Choosing Your Components

Choosing components which are not specified.
Some people find it difficult to select a suitable component where the exact component is not specified or where an alternative has to be selected, so perhaps a few general pointers might be helpful. The reason why component values are omitted may well be because a very wide range of alternative values can be used and if one particular is specified, the newcomers to electronics feel that they have to use that one value or the circuit will not work, (which is almost never the case). For example, I have been asked if a capacitor rated at 25V could be used instead of the same value capacitor rated at 16V shown in the circuit, to which the answer is ‘yes, most definitely’. The lower voltage rating is adequate and the component cheaper to buy, but if one of a higher voltage rating is available, then it can be used.

With capacitors, you need to consider the physical size and wire connections, the capacitance, the voltage rating, and the leakage. The cost and size of a capacitor is directly related to it’s voltage rating, and once the voltage rating exceeds that normally used, the price shoots up rapidly as the sales volume reduces rapidly, which in turn, discourages further sales. This sometimes causes circuit builders to connect chains of cheaper capacitors together to make a smaller-capacity high-voltage capacitor. In the case of Tesla Coil builders, they then may connect several of these chains in parallel to boost the capacitance.

If the voltage rating is exceeded (usually by a very large amount), the capacitor will be damaged and become either a short-circuit, or more likely, an open circuit. Either way, it will never work as a capacitor again. In a household circuit, where the capacitor is being used as part of the power supply to the circuit, the voltage rating does not need to be much higher than the supply voltage, with say, 16V being used for a 12V circuit. You could use a capacitor rated at 25V, 40V, 63V, 100V or 400V and it would work perfectly well, but it will be much larger and have cost much more. But, if you have one sitting around and not being used, there is no reason why you should not use it rather than paying to buy another one.

If the capacitor is being used in a timing circuit where a high-value resistor is feeding current to it, then the leakage current of the capacitor becomes very important. Electrolytic capacitors are seldom suitable for such an application as they have a small, unpredictable leakage current which will vary with the age of the capacitor. For accurate timing with a capacitor, ceramic, polypropylene, mylar or tantalum should be used.

The voltage rating for an electrolytic capacitor is for DC, so if you use it for limiting current in an AC power supply, that is, where the current flows through the capacitor rather than the capacitor being placed across the supply and is acting to combat ripple, then great care is needed. The capacitor will heat up due to the power flowing through it, and it is possible for an electrolytic capacitor used in that way to rupture or ‘explode’ due to the electrolyte boiling. Instead, you need to use the very much more expensive oil-filled can capacitors (as shown near the end of chapter 10). That style of usage is unusual for home constructors.

With bi-polar transistors, you need to use commonsense. Suppose a 555 timer chip is required to power a transistor which controls a relay:

For the moment, we will ignore the fact that the 555 could drive the relay directly without the need for a transistor. let’s say that the relay draws a current of 30 mA when connected to a 12V supply. Therefore, the transistor needs to be able to handle a current of 30 mA. Any small switching transistor such as the BC109 or 2N2222 can easily handle that current. The transistor also needs to be able to handle 12 volts. If in doubt, look up the characteristics of your choice of transistor at alldatasheet by entering the transistor name ‘BC109’ or whatever in the entry box at the top of the screen and clicking on the button to the right of it. Eventually, it will let you download a pdf document specifying the transistor, and that will show you the voltages which the transistor can handle. Both of the above transistors can handle far more than 12V.

The next question is, ‘can the transistor switch fast enough to work in this circuit?’ and the data sheet will show that they can switch on and off a million times per second. As the relay can only switch on and off a few times per second, the transistor can easily operate fast enough to handle the switching.

Next, we need to know what size of resistor would be suitable. The data sheet will also show the DC current gain of the transistor. This is usually marked as “hfe” and for these transistors is likely to be a minimum of, say, 200. This means that the current flowing into the base of the transistor needs to be one two-hundredth of the relay’s 30 mA which is 0.15 mA. The resistor will have about +11 volts at pin 3 of the 555 timer and around +0.7 volts at the base of the transistor when it is switched fully on. That means that the resistor will have about 10.3 volts across it when the relay is switched on:

So, what size of resistor will have 0.15 mA flowing through it when there is a 10.3-volt drop across it? We know that a 1K resistor passes 1mA per volt and so would pass 10.3 mA with 10.3 volts across it. That is far more than we need. A 10K resistor would pass 1.03 mA which is still far too much but certainly could be used. As it is a resistor, we can use Ohm’s Law: R = V / A (Ohms equals Volts over Amps), or R = 10.3 / 0.00015 which is 68K. So, any resistor between 68K and perhaps 15K should work well.

The diode is there to protect the transistor from excessive voltage caused by the coil of the relay. When a coil is switched off suddenly, it generates a reverse voltage which can be hundreds of volts, pulling the collector of the transistor far above the +12V power supply line. When that starts to happen, it effectively reverses the diode direction, allowing it to conduct and short-circuit that big voltage spike:

Due to the short-circuiting, the voltage can’t get any higher and the current through the diode is not large, so most diodes such as the popular and cheap 1N4001 or 1N4007 types can be used.

When a transistor is connected like that and switched on, it is effectively a short-circuit between it’s collector and emitter, and that places the full 12 volts across the relay, powering it very solidly. This connection method is called a “common-emitter” circuit because all of the transistors used have their emitters all wired in common to the 0V line. An alternative arrangement is the “emitter-follower” circuit:

With this circuit arrangement, the emitter of the transistor “follows” the voltage on pin 3 of the 555 timer. It is always a constant voltage below it, typically about 0.7 volts. The output of the 555 timer has a maximum of about 0.7V below the supply voltage, and so it’s maximum value is about 11.3V in this circuit. The transistor drops that by a further 0.7V, which means that the relay only gets about 10.6V across it instead of the full 12V of the supply, which means that it should be a 10-volt relay rather than a 12-volts relay.

Those are the easy cases because the 555 timer can supply at least 200 mA through it’s output pin, while keeping the output voltage steady. That is not the case with simple transistor circuits. Take a situation like this:

For audio work – microphone pre-amplifiers and the like – the rule of thumb is that the current flowing through the first transistor should be at least ten times the current required by the base of the second transistor in order not to drag down and distort the audio waveform.

Relay switching is not so critical but the same general principle applies and attention needs to paid to the collector resistor of the preceeding transistor. For example, if the current flowing through the preceeding transistor is small, say, 0.5 mA and the output transistor needs 1.5 mA flowing into it’s base, then there can be a problem. In this circuit, for example:

Here, the voltage at point “A” goes high because the first transistor switches off and so becomes the same as a resistor of 1Meg or more. Normally, that resistance is so much greater than the 27K of it’s resistor, that the voltage at point “A” would be nearly +12V, but if you were to connect the resistor “R” of just 1K in value, then the situation is changed completely. the base of “Tr” can’t rise above 0.7V. The first transistor can be ignore due to its very high resistance. That leaves a voltage-divider pair of resistors, the 27K and the 1K, with 11.3 volts across them, stopping the voltage at point “A” from rising above 1.13V instead of the original 12V and transistor “Tr” will only get 0.43 mA instead of the 1.5 mA which was wanted. The transistor “Tr” has effectively a 28K resistor feeding it current from the +12V rail.

One solution would be to raise the current through the first transistor by using a resistor a good deal smaller than the present 27K. Another option is to lower the input current requirement of transistor “Tr” by making it a Darlington pair or by using a transistor with a much higher gain.

Electronics Tutorial

    • Leslie Risner
    • November 14th, 2012

    You really make it seem so easy with your presentation but I find this matter to be actually something which I think I would never understand. It seems too complicated and very broad for me. I am looking forward for your next post, I’ll try to get the hang of it!

  1. Hello there! Would you mind if I share your blog with my facebook
    group? There’s a lot of people that I think would really enjoy your content. Please let me know. Thanks

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