Light up the running light, rough notes for beginners (1)

Turn on the running water lamp , Rough notes for beginners （1）

As a beginner, what I want most is to be able to be taught in detail without professional terms , But I often don't know where to start , So here is a summary of useful and understandable ways I have heard , Mixed with my learning ideas , Also record my learning process , Maybe in the end , The more you learn, the easier it is to remember , But at first, I always like to record in detail .
For a simple description, the following low level is used “-” Express . For high level “+” Express

Lighten up led Code

Latch open , And turn it off after finishing the operation .
It's a matter of habits , It is also to prevent programming logic errors later .
No addition “while（1）;”
You write a piece of code, and the single chip computer runs super fast all at once , You can't see the light on .
And it can be seen on the board with a latch without adding it, because there is a latch .
（ It means , No addition while For example, the lamp delay setting is 1 millisecond , If there is no latch at the same time , You can't see it bright, but you can't see the result with the naked eye .）
added while（1）, It will be executed until while It has been implemented here while.while Except for conditional restrictions .

Schematic diagram used

①

When Y4c For high voltage Q1-Q8 The level of will follow
D1-D8 Change by change .
（D1+,Q1 also +）;
If at this time Q1 by “+”
Y4c Change it to “-”
Well, then D1 Change it to “-”, here Q1 Will only maintain the previous state , That is to say “+”, Not with me D1 Change by change .
（ So we should make Q1-Q2 along with D1-D2 Change and change requires Y4c by “+”）.
②

It's up here 4 One is concentrated on the following chip

The above one is counterproductive , So be Y4c take “+” that Y4 Just take “-”.
③

*** The principle of the above diagram is to convert binary to decimal

Here is an example

If you want to light led L1, You need to look at the flow chart like this

①

y4c by “+” Can we make Q1 along with D1 Change and change

②
Y4c for “+” be
Y4 for “-”

③
Y4 for “-” be Y4 for “0”

therefore P2 The first three are 100
Back 5 The bit hypothesis is 00000.
So point L1 The code is ,P2=0x80;P0=0x01;

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（ Lighten up L1——Y4c for “+” Can we make ——D1 by 0——Q1 Also for the 0

Y4c for “+”——Y4 for “-”——Y4 Corresponding P27P26P25 From binary to decimal, the result is 4, So that Y4=0——Y4=0——Y4“-”——Y4c“+”——Q1 Can follow D1 Change and change —— At this time make D1=0 be ——L1 On one side “+” On one side “-” The light is on .）

About P2=0x80;P0=0xff;P2=0x00;

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P2=0x80 open Y4 Latch ;
P0=0x00 Close all led The lamp ;
P2=0x00 close Y4 Latch ;

therefore P2=0x80;

About P2=0xa0;P0=0x00;P2=0x00;

Both sides of the buzzer are regular off
N BUZZ To be regular

therefore P0=0x00;// Turn off the buzzer

P2=0xa0;// open Y5 Latch ;
P2=0x00;// Close the latch ;
P2 For binary order P27 P26 P25 P24 P23 P22 P21 P20
P0 For binary order P07 P06 P05 P04 P03 P02 P01 P00

How to only give P27P26P25 Assign values without affecting the remaining .

1 It's true 0 For false ;

use “|“”&“

P2=0x9f&（P2|=0xe0） ;// open Y4

|： Self understanding is equivalent to “∪”

Two true is true ,
Two false is false ,
One true and one false is true .

&： amount to “∩”

Two true is true ,
Two false is false ,
One true and one false is false .
Suppose at the beginning P2 All eight don't know
？？？？ ？？？？

or “|” On “e0” The process

？？ ？？ ？？？？
1 1 1 0 0 0 0 0（e0）

Or go up 1 1 1 ？ ？？？？

The process is as follows

111？？？？？ And again 10011111（9f） And “&” The process

100？？？？？
The process is as follows

In this way, only the first three mouths are changed , And make the remaining several mouths unchanged .

Relay

The same is true of the way the buzzer is used

About simple usage

In short , About lighting , You only need to know the function of the schematic diagram of the lamp , If you take the opposite , What will change with what changes , Common Yang or common Yin ？ When one end is “+” one end “-” Then it's all right .