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00:00:02

electronic devices, and recently there are

00:00:06

more and more such devices, it is necessary to ensure

00:00:09

uninterrupted power supply, that is, in the event of a

00:00:11

loss of the main power supply, an

00:00:14

auxiliary

00:00:15

additional power supply of the electronic

00:00:18

device must be connected; such a device, for example,

00:00:20

can serve as an electronic watch, but not a wrist watch, but a

00:00:23

large wall clock which by

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default is powered by a power supply

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connected to a 230 volt network, or for those who are

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more accustomed to 220 volts, and if

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the voltage disappears even for a short time,

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then naturally such a clock simply

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goes wrong, but this is not the worst thing, but for

00:00:40

example, if there is already some kind of burglar

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alarm or fire alarm,

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the work is still only from a 230 volt network

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suddenly loses power,

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then there will be more serious

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consequences, for example, to disarm

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such an alarm it would be enough to

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turn off the voltage or cut

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some relevant wires, but

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of course, no one does it directly and powering

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security systems from a 220 or 230

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volt network is always

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provided additional

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backup power is most often from a

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battery;

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however, in most simple electronic

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devices, there is most often no need to

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use any complex circuits to

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ensure uninterruptible power supply, and

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novice radio amateurs or

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electronics engineers, not very familiar with

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this topic, begin to fence off

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all sorts of complex circuits, although as we will see

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later, it will be enough the use of

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just two diodes, let's look at this

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first theoretically in a more simplified way and

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then put together a simple circuit,

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so let's imagine that we have

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some kind of load, let's designate it simply as a

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resistor r m which needs

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to provide constant uninterrupted

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power supply for greater clarity, let's

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replace the resistor rn with an LED

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Only the limiting resistor

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and LED are designated as

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VD.

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and the resistor r in this case can

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be any other load, up to the

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alarm system

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and you here we will have a plus minus and

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we need to power this load all the time

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by default, of course, if this

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load works for us 24 hours a day,

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7 days a week and so on all year round,

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then powering such a load from a battery

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or from a rechargeable battery

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is ineffective, especially if it is

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possible to connect a power supply

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powered from a 230 volt network, let's just

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draw the power supply as

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a square, here it

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will receive 230 volt 50 hertz power at the input It’s

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more common to let it be 220

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volts, but in most cases everyone has

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switched to the 230 volt standard. I

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already talked about this. At the output of such a

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power supply, we get let’s say a

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voltage plus minus a constant

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voltage of four and a half volts,

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although there may be a completely different

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voltage here, but this circuit We

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will collect it with you later, so we will immediately

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write down the values close to our further

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circuit, and we need to connect these plus or minus volts

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to the

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plus and minus terminals of the load, and if we have voltage at the input of the

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power supply, then

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accordingly there is also at the output and

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the load receives power,

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but as soon as the power supply to the 230 volt network is lost, the

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power to

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the load naturally disappears, so it provides an

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additional battery

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plus the battery and a minus and

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simply connects the minus to the minus of

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the load and to the minus of

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the power supply, and the plus naturally to the plus,

00:03:38

that is, they connect it in parallel to the main power supply the

00:03:42

battery will indicate its death, but

00:03:45

directly connecting

00:03:47

two power sources in parallel in this case

00:03:49

will be ineffective,

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since in the event of excess voltage

00:03:53

on one of the power sources, a

00:03:56

balancing current will occur with subsequent

00:03:59

consequences. I will depict these two

00:04:01

power sources and the load separately

00:04:04

here to make it more clear what what

00:04:05

happens is that we will have a plus and a minus of

00:04:08

the bp power supply. the output voltage is

00:04:12

four and a half volts, here we

00:04:14

will have a battery connected in parallel,

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labeled exactly the same as

00:04:21

GB, and in parallel with this

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power source, our video

00:04:26

resistor and LED load are connected, so when

00:04:29

the voltage is higher, for example, on the

00:04:31

power supply we always let the voltage

00:04:33

be higher than

00:04:34

4.5 volts, and here let’s say 4 volts, then

00:04:38

due to the potential difference, the current from the

00:04:41

power supply will flow not only through

00:04:43

the load but also through the battery itself, the

00:04:46

opposite situation will be

00:04:48

observed if there is a voltage here,

00:04:49

let’s say 5 volts on the battery, it’s

00:04:52

power supplies 4 and a half then the current

00:04:53

will flow

00:04:54

and through the power supply and through the load,

00:04:57

by the way, you can personally verify this

00:04:59

by placing

00:05:00

an ammeter here and placing an ammeter here in the same way,

00:05:03

or a milliammeter, and

00:05:08

making the voltage of one of the power sources higher and then you will see that when the

00:05:12

power supplies are connected in parallel,

00:05:13

reverse current will begin to flow through that

00:05:15

battery the voltage will be lower,

00:05:18

so in order to eliminate this

00:05:20

drawback it is enough to

00:05:24

put a diode here instead of an ammeter in our case,

00:05:26

then in the case when the voltage

00:05:28

here is higher than here, that is, on the

00:05:30

power supply it will be higher than on the

00:05:32

battery,

00:05:34

the diode will simply be locked higher

00:05:37

potential coming from the power supply,

00:05:39

it looks more clearly like this, this is how

00:05:42

our diode here has an anode-cathode

00:05:45

current flows in this direction and there

00:05:48

will be a plus here and a minus here, so the current

00:05:50

flows only from the anode to the cathode in the

00:05:52

opposite direction, it does not flow like

00:05:55

we do we know, but it flows not only

00:05:58

when there is 5 volts here and zero volts here,

00:06:00

but also when let’s say there will be 10 volts here

00:06:03

and there will be, for example, 8 volts, that is,

00:06:05

when the potential on the side

00:06:07

will be higher than the potential on the

00:06:09

cathode side, now all the current coming from the

00:06:13

power supply will already flow

00:06:15

exclusively through the load, it will

00:06:17

no longer pass here, but in the case when the voltage of

00:06:19

the battery becomes higher than the

00:06:22

voltage of the power supply for some

00:06:24

reason, then the top from the

00:06:27

battery can still flow through the power

00:06:30

supply and through the load; for this,

00:06:32

another diode is installed and then the current

00:06:35

from the battery

00:06:37

will be able to flow exclusively through the

00:06:39

load on our circuit it will

00:06:41

look like this here we will have a

00:06:45

positive diode and here we

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will have another diode in the same way everything is such a

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simple circuit now we will see how

00:06:52

it works in practice for For this we

00:06:54

will need any two diodes since we

00:06:56

will use an LED; its power is

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naturally low, so a

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1 n 41 48 diode will suit us, or I already have two

00:07:05

soldered 1 n 4007 diodes, they are of course more

00:07:10

powerful than these diodes, but this is

00:07:12

not so important for us now, please note

00:07:15

that both diodes are connected to each other,

00:07:17

in this case they are soldered with cathodes, our diodes

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exactly correspond to this figure,

00:07:22

but who doesn’t remember how to determine the anode and

00:07:24

cathode of a diode, to do this, just turn on the

00:07:26

multimeter and turn on the continuity mode;

00:07:28

the result should flow only in

00:07:30

one direction, for example, we apply a black probe

00:07:32

to one terminal of the diode and the

00:07:34

red one, which was just the probes,

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came off, the historical moment

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is applied to the second terminal and we

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see the diode conducts current if we swap

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the spikes

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the diode will not conduct, this is how

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we see the diode does not conduct current,

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and the easiest way to navigate is by the

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marking strip in this case,

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here it’s black

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and here it’s gray, it’s just applied

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closer to the output side indicating the

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cathode, and so our load here we

00:08:07

’ll supply plus from the battery and the

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connection point of the two diodes will be connected

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to a

00:08:13

resistor, check the circuit, connect

00:08:16

the battery and see the LED lights up,

00:08:19

that is, voltage the voltage of the batteries is supplied to the load;

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we have 3 volts.

00:08:23

Now we will connect the power from the

00:08:25

power supply 4 volts plus we

00:08:28

connect the diode to the anode

00:08:31

and we have a common minus and we see the LED is

00:08:34

lit. We also connect the plus from the

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battery to the anode of 2 diodes and we see everything

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we are working here, we have a

00:08:43

plus coming from the power supply and powering our

00:08:47

LED, but at this time the

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battery voltage is not supplied because

00:08:51

the voltage supplied from the power supply is

00:08:53

higher than the battery voltage, therefore

00:08:56

no equalizing current flows in this circuit

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and does not consume the rock of the batteries, now

00:09:01

we remove voltage of the power supply and

00:09:04

we see the LED is still lit, having

00:09:06

received power from the battery, here is such a

00:09:10

simple circuit, but here you should also

00:09:11

remember that a voltage of the order of half a

00:09:14

volt will be lost on the dione, as when

00:09:17

powered by batteries such when powered by a

00:09:19

power supply, and here you need to remember the

00:09:22

following: the voltage coming from the

00:09:24

power supply should be higher than

00:09:26

the voltage of the battery or, in this

00:09:28

case, batteries by the amount of the

00:09:31

voltage drop of the diodes, that is, it is desirable

00:09:33

that the minimum difference be of the order of

00:09:35

one volt, but it is better to make a difference of

00:09:38

two volts; in addition, such circuits

00:09:40

can not immediately power the load and then the

00:09:42

stabilization unit and then the load, well, that’s

00:09:45

all, I hope the circuit will be

00:09:48

useful to those who didn’t know it or to those who

00:09:50

forgot, in the next video we will continue to

00:09:52

study the inductance of an electric

00:09:55

filter and so on, but that’s all,

00:09:58

subscribe to the channel and bye

Description:

Надежная схема бесперебойного питания состоит всего из двух диодов, соединенных катодами. К аноду одного диода подключается положительный потенциал блока питания. А к аноду другого – положительная клемма аккумулятора, что позволяет избежать уравнивающих токов, вызванных неравенством напряжений. Напряжение блока питания должно быть выше минимум на 1…2 В от аккумуляторной батареи. Такая схема бесперебойного питания от аккумулятора обеспечивает резервное питание электронного устройства напряжением основного источника тока. Добро пожаловать на мой сайт: https://diodov.net/

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