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https://www.ikalogic.com/8-bit-digital-to-analog-converter-dac/


아래글을 읽기전에

저항의 분배법칙

Op Amp의 기능중 Buffer Solution을 알아야한다.

8-BIT DIGITAL TO ANALOG CONVERTER (DAC)


This article aims to introduce to beginners and intermediate readers 

a simple solution to build a digital to analog converter, 

based on the famous r/2r resistors network.

This article also discuss a problem encountered 

by many beginners 

while trying to build their own DAC, 

and proposes some very simple solutions to that problem.

Learn how to build an Analog to digital converter using the same simple technique explained in this page.

Through this article, 

I am going to explain how to build an 8-bit digital to analog converter 

with parallel input. 

If you don’t know what this means, well 

its simply a circuit that will take 

as input a digital 8-bit number 

from 0 (00000000) to 255 (11111111), 

and output the relative value on a scale from 0 to 5v.

The maths that describe this process is very simple, 

an 8 bit converter will divide the 5 volts into 255 steps, each step having a value of:

5/255 = 0.019 V

한비트당 0.019V가 할당된다는 의미이다.

Then the output voltage for the converter 
should be equal to the binary input multiplied 
by the step value, 
e.g. for an input of 129 (1000 0001 in binary) 
the output voltage should be:

129 X 0.019 = 2.451V

1000 0111 = 2^7 + 2^0 = 128 + 1 = 129

한 단계당 0.019V 이므로 두값을 곱하면 2.451V.

Here is a simplified functional diagram of an 8-bit DAC.


Some vocabulary:

DAC: Digital to Analog converter.
D0, D1, D..: Data lines.
Analog: Continuous electrical signals.
Digital: Method of representing information using “1” and “0” (usually 5v and 0V).
LSB: Less significant bit.
MSB: Most significant bit.



The R/2R resistor network.


The digital data entering thought the 8 lines (D0 toD7

are going to be converted 

to an equivalent analog voltage (Vout

by the mean of the R/2R resistor network. 

Actually a lot of commercial Digital to Analog converter ICs 

are based on this same principle.

The R/2R network is build by a set of resistors of two values, 

with one of them double the other (example 10K and 20K), 

in on of my circuits 

I used 1M ohm and 470K ohm resistors, 

which is quite near to the R/2R ratio, 

and this small difference didn’t cause any detectable errors in most applications.

However, 

if you want to build a very precise DAC, 

be precise when choosing the values of the resistors 

that will exactly match theR/2R ratio

Note that you can build a DAC with any number of bits you want, 

simply by enlarging the resistor network, 

by adding more R/2R branches (like the one shaded in green)

BUT you must keep the 2R resistance connected to ground (shaded in light red)

Going through the mathematical proof for the operation of 

this converter can be a pain for some of us, and I am only intending to keep things simple.

Now, in order to use this Resistor Network (also called R/2R Ladder

for real applications, 

you will have to build a very simple voltage buffer circuit, 

which will be explained in the next section.


The applied circuit


All the components are labeled on the circuit, 
so i’ll start directly to explain how it works. 
To simplify this task, 
i’ll split the circuit into two main stages: 
the Digital to analog converter 
and 
the Voltage buffer stage.



Stage 1: the Digital to analog converter (The R/2R network)

This part have been explained in detail in the previous section, 

its purpose is to create the voltage V1 

which is equivalent to the weight of the binary number 

on the lines (D0 to D7).

Now that this is a resistor network, 

if we apply any load on the output of the first stage

this load will be considered 

as an additional resistor in the network, 

and thus 

will disturb the network which will no longer provide the correct & desired output voltage. 

Therefore, to overcome this problem, 

we need a voltage buffer, here is where the next stage comes…

출력에 부하를 연결하지 않는걸 권한다.


Stage 2: the voltage buffer

This stage will isolate the point V1 

from the final output V2

while always keeping the voltageV2 at the exact same value of V1

This is what we call a voltage buffer.

for the voltage buffer 

we use an opamp with the output 

connected to the inverting input 

(this special configuration of the Op Amp is also called Voltage Follower).

The most important things to note are:

1.No current (almost 0A) 

will flow from the point V1 into the OpAmp, 

    so we wont be disturbing the resistor network configuration.

네거티브 FeedBack에 의해서 이다.

2.V2 will always equal V1 (theoretically, see the rest of this document).

3.The current going out from the point V2 

to any other stage 

is sourced from the power supply of the OpAmp.

V2로 부터의 출력전류의 근원은 OpAmp의 공급전류이다.


The most encountered problem & some solutions

A quick look on those two graphs 

can be sufficient to understand the problem: 

the output of the op-amp is not linear on the full 0-to-Vcc scale. 

actually an OpAmp, depending on its type, 

will deliver a maximum voltage of (Vcc – 0.5V), 

where Vcc is the supply voltage of the OpAmp.

So, in our application, the OpAmp will only deliver 4.5V 

even if theoretically it should deliver 5V.

이론상으로 출력이 5V지만, 실제로는 0.5V가 빠진 4.5V가 출력된다.

You may think this caused by the resistor network, 

but it’s not! this is a limitation in the op-amp itself.

Lets get a little deeper into the problem, 

the actual output curve in red should be linear, 

but actually it begins loosing its linearity beginning from 3.9 volt.

3.9V이상부터는 선형성을 읽는다.(불규칙해진다.)

(Again this depends on the type of OpAmp, those results a based on my own tests on a LM350 OpAmp

The red ‘Error zone’ is where the output of the DAC no longer math the relative binary input.

This is the error we will be trying to overcome in the next part, 

through two very simple solutions.

Solution 1 :


The first solutions – shown in the red shading – is to increase the supply voltage of the Op-Amp, as shown in the schematic. 

This will totally solve the problem, 

and, whether you are supplying 6.5 volts or more, 

you will get neat linear output from 0V to 5V.

공급전압을 6.5V이상으로 하면, 0~5V의 선형값을 얻을 수 있다.



Solution 2 :

The second solutions – shown in the red shading – is to reduce the range of the input to [0 to 127] from the original range of [0 to 255]. 

This will result on a voltage swing of 0 to 2.5 volt at the output, 

which will be in the linear operating area of the Op-Amp 

(this done by attaching the MSB line to ground, this way you only control the 7 other lines, and a 7 bit value can swing from 0 to 127).

I hope this introduction was interesting, and that it will help you to build simple but reliable DACs to suit your application.

비트를 하나 출여서 변환시킨다.


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PIC16C73

스마트 비닐하우스 (1).zip

OUSKI-IP6 (웹캠어플)


T:온도 H:습도

온도가 세팅온도와 안맞으면 맞을때 창문열림

습도가 세팅습도와 


▼보고서 분할압축

it 거의 최종.vol1.egg

it 거의 최종.vol2.egg




박성진하우스제어.zip


▼설명

1.과제의 필요성 (3페이지 이내)

Abstract를 작성한 후에, 서론을 작성합니다.

서론에는 왜 이 작품(과제, 논문, 연구)를 하면 좋을지?

2.선행연구 및 기술현황 (2페이지 이내)

작품/논문과 관련된 연구와 기술 현황을 작성합니다.

과제 제안목표와 방향을 기술합니다.

스마트폰을 이용해 원격으로 물공급, 창문개폐가 가능하다.

온,습도 Setting값에 따라 자동으로 동작한다.

3.작품/논문 전체 진행계획 및 구성 (2페이지 이내)

전체 진행해야 하는 item들을 대체적으로 나열합니다.

4.기대효과 및 개선방향 (1페이지 이내)

5.기타 (1페이지 이내)

팀원들간의 역할

팀원들간의 역할 분담을 명확히 기술하고, 실행계획을 기술합니다.


소스.C


'졸업작품 > 스마트하우스' 카테고리의 다른 글

스마트하우스 (공개용)  (0) 2018.11.03
ㄱㅂㅅ(ㄱㅎㄷ)  (0) 2017.05.16
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Naver

[boost ic]

[ac모터 back emf] :: ▶LINK 여기로 와주세요~

[diode cathode voltage]



★ 질문 사항은 댓글 남겨주세요

http://www.learnabout-electronics.org/PSU/psu32.php


Boost Converter

Switched mode supplies can be used for many purposes 

including DC to DC converters.

Often, although a DC supply, such as a battery may be available, 

its available voltage is not suitable for the system being supplied.

DC공급기라도, 그 전압이 시스템에 적합하진않다.

For example, 

the motors used in driving electric automobiles require much higher voltages, 

in the region of 500V, than could be supplied by a battery alone.

전기차 운용에 쓰이는 모터들은 요구한다. much higher 전압들을

500V의 일부는 could be 공급될수있다. by a 베터리 자체에의해.

Even if banks of batteries were used, 

the extra weight and space taken up 

would be too great to be practical.

The answer to this problem 

is to use fewer batteries 

and 

to boost the available DC voltage 

to the required level 

by using a boost converter.

이 문제에 대한 대답

은 to 사용하는거다. 더 낮은 베터리들을

and

to 부스트하는거다. the available DC 전압을

to the 요구하는 레벨까지

by 사용함으로써 a 부스트 컨버터를.

Another problem with batteries, large or small, 

is that their output voltage varies as the available charge is used up, and at some point the battery voltage becomes too low to power the circuit being supplied.

다른 문제

는 that their 출력 전압은 변한다. as the available한 charge를 다만큼,

and at 어떤 의미에선 the 베터리 전압은 becomes 너무 낮게된다. to 파워공급하기에 the 회로가 being 공급되게끔.

However, if this low output level can be boosted back up to a useful level again, 

by using a boost converter, 

the life of the battery can be extended.

하지만 만약 this 낮은 출력 레벨이 can be 부스티드되면 back upto쓸모있는 레벨로 다시,

by 사용함으로써 a 부스트 컨버터를,

the life of the 베터리는 can be 확장된다.




The DC input to a boost converter 

can be from many sources as well as batteries, 

such as rectified AC from the mains supply, 

or DC from solar panels, fuel cells, dynamos and DC generators.

The boost converter is different to the Buck Converter in that it’s output voltage is equal to, or greater than its input voltage. 

However it is important to remember that, as power (P) = voltage (V) x current (I), if the output voltage is increased, the available output current must decrease.



Fig. 3.2.1 illustrates the basic circuit of a Boost converter.

However, in this example the switching transistor is a power MOSFET, both Bipolar power transistors and MOSFETs are used in power switching, the choice being determined by the current, voltage, switching speed and cost considerations.

The rest of the components are the same as those used in the buck converter illustrated in Fig. 3.1.2, except that their positions have been rearranged.


▲MOSFET이 궁금하면 클릭 ☜


Boost converter Operation



Fig 3.2.2 illustrates the circuit action 

during the initial high period of the high frequency square wave 

applied to the MOSFET gate at start up.

그림3.2.2는 설명한다. the 회로 액션을

during the 초기 high 주기동안 of the high frequency 사각파의

적용된 to the MOSFET 게이트에 at 스타트업에서.

During this time 

MOSFET conducts, placing a short circuit 

from the right hand side of L1 to the negative input supply terminal.

Therefore a current flows between the positive and negative supply terminals through L1, which stores energy in its magnetic field.

There is virtually no current flowing in the remainder of the circuit 

as the combination of D1, C1 and the load 

represent a much higher impedance than the path directly through the heavily 

conducting MOSFET.





Fig. 3.2.3 shows the current path during the low period 

of the switching square wave cycle. 

그림 3.2.3은 보여준다. the 전류 경로를 during the low 주기 동안

of the 스위칭 사각파 사이클의.

As the MOSFET is rapidly turned off the sudden drop in current causes L1 to produce a back e.m.f. in the opposite polarity to the voltage across L1 during the on period, to keep current flowing.

This results in two voltages, the supply voltage VINand the back e.m.f.(VL) across L1 in series with each other.

This higher voltage (VIN +VL), now that there is no current path through the MOSFET, forward biases D1. The resulting current through D1 charges up C1 to VIN +VL minus the small forward voltage drop across D1, and also supplies the load.





Fig.3.2.4 shows the circuit action during MOSFET on periods after the initial start up. Each time the MOSFET conducts, the cathode of D1 is more positive than its anode, due to the charge on C1. D1 is therefore turned off so the output of the circuit is isolated from the input, however the load continues to be supplied with VIN +VL from the charge on C1. Although the charge C1 drains away through the load during this period, C1 is recharged each time the MOSFET switches off, so maintaining an almost steady output voltage across the load.

The theoretical DC output voltage is determined by the input voltage (VIN) divided by 1 minus the duty cycle (D) of the switching waveform, which will be some figure between 0 and 1 (corresponding to 0 to 100%) and therefore can be determined using the following formula:

Example:

If the switching square wave has a period of 10µs, 

the input voltage is 9V and the ON is half of the periodic time, i.e. 5µs, 

then the output voltage will be:

만약 the 스위칭 사각파가 갖고있으면 주기를 of 10us의,

the 입력 전압은 9V이다. and the ON이 절반 주기이다. i.e. 5us,

그럼 출력전압은 will be:

VOUT = 9/(1- 0.5) = 9/0.5 = 18V (minus output diode voltage drop)

Because the output voltage is dependent on the duty cycle, 

it is important that this is accurately controlled.

For example if the duty cycle increased from 0.5 to 0.99 

the output voltage produced would be:

because the 출력 전압이 의존하기때문에. on the 듀티 사이클에

이건 중요하다. that 이건 정확히 컨트롤된다.

For example, 만약 듀티 사이클 증가했다면 from 0.5 ~ 0.99까지

그 출력 전압 만들어진 would be:

VOUT = 9/(1- 0.99) = 9/0.01 = 900V

Before this level of output voltage was reached however, there would of course be some serious damage (and smoke) caused, so in practice, 

unless the circuit is specifically designed for very high voltages, the changes in duty cycle are kept much lower than indicated in this example.



Boost Converter Animation

Click play to start.

See the current paths during the on and off periods of the switching transistor.

See the magnetic field around the inductor grow and collapse, and observe the changing polarity of the voltage across L.

Watch the effect of ripple during the on and off states of the switching transistor.

See the input voltage and the back e.m.f. of VL add to give an output voltage greater than the input voltage.

Click pause to hold the animation in either the on or the off state.

Click back to reset the animation.





I.C. Boost Converter

In this circuit,

an appropriate fraction of the output voltage (VOUT), 

dependent on the ratio of R2:R3 

is used as a sample 

and 

compared with a reference voltage within the I.C.

이 회로에서,

an 근사치 fraction of the 출력전압(Vout), 

의존하는 on the 비율에 of R2:R3

는 사용된다. as a 셈플로 

and 

비교된다. with a 참조 전압에 within the IC내의.

This produces an error voltage that is used to alter the duty cycle of the switching oscillator, 

enabling a range of automatically regulated boost voltages between 5V and 28V to be obtained.

Because of the ease with which boost converters can supply large over voltages, they will almost always include some regulation to control the output voltage, and there are many I.Cs. manufactured for this purpose A typical example of an I.C. boost converter is shown in Fig. 3.2.6, in this example the LM27313 from Texas Instruments

This chip is designed for use in low power systems such as PDAs, cameras, mobile phones, and GPS devices.

The LM27313 contains an internal oscillator 

operating at a fixed frequency of about 1.6MHz.

The FET switching transistor is also internal 

and switches the current through L1 via the SW terminal. 

Notice also that 

a Schottky diode with an appropriate voltage 

and 

current rating is used for D1 to keep losses 

due to the forward voltage drop of the diode as small as possible, 

and to enable high switching speeds to be achieved.

The I.C. also has a shut down (SHDN) facility, operated by external logic, by which the boost converter may be disabled when not required, to save battery power.


▼사용된부품

LM27313 (부스트 컨버터)
인턱터,저항,커페시터
schottky Power Recifier



Protection Circuits

Other safety features provided by the I.C. are over current shut down, which disables the switch on a cycle-by-cycle basis if too much current is sensed, and an over temperature shut down facility.

Stability

Another problem facing designers of high frequency boost converters is that of stability, as at MHz frequencies both negative and positive feedback can occur simply due to electromagnetic fields radiating between components within the circuit, especially where the circuit components are in very close proximity as in surface mount layouts. C2 is therefore added to improve stability and prevent possible oscillation due to unwanted positive feedback occurring.

 


https://reibot.org/2011/08/07/intro-to-boost-converter/


BuckConverter VS BoostConverter









DC-DC Converter


용도 :: AC-DC컨버터를 거쳐서 나온 DC전압을 각 용도에 맞춰서 낮춰야 할경우

DC-DC컨버터를 사용하게 된다.

전압을 낮추는 방법에는 여러가지 방법이 있겠지만 대체로 PWM을 이용한 High/Low 빈도를 이용하여 출력 전압을 내보낸다.

Pspice로 설계시 자료(영문 PDF)

4763126.ppt



http://www.circuitstoday.com/12v-to-120v-dc-dc-converter


555timer는

PSpice에서 anl_misc.olb 에 있다. 여기서 555B를 사용한다.



▼Buck Converter (스위치에 따른 상태변화)

Buck Converter is controlled by Active Switch S1.


▼Pulse 주기는 아래 공식에 따라 결정된다. (Duty비 라고 불린다.)

ON Time divided by the Total Period as shown in this equation.


ΔiL

Δvc


What is the voltage over the inductor during the switch is ON.

Vi는 initial Volate라는 뜻이다. (초기전압)



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