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For the Ham, a power transformer is a device used to step down, or step up the voltage, typically coming from the mains, to power equipment. They are also used at various stages in the electrical grid which gets power from the power station to the consumer. Other transformers handle audio and radio frequency signals, audio output to a speaker, or RF power to an antenna.
If a power transformer is placed across the mains (or a generator), a small "magnetising" current induces an alternating magnetic flux in the core material, typically laminated steel. This causes an alternating voltage across the output terminals. This transfer of energy is the result of "mutual inductance". The amount of current flowing in the secondary is governed by the output voltage, and the resistance of the load. As discussed below, this is reflected in the current in the primary.
These consist of two or more coils of wire, usually with enamel as the insulation. These are would on a core of laminated steel. One shape is the "E-I" core, with interleaved layers on a E shaped leaf facing an I, then a I above facing an inverted E, with the windings on a bobbin. Modern ones have the primary and secondary separated for safety, while older ones had the secondary wound over the primary. The latter should not be used when a person can come into contact with the secondary voltage such as in a lab or bench supply. C cores were popular in the past. Toroidal core or "Toroids" are a popular modern transformer.
The turns ratio is what determines the output voltage for a particular input. If we have a 120 volt supply and want 12 volts out we use a 10 to 1 ratio. For 240 to 12 volts it is 20:1. To supply a lower power valve circuit (from 120 volts) we might use a 200 volt AC output transformer for generate 280 volts DC. The ratio is thus 3:5, with more turns on the secondary that the primary. For a linear amp we might use one with a 1400 volts out to generate just under 2000 volts. Big transformers tend to be run from 240 volts, between the two "hots". Thus the ratio is 3:35.
For a low voltage output the input current is lower than the output current, but for a step-up voltage, the current in the primary is higher than that in the secondary. Also, if rectification and filtering increases the output voltage, then the charging current for the capacitors is higher than the output, or we are creating energy! Anyway, if the 2000 volt supply is supplying 1 amp then the 240 volt input would be a little over 8.33 amps, taking losses into account. If the transformer had a 120 volt primary the current would be a fairly hefty 16.7 amps. Thus, if you are using a step-up transformer the primary wire is often heavier than the secondary. This is the opposite of a step-down transformer.
In almost all cases related to Amateur Radio, some form of rectification is required. Most often this is a rectifier bridge consisting of 4 diodes, used at low voltages; or strings of series connected diodes at high voltages. Smoothing capacitors follow this, and potentially some form of regulation.
The exception is when a transformer is used to power a 120 volts device from 220 or 240 volts, or a 240 volts one from 120 volts; or converting industrial voltages such as 277 or Canada's 347 volts to 120 volts.
It is possible to drive a small commercial step-down transformer backwards, using a transistor switched low-voltage source, such as if we need 60 Hz power for a clock in a 50 Hz country, or vis-a-versa. However, we must NOT use such a transformer with the secondary connected to the mains, to generate a high voltage, as the LV side insulation may not be rated for mains voltages. Likewise, the mains rated insulation is likely not rated for the many hundreds or thousands of volts which the new secondary now has on it. Also, the impedance may be low on the now mains side, resulting in high current in this winding, and overheating. The previous version of the question relates to doing this by accident, as someone I met at a college did, destroyed her digital clock project with (a perhaps theoretical) 4800 volts AC applied to the circuitry, rather than the required 12 volts.
A signal transformer can also be used in either direction.
Off the exam, the alternative is a switchmode supply, where the mains is rectified to a voltage such as 170 or 340 volts DC, then fed via a high frequency oscillator to a compact transformer, and into a rectifier. A typical frequency is 25 kHz. In a low quality supply this will generate harmonics at 25 kHz spacings across HF amateur bands. Jaycar's lower cost one is an example, certainly only suitable for VHF and UHF, etc. despite being advertised as being suitable for transceivers, without qualification.
Caution! The output of transformers or power supplies beyond maybe 60 volts AC or 100 volts DC can be dangerous. At several hundred volts the chance of fatality becomes high, should you touch a live connection. One reason is that beyond a transformer GFCIs / RCDs / Safety Switches do no see the out-of-balance current they need to trip, to protect you. The other is that these supplies can supply large currents. There is an insane (in a bad way) new hobby involving using a microwave oven transformer to burn patterns into wood. Many practitioners have required the services of the undertaker. With 2000 volts, the maxim, "One flash, and you're ash" very much applies.
Large, high current, low voltage transformers can generate large, welder-like currents (considering that is what a traditional stick welder is - a big transformer), which may cause damage, or thermal burns.
One means to match signals to a load where the impedance is different is to use a transformer. Depending on the frequency, laminated steel, iron powder, ferrite, or air is used for the core.
Valves (tubes) are high impedance devices, and to match these to a speaker a transformer is needed. In the days of early transistors, especially germanium ones, transformers were also used. In several amplifier configurations, instead of a complimentary pair of PNP and NPN transistors, two transistors of the same kind were used, alternatively pulling current through one winding, then the other, as the wave is positive or negative. This was does because NPNs were once far more expensive than PNPs.
This arrangement is also used in the home-made inverters mentioned above, although typically with now affordable NPNs, or perhaps MOSFETs.
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| This circuit uses two transistors of the same type. When X is positive, current flows from the positive (+), via the upper winding, through the upper transistor, and to ground. When the other input is high, current flows via the bottom winding and the lower transistor. You will notice that the current flows upwards in one case, and downwards in the other. Thus there is an alternating flux in the transformer, and so an alternating voltage and current generated in the secondary winding. If instead of a split phase audio signal, we put a square-wave into the top input, and an inverted version into the lower input, and the output winding has a large number of turns, we could get an alternating 120 or 240 volt output. Some modification to the driving signal to provide a gap between the positive pulses helps avoid excessive currents at the transitions. |
Transformer output arrangements also allow lower voltage, high current (low impedance) to drive high power into relatively high impedance load, useful for running 100 watts into 50 ohms from a 13.8 volt supply (which would otherwise only supply a few watts).
The relationship between the turns ratio and the impedance ratio is a square. If a transformer has a 2:1 turns ratio then the impedance ratio is 4:1. A small version is used as a 300 to 75 ohm balun, a balanced to unbalanced converter to connect 300 ohm twin-lead / ribbon to a 75 ohm coaxial socket. The same ratio can be used for a 50 ohm to 200 ohm connection. A 1:3 winding ratio gives a 1:9 impedance ratio, suitable for a 50 ohm to 450 ohm connection.
There are two kinds of transformer / balun: one is an isolation transformer, with separate windings; the other is the auto-transformer variety, with a single tapped winding.
A small TV balun is wound on a small bead of ferrite, but for a transmitter a large toroid, or several glued together, are needed.
One way to think about the impedance transformation is this: A one volt RMS AC signal feeding a one ohm resistor means one amp, and one watt. If we put that into a 1:3 transformer there will be 3 volts, but for the same power the current needs to be only ⅓ amp, or 0.33333 A. To have this flow we need a resistance of 3 / 0.33333 = 9 ohms. Thus we have a 1:9 impedance ratio.
In various cases we wish to remove a direct electrical connection between two circuits. One way to do this is to use a transformer. This may be a 600:600 ohm transformer in a radio to telephone "patch" box, an audio transformer to isolate a PC soundcard from a radio. While safety may be one motivation, preventing ground loops, which often result in hum, can be another.
At one point transformers were used between amplifier stages.
Off the exam, suitable transformers can also be used at mains voltages, so that even if a person touches one wire beyond the transformer they should not receive a shock, even if they are grounded, as the voltage on the wire is not referenced to ground. An example is illuminated Christmas trees in a department store, and things like lights for working under cars, or lights in swimming pools.
For various reasons we place components in series and parallel. This can be due to commonly available parts not having the desired value, voltage rating, power rating, or current rating. We can also do this to adjust a timing circuit. We might, for example place a high value resistor in parallel with one in a timing circuit to speed it up (increase frequency), or a small capacitor in parallel with the main capacitor in the same circuit to slow the charging and discharge (decrease frequency).
In other cases we might have a component with a specific function, and a characteristic resistance, inductance, etc, and we either place a second such component in series or parallel, or we place another component in series or parallel with it.
For resistors in series, the resistances just add together. Thus a 100 ohm and a 220 ohm resistor makes a 320 ohm one.
Resistors in parallel can be more complex to calculate, unless the values are equal. For equal values, just divide the value by the number of resistors. Five 10 ohm resistors in parallel gives 2 ohms.
For different values we must find the reciprocal (1/x) of each one, add these, and then find the reciprocal of this value, thus:
RT = 1 / (1/R1 + 1/R2 + 1/R3)
Suppose we have a 2 ohm, a 5 ohm, and a 40 ohm resistor in parallel:
RT = 1 / (1/2 + 1/5 + 1/40) = 1 / (0.5 + 0.2 + 0.025) = 1 / 0.725 ≈ 1.37931 ohms
If paralleling two resistors note that the result will not be less than half the value of the lowest value resistor. This acts as a sanity check for your calculations.
For Inductors the same rules apply, in series these add; and in parallel, the simple division, or more complex reciprocal operations above apply.
Adding inductors in series does increase the impediment to AC current flow, while placing them in parallel eases the passage of such current.
We can place speakers in series or parallel, and apply the formulas above, provided we don't overload the amplifier.
Capacitors is parallel are simple, just add them together. Note that you need to take into account the sub-multipliers (nano, pico, etc).
Capacitors is series gain a greater voltage rating, at the cost of capacitance. For identical values these are divided, so if we have three 330 μF items, we get only 110 μF, but if these are 400 volt electrolytics then we can theoretically have 1200 volts on the string, although it would be wise to de-rate them.
CT = 1 / (1/C1 + 1/C2 + 1/C3)
One of the questions asks about increasing resistance, which is done by lifting one connection from the PCB, or otherwise cutting the circuit, and inserting a second resistor. One of the spoilers is to put a capacitor in parallel, but this is NOT done to the resistor. It is only done in a timing circuit or maybe a filter, and only to the other capacitor (unless all are in parallel). This is simply because an extra component in parallel can be easier.
The keystrokes depend on the type of calculator you have:
If you have a low cost "4 function" unit you type "1 ÷ 470 =", and note the result, then clear and entter "1 ÷ 270 =". This will give you 0.0037037. Type "+ 0.0021277 =" and note the answer 0.0058314. Clear, and type "1 ÷ 0.0058314 =". This will give "171.48541", being the resistance in ohms.
Scientific calculators should allow something like 470 [1/x] + 270 [1/x] =" which gives 0.00583136327. Taping [1/x] gives 171.486486727.
Fancy calculators allow entering full formulas, "1 ÷ ( 1 ÷ 470 + 1 ÷ 270 ) =", giving the answer above, noting that such devices perform the calculations the "order of operations" rules, evaluating the division parts withing the brackets ahead of the addition, then doing the final division.
For RPN use "470 [1/x] 270 [1/x] + [1/x], noting only the oldest RPN calculators, or those of Soviet origin comply with the prohibition against programmable units during the exam, or the use of emulations on 'phones or PCs.
A circuit published periodically in hobby electronics or ham magazines involves a string of high value resistors, typically totalling 90 MΩ, or perhaps 990 MΩ. This is placed inside a high voltage probe shell, and connected to a multimeter with 10 MΩ input impedance. Thus the voltage is divided by 10 or 100.
The danger is that the author often fails to mention that special resistors with a high voltage rating must be used, or one of the resistors might "break down". Once one breaks down, this increases the stress on the rest of the string, and another might fail, and so on, until you are shorting a high voltage, possibly high current, supply directly to a meter which you may be holding, and which might have a rating of only 600 volts. At best this will destroy the meter; at worst, it may add you to the coroner's case list.
As you may remember, metric uses multiples, and sub-multiples as a prefix to units. Thus we have millimetres, and some medications in micrograms, μg.
This table below reminds you of the sub-multiples relevant to this paper:
| Symbol | Prefix | Meaning |
| m | milli | ÷ 1 000 |
| μ | micro | ÷ 1 000 000 |
| n | nano | ÷ 1 000 000 000 |
| p | pico | ÷ 1 000 000 000 000 |
In the questions below these relate to capacitance and inductance. While removed from the exam, to move a line up the table, we divide by 1000, from 1000 pF to 1 nF, or 2200 nF to 2.2 μF.
Centi (c) is not used with electrical units, as it is not a power which is a multiple of 3. The main uses are in the centimetre band names, and in some cases where a HF or similar antenna is being described, although millimetres are preferable. "deci" is only used in decibels, dB. Atmospheric pressure has changed from millibars to hectopascals. French soft-drinks (sodas) are marked in cl, or centilitres.
For some reason the examiner has decided to use a dash between the number and the unit for a component value, but not for the calculated value resulting from placing the components in combination. Normally only a space is used, even when discussing the marking vs actual value of a component, such as a 470 ohm resistor with an actual value of 472.7549 ohms. Perhaps it is like Subway somehow dodging persecution in Australia for selling products in imperial units, which is generally illegal, by perhaps selling a product called a "6-inch" rather than selling saying they are selling what should be 15.24 cm of sandwich. Interestingly, they have been sued over undersized "Footlong" items.
These are actual questions from the General exam pool.
G5C01
What causes a voltage to appear across the secondary winding of a transformer when an AC voltage source is connected across its primary winding?
A. Capacitive coupling
B. Displacement current coupling
C. Mutual inductance
D. Mutual capacitance
Given transformers consist of coils, or inductors, this is mutual inductance, answer C.
G5C02
What is the output voltage if an input signal is applied to the secondary winding of a 4:1 voltage step-down transformer instead of the primary winding?
A. The input voltage is multiplied by 4
B. The input voltage is divided by 4
C. Additional resistance must be added in series with the primary to prevent overload
D. Additional resistance must be added in parallel with the secondary to prevent overload
The voltage is multiplied by 4, answer A.
Assuming a 120 volt to 30 volt transformer, we get 480 volts on the output, 4 times the primary voltage. It is thus likely that the connected electronics will be damaged, and electrolytic capacitors may rupture. A 240 to 60 volt jobbie will put out 960 volts; and a 480 to 120 volt one (used to convert industrial power for office equipment), will output a scary 1940 volts! Contact with these may prove fatal.
This demonstrates that a transformer has a bi-directional function, so an audio or RF transformer can work in either direction. In the latter case, a received signal can pass through an antenna matching a high impedance antenna such as an end-fed dipole to 50 ohm coax, and into the radio; and from the radio in transmit mode via 50 ohm coax, through the transformer, and into the high-Z antenna. This also applies to baluns.
G5C03
What is the total resistance of a 10-, a 20-, and a 50 ohm-resistor connected in parallel?
A. 5.9 ohms
B. 0.17 ohms
C. 17 ohms
D. 80 ohms
The only one not stupidly high or stupidly low, is answer A, but better prove it: R = 1 / (1/10 + 1/20 + 1/50) = 1 / (0.1 + 0.05 + 0.02) = 1 / 0.17 = ≈ 5.88235 ohms, yep, A.
The value must be below the lowest value, and for three different resistors, larger than one-third that value.
G5C04
What is the approximate total resistance of a 100- and a 200-ohm resistor in parallel?
A. 300 ohms
B. 150 ohms
C. 75 ohms
D. 67 ohms
The value must be less that the lowest value. R = 1 / (1/100 +1/200) = 1 / (0.01 + 0.005) = 1 / 0.015 = ≈ 66.66666667 ohms, or around 67 ohms, answer D.
G5C05
Why is the primary winding wire of a voltage step-up transformer usually a larger size than that of the secondary winding?
A. To improve the coupling between the primary and secondary
B. To accommodate the higher current of the primary
C. To prevent parasitic oscillations due to resistive losses in the primary
D. To ensure that the volume of the primary winding is equal to the volume of the secondary winding
To obtain one amp output at high voltage, the input may exceed 10 amps, so the winding needs to be quite heavy due to this current, answer B.
This may well exceed 1 mm in diameter, or be even heavier for a 120 volt input transformer.
G5C06
What is the voltage output of a transformer with a 500-turn primary and a 1500-turn secondary when 120 VAC is applied to the primary?
A. 360 volts
B. 120 volts
C. 40 volts
D. 25.5 volts
The ratio is 1:3 so we multiply 120 by 3 to get 360, answer A.
The formula, more useful when there isn't a simple integer relationship, is: 1500 / 500 x 120 = 360 volts.
If the output was rectified this would provide up to 500 volts DC, a voltage useful for a tube / valve transmitter of several tens of watts.
G5C07
What transformer turns ratio matches an antenna's 600-ohm feed point impedance to a 50-ohm coaxial cable?
A. 3.5 to 1
B. 12 to 1
C. 24 to 1
D. 144 to 1
The impedance ratio is 600:50 = 12:1, so the turns ratio is √12 ≈ 3:464, or about 3.5 to 1, answer A.
G5C08
What is the equivalent capacitance of two 5.0-nanofarad capacitors and one 750-picofarad capacitor connected in parallel?
A. 576.9 nanofarads
B. 1,733 picofarads
C. 3,583 picofarads
D. 10.750 nanofarads
Capacitors in parallel add, so the 2 x 5 nF give us 10 nF, and the 750 pF is 0.750 nF, giving a total of 10.750 nF, answer D.
G5C09
What is the capacitance of three 100-microfarad capacitors connected in series?
A. 0.33 microfarads
B. 3.0 microfarads
C. 33.3 microfarads
D. 300 microfarads
Divide 100 by 3, and you get 33.3 μF, answer C
G5C10
What is the inductance of three 10-millihenry inductors connected in parallel?
A. 0.30 henries
B. 3.3 henries
C. 3.3 millihenries
D. 30 millihenries
Divide to 10 by 3, and get 3.3, and check that it is mH, so answer C.
G5C11
What is the inductance of a 20-millihenry inductor connected in series with a 50-millihenry inductor?
A. 7 millihenries
B. 14.3 millihenries
C. 70 millihenries
D. 1,000 millihenries
In this case you add them, so 70 mH, answer C.
G5C12
What is the capacitance of a 20-microfarad capacitor connected in series with a 50-microfarad capacitor?
A. 0.07 microfarads
B. 14.3 microfarads
C. 70 microfarads
D. 1,000 microfarads
C = 1/(1/20 + 1/50) = 1/(0.05 + 0.02) = 1 / 0.07 ≈ 14.2857 μF, nearest to answer B.
To do this without the exact maths, the value has to be less than 20 μF, but the large value in series means the value is not going to be halved, as it would be if the second one was 20 μF. Thus 0.07 is way too small, and 14.3 the only sensible value.
G5C13
Which of the following components should be added to a capacitor to increase the capacitance?
A. An inductor in series
B. An inductor in parallel
C. A capacitor in parallel
D. A capacitor in series
Add an extra capacitor in parallel, answer C.
If we have a timer which is too fast, or an oscillator operating at too high a frequency, the extra capacitance with slow it, or reduce the frequency. This can also apply to a power supply filter capacitor.
G5C14
Which of the following components should be added to an inductor to increase the inductance?
A. A capacitor in series
B. A capacitor in parallel
C. An inductor in parallel
D. An inductor in series
It has to be an inductor, and it is added in series, answer D.
Sometimes we need to work out what resistors in parallel make the value we need. In the real world we may determine we need a 450 ohm resistor, but we don't have anything near (430 or 470 ohms are the nearest at Jaycar, and 442 or 453 are a week away at Farnell / E14 in the UK), or it might be 2 am, and our parts collection is sparse. 448 and 450 are available in the US, potentially with long lead times, with large minimums or high prices. Anyway, the simplest solution may be to divide the value to see what suitable values are available. 150 ohms resistors are common, so three in series area a good option. If we used this resistor to drop 9 volts to operate an LED at 20 mA it would dissipate 180 mW, making a 250 mW item (a standard quarter watt product) quote warm. If 3 resistors were used, the disipation would be a very comfortable 60 mW.
If we have an actual resistor (RA) higher than desired RD, we can determine the parallel value (RP) required:
RP = 1 / (1/RD - 1/RA)
RP = 1 / (1/450 - 1/451) = 202,950 ohms, or 202.95 kΩ.
I've once watched a video re tracing faults in a car, and while the wire in question had continuity between the two ends, it had shorted to ground, meaning that the signal could not travel from end to end. The culprit was damage within a multi-pin connector. Don't trust a simple continuity test! Check for unwanted connections to ground, and unwanted connection to power, or other signals.
A second trap is that a wire may test at (say) 12 volts or at ground with a digital meter, but these will be via a high impedance path, so will not provide any current. A test lamp can be useful, as these draw some current. This can also be the case with mains voltages.
You can end up with half the mains voltage on the braid of TV coax, thanks to capacitive leakage within the TV, making grounding the coax shield to your station earth a good idea, in order to prevent unpleasant tingles from it, as well as providing some protection from nearby lightning strikes, etc. Metal brackets with an F-connector joiner and a hole with a screw which accommodates a wire of several square millimetres are available to do this. Metal cased F-connector splitters may also have an earthing point.
If you have already read and absorbed the Components pages, head to: Circuits 1
On to: Components 1 - Batteries, Diodes and Transistors
You can find links to lots more on the Learning Material page.
Written by Julian Sortland, VK2YJS & AG6LE, October 2024.
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