Which current should I use to determine the Watt?

Question

I am making a heated blanket and I have some troubles to determine the heat that will be generated.

Resistance: Let say, I use this heated cable with a resistance of 0.3 Ohms/m*. If I am using 10m of this cable the resistance will be 3 ohms.

Voltage: I will use a AC adaptator with a voltage of 19V.

Current: So the current that will be drawed is 19/3 = 6.3A

But my adaptator delivers a current of 3.16A, so which current should I use to determine the Watts that my blanket will output? 3.16A or 6.3A?

My guess is that it will be limited to 3.16A but that's where I get confuse because it seems that the length of the cable doesn't count anynore (if it's below 20m) :

• a 1 m cable should draw 63.3A (at 19v and 0.3 ohms/m) but with a input current of 3.16A, it should only draw 3.16A.

• a 15 m cable should draw 4.22A (at 19v and 0.3 ohms/m) but with a input current of 3.16A, it should only draw 3.16A.

• a 20 m cable should draw 3.16A (at 19v and 0.3 ohms/m) but with a input current of 3.16A, it should only draw 3.16A.

• a 30 m cable should draw 2.11A (at 19v and 0.3 ohms/m) but with a input current of 3.16A, it should only draw 2.11A.

This seems to mean that A 1m cable or a 20m cable will generate the exact same heat but a cable longer than 20m will start to generate less heat.

Could that be correct because it doesn't make sense to me and I can't find where is my reasoning mistake.

Thanks a lot for shedding some light on this!

*I could only find the resistance per m of this cable on a forum. But for all the other cables I looked for, I've never found the resistance/m. Why the sellers never give this info? It is a specification too obvious to notify?

Answer 1

Your guess that "it will be limited to 3.16A" is not the full story because you are trying to draw more current from the power supply than it can deliver.

There are a few scenarios for how over current will be handled depending on the power supply you're using (I'm assuming a standard CV -continuous Voltage- power supply):

1. Power supply with Constant Current (CC) Limit: The current supplied will be ~3.16A but the voltage will drop to 3.16A x 3Ohms = 9.48V (for you 3 Ohm example), reducing the power accordingly .

   2. The Power supply will Fault because of over-current, setting the output voltage to zero whilst the over-current condition persists.

   3. The power supply does not have any over-current protection and will try to supply the power needed, but this causes lower output voltage, higher ripple voltage, possible over heating of the power supply components and eventual failure.

Answer 2

Chances are your power supply will either shut itself down, or overheat and break. Trying to draw more than 3.16A from it is a Bad Idea™.

Assuming the supply is robust enough to survive, and the current draw is the overriding factor, and it doesn't shut itself down:

• The supply has a maximum power of 19*3.16 = 60.04W.
• You try drawing 6.3A from it, the voltage would drop to V=P/I = 60.04/6.3 = 9.53V

To keep the voltage at 19V you will need a power supply capable of providing 6.3A, which equates to ~120W power supply.

Answer 3

If your resistance is such that it would draw more current than your power supply can deliver, you have the following possibilities depending on the design of your supply:

1. The power supply output voltage drops until the current is some safe limited value. In this case your power is the current limit X the actual output voltage.

2. The power supply shuts down immediately and delivers no current.

3. The power supply delivers more current than it's designed for, eventually destroying itself or shutting down from over temperature.

4. The power supply hiccups turning on for a short time and then off if the load is still higher than the current limit.

In any of these cases the (instantaneous) power will be the actual current times the actual voltage. It may have no relationship to the rated output voltage and current of the supply, because the supply is operating outside its design limits.