Aspect #07 07 марта 1998

Железо

Let's start with the basics - Electrical network and Ohm's law.

 We are asking for quite some time to begin publishing articles 
for beginners "zhelezyachnikov. The desire of our readers a law 
for us.:) Well, nachem from scratch ... 



     Electrical circuit and Ohm's law.


   The electrical circuit is the foundation of any wireless 
devices, including including those of low frequency amplifiers,

receivers and other devices are you going to design. In the 
meantime, figure it out in a very simple circuit and its 
Dielectric laws in the calculations of some of its elements.

Thus, the simplest electrical circuit (Fig. 2). Her
can be composed of DC (GB),
its load (H), ie
consumer power, the switch (S) and connecting
conductors. The current source may be battery
3336L, consumer - bulb figured on voltage 3,5 V and
current of 0.26 A (or a resistor -
RADIODETAL having
specific resistance), switch - switch or bell button
connecting conductors - segments isolated
wire.

   Draw up a chain by expanding its elements directly on the 
table. It should remind your chain of electric lamps. All points

compound preferably propayat. If the battery is fresh (new), 
filament bulbs serviceable, all connections are secure, then

contact closure switch S in the chain
current flows and the bulb will glow brightly. Check whether 
this is so. 

   Of these circuits, only
with other elements, will take shape all
your future raditehnicheskie device.

   Remember: the current across the outer part
chain flows from positive to negative battery terminal.

   For those who do not Zaneta what Parallel and serial 
connection try to explain this with an example

connection of resistors:


                 - Is consistently

                   connected three

                       resistor.


       - A parallel connected two

         resistor.



   For the connection, the current in the
vsy chain, and each of its sections are identical. Check it out 
you can with the help of Ammeter DC. Turn it on,

For example, in the open circuit between the positive pole of 
the battery and light bulb. The scheme shown in Figure 2, this 
point of inclusion ammeter indicated a cross. Then

ammeter inserted between the switch and
the negative pole of the battery. Everywhere,
Wherever the chain you did not include the meter, the needle 
will record the same value of current - about 0.2 A. As the 
battery discharge current in the circuit decreases and the 
light bulb dims. 

   We now carry out such an experience:
Open the circuit breaker
Plug the battery voltmeter PU
(Fig. 3) to measure the voltage on it, and then, without 
disconnecting the voltmeter on the battery, re- closed circuit. 
There is a difference in the testimony of a voltmeter? 

   After the closing circuit voltmeter should show a slightly 
lower voltage: it shows the tension developed by battery at the 
ends of the external circuit, which always less than "idle" 
battery voltage. Part of the same voltage drops (decreases, 
goes out, lost) on its domestic resistance. As a low battery, it

internal resistance and voltage drop across it increases.

   The following experiment. Turn
in series has
one the same bulb (Fig. 4). How to burn a light bulb? In 
polnakala. So it should be. Why?


   If you do not take into account the resistance of connecting
conductors and contacts
switch Cator small
by sravnieniyu resistance of the filaments of bulbs, the 
resistance outer region of the chain will increase by 
approximately twice. Now, the battery voltage is applied to the 
threads of the two filament bulbs. For each of them has

half the power than the earlier
one. Respectively, decreased the current through the bulb, and 
the heat of their threads. 

   In a closed circuit relationship between the
its existing voltage, amperage,
developed by this strain, and the resistance of the circuit 
determined by Ohm's law: Current I just proportsionaleni 
voltage U, and inversely proportional to the resistance R.

Mathematically, this law is an electrical circuit is as follows:


     U U
 I = or U = I * R or R =.

     R I


   Bear in mind: the current I, voltage U, and the resistance R 
in the formulas of this Act shall vyrazhatsya in basic 
electrical quantities - amperes (A), voltage (V) and Ohm (Om).


   This law is valid for
subcircuit, for example,
bulb or resistor, included in a closed circuit. In this case 
you can see immediately, reaching

the same circuit as that circuit
which is shown in Figure 5.
Battery voltage GB = 4,5 V
a resistor
R = 10 Om, then the ammeter will show the PA current equal
0.45 A (450 mA), a voltmeter PU - about
4.5 V. In this case, all the battery voltage through the 
ammeter, the internal resistance of which little is applied to 
the resistor R, so it falls by almost all voltage current 
source. 

   Replace another resistor resistor with a nominal (indicated 
on the package) resistance 20 ... 30 Ohm. Voltmeter

connected to a resistor, should show the same voltage. A 
ammeter? The ammeter will show the current value less

than in the previous case. If, for example,
30 ohm resistor, the ammeter will show the current 0.15 A (150 
mA). However, knowing the resistance of the resistor and the 
fall voltage across it, the value of current in the circuit you

you can see the arrow neglyadya ammeter. To do this we need 
only divide by instructions voltmeter (volts) on the resistor 
(in ohms), that is to solve the problem, using the formula of 
Ohm's law: 


     U
 I =

     R


   Receiver or amplifier - it is not just
electrical circuit and interconnected
circuit where a circuit controlled by another,
electrical circuit is passed from one to
another.

   Graphic illustration
This may be, for example, the experience (Fig. 6):
Plug the battery 3336L
Wirewound variable
resistor, resistance
10 .. 15 ohms, and between one
from its extreme conclusions and
engine (the role of such a resistor can do
small portion of the helix
electric), including those
the same bulb.
The engine is put in the middle
position with respect to
extreme conclusions.

   How to burn a light bulb?
Half-heartedly. Pass
cursor to bottom (by scheme) conclusion. How now? Do not burn. 
And if engine will be in the uppermost (again

scheme) position? The lamp will burn a full incandescence. As 
you see, by variable resistor can smoothly increase or decrease 
heat bulbs. 

   In this experiment, two interlinked chain.
First strand form a battery and resistor GB
R, the second - light H and that portion of the resistor 
between the lower (scheme) O and engine to which the lamp is 
connected. Throughout the resistor falls (decreases) all

battery voltage. And that part of the stress that is 
attributable to lower plot resistor, through the engine is fed

on filament light bulbs. And the more
plot resistor is inserted into the second chain,
the greater the stress on the filament bulb, the brighter it 
glows. 

   Variable resistor that is used in such
way, acts as a voltage divider battery, or as they say, a 
potentiometer. In this case it divides the voltage of the 
battery into two parts and one of its portion that can govern, 
transmitted to the control of a second chain. Running forward, 
we say that it is fundamentally so is the volume control in

receivers and amplifiers, low frequency.

   With the voltage divider the same
light bulb can be powered from the battery, the voltage which 
is much more than the voltage at which the filament 
rastschitana bulbs.


   The role of divider can perform and two fixed resistors, as 
shown in Fig. 7. Here the resistor R2 should be such that

to in this section divider voltage falls,
corresponding naminalnomu voltage bulbs H. In that case,
If the battery voltage is twice the voltage, which should bring 
to the lamp, the resistance divider R1R2

should be roughly equal.

   Such voltage dividers you can
seen in any wireless devices. They will be integral to and
your designs.

   However, there is another way to supply that
the same bulb from the battery voltage greater
- By including the quenching resistor in the circuit, ie 
resistor, which will extinguish some of the source voltage 
supply. Match consecutively two

3336L battery - it will
battery voltage of 9 V.
Connect to it the same bulb (3,5 X 0,26 A), but
so as shown in Scheme
Fig. 8, - through a resistor
R with a resistance 20 .. 25
Ohms, calculated on the power dissipation of at least
1 watt. (The diameter of the resistor in the center of about 
5,5 mm). Resistor such resistance and the same power

can be composed of two resistors rated at 0.5 watts, (the 
diameter of the resistor in the center of about 3 mm), ie, 
resistors, such as MLT-0, 5, with face values ​​of 39 ... 51 
Ohm, connecting them in parallel. Lightbulb, you see, is lit 
properly, just maybe, the resistor (s) a little heated. 

   In this experiment, the resistor and the filament
bulbs, too, in fact, form a voltage divider. On the filament 
falls voltage (3,5 V), corresponding

its resistance (about 13 ohms), so
she glows. The rest is the battery voltage
drops across the resistor. Resistor, thus extinguishes 
(absorbs) the excess voltage of the battery, so it is usually 
identified as blanking.


   From another point of view, the resistor limits the current 
in the circuit, and hence the current through the filament of 
the bulb. Therefore, it can be called restrictive. Problem

is it - create a light bulb conditions
under which it would be fine filament
shone and burned.

   Resistance to blanking (restrictive), the resistor is 
calculated from the fact excessive stress, which they must

repay, and the current needed to supply
power of the payload. In our
experiment payload was light,
filament which is nice for a voltage 3,5 V and current of 0.26 
A. And once the battery voltage is 9V, then the resistor is the 
site of the chain, should extinguish the voltage 5.5 V at 0.26 
A. 

   What should be the resistance of this
resistor? According to Ohm's law - about 20 ohms.


    U 5,5 B
R = = = 21,15 Ohm.

    I 0,26 A

When the battery voltage 9V, this resistor
resistance will not pass through itself to
load current is more than 0.26 A.

   And what should be the power resseevaniya this resistor? 
Calculate it by such may already be familiar to you the 
following formula: P = U * I. In this formula, U - voltage

volts, which resistor should pay,
and I - current in amperes, which must be
load. Consequently, for our example power expressed in watts (W)
dissipated quenching resistor is: P = 5,5 * 0,26 = 1,43 Tues 
Hence, the resistor must be rated for power dissipation is not 
less than 1,5 Tues This may be For example, a resistor-type 
MLT-2, 0, or wire. If the resistor will be less power 
dissipation, such as MLT-1, 0 or MLT-0, 5, and it always will 
be warm, that may have been in your experience, and may even 
burn. 

   Extinguishing resistors are very many elements of electrical 
circuits your future designs.


   You will also need to count and power consumed by the 
designs of power sources. This is in order, for example, to 
know how long the work receiver or amplifier will be enough 
electrical capacity to supply batteries. Power consumption from 
the current source learn multiplying the voltage at the ends of 
the chain on the current in the circuit. For example, the power 
consumed by an incandescent lamp, you used for the experiments, 
composes about 1 W (P = U * I = 3,5 * 0,26 = 0,91 W).


   Electric battery capacity 3336L
equal to 0.5 Ah (ampere-hour). Divide this
capacity to power consumption lamp, and you will know at what 
time (in hours) of battery power is enough for food bulbs. Yes, 
just half an hour. And if the battery is weak, and even less. : 
( 

   Finally - a little advice,
having direct relevance to the topic of this
workshop. The fact that the principal circuit diagrams and 
explanations of radoapparatury ratings the resistance will mark 
made in ohms. For example: (220) - in ohms, (5,1 k) -

in kiloomah (1 kohm = 1000 ohms), (1,5 M) - in
megaomah (1 MW = 1000 ohms). At the same time,
on the small resistors, manufactured
our industry, their nominal resistance marked by another 
conditional system: the unit of ohms of resistance denoted by 
the letter E, kOhms - K, mega - M. Resistance of resistors from 
100 to 910 ohms expressed as a fraction of kilooma, and 
resistance from 100 ohms to 990 ohms - in fractions of megaoma.


   If the resistor express
integer, then the letter of
units are placed after this number, for example: 27E (27 ohms), 
51K (51 ohms) 1M (1 M). If the resistor

expressed as a decimal fraction less than unity,
then the letter of the unit disposes before the number, for 
example: K51 (510 ohms) M47 (470 ohms).


   Expressing resistor by a
number to a decimal, integer
put before the letter, and a decimal fraction -
the letter that symbolizes the unit. For example: 5E1 (5,1 
Ohm), 4K7 (4,7 k) 1M5 (1,5 MW).


Well, as it were on today, (very much
I had enough to fill the text:). If you
something is not understood please contact us. Your
questions, we need the following razesnim nomirah our 
newspaper. In the near future I plan to Tell you that it 
represents a diode, and how it works. A I now advise you first 
practice with a soldering iron. 

      Know: skill comes with experience!


                            (C) Fedy Savin