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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. 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, or audio or RF power to a speaker or antenna.

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, 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.

Note that while it is possible to drive a 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, 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, and 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 question relates to doing this by accident. Certainly, someone I met at a college destroyed her clock project this way, with, at least theoretically, 4800 volts AC applied to the circuitry.

One means to match signals to a load where the impedance is different is to use a transformer. Depending on the frequency, steel laminations, 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, having two transistors of the same kind, alternatively pulling current through one winding, then the other as the wave is positive or negative. This arrangement is also used in the home-made inverters mentioned above.

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. |

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 varions 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.

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, then add these, and finally 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.

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** uses the formulas above. 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 these.

`CT = 1 / (1/C1 + 1/C2 + 1/C3)`

One of the questions ask 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! This is simply because an extra component in parallel can be easier.

A circuit published periodically involves a string of high value resistors, typically totalling 90 MΩ, or perhaps 990 MΩ. This is placed inside a high voltage probe, 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 fails to mention that special resistors with a high voltage rating must be used, or the resistor 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 through a meter you may be holding, and which is only rated for maybe 600 volts.

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. To move a line up the table, we divide by 1000, from 1000 pF to 1 nF, or 2200 nF to 2.2 μF.

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 happens if you reverse the primary and secondary windings of a 4:1 voltage step down transformer?
A. The secondary voltage becomes 4 times the primary voltage
B. The transformer no longer functions as it is a unidirectional device
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`

Assuming a 120 volt to 30 volt transformer, we 480 volts on the output, or at least, that is what the transformer attempts to supply, answer A.

It is quite possible that something is going to be damaged, either the connected electronics, or the electrolytic capacitors, to the extent that they 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), you get a scary 1940 volts! These voltages can be seriously dangerous, especially if you were expecting a lower voltage, and touched the terminals while testing.

You should NOT just reverse a transformer to get a high voltage for valve amplifiers, for example, but use one designed for the output voltage desired.

`G5C03
Which of the following components increases the total resistance of a resistor?
A. A parallel resistor
B. A series resistor
C. A series capacitor
D. A parallel capacitor`

An additional resistor must be placed in series, answer B.

`G5C04
What is the total resistance of three 100 ohm resistors in parallel?
A. 0.30 ohms
B. 0.33 ohms
C. 33.3 ohms
D. 300 ohms`

For resistors of the same value in parallel, just divide the value by the number of resistors, 100 / 3 being 33.3 ohms, answer C.

`G5C05
If three equal value resistors in series produce 450 ohms, what is the value of each resistor?
A. 1500 ohms
B. 90 ohms
C. 150 ohms
D. 175 ohms`

450 / 3 is 150 ohms, answer C.

`G5C06
What is the RMS voltage across a 500-turn secondary winding in a transformer if the 2250-turn primary is connected to 120 VAC?
A. 2370 volts
B. 540 volts
C. 26.7 volts
D. 5.9 volts`

The only answer anywhere near sensible answer is C, but better prove it: 500 / 2250 x 120 is 26.666667 volts, answer C.

`G5C07
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?
A. 12.2 to 1
B. 24.4 to 1
C. 150 to 1
D. 300 to 1`

This would be, in all likelihood, a valve amplifier. The impedance ratio is 150:1, so the **turns ratio** is √150 ≈ 12.247, about 12.2 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. 1733 picofarads
C. 3583 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. 0.30 microfarads
B. 0.33 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 henrys
B. 3.3 henrys
C. 3.3 millihenrys
D. 30 millihenrys`

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. 0.07 millihenrys
B. 14.3 millihenrys
C. 70 millihenrys
D. 1000 millihenrys`

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. 1000 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. A resistor in series
C. A capacitor in parallel
D. A capacitor in series`

Adding an extra capacitor in parallel, answer C.

`G5C14
Which of the following components should be added to an inductor to increase the inductance?
A. A capacitor in series
B. A resistor in parallel
C. An inductor in parallel
D. An inductor in series`

It has to be an inductor, and it is in series, answer D.

`G5C15
What is the total resistance of a 10 ohm, a 20 ohm, and a 50 ohm resistor connected in parallel?
A. 5.9 ohms
B. 0.17 ohms
C. 10000 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.

`G5C16
Why is the conductor of the primary winding of many voltage step up transformers larger in diameter than the conductor 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.

`G5C17
What is the value in nanofarads (nF) of a 22,000 pF capacitor?
A. 0.22
B. 2.2
C. 22
D. 220`

Lop off the three zeros, and get 22 nF, answer C.

Pico is 10^{-12}, and nano is 10^{-9}, so there are 1000 pF in 1 nF, or 22,000 pF in 22 nF.

`G5C18
What is the value in microfarads of a 4700 nanofarad (nF) capacitor?
A. 47
B. 0.47
C. 47,000
D. 4.7`

There are 1000 nF in 1 μF, so 4700 nF is 4.7 μF, answer D.

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, March 2022.

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