Do-it-yourself mirror heating

FROMwinter cold snaps, decided to make self-heating rear-view mirrors, because last winter I got tired of scraping them from frozen ice and snow before each trip. And besides, after these actions, I noticed that I scratched the mirrors themselves with a brush, even if the scratches are small and they are not very visible, it’s still unpleasant. And besides, it is very good in the rain, the drops that fall on the mirror immediately dry up and the mirror is constantly dry!

Do-it-yourself steering wheel heating

Do-it-yourself steering wheel heating

In winter, it’s not very comfortable, especially in cool areas, when everything in the car is at minus, including the steering wheel, which sometimes you have to wear gloves. This issue has been resolved DIY steering wheel heating.

From several options, I chose the best in my opinion. Using carbon tape (12mm*0.6mm).

Electronic relay for turning on the cooling fan

Do-it-yourself electronic relay for turning on the cooling fan.

In hot weather, the temperature sensor that controls the radiator cooling fan has to be changed very often. And the turn-on temperature is not how not to adjust. All these shortcomings can be solved with just do-it-yourself electronic relay. In which car you will use it, the question is not fundamental, vaz, gas, UAZ and other brands.

DIY police siren

DIY police siren

I’ll go straight to what it is, and what kind of sounds we get. it homemade police siren made on a microcontroller PIC16F628. If you want to assemble a police cracker with your own hands, it will not take much effort. There are two sounds in this assembly, the first is a siren, the second when a button is pressed is a kind of police "quack". Let's move from theory to practice.

DIY strobe lights

Do-it-yourself strobe lights for a car

I think it’s clear from the picture what strobe lights are, and what we visually see when they work, I think they know without any explanation. Found a solution how to do it simple do-it-yourself strobe lights.

How to connect a 12 volt fan to 24 volts

How to connect a 12 volt fan to 24 volts

Each owner of a heavy vehicle (truck, bus, etc.) with on-board network voltage 24 volts at least once faced with a problem when it is necessary connect a 12 volt consumer.

One of the simplest solutions to this is to connect this consumer (radio, radio, kettle or something else) to one of the batteries that are connected in series in such machines. But this solution has one very big drawback: the battery to which the 12 volt consumer is connected will be undercharged all the time, and the second battery may be overcharged. >Both of these cases will lead to reduced battery life. The second, most correct way to connect 12-volt consumers to a 24-volt network is to use a 24-volt to 12-volt converter.

A simple do-it-yourself converter 12-220 Volts

Do-it-yourself converter 12-220V

Recently, more and more people are interested in assembling do-it-yourself inverters (converters). The proposed assembly is capable of delivering power up to 300W.

An old and good multivibrator is used as a master oscillator. Of course, such a solution is inferior to modern high-precision chip generators, but let's not forget that I tried to simplify the circuit as much as possible so that I ended up with an inverter that would be available to the general public. A multivibrator is not bad, it works more reliably than some microcircuits, it is not so critical to input voltages, it works in harsh weather conditions (remember the TL494, which needs to be heated, at sub-zero temperatures).

The transformer is used ready-made, from UPS, the dimensions of the core allow you to remove 300 watts of output power. The transformer has two primary windings of 7 volts (each arm) and a network winding of 220 volts. In theory, any transformers from uninterruptible power supplies will do.

The diameter of the primary winding wire is about 2.5mm, just what you need.

Car battery charger

Car battery charger

In this article I want to give a simple assembly of do-it-yourself car battery charger. Even very simple, it does not contain anything superfluous. After all, often complicating the scheme, we reduce its reliability. In general, here we will consider a couple of options for such simple car chargers that can be soldered to anyone who has ever repaired a coffee grinder or changed a switch in the corridor. For a long time I was visited by the idea of ​​​​assembling the simplest charger for the battery of my motorcycle, since the generator sometimes simply cannot cope with charging the latter, it is especially difficult for him on a winter morning when you need to start it from the starter. Of course, many will say that it is much easier with a kick starter, but then the battery can be thrown out altogether.

car battery charger

Do-it-yourself car battery charger

In the winter season, more and more often we pay attention to car battery charging, due to its discharge, and poor performance. But the prices for battery chargers are not very small, and sometimes it's easier to do Do-it-yourself memory, which will be discussed further.

The proposed scheme is very high quality will charge your battery and it will extend its service life.

Do-it-yourself stroboscope for setting the ignition timing

Do-it-yourself stroboscope for adjusting the UOZ

When replacing a distributor, or repairing the ignition of the mixture, whether it is a carburetor change, we are faced with the need to adjust the ignition timing.

What is Ignition Advance (Ignition)the angle of rotation of the crank from the moment at which voltage begins to be applied to the spark plug to breakdown the spark gap until the piston reaches TDC.

To set up the UOS, most masters use the so-called car strobe light, which flares up at the moment when a spark runs through the spark plug. Details on how to use a stroboscope to adjust the UOS can be seen on the Internet. The same article provides simple car strobe circuit, which the do it yourself can be assembled by almost any novice radio amateur.

In this article, we will consider a stabilized power supply with a continuously adjustable output voltage of 0 ... 24 volts and a current of 3 amperes. The power supply protection is implemented on the principle of limiting the maximum current at the output of the source. Adjustment of the current limit threshold is made by resistor R8. The output voltage is regulated by a variable resistor R3.

The schematic diagram of the power supply is shown in Figure 1.

Item List:

R1........................180R 0.5W
R2, R4................. 6K8 0.5W
R3.......................10k (4k7 – 22k) reostat
R5........................7k5 0.5W
R6........................0.22R at least 5W (0.15- 0.47R)
R7.............20k 0.5W
R8.......................100R (47R - 330R)
C1, C2...................1000 x35v (2200 x50v)
C3............1x35v
C4............470 x 35v
C5......................100n ceramic (0.01-0.47)
F1......................5A
T1......................KT816 (BD140)
T2......................BC548 (BC547)
T3......................KT815 (BD139)
T4......................KT819 (KT805,2N3055)
T5......................KT815 (BD139)
VD1-4................KD202 (50v 3-5A)
VD5............ BZX27 (KS527)
VD6..............AL307B, K (RED LED)

Let's start in order:

A step-down transformer such a power is selected so that it is able to supply current to the load of the required value for a long time, and the voltage on the secondary winding is 2 ... 4 volts more than the maximum voltage at the output of the power supply. Accordingly, the rectifier bridge is selected with a current margin, so that later the bridge diodes or the diode assembly do not have to be molded onto the radiator.

How to estimate the power of a transformer? For example: there should be 25 volts on the secondary at a current of 3 amperes, which means we have 25 * 3 = 75 watts. In order for the transformer to be able to supply 3 amperes to the load for a long time, increase this percentage value by 20 ... 30, i.e. 75 + 30% = 97.5 watts. It follows that a 100 watt transformer must be selected.

The maximum voltage at the output of the power supply depends on the Zener diode VD5, which is in the collector circuit of the transistor T1. For example: when using a KS168 zener diode, we get a maximum voltage of about 5 volts at the output, and if we put KS527, we get a maximum voltage of 25 volts at the output. Information on zener diodes can be found in the article:

What rating should be the filter capacity standing after the diode bridge? In our case, according to the scheme, there are two capacities in parallel C1 and C2 of 1000 microfarads each. In general, the capacitance of this capacitor is selected based on the order of 1000 microfarads per 1 ampere of output current.
Electrolyte C4, standing at the output of the power supply, is selected in the region of 200 microfarads per 1 ampere of output current.

What voltage should the electrolytes C1, C2 and C4 be set to? If you do not go into abstruse calculations, then you can use the formula: ~Uin:3×4, i.e. the voltage value that the secondary winding of the step-down transformer produces must be divided by 3 and multiplied by 4. For example: on the secondary we have 25 volts of change, hence 25: 3 * 4 \u003d 33.33, therefore capacitors C1, C2 and C4 are selected for Uwork \u003d 35 volts. You can put containers with a higher operating voltage, but not less than the calculated value. Of course, this calculation is rough, but nevertheless ...

A current limiter is assembled on T5. The limit threshold depends on the value of the resistor R6 and the position of the variable resistor R8. In principle, the R8 variable may not be set, and the limit threshold can be made fixed. To do this, we connect the base of the transistor T5 directly to the emitter T4, and by selecting the resistor R6 we set the required threshold. For example: with R6 = 0.39 ohms, the limitation will be about 3 amperes.

Limiting current adjustment. Without load, set the potentiometer R3 to Uout about 5 volts. Connect an ammeter and a 1 ohm resistor connected in series to the output of the power supply unit (the power of the resistor is 10 watts). Adjust R8 to the required current limit. We check: we gradually unscrew R3 to the maximum, while the readings of the control ammeter should not change.

In the process of operation, the T1 transistor heats up slightly, put it on a small radiator, but T4 heats up thoroughly, decent power is dissipated on it, you can’t do without an impressive radiator, and even better to adapt a computer cooler to this radiator.

How to estimate the dissipation power T1? For example: the voltage after the diode bridge is 28 volts, and the output is 12 volts. The difference is 16 volts. Let's estimate the power dissipation at a maximum current of 3 amperes, i.e. 12*3 = 36 watts. If we set the output voltage to 5 volts at a current of 3 amperes, then the transistor will dissipate power (28 - 5) * 3 = 69 watts. Therefore, when choosing a T4 transistor, do not be too lazy to look into the transistor reference book, look at what dissipation power it is designed for (in the table column Pk max). Reference material on the transistor, see the figure below (click on the image to enlarge the image):

The printed circuit board of the power supply is shown in the following figure:

What value should the fuse be? In this circuit, there are two fuses: on the primary winding of the transformer (selected 0.5 ... 1 amperes more than the maximum current of the primary winding), and the second in front of the rectifier bridge (selected 1 ampere more than the maximum limiting current of the PSU).

You can squeeze out much more than 3 amperes from this circuit, for this you need to have a trans-r capable of delivering the required current, put a diode bridge with a current margin, recalculate the filter capacitances, reinforce the tracks on the board through which a large current will flow with a thick wire, and apply parallel connection of transistors as T4 as shown in the following figure. Transistors are also placed on a radiator with forced airflow by a fan.

If you are going to use this PSU as a charger for a car battery, set it to no load (no battery connected) with a voltage regulator of about 14.6 volts at the output and connect the battery. As the battery charges, the density of the electrolyte increases, the resistance increases, and the current will decrease accordingly. When the battery is charged and there are 14.6 volts at its terminals, the charging current will stop.

The appearance of the printed circuit board and the assembled power supply, see below:

Every motorist dreams of having a rectifier for charging the battery. Without a doubt, this is a very necessary and convenient thing. Let's try to calculate and make a rectifier for charging a 12 volt battery.
A typical battery for a passenger car has the following parameters:

• the voltage in the normal state is 12 volts;
• Battery capacity 35 - 60 amp hours.

Accordingly, the charge current is 0.1 of the battery capacity, or 3.5 - 6 amperes.
The rectifier circuit for charging the battery is shown in the figure.

First of all, you need to determine the parameters of the rectifier device.
The secondary winding of the rectifier for charging the battery must be rated for voltage:
U2 = Uak + Uo + Ud where:
- U2 - voltage on the secondary winding in volts;
- Uak - battery voltage is 12 volts;
- Uo - the voltage drop across the windings under load is about 1.5 volts;
- Ud - the voltage drop across the diodes under load is about 2 volts.

Total voltage: U2 = 12.0 + 1.5 + 2.0 = 15.5 volts.

We accept with a margin for voltage fluctuations in the network: U2 \u003d 17 volts.

We take the battery charge current I2 \u003d 5 amperes.

The maximum power in the secondary circuit will be:
P2 = I2 x U2 = 5 amps x 17 volts = 85 watts.
The power of the transformer in the primary circuit (the power that will be consumed from the network), taking into account the efficiency of the transformer, will be:
P1 = P2 / η = 85 / 0.9 = 94 watts. where:
- P1 - power in the primary circuit;
- P2 - power in the secondary circuit;
-η = 0.9 is the efficiency of the transformer, efficiency.

Let's take P1 = 100 watts.

Let us calculate the steel core of the Ш - shaped magnetic circuit, the transmitted power depends on the cross-sectional area of ​​\u200b\u200bwhich.
S = 1.2√P where:
- S core cross-sectional area in cm2;
- P \u003d 100 watts is the power of the primary circuit of the transformer.
S \u003d 1.2 √ P \u003d 1.2 x √100 \u003d 1.2 x 10 \u003d 12 cm.sq.
The section of the central rod, on which the frame with the winding will be located S = 12 cm.sq.

Let's determine the number of turns per 1 one volt in the primary and secondary windings, according to the formula:
n = 50 / S = 50 / 12 = 4.17 turns.

Take n = 4.2 turns per volt.

Then the number of turns in the primary winding will be:
n1 \u003d U1 n \u003d 220 volts 4.2 \u003d 924 turns.

Number of turns in the secondary winding:
n2 = U2 n = 17 volts 4.2 = 71.4 turns.

Let's take 72 turns.

Let's determine the current in the primary winding:
I1 = P1 / U1 = 100 watts / 220 volts = 0.45 amps.

Current in the secondary winding:
I2 = P2 / U2 = 85 / 17 = 5 amps.

The wire diameter is determined by the formula:
d = 0.8 √I.

Wire diameter in the primary winding:
d1=0.8 √I1 = 0.8 √ 0.45 = 0.8 0.67 = 0.54 mm.

Wire diameter in the secondary winding:
d2 = 0.8√ I2 = 0.8 5 = 0.8 2.25 = 1.8 mm.

The secondary winding is wound with taps.
The first withdrawal is made from 52 turns, then from 56 turns, from 61, from 66 and the last 72 turns.

The conclusion is made with a loop, without cutting the wires. then the insulation is peeled off the loop and the outlet wire is soldered to it.

The charging current of the rectifier is adjusted in steps by switching taps from the secondary winding. A switch with powerful contacts is selected.

If there is no such switch, then you can use two toggle switches for three positions rated for current up to 10 amperes (sold in an auto store).
By switching them, it is possible to sequentially issue a voltage of 12 - 17 volts to the output of the rectifier.

The position of the toggle switches for output voltages 12 - 13 - 14.5 - 16 - 17 volts.

Diodes must be designed, with a margin, for a current of 10 amperes and each stand on a separate radiator, and all radiators are isolated from each other.

The radiator can be one, and the diodes are installed on it through insulated gaskets.

The radiator area for one diode is about 20 cm2, if there is one radiator, then its area is 80 - 100 cm2.
The charging current of the rectifier can be controlled with a built-in ammeter for current up to 5-8 amperes.

You can use this transformer as a step-down transformer to power a 12-volt emergency lamp from a 52-turn tap. (see diagram).
If you need to power a light bulb at 24 or 36 volts, then an additional winding is made, based on for every 1 volt 4.2 turns.

This additional winding is connected in series with the main one (see the top diagram). It is only necessary to phase the main and additional windings (beginning - end) so that the total voltage develops. Between points: (0 - 1) - 12 volts; (0 -2) - 24 volts; between (0 - 3) - 36 volts.
For example. For a total voltage of 24 volts, you need to add 28 turns to the main winding, and for a total voltage of 36 volts, another 48 turns of wire with a diameter of 1.0 mm.

A possible variant of the appearance of the rectifier case for charging the battery is shown in the figure.

How to make a frame for transformer on W - shaped core.

Let's make a transformer frame for the article"How to Calculate a Power Transformer"

To reduce losses due to eddy currents, the cores of the transformer are recruited from plates stamped from electrical steel. In low-power transformers, “armored” or W-shaped cores are most often used.

The transformer windings are on the frame. The frame for the W-shaped core is located on the central rod, which simplifies the design, allows better use of the window area and partially protects the windings from mechanical influences. Hence the name of the transformer - armored. .

To assemble armor cores, W-shaped plates and jumpers to them are used. To eliminate the gap between the plates and jumpers, the core is assembled, in an overlap,.

The cross-sectional area of ​​\u200b\u200bthe W-shaped core S, is the product of the width of the central rod and the thickness of the set of plates (in centimeters). Suitable inserts for the core must be selected.

For example, from the article "How to calculate a 220/36 volt transformer":

- transformer power P = 75 watts;
- cross-sectional area of ​​the magnetic circuit S = 10 cm.kv. = 1000 mm.kv.

Under such a section of the magnetic circuit, we select plates:

— width b = 26 mm. ,
- height of the plate window c = 47 mm,
- window width - 17 mm.,

If there are plates of a different size, you can use them.

The thickness of the plate pack set will be:

S: 26 = 1000: 26 = 38.46. Let's take: a \u003d 38.5 mm.

There are many ways to make frames for a W-shaped core from different materials: electric cardboard, pressboard, textolite, etc. Sometimes frameless winding is used. For low power transformers up to 100W. frames glued together from cardboard and paper work well.

Frame manufacturing.

How to calculate a 220/36 volt transformer.

In the household, it may be necessary to equip lighting in damp areas: basement or cellar, etc. These rooms have an increased degree of danger of electric shock.
In these cases, you should use electrical equipment designed for reduced supply voltage, no more than 42 volts.
You can use a battery-powered electric flashlight or use a step-down transformer from 220 volts to 36 volts.
We will calculate and manufacture a single-phase power transformer 220/36 volts, with an output voltage of 36 volts, powered by an AC electrical network with a voltage of 220 volts.

To illuminate such areas suitable light bulb at 36 volts and a power of 25 - 60 watts. Such light bulbs with a base for an ordinary electric cartridge are sold in electrical stores.
If you find a light bulb for a different power, for example 40 watts, it's okay - it will do. It's just that the transformer will be made with a power margin.

Let's make a simplified calculation of the 220/36 volt transformer.

Power in the secondary circuit: P_2 \u003d U_2 I_2 \u003d 60 watts

Where:
P_2 - power at the output of the transformer, we set 60 watts;
U _2 - voltage at the output of the transformer, we set 36 volts;
I _2 - current in the secondary circuit, in the load.

The efficiency of a transformer with a power of up to 100 watts is usually equal to no more than η = 0.8.
Efficiency determines how much of the power consumed from the network goes to the load. The rest is used to heat the wires and the core. This power is irretrievably lost.
Let's determine the power consumed by the transformer from the network, taking into account losses:

P_1 = P_2 / η = 60 / 0.8 = 75 watts.

Power is transferred from the primary winding to the secondary winding through the magnetic flux in the magnetic circuit. Therefore, from the value R_1, power consumed from a network of 220 volts, depends on the cross-sectional area of ​​the magnetic core S.

The magnetic circuit is a W - shaped or O - shaped core, assembled from sheets of transformer steel. The primary and secondary windings of the wire will be located on the core.

The cross-sectional area of ​​the magnetic circuit is calculated by the formula:

S = 1.2 √P_1.

Where:
S is the area in square centimeters,
P_1 is the power of the primary network in watts.

S \u003d 1.2 √75 \u003d 1.2 8.66 \u003d 10.4 cm².

The value of S determines the number of turns w per volt by the formula:

w = 50/S

In our case, the cross-sectional area of ​​the core is S = 10.4 cm2.

w \u003d 50 / 10.4 \u003d 4.8 turns per 1 volt.

Calculate the number of turns in the primary and secondary windings.

The number of turns in the primary winding for 220 volts:

W1 = U_1 w = 220 4.8 = 1056 turns.

The number of turns in the secondary winding at 36 volts:

W2 = U_2 w = 36 4.8 = 172.8 turns,

round up to 173 turns.

In load mode, there may be a noticeable loss of part of the voltage across the active resistance of the secondary winding wire. Therefore, for them it is recommended to take the number of turns by 5-10% more than the calculated one. Take W2 = 180 turns.

The magnitude of the current in the primary winding of the transformer:

I_1 = P_1/U_1 = 75/220 = 0.34 amps.

Current in the secondary winding of the transformer:

I_2 = P_2/U_2 = 60/36 = 1.67 amps.

The diameters of the wires of the primary and secondary windings are determined by the values ​​of the currents in them based on the permissible current density, the number of amperes per 1 square millimeter of conductor area. For transformers current density, for copper wire 2 A/mm² is accepted.

With such a current density, the diameter of the wire without insulation in millimeters is determined by the formula: d = 0.8√I.

For the primary winding, the wire diameter will be:

d_1 = 0.8 √1_1 = 0.8 √0.34 = 0.8 0.58 = 0.46 mm. Take 0.5 mm.

Secondary wire diameter:

d_2 = 0.8 √1_2 = 0.8 √1.67 = 0.8 1.3 = 1.04 mm. Let's take 1.1 mm.

IF THERE IS NO WIRE OF THE REQUIRED DIAMETER, then you can take several, connected in parallel, thinner wires. Their total cross-sectional area must be at least that which corresponds to the calculated one wire.

The cross-sectional area of ​​the wire is determined by the formula:

s = 0.8 d².

where: d is the diameter of the wire.

For example: we could not find a wire for the secondary winding with a diameter of 1.1 mm.

The cross-sectional area of ​​the wire with a diameter of 1.1 mm. is equal to:

s = 0.8 d² = 0.8 1.1² = 0.8 1.21 = 0.97 mm².

Rounded up to 1.0 mm².

Fromselect the diameters of the two wires, the sum of the cross-sectional areas of which is 1.0 mm².

For example, these are two wires with a diameter of 0.8 mm. and an area of ​​​​0.5 mm².

Or two wires:
- the first with a diameter of 1.0 mm. and a cross-sectional area of ​​​​0.79 mm²,
- the second diameter is 0.5 mm. and a cross-sectional area of ​​0.196 mm².
which in total gives: 0.79 + 0.196 = 0.986 mm².

The article explains how to convert a conventional computer power supply to a voltage of 24 volts.

In some cases, there is a need for powerful power supplies for various equipment designed for 24 volts.

In this article I will tell you how you can convert a conventional computer power supply, both ATX and AT, to a voltage of 24 V. Also, from several of these blocks, you can combine any voltage to power all kinds of devices.

For example, to power a local PBX UATSK 50/200M, designed for a voltage of 60 V and a power of about 600 watts, the author of the article replaced the usual huge transformer units with three small computer power supplies that neatly fit on the wall next to the power switch and almost without creating any noise.

The alteration consists in adding two power diodes, a choke and a capacitor. The circuit is similar to the +12V power bus after the pulse transformer, only the diodes and the polarity of the capacitor are reversed, as shown in the figure (filter capacitors not shown).

The beauty of this alteration is that the protection and voltage stabilization circuits remain intact and continue to operate as before. It is possible to get a voltage different from 24 volts (for example, 20 or 30), but for this you will have to change the parameters of the reference voltage divider of the control microcircuit and change or disable the protection circuit, which is more difficult to do.

Additional diodes D1 and D2 are mounted through the insulation on the same radiator as the others, in any convenient place, but with a full contact patch with the radiator.

Choke L1 is mounted in any place available on the board (can be glued), but it should be noted that in different models and brands of power supplies it will heat up differently, perhaps even more than already standing in the + L2 circuit (depending on the quality of the power supply) . In this case, you must either select the inductance (which should not be less than the standard L2) or mount it directly on the case (through insulation) to remove heat.

You can check the block at full load or at the load for which it will work for you. In this case, the case must be completely closed (as expected). When checking, it should be observed whether the radiators on which the semiconductors and the additionally installed inductor are installed along the -12v circuit are overheated. For example, a power supply designed for 300 watts can be loaded with a current of 10-13A at a voltage of 24V. It will not be superfluous to check the output voltage ripple with an oscilloscope.

It is also very important to note that if you have two or more blocks connected in series working together, then the case (ground) of the circuit must be DISCONNECTED from the metal case of the power supply (I did this by simply cutting the tracks at the places where the board was attached to the chassis). Otherwise, you will get a short circuit either through the ground wire of the power cords or through the bodies touching each other. To visualize the correct operation of the unit, you can bring a light bulb or LED out.

The difference between the alteration of the AT and ATX standards is only in the launch of the block. AT starts working immediately after being connected to a 220 V network, and ATX must either be started with a PS-ON signal, as is done on a computer, or ground the wire of this signal (usually it goes to the control leg of the microcircuit). In this case, the unit will also start when plugged into the network.