A PWM Algorithm For Software And HardwareRealized With Only One Add Instruction

- A PDM Algorithm Based On Addition (1)

现在出的微控制器，几乎没有不带PWM输出的。PWM一般都用特殊的定时器电路实现。

PWM的本质是在脉冲总周期Nt内输出累积时间为Ht的正脉冲。

本人提出过一种算法，用C语句表达如下：

CNT += H ;

if ( CNT >= N )

{ CNT %= N ; Pout = 1 ; }

else Pout = 0 ;

// 每个t计算一次；

// N为任一正整数，H为0 ~ (N-1)，CNT为累计数并保存模除余数。

严格讲，本算法应称为“基于累加的PDM(Pulse Density Modulation)算法”。

本算法可以用FPGA/CPLD或PSoC UDB的Datapath硬件实现，资源占用恐怕比常规PWM还要少。而软件实现，在任何单片机上都可以做，只要它有一个最简单的定时器，就可做出任意多个PDM输出。

本文附上一个用PSoC4 Pioneer Kit实现的实例，软硬并用。其中，红色LED用UDB的Datapath调光；简单到只用一条指令：A0 = A0 + D0，进位co_msb通过引脚输出给红灯。而绿色、蓝色二只LED在SysTick中断里用软件进行调光。三灯都用10KHz时钟，即每100uS计算一次。为便于看程序，没有用汇编。

实例：ZhuPDM8(PSoC Creator 2.2 SP1).rar

http://www.eeskill.com/file/id/26301。

用PSoC4 Pioneer Kit做实例，也是表示我对AET“赛普拉斯PSoC4”主题季活动的支持。更主要的是PSoC（从PSoC3起）UDB的Datapath是我将本算法在MCU芯片内用硬件实现的第一种芯片，而且是目前唯一的一种MCU。这也是我推崇PSoC UDB的原因，而且爱“PSoC Datapath”及“PSoC葡萄酒”。

二十多年前，单片机内还没有PWM，这种算法是我自己逼出来的，算是一种“独门绝技”，我只告诉过几个朋友。如果什么人发现有谁谁谁在什么地方使用这种方法，请告诉我，我会把“独门”二字去掉。

应该说，一个表达式和一个实例，已经把算法叙述清楚了。

有人说当N=2、H=1时占空比不是50%，我以实例说明占空比是50%。录下回应如下：

Re: A PWM Algorithm Realized With Only One Add Instruction

H Lposted on 21 Jul 2013 09:48 PM PST

A clever approach. However, with N =2 and 1 =1, the output duty cycle would not be 50% as there is more CPU cycle in one half of the operation. The slower the CPU the higher the error.

Re: A PWM Algorithm Realized With Only One Add Instruction

fyzhuposted on 22 Jul 2013 09:28 PM PST

What's wrong?

MATLAB shows the duty cycle is 50% for N=2 and H=1:

>>

npulse = 8;

N = 2;

H = 1;

CNT = 0;

disp( ' k Pout for init. CNT=0' )

for k = 1:npulse

CNT = CNT + H;

if CNT >= N

CNT=mod(CNT,N) ;

Pout = 1 ;

else

Pout = 0 ;

end

disp( [k Pout] )

end

k Pout for init. CNT=0

1 0

2 1

3 0

4 1

5 0

6 1

7 0

8 1

>> ...

k Pout for init. CNT=1

1 1

2 0

3 1

4 0

5 1

6 0

7 1

8 0

Re: A PWM Algorithm Realized With Only One Add Instruction

H Lposted on 23 Jul 2013 10:36 PM PST

The difference between the two condition is onewith the % instruction, and the other does not.

My answer is runing it froma 'CPU'. The CPU needs to process the % instruction.

You can compile the code and check the difference of instructions.

我继续回应：

Hi H L,

I don't think there is any difference between the mod() and the %-operator.

The debug result (see attached screen copy) of following programm runing in a real CPU with the %-operator shows the same duty cycle of 50%.

void main()

{

uint16 N = 2;

uint16 H = 1;

uint16 CNT = 1;

uint8 Pout[8];

uint16 k=0;

do {

CNT += H;

if (CNT >= N)

{ CNT %= N ; Pout[k] = 1 ; }

else Pout[k] = 0 ;

k++;

}while(k<8);

for(;;) { }

}

May the defferent compiler cause the defferent result of %-operator ?

附图：

FY_ZHU

2013-07-18 BOS-MA