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FLOWCODE

PROTEUS

ISIS

ARES

УРОК

HIASM

ARDUINO

00:00:00

and one more important point that I will

00:00:02

probably take into a separate lesson is the

00:00:04

use of transistors to

00:00:06

connect, in this case, a

00:00:07

seven-segment

00:00:09

indicator. I have already done a lesson on

00:00:11

connecting Well, for

00:00:14

beginners on connecting transistors to a

00:00:17

microcontroller, but now let’s look

00:00:19

at this example. So why do we

00:00:21

use transistors the output current of

00:00:24

microcontrollers at each pin is

00:00:26

limited; this current value is written in

00:00:29

Dashi and the maximum current that can be

00:00:31

obtained from one pin without destroying it

00:00:35

is several tens of miami.

00:00:38

Therefore, if we control some

00:00:40

powerful load, then we need to use

00:00:43

transistors, in any case, I try my best to

00:00:45

practice more than 10

00:00:47

worlds do not load the controller outputs So

00:00:51

let’s say we have some kind of powerful

00:00:53

indicator and to control it from the

00:00:55

microcontroller we will use

00:00:57

transistors. What options could there be, or a

00:01:00

bipolar transistor or a

00:01:03

field-effect transistor in our time if there is a

00:01:06

radio parts store nearby and others

00:01:08

available at approximately the same

00:01:10

price but at low frequencies, here I am. Well,

00:01:15

since I work with digital technology most of

00:01:17

all, I give clear preference to

00:01:20

field-effect transistors because they have

00:01:23

easier control and better characteristics.

00:01:26

Well, we’ll start with bipolar ones.

00:01:30

If you assemble a circuit using

00:01:32

this circuit, of course you need

00:01:37

to limit the current from the current flowing to the

00:01:40

indicator, for this you

00:01:44

need to put resistors on the terminals of the segments. Well, approximately like

00:01:46

this, here is a resistor on each terminal,

00:01:50

its resistance

00:01:52

is calculated by the color of the indicator light,

00:01:55

let’s say we have a red

00:01:57

voltage drop on the Red LED

00:01:59

is 2 2 s. Well, approximately Volts

00:02:02

maximum glow current Well,

00:02:05

let’s say 10 mA, let it be 5 mA,

00:02:08

enough for there to be a glow that is

00:02:11

Well, not very high brightness, but convenient

00:02:14

for

00:02:16

perception. Totally, the controller power supply is

00:02:19

5 V minus 2 V drop on the

00:02:23

LED, we get 3 V and divide by the current

00:02:27

that we need to get by we

00:02:30

get how much there, like 600 ohms,

00:02:34

we choose the closest of the available

00:02:38

resistors, let it be 560 ohms, these are

00:02:41

the resistors we put on each

00:02:44

segment to limit the current Well, again, I

00:02:47

retreated and again let’s return to our

00:02:51

transistors, we have an indicator with a common

00:02:53

anode. I remind you that in the flo

00:02:56

code for a four-digit indicator you

00:03:00

cannot change common anode or common

00:03:03

cathode, if

00:03:06

you put the usual one symbol, then here

00:03:10

you can already

00:03:11

change what we will have as a common anode

00:03:14

or a common

00:03:16

cathode. Well, in this case, we have a

00:03:20

common anode with this indicator, that is,

00:03:24

supplying PSM from bipolar transistors to the etivo wo

00:03:30

n’t even work, well ok now let's

00:03:32

try So a bipolar transistor

00:03:34

let's say the current we have a small

00:03:37

power there is small Well let's put an SMD

00:03:39

small package

00:03:41

bc847 these are the transistors in the

00:03:45

c23 package the maximum voltage there I don't remember

00:03:48

something 20 from 20 to 100 V and the

00:03:51

maximum current is 100 mA we select and they

00:03:54

should give us Plus That is, we

00:04:00

will connect these indicator anode pins

00:04:02

to the emitter circuit, here we

00:04:07

will have a power bus plus 5

00:04:14

V and we will break these 5 V with these

00:04:19

transistors like this. We break the circuit from here.

00:04:23

Okay, now I’ll draw there, let’s

00:04:28

continue. So, so big The question is how

00:04:31

to calculate the resistors that are in

00:04:34

the base and emitter circuits of the transistor, we

00:04:36

consider the base

00:04:42

circuit which resistor should be placed

00:04:45

between the output of the microcontroller and the base of

00:04:48

the transistor if our circuit is built in

00:04:52

such a way that the emitter sits directly on

00:04:55

the ground, that is, there is no power here, there is

00:04:57

a load in the collector then if we supply the

00:04:59

voltage output directly to

00:05:02

the base, then due to the low drop at the

00:05:07

transition, well, roughly speaking, a

00:05:09

short circuit will result, that is, all the current

00:05:12

will go to the ground, the microcontroller

00:05:15

will fail, for this it is necessary to

00:05:18

limit it, that is, in theory, the greater

00:05:20

the resistance, the better the less

00:05:22

the load on

00:05:24

the microcontroller, on the other hand, if

00:05:27

the resistance is very large, then we will not

00:05:30

get the required current at the output. We remember

00:05:34

that a bipolar transistor has such an

00:05:36

indicator as the current transfer coefficient,

00:05:39

that is, roughly speaking, how much it

00:05:41

amplifies the signal if it has a gain coefficient,

00:05:45

for example, these transistors 100 Well, the average coefficient And

00:05:48

if we supply 1 ma to the base, then with

00:05:51

the collector we will not be able to remove more than

00:05:53

100 ma Based on these considerations,

00:05:57

this resistor is calculated

00:06:01

Let's open the SM

00:06:05

Studio program, we need to know approximately the

00:06:08

gain coefficient, it is designated as

00:06:10

H Well, let us will be on

00:06:13

average

00:06:16

100 you need to

00:06:18

know the

00:06:21

load current Well now, of course, I do a very

00:06:25

detailed calculation, then if with experience comes

00:06:28

what to

00:06:30

estimate by eye What resistance is

00:06:32

needed So we have a load current equal to

00:06:36

how much Well, let’s say let it be 100 mi

00:06:38

am consumes the indicator this is of course

00:06:41

a lot

00:06:42

but let it will be like this 01

00:06:49

ampere

00:06:54

load current and now we find the required

00:06:58

current that is needed in the base circuit of

00:07:02

the transistor it turns out

00:07:04

and the

00:07:07

base will be

00:07:09

equal to the load current

00:07:12

and we divide by our gain factor

00:07:15

divided by

00:07:18

H from here

00:07:20

and the base

00:07:23

equals we get this current 1

00:07:30

should flow through the

00:07:32

base we calculate the resistance of this

00:07:37

resistor resistance will be

00:07:39

equal to Supply voltage, that is, the

00:07:42

voltage that comes from the

00:07:44

microcontroller port, it is approximately equal to the

00:07:46

supply voltage, let it be 5 V 5 so

00:07:51

parentheses 5 V from it we subtract the

00:07:55

voltage drop at the junction of the base emitter for

00:07:59

transistors for modern ones for

00:08:01

most this is where - then 05-07 in let

00:08:04

it be 05 and divide by the resulting

00:08:16

base current equals 4.5 Coma That is,

00:08:21

as you can see,

00:08:22

we need to put quite a large resistance to

00:08:25

provide the required current for the indicator to the

00:08:27

microcontroller

00:08:29

no load there is about 1

00:08:32

mi and the indicator will flow required

00:08:35

current Well, in order to take it with a margin, we take

00:08:37

the resistance a little lower, let’s say there,

00:08:40

well, usually 1 Kim is taken, that is,

00:08:44

I will say again that the minimum possible

00:08:46

resistance is calculated based on the

00:08:49

maximum output current of

00:08:51

the microcontroller, and the

00:08:54

maximum maximum possible

00:08:56

resistance of the base circuit is calculated

00:08:58

Based on the required load current I hope I

00:09:02

explained it clearly in this case we take

00:09:04

Well, let there be 1

00:09:11

com Let it go

00:09:14

from the output of the microcontroller and from the

00:09:19

emitter output we draw a

00:09:23

bus Let

00:09:26

it be like this, okay Let it be like

00:09:29

this from the output of the emitter we will get

00:09:32

the voltage to the first leg of the

00:09:37

indicator sign

00:09:41

one but again here we have there is no

00:09:44

flow limitation for the indicator,

00:09:46

that is, it will receive 5

00:09:48

V; if there are red LEDs and it is not

00:09:51

designed for such a voltage, then

00:09:54

the indicator will fail. Therefore, we

00:09:56

need to limit the current; this can be done by

00:09:58

including a resistance in the collector

00:10:00

or emitter circuit of the transistor. Well,

00:10:03

or if we already have resistors at these terminals,

00:10:06

then we do

00:10:08

n’t need to add anything. Well, let’s assume that there are

00:10:11

no resistors here. And we need to add

00:10:13

resistors. Well, in the emitter

00:10:19

circuit, that is, we put a

00:10:24

resistor here and calculate its

00:10:26

resistance.

00:10:29

So, this is our emitter

00:10:32

follower. If anyone

00:10:34

knows the theory bipolar transistors, then

00:10:37

we will have the same voltage at the emitter

00:10:40

as at the base minus the

00:10:42

voltage drop at the

00:10:44

junction, that is, U at the

00:10:50

emitter will be equal to the power supply, that is, 5

00:10:54

V

00:10:56

-05 conventionally, so we

00:11:03

remove 4.5 V from the emitter and we need limit the

00:11:08

current through the indicator at a level of, say,

00:11:10

20

00:11:12

Ma. How is this calculated, that is, here

00:11:15

we have a plus power transistor

00:11:18

resistor and here the indicator goes to ground,

00:11:22

we find out the voltage drop on our

00:11:24

LED on the

00:11:26

indicator, let’s say it’s 2.5 V

00:11:31

and knowing the load current we calculate the

00:11:36

resistor resistor will be equal, open the

00:11:39

brackets, the voltage on the

00:11:42

emitter minus the voltage on the LED,

00:11:46

let's say this is well, 2 s to 2,

00:11:49

this is the drop voltage on the LED

00:11:53

for red ones. This is from poto to 2 s, according to conventional

00:11:57

rules, for blue and

00:12:01

white, it’s the largest, about 3- 35

00:12:05

V That is, how does the spectrum of colors go there

00:12:09

Every hunter wants to know where the

00:12:11

pheasant sits from red and so on green

00:12:14

blue purple red - the

00:12:17

smallest voltage drop purple

00:12:19

color - the largest voltage drop

00:12:22

purple ultraviolet white color

00:12:25

Who doesn’t know the white color in our

00:12:27

LEDs turns out using a

00:12:30

phosphor that is irradiated with

00:12:32

ultraviolet light So 4.5 -

00:12:38

2.5 and we divide all this by the current that

00:12:41

should go through the indicator, let’s say 10

00:12:45

mA

00:12:48

0.01, so our resistance is

00:12:51

200

00:12:53

ohms, set

00:12:56

here 200 ohms,

00:13:01

so we calculated everything,

00:13:06

add the necessary ones the cells are the

00:13:24

same, so

00:13:26

we sign 2, so

00:13:32

everything is connected, well,

00:13:34

we have the properties of the indicator. And here there are no

00:13:37

properties,

00:13:38

that is, it won’t work, but then

00:13:41

I’ll show it on

00:13:43

the LED. So, we launch the program and

00:13:47

see that everything is displayed. The

00:13:49

indicator is controlled

00:13:51

through key

00:13:54

transistors. Let’s do it

00:13:57

now instead of indica, I’ll put

00:14:01

LEDs here, that is, the indicator has the

00:14:04

same ordinary

00:14:12

LEDs, we adjust the

00:14:15

voltage drop. Let it be like this: as much as I

00:14:17

took 2 s to Volt and a current of 10 mim, this is at

00:14:22

which it is completely

00:14:25

lit 2 s polo

00:14:29

10

00:14:30

mA and now let’s see the

00:14:34

currents, put a probe

00:14:37

current here Well, they are the same circuits Let's

00:14:40

just put it on one circuit, that is,

00:14:42

on the first transistor, what we have in the

00:14:44

base circuit and what we

00:14:47

have Through our indicator through the LED

00:14:50

If I turn it on now, we won’t

00:14:52

see anything because there is a high

00:14:56

frequency, so I set up a graph, an

00:15:00

analog

00:15:02

graph, transfer

00:15:07

the indicators to it and

00:15:12

run it, okay, better

00:15:16

one at a time,

00:15:20

oh, that is, now we will

00:15:24

look at this probe, what we have

00:15:26

flowing from the base, we look at the

00:15:29

amplitude of our current 1.6

00:15:34

mA, as I will remind you here is calculated from there

00:15:37

were 4.5 of us Coma I took less than 1

00:15:41

Coma here the maximum current is less than 2 Mi

00:15:46

goes from the microcontroller to each

00:15:48

transistor this is safe for the

00:15:50

microcontroller and it can control a

00:15:52

large load

00:15:54

now

00:15:57

we remove this and install the probe that we

00:15:59

have going through the

00:16:01

indicator we see we see from zero

00:16:06

to almost 9 miam, the current flows through the indicator,

00:16:12

approximately the one that was calculated. This is how the

00:16:15

resistors in the

00:16:17

base and collector or emitter

00:16:20

circuits for bipolar transistors are calculated. I hope I explained it.

00:16:25

Clearly. But the only difference is if

00:16:29

let’s say here the emitter goes to the

00:16:32

ground and the collector goes to the circuit. you have

00:16:35

some kind of

00:16:37

load turned on. Well, let’s say the same

00:16:43

indicator. That is, here

00:16:47

the resistor and here the

00:16:50

LED, the calculation will be exactly the same with

00:16:54

the only difference being that the

00:16:56

voltage drop across the base of the emitter we will not do

00:16:59

for this transistor, that is, here we

00:17:01

imagine a completely

00:17:03

open key and calculate

00:17:05

Supply voltage minus the voltage drop on

00:17:08

the LED Well, in fact, there is a

00:17:10

voltage drop on the transistor. Well,

00:17:12

you can ignore it in this

00:17:15

case and divide by the current needed by the LED,

00:17:19

we get a resistor and Let’s

00:17:24

quickly show an example on a field-effect

00:17:26

transistor; it wo

00:17:29

n’t be possible to use it because

00:17:35

we are an indicator we plug it into the emitter circuit,

00:17:38

or into the circuit from the field-effect

00:17:42

transistor current, let’s take

00:17:44

72 2n

00:17:49

72 this is a field-effect transistor in exactly the

00:17:52

same

00:17:55

case, it seems that it also has 100

00:17:58

maximum Well, or something

00:18:02

like that Why won’t it work because we

00:18:06

also see that it’s

00:18:08

included in as Tori sources, that

00:18:12

is, the voltage at the sources here

00:18:15

below will repeat the voltage at the

00:18:17

gate minus the gate-source voltage. And

00:18:22

for field-effect transistors it is quite

00:18:24

large. Well, that is, most of the

00:18:26

ones I work with have this voltage from up to

00:18:29

5 V, that is, in this situation it is not

00:18:32

very applicable And besides,

00:18:35

the transistor will be useless, it will

00:18:37

heat up, there will be no special advantages of the field-effect

00:18:40

transistor.

00:18:41

If we turn on the load in the

00:18:44

drain circuit of the field-effect transistor, then here

00:18:49

everything starts to look much

00:18:54

better; powerful field-effect transistors,

00:18:57

for example,

00:18:59

which are not very expensive there, but at the

00:19:01

same time have very cool

00:19:04

characteristics roughly speaking, by opening it,

00:19:07

feeding it from the controller 5 V through a

00:19:10

resistor. Well, let’s say 10

00:19:12

kWh, that is, the microcontroller doesn’t

00:19:15

feel this load at all. And we

00:19:19

can control the transistor, let’s say a heater

00:19:23

that consumes 20 A from a car battery.

00:19:32

These field-effect transistors have the peculiarity

00:19:34

that the gate is isolated from the rest of

00:19:39

the channel. It turns out that there is a very large

00:19:42

resistance here and it does not need to be taken into account at

00:19:47

low frequencies of the order. Well, somewhere, let’s

00:19:50

say conditionally up to 1 kHz, that is, here I

00:19:53

can just turn on directly like this, the

00:19:59

output of the microcontroller to the gate and it

00:20:01

will not see any load at all

00:20:04

when the zero transistor is closed

00:20:07

and the load flows. Well, almost no

00:20:11

current flows, roughly speaking. If you open it and

00:20:15

apply a logical one, then it

00:20:17

will open and the resistance of this

00:20:20

junction

00:20:22

in some types of transistor will

00:20:25

reach several

00:20:27

millimeters then it’s as if there’s just a

00:20:29

wire here, these are their properties: high

00:20:31

input resistance And very low

00:20:34

transition resistance I like them; they’re

00:20:37

easy to control and the circuits turn out to be

00:20:39

much simpler and more economical; less

00:20:42

heating, that is, I recommend it for use;

00:20:51

for example, here is the problem: How to remember where

00:20:53

what terminals are on a transistor, like everyone else, I

00:20:56

probably started with bipolar

00:20:57

transistors and only then learned all the

00:21:00

delights of field-effect transistors and it

00:21:03

was quite difficult to remember where what terminals are,

00:21:05

how they are designated. If you know the

00:21:08

designation of the terminals of a bipolar

00:21:10

transistor, then I will show you how easy it is

00:21:12

remember the outputs of the field-effect transistor, we

00:21:15

remember this is the base. The central terminal is the

00:21:19

base; the output with an arrow is the

00:21:23

emitter and the output that is empty, the remaining one, is the

00:21:29

base; the emitter; the collector of the field-effect

00:21:32

transistor; the conclusions are called gate-source;

00:21:38

drain; how can you remember them? I discovered

00:21:41

such a pattern and remember them by

00:21:43

the first letters, that is, we look Well, the shutter

00:21:46

is clear That is, this is the middle pin

00:21:49

that is controlled by the shutter, it is easy

00:21:51

to remember That is, it is like a

00:21:53

base But then we remember by the first

00:21:57

letters here below that the emitter

00:22:00

begins with a vowel letter

00:22:04

e, which means that here the pin that goes down

00:22:08

begins with a vowel and

00:22:11

the source here the collector begins with k with a

00:22:15

consonant letter means and Here it

00:22:18

begins with a consonant letter this is Stock

00:22:21

Well, that seems to be all I wanted to say

00:22:25

I hope I helped everyone good luck

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The browser/computer should not freeze completely! If this happens, please report it with a link to the video. Sometimes videos cannot be downloaded directly in a suitable format, so we have added the ability to convert the file to the desired format. In some cases, this process may actively use computer resources.

How can I download "PROTEUS УРОК 23 Расчет транзисторного ключа" video to my phone?

You can download a video to your smartphone using the website or the PWA application UDL Lite. It is also possible to send a download link via QR code using the UDL Helper extension.

How can I download an audio track (music) to MP3 "PROTEUS УРОК 23 Расчет транзисторного ключа"?

The most convenient way is to use the UDL Client program, which supports converting video to MP3 format. In some cases, MP3 can also be downloaded through the UDL Helper extension.

How can I save a frame from a video "PROTEUS УРОК 23 Расчет транзисторного ключа"?

This feature is available in the UDL Helper extension. Make sure that "Show the video snapshot button" is checked in the settings. A camera icon should appear in the lower right corner of the player to the left of the "Settings" icon. When you click on it, the current frame from the video will be saved to your computer in JPEG format.

What's the price of all this stuff?

It costs nothing. Our services are absolutely free for all users. There are no PRO subscriptions, no restrictions on the number or maximum length of downloaded videos.