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Topic: batteries (Read 648 times) previous topic - next topic

LIAM_MAD

Hi Folks,

I'm looking to power two heat pads, the pads require 600ma each, so that 1.2amps in total as they are joined together. I bought 2 x 7.4v, 2250mah lithium batteries and set them up in series, which gives a voltage of . The pads powered on really quickly (which is what I want them to do), they got very hot after about 10 seconds but unfortunately the batteries drained very quickly, (I think). I'm just wondering is there a way that I can power the pads with these batteries in series, without them draining so quickly. I'm not sure but If I put some more resistance into the circuit, this would resist the flow of current and prolong the battery life?

Thanks

slipstick

Provide full specification of the heating pads, e.g. 600mA at what voltage? How are they joined together, series or parallel?

At 1.2A a charged 2250mAh battery would last almost 2 hours so I think there's something going on that you have missed.

Steve

LIAM_MAD

Hi Steve,

This is the schematic I'm working off, http://etextile-summercamp.org/2013/wp-content/uploads/2013/06/heatcontrolling_breadboard_example2-1024x682.jpg

The specs for the heat pads suggest that the draw 600ma at 5v

Thanks


westfw

If they draw 600mA at 5v, that probably means 3times that with the series connected batteries, and about one hour of runtime?

ElCaron

The specs for the heat pads suggest that the draw 600ma at 5v
So what made you think connecting them to 14.4V (actually, 16.8V, if your Lithium battery was fully charged) was a good idea?
Maybe, an overcurrent protection of your batteries kicked in, you drew 4A. Fucking around with LiPos is pretty dangerous.

LIAM_MAD

El Caron, the total voltage with the lipos in series was 14.8v, according to ohms law if the resistance of each pad is 8.3ohms, I=v/r, I would  work it out as I= 14.8 / 8.3, the current being 1.78amps. How did I draw 4Amps, excuse my lack of knowledge on this but I'm new to creating circuits. thanks.

slipstick

You said you had 2 heating pads in parallel so that would be 1.78A x 2 = 3.56A at 14.8V. More at the full lipo voltage of 16.8V.

So you have 3 times the normal voltage and 3 times the normal current so that's 9 times as much power as the poor heating pads are expecting. And then there's the problem that your current is way too much for any breadboard. Even 600mA would be more than I'd ever use. Why don't you use the lipos in parallel for 7.4V which would at least be a bit more sensible?

With your existing set up it's a question of whether the heaters or the breadboard are going to burn up first. Or it might be the transistors because I bet they're getting quite hot too.

So what code do you have in the UNO driving this set up?

Steve

LIAM_MAD

Hi Steve,

Thanks for getting back to me, I see what you mean by putting the batteries in parallel. I think the max current output is 2200ma, so by putting them in parallel, they remain at 7.4v and the max current is doubled to 4400ma, is that correct?

Below is the code I'm using with the Arduino nano, I forgot to mention that I have substitued the button in the schematic for a flex sensor:

// constants won't change. They're used here to
// set pin numbers:
const int heatPin1 =  6;      // the number of the LED pin
const int heatPin2 =  5;      // the number of the LED pin

void setup() {
  // initialize the LED pin as an output:
  pinMode(heatPin1, OUTPUT);
  pinMode(heatPin2, OUTPUT);

  Serial.begin(9600);
}


void loop() {

  //make sure pads are off


  //Serial.println("I'm off, thanks.");
 

  int flexSensorReading = analogRead(0);
  Serial.println("Flex");
  Serial.println(flexSensorReading);
  delay(1000); //just here to slow down the output for easier reading
/*
  //make this number between 1 and 100
int flex0to100 = map(flexSensorReading, 0, 540, 0, 100);
 Serial.println(flex0to100);

  delay(1000); //just here to slow down the output for easier reading
*/
  if (flexSensorReading < 990)
  {
    //turn on
    //Serial.println(flex0to100);

    digitalWrite(heatPin1, HIGH);
    Serial.println("1 I'm on, yay");
    delay(0); // you can change the timing as you like, 1sec=1000
    digitalWrite(heatPin2, HIGH);
    delay(0); // you can change the timing as you like, 1sec=1000
    Serial.println("2 I'm on, well done");
  }
  else {
    // turn both the heating off
    digitalWrite(heatPin1, LOW);
    digitalWrite(heatPin2, LOW);
    Serial.println("i'm off, now, bye");
  }


}

ElCaron

#8
Dec 11, 2017, 04:31 pm Last Edit: Dec 11, 2017, 04:32 pm by ElCaron
El Caron, the total voltage with the lipos in series was 14.8v, according to ohms law if the resistance of each pad is 8.3ohms, I=v/r, I would  work it out as I= 14.8 / 8.3, the current being 1.78amps. How did I draw 4Amps, excuse my lack of knowledge on this but I'm new to creating circuits. thanks.
No, parallel resistors combine according to 1/R = 1/R1 + 1/R2. For R1=R2, that means R=R1/2. So that is 4.15Ohm

Also, use code tags. What is your issue with the code? What is your question right now?

slipstick

Thanks for getting back to me, I see what you mean by putting the batteries in parallel. I think the max current output is 2200ma, so by putting them in parallel, they remain at 7.4v and the max current is doubled to 4400ma, is that correct?
No. The voltage is correct. But the CAPACITY of the battery is 2250 mAh. The maximum current could be anything. E.g. many lipos 2250mAh lipos are 20C or 30C so the max current will 45A or 67.5A. Capacity and current are NOT the same thing. Two packs in parallel doubles the capacity.

If you mess about with lipos without any understanding of what you're doing you can get in real trouble. Houses have burnt down through people charging lipos incorrectly.

Steve

TomGeorge

No, parallel resistors combine according to 1/R = 1/R1 + 1/R2. For R1=R2, that means R=R1/2. So that is 4.15Ohm


Yes and   I = V / R
so
I = 14.8 / 4.150
I = 3.56 Amps
1.78Amps in each heater.
Tom.. :)
Everything runs on smoke, let the smoke out, it stops running....

MarkT

I see no datasheets for these heaters still...

Also its very important to use fuses with high current batteries like LiPos and SLA, the fuse
will stop bad things happening (like the wiring burning up and starting a fire).  That
eTextile link shows no fuse - the very thought of eTextiles and heater elements run at 3 times
their rated voltage should also produce the thought "No, that's dangerous, that will start a fire".
[ I will NOT respond to personal messages, I WILL delete them, use the forum please ]

MarkDerbyshire

#12
Dec 12, 2017, 09:20 am Last Edit: Dec 12, 2017, 09:21 am by MarkDerbyshire
It's only a battery with a low voltage - what harm can it do!!!  If you can't or not willing understand the basics of Ohms Law and LiPo care take up knitting


TomGeorge

"I didn't do it, it was like that when we got here!!!!!!"
Everything runs on smoke, let the smoke out, it stops running....

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