Electronic – Dallas 1-wire Digital Thermo – issue with strong pullup

1-wireatmegapullup

I have 2 Dallas/Maxim DS18b20 digital thermometers connected in parasitic-power mode. If I include a strong-pullup as shown on p.6 of the datasheet, the nearer device gives good data; the further one reads back 85C (== the reset value). If I disconnect the strong pullup, I get valid temperature readings on both devices.

I omitted the MOSFET switch on the pullup and connected the strong-pullup i/o pin directly to the one-wire bus, floating it by making the pin an input with no pullup.

Why does the strong pullup not work as expected?
Why do the devices work OK without one?

schematic

^{simulate this circuit – Schematic created using CircuitLab}

Sensor read function:

 int8_t DSSensorsSample(TempSens_t t[SENSORS_REQD], uint8_t nsens){
   for( TempSens_t *p = t;  nsens-- > 0;  ++p ){

      p->type = p->addr[0];

      ds.reset();
      ds.select(p->addr);
      ds.write(0x44, 1);        // start conversion, with parasite power on at the end
      pinMode(DSSTRONGPULLUPPIN, OUTPUT);
      digitalWrite(DSSTRONGPULLUPPIN, HIGH);

      delay(1000);     // maybe 750ms is enough, maybe not
      // we might do a ds.depower() here, but the reset will take care of it.

      pinMode(DSSTRONGPULLUPPIN, INPUT);
      digitalWrite(DSSTRONGPULLUPPIN, LOW);

      p->is_present = ds.reset();
      ds.select(p->addr);
      ds.write(0xBE);         // Read Scratchpad

      for ( uint8_t j = 0; j < 9; j++) {           // we need 9 bytes
         p->data[j] = ds.read();
      }
   }
   return( 0 );
}

Best Answer

Why are you switching DSSTRONGPULLUPPIN high/low each time? Just leave it high all the time and change the pin mode only. Even if this leaves the ATMEGA's internal 20kΩ pullup enabled, this is insignificant with respect to the 4.7kΩ fixed pullup you have.

In particular, the sequence

pinMode(DSSTRONGPULLUPPIN, OUTPUT);
digitalWrite(DSSTRONGPULLUPPIN, HIGH);

is causing the the pin to be a strong pull-down briefly. This is probably creating the errors you are getting.

Your problem:

Your DDR setting is incorrect: You've written

DDRC |= _BV(PC0) | _BV(PC1) | _BV(PC3) & _BV(PC2);

but you want

DDRC |= _BV(PC0) | _BV(PC1) | _BV(PC3)
DDRC &= ~_BV(PC2);

You say you want more detail because you just kinda emulated other DDRC assignments and don't really understand what you wrote? OK, here goes:

How to set registers with bitwise operators on the AVR.

Let's unpack the _BV() macro and PCx definitions for a moment. _BV(x) is simply 1 << x. If you write _BV(5), the macro will evaluate to the number 0b 0001 0000. PCx is simply a macro defined to the value of x, so _BV(PC5) will evaluate to the same thing as _BV(5). You're probably expecting something like the following to happen:

   0b 0000 0001  // PC0
|  0b 0000 0010  // PC1
|  0b 0000 1000  // PC3
&  0b 0000 0100  // Misconception: AND does not cause this to output a zero at the set bit
|= ------------
   0b 0000 1011

However, you're not clearing PC2 correctly. You're applying a binary AND at the end. AND has a higher precedence than OR in C, so your operation is actually

DDRC |= _BV(PC0) | _BV(PC1) | ( _BV(PC3) & _BV(PC2) );

PC3 is 0b 0000 1000, and PC2 is 0b 0000 0100, so the parenthetical section evaluates to zero. This is actually fortunate for PC2, but unfortunate for PC3, which remains clear when you wanted it set. To clear a bit, you do use &, but you need to take the complement of the number using the ~ operator. In the following equations, I've used ? for bits which we don't want to change:

   0b ???? ???1  // Set PC0
|  0b ???? ??10  // Set PC1 
|  0b ???? 1?00  // Set PC3
|= ------------
   0b ???? 1?11

   0b ???? 1?11
&  0b 1111 1011  // Clear PC2
&= ------------
   0b ???? 1011

In code, these equations are:

DDRC |= _BV(PC0) | _BV(PC1) | _BV(PC3)
DDRC &= ~_BV(PC2);

as shown above. You could either do this in two steps as I did, or apply the previous value of DDRC with a bit mask:

DDRC = (DDRC & 0xF0) | ( ( _BV(PC0) | _BV(PC1) | _BV(PC3) ) & ~_BV(PC2) );

0xF0, or 0b 1111 0000, when AND'ed with the previous value of DDRC, returns 0b ???? 0000, to which we can OR our lower four bits. I think the two-step method is clearer: Your inputs and outputs are on different lines (note that you can add additional inputs with, for example, DDRC &= ~_BV(PC2) & ~_BV(PC4)). Sometimes, I'll even break it up onto more lines:

DDRC |= _BV(PC0)   // RELAY STATUS LED - O/P
DDRC |= _BV(PC1)   // RELAY CTL LINE - O/P
DDRC &= ~_BV(PC2)  // PUSH BUTTON - I/P
DDRC |= _BV(PC3)   // SPARE - O/P

That, to me, is the most clear mechanism possible, and the easiest for future readers to understand. It's also the most verbose, but who cares? The compiler will optimize all three mechanisms to the same code.

Electronic – DS1822 1-Wire Sensor, Parasitic Power, and Strong Pull-Up Circuit

As shown here, there are two things that can be done.

enter image description here

The Strong Pullup through the Mosfet, should have a appropriately sized Current Limiting Resistor, allowing the tens of milliamps required. If those tens of milliamps are more than your ic's pin can sink, then a secondary in-line resistor (Rt) can be used on the line, for protection.

Best source of information regarding one-wire is of course, Maxim-Dallas's app notes.

Best Answer

Related Solutions

Electronic – ATMega8 – PORT C – Digital Output Issue

Your problem:

How to set registers with bitwise operators on the AVR.

Electronic – DS1822 1-Wire Sensor, Parasitic Power, and Strong Pull-Up Circuit

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