer is b) current shunt.

Which one of the following is a disadvantage from which an absorption frequency meter suffers?

a) Low sensitivity
b) High cost
c) It only responds to DC
d) Its power supply must be stable.

The answer is a) Low sensitivity.

Tuesday, September 15, 2009

Three weeks to go - and much to be done !

Countdown Timer: 3 weeks 0 days to go

Well, there's just three short weeks left to go until the exam, and I'm a bit anxious. I feel that I've made major progress, of course, but just fear that on the day a whole heap of questions will be thrown at me that I can't deal with.

That's fear of the unknown for you.

Looking at the facts as they stand, I've come a long way in seven weeks. I have covered a lot of topics which were previously pretty alien to me, and have even managed to convince myself of something I thought I'd never say - the maths ain't so difficult after all. There you go, I said it !

So in reality, I've put myself in a reasonable position to pass the exam. I know there's a chance that questions will come up that I won't be able to answer, but that has been the case since the start. What I need to do now, and what I would urge all of you to do, is to use the time left (21 days in total!) to cover over the things you are weak on.

In my case, these topics would include:

Power (haven't learned the formulae yet)
Transistors and Diodes (not enough done here)
Circuit diagrams (need to learn filters, baluns etc)
Oscilloscopes and sine waves (not a tap done there yet!)

Although I have done a lot of study this week on Elecromagnetic Compatibility (EMC), I still feel there is room for improvement.

To aid me in further study, I am going to concentrate on the best sources available to me. These would be:

Towards the Radio Amateurs' Examination by John Bowyer


The IRTS course study CD.

Both are invaluable. The Bowyer book because it has many sample questions, plus the answers, plus an explanation of the answers. The IRTS disc because it is relevant to the Irish exam.

Oh, and of course I will continue doing the QADV questions like a maniac.

Anyhow, I need at this point to give you a few words of encouragement. Michael from Kilkenny is studying hard having failed the exam in June. He has come a long way. He gave me a DVD which contains material on the maths for the exam which I started watching last night and it is amazingly beneficial. Keep up the good work Michael - you'll get there.

Just by way of testing myself, I did the two sample papers again today, mainly to see if I have kept all the information in my head. Here are my results:

Sample Paper Revision 3:

Section A: 100%
Section B: 93.3%
Section C: 80%

Result: PASS !!

Sample Paper Revision 4:

Section A: 90%
Section B: 100%
Section C: 90%

Result: PASS !!

Look, it's as simple as this. If I can get this far, so can you . . . !! :D

Monday, September 14, 2009

Electromagnetic Compatibility (EMC) - More Questions and points

I'm finding it a useful exercise, when doing the QADV questions on Electromagnetic Compatibility, to try to draw the circuit diagrams for the following, in order to help me better remember them:

Band-stop filter
Band-pass filter
High-pass filter
Low-pass filter

Perhaps you might benefit from the same exercise?

Here's a point which you might need to know for the EMC questions:

What is another name for a notch filter? The answer is a Band Stop Filter.

Band pass filters are constructed using:

a) capacitors alone
b) transistors and resistors
c) capacitors and inductors
d) diodes and inductors

The correct answer is c) capacitors and inductors.

Another name for 'blocking' (something that a very strong signal can cause in a receiver) is:

a) de-sensing
b) decoupling
c) attenuating
d) supressing

The answer is a) de-sensing.

If you thread a ferrite bead onto a wire, it increases the inductance of the wire.

Saturday, September 12, 2009

Some Electromagnetic Compatibility questions

OK, we're going to start on EMC with some questions from the QADV program:

If a TV picture contains a 'ghost' in the form of a faint image of the same picture displaced to the right, this CANNOT be caused by

a) large buildings
b) trees
c) tower cranes
d) transmissions by radio amateurs

The answer is d) transmissions by radio amateurs. (By the way the image is known as "Ghosting").

In amateur radio, a low-pass r.f. filter is used to

a) decrease the impedance of an antenna
b) suppress harmonics
c) increase audio intelligibility
d) remove intermodulation products

The correct answer is b) suppress harmonics.

Which one of these actions would reduce breakthrough from a radio transmission onto a TV?

a) Use a speech processor in the transceiver
b) Lower the height of the transmitting antenna
c) Move the transmitting antenna away from the TV
d) Operate using SSB exclusively.

The answer is c) Move the transmitting antenna away from the TV.

Remember that d) is incorrect because SSB is more likely to cause breakthrough than AM or FM.

Which of the following should be fitted in the downlead of a TV antenna to prevent breakthrough from a 2 metre amateur signal?

a) Band-stop filter
b) Mains filter
c) Low-pass filter
d) Balun transformer

The correct answer is a) Band-stop filter.

The occurrence known as the "rusty bolt effect" results in

a) image breakthrough onto the i.f. of a receiver
b) 'ghosting' on analogue TV signals
c) an increase in the maximum usable frequency (MUF)
d) the generation of passive intermodulation products

The answer is d) the generation of passive intermodulation products.

Which one of the following helps to prevent the second harmonic of a transmission being radiated?

a) A key-click filter
b) An output stage with valves or transistors in push-pull
c) A vertical antenna
d) A screened mains lead to the transceiver.

The correct answer is b) An output stage with valves or transistors in push-pull.

To avoid EMC problems, the transmitting antenna should NOT be

a) mounted high up
b) balanced
c) connected directly to the transceiver
d) located at a distance from the house

The correct answer is c) connected directly to the transceiver.

The gain of a receiver is being affected by a strong s.s.b. signal on another frequency. This effect is called

a) cross modulation
b) image pick-up
c) ghosting
d) fading

The correct answer is a) cross-modulation.

Wednesday, September 9, 2009

Less than four weeks to go now

Countdown Timer: 3 weeks 6 days

I was told by Brian EI7GVB, or should I say warned, that I would not be able to study every day for the exam, and that there would be times when I would have to take a break from it and clear the head. Well, the past week has been that time for me. I have not touched a book, looked at a question, or updated the blog in over a week now. Probably a bad idea!

Anyhow, I'm hoping to get back into the swing of it this week. I need to do Mosfet and Jugfet transistors, and also diodes. And then I want to move onto EMC - otherwise known as Electromagnetic Compatibility. Sounds daunting, doesn't it? Indeed, such things as chirp and frequency drift come into play. I will, of course, properly define what EMC is. But for the moment, I will just quote from the exam syllabus if you wish to Bing or Google some of the search terms and do a bit of homework.

Or I would, if I could find it on the syllabus!! Indeed, I have checked both revision 3 and revision 4 of the exam notes and sample paper, and it does not seem to be there!

OK, for now, we'll just have to do with a basic definition of EMC from Wikipedia:

Electromagnetic compatibility (EMC) is the branch of electrical sciences which studies the unintentional generation, propagation and reception of electromagnetic energy with reference to the unwanted effects (Electromagnetic interference, or EMI) that such energy may induce. The goal of EMC is the correct operation, in the same electromagnetic environment, of different equipment which use electromagnetic phenomena, and the avoidance of any interference effects.

In order to achieve this, EMC pursues two different kinds of issues. Emission issues are related to the unwanted generation of electromagnetic energy by some source, and to the countermeasures which should be taken in order to reduce such generation and to avoid the escape of any remaining energies into the external environment. Susceptibility or immunity issues, in contrast, refer to the correct operation of electrical equipment, referred to as the victim, in the presence of unplanned electromagnetic disturbances.

Interference, or noise, mitigation and hence electromagnetic compatibility is achieved primarily by addressing both emission and susceptibility issues, i.e., quieting the sources of interference and hardening the potential victims. The coupling path between source and victim may also be separately addressed to increase its attenuation.