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ENCYCLOPEDIA OF RADIO ELECTRONICS AND ELECTRICAL ENGINEERING Why do we need amateur radio calculations. Encyclopedia of radio electronics and electrical engineering
Encyclopedia of radio electronics and electrical engineering / Beginner radio amateur Amateur radio is art. When starting to manufacture any design, even described in detail somewhere (whether it be an amplifier, a radio receiver, a power supply, a TV set-top box, etc.), most often it is not possible to repeat it exactly, because there are no necessary parts, no satisfied with some constructive or circuit solutions, I want to get slightly different parameters and results, something to be finalized and improved. You can, of course, act by trial and error, selecting structural elements blindly, but isn’t it easier to arm yourself with a pen, a piece of paper and figure out what needs to be changed, what should happen, in which direction to act and what kind of details are needed? Let us make a reservation right away that experimental refinement may still be required, but its volume will be immeasurably less. Having mastered and repeated well-known designs, an amateur rarely stops there and begins to develop something of his own, original and unique. Here you can’t do without elementary calculations! How to set the transistor mode correctly, what value and power to install resistors, how much power will be dissipated by transistors and diodes, whether the bandwidth will be wide - these and many, many other questions can be answered by doing elementary calculations. I'm not talking about the calculation of circuits, the number of turns of coils and transformers - no one has yet been able to guess by eye the optimal data of these elements. Graphical representations are extremely useful and carry a lot of information - it is not for nothing that the characteristics of transistors and many other elements are given in the reference books in the form of graphs. Now suppose that in some calculation you come across the formula V (a + b2), in which you need to substitute a \u6,3d 0,3 and b \u1d 3. Come up with a geometric analogue of this formula and get the answer. The example was taken by no means by chance, this is how active and reactive resistances are added. While you are thinking, let's discuss the question: with what accuracy should we count? If you have already taken out a calculator to calculate the answer in the proposed example, then do not do this, but divide XNUMX by XNUMX. The calculator will fill in all the digits after the decimal point in triples. Do they all need to be rewritten in response? You are smarter than a calculator and will not do empty work. The result of the calculation must be rounded off, but what should be written - 0,3 or 0,33? It depends on the accuracy with which you make the calculations. The last digit is discarded if it is less than 5, and if it is greater, then 1 is added to the previous one. For example, 0,33 is rounded to 0,3, and 0,37 to 0,4. In both cases, the error can reach half of the unwritten digit, i.e. 0,05. The accuracy of the answer (relative error) will be 0,05 / 0,3 \u17d 0,3% in the first case (when you wrote down 1,5 in the answer) and only 0,33% - in the second (when you wrote down XNUMX) Very often, well-written source data already contains information about their accuracy. I have a quartz resonator in front of me that says 27,000 MHz, and although the frequency is given in megahertz, I am sure that the crystal is ground to an accuracy of 0,5 kHz, and the relative error is less than 0,002%. If it has an inscription of 27 MHz, it is difficult to expect the same accuracy. High accuracy is needed to get to the standardized frequency of the CB channel, but is it needed, say, when calculating the resistance of a resistor? Of course not, because the resistors themselves are mainly produced with tolerances of 5, 10, or even 20%. The same applies to capacitors, and the spread of characteristics of transistors is even greater. I will take the liberty of saying that in the vast majority of radio engineering calculations, two significant figures can be dispensed with and an accuracy of 5 ... 10% is quite enough. When something needs to be adjusted more precisely, trimming resistors and capacitors are installed, and the coils are supplied with adjustable magnetic circuits with "cores" - trimmers. Now let's answer the above problem. Its geometric analogy is a right triangle (Fig. 1) and the Pythagorean theorem. The lengths of the legs are a and b, the answer is the length of the hypotenuse. It is even impossible to draw a triangle with the given data to scale - it is too sharp! And it is quite clear that the length of the hypotenuse c differs very little from the length of the long leg a. If one of the impatient readers has already solved the problem on a calculator, then they saw the answer: 6,3071388, and this number needs to be rounded off. We will not solve this problem at all, since it is now clear to us that in the answer 6,3 with an accuracy better than 1%.
There is also an algebraic method that simplifies the calculation. Let's take a as a unit of measurement. And why not, because it's all the same how to measure the length of a boa constrictor - in meters, in yards or in parrots, you just need to know the coefficients for converting one unit to another. So, a measured in a is equal to one. But b measured in a is b/a = 0,3/6,3 = 0,05 (round up). This is a small value compared to unity, let's denote it x = b/a. Now it is convenient to present the formula side by side and limit ourselves to only the first two terms: (1 + x2)1/2 = 1 + x2/2. It is easy to calculate in your mind that the second term is only 2,5 10-3, and it can also be neglected. So, the answer in a is one, and in the previous values \u6,3b\uXNUMXb- XNUMX. Question for self-test, what is the duration of single pulses (in relation to the period) at the output of the logic element (Fig. 2), if it switches at a voltage of 2 V, and a sinusoidal signal with an amplitude of 4 V is applied to the input?
Author: V.Polyakov, Moscow
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