Hardware ESC 8xSERVO CONTROL on PIC (Oshonsoft BASIC)

Pommie said:

Did I not write that?

Mike.

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Pommie said:

So, I did write that. The use of but was that I believe 1 chip can do everything.

Mike.

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Pommie said:

No, just 1 microcontroller but still needs the 164.

Mike.

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• The output of the 74HC595 has a latch, so it can remain unchanged during the shift; the output of the 74HC164 does not have a latch, so it changes every time a shift clock is created. This is the most significant distinction between the two.
• To achieve a multi-chip cascade, the 74HC595 utilizes a specific Q7 pin; the 74HC164 cascades directly using output pin Q7.
• When the enable OE is faulty, the output pin of the 74HC595 is a high impedance, but the 74HC164 has no enable pin.
• The shift register is reset by the 74HC595. If you want to reset the LATCH register, you must load the contents of the shift register into the latch register on the rising edge of ST CP; that is to say, the 74HC595 reset is synchronous, but the 74HC164 reset is asynchronous.
• The parallel-to-serial chip 74HC165 corresponds to the 74HC164.

camerart said:

So, is the 74HC164 the best choice?

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rjenkinsgb said:

Yes, absolutely - for this specific application.

The idea is to shift one bit through, so it sets each output high in turn. The "flicker" if you like, but that is a sequence of servo pulses.

The clock input to shift it will come from the output compare pin so the timing is then purely hardware, and software delays cannot cause jitter - the software has plenty of time just to set up the next compare each time.


The 595 is ideal as as "output expander" for normal things like switching LEDs or relays, as you can shift data in without affecting the pins then transfer it to the latch once it's in the correct bit alignment.

It needs more control signals though so it can only be driven via a software routine, which means the likelihood of jitter again.

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rjenkinsgb said:

In a way, yes!

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camerart said:

One possible issue with the 74HC164 is as the BITs are filling the BYTE, if they don't fill fast enough, they have a momentary voltage, I am assuming that the PIC that clocks them is too fast to worry about.

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rjenkinsgb said:

That's not a problem, it is essential for how this setup works!

Look at the actual RC signal waveforms I posted earlier on:

View attachment 140109

The shift register moves a single bit through to each output in turn - like the "Channel x Servo signals"

The PPM waveform at the top is from the PIC compare hardware rather than a radio receiver, shifting the bit each 1 - 2 mS as appropriate for each respective servo position.

Unless the outputs changes as the SR was clocked, there would not be any servo outputs!

For decades, Radio control receivers used to use a shift reg IC for the conversion back to servo pulses, before MCU based ones took over.


ps. If you need to translate to 5V power for the shift register, it should have been a 74HCT164
The HCT ones take TTL levels and 3.3V logic is perfect to feed those with the HCT device working on 5V.

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Pommie said:

This is code I think will work with a 164. RC0 should go to the data input (pins 1&2) and RC1 to clock (pin 8). The servos go to Qa to Qe.
Not tested as no hardware, comments welcome.

Code:

#include <xc.h>
#include "config.c"
#include <stdint.h>
#define _XTAL_FREQ 32000000
#define NUM_SERVOS 10

uint16_t servoPos[NUM_SERVOS];
uint32_t ms;
uint8_t count=0,servoCount;


void main(void) {
    OSCCON=0b01110000;      //8MHz
    PLLEN=1;                //x4=32MHz
    //setup 1mS interrupt = 8,000,000/16 = 500,000/10 = 50,000 set PR2=49 = 50,000/50 = 1000 = 1mS
    T2CON=0b01001110;       //pre=16 post=10
    PR2=49;
    TMR2IE=1;               //timer 2 interrupts enable
    T1CON=0;                //timer 1 stopped
    for(uint8_t i=0;i<NUM_SERVOS;i++){
        servoPos[i]=i*1000+8000; //1ms(8000) to 1.875(7/8ths - 15000)ms in 1/8th mS steps
    }
    TRISC=0b11111100;           //CCP0 & 1 output
    PEIE=1;
    GIE=1;
    while(1){
            //adjust servo positions here
    }
}


void __interrupt() inter(void){
    if(TMR2IE && TMR2IF){
        ms++;
        count++;
        if(count>=20){          //start every 20mS
            TMR1=0;             //zero timer 1
            T1CON=1;            //start timer 1
            count=0;
            CCP1CON=0b1000;     //CCP1 pin low and high on match - will be first pulse
            CCPR1=2000;         //start pulse in 0.25mS (2000 clock cycles)
            CCP1IE=1;           //enable CCP1 interrupts
            CCP1IF=0;           //ensure interrupt flag is clear
            servoCount=0;       //reset servoCount
            LATC0=1;            //connected to data in of shift register will clock in a high when CCP1 goes high
        }
        TMR2IF=0;
    }
    if(CCP1IE && CCP1IF){
        LATC0=0;                //clear the data in pin
        if(servoCount==9)       //have we done all servos?
            CCP1IE=0;           //yes so no more CCP1 interrupts 
        if(CCP1CON==0b1000){    //have we started the 4000 cycle pulse
            CCP1CON=0b1001;     //yes so end the pulse after 0.5mS
            CCPR1=CCPR1+4000;   //4000 cycles=0.5mS
        }else{
            CCP1CON=0b1000;     //No so output the timed gap        
            CCPR1=CCPR1+servoPos[servoCount++]-4000;   //will generate an interrupt when servo time is up
        }
        CCP1IF=0;
    }
}

It (I hope) outputs a 0.5mS pulse every servoTime.

Mike.
Edit, added a millisecond interrupt as it's useful.

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Pommie said:

C can you confirm that doing CCPR1 = 0x55AA writes 0x55 to CCPR1H and 0xaa CCPR1L?

Mike.
Edit, can Ian Rogers or ericgibbs confirm this?

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camerart said:

Hi M,
I tried this test, and it looks as if CCPR1H is not being written to.
C
================================
Dim test As Word
test = 0x55aa

CCPR1 = test
=================================

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jjw said:

Write:
CCPR1H = test.HB
CCPR1L = test.LB

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Pommie said:

For
CCPR1=CCPR1+servoPos[servoCount++]-4000; //will generate an interrupt when servo time is up

You can do,

Code:

test.HB=CCPR1H
test.LB=CCPR1L
test=test+servoPos(servoCount)
test=test-4000
CCPR1H=test.HB
CCPR1L=test.LB
servoCount-servoCount+1

OR, much better solution,
switch to C.

Mike.

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