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?

  • \$\begingroup\$ Use twice as much wire, or a different supply. \$\endgroup\$ – Brian Drummond Dec 10 '14 at 12:01

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 .

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

    2. 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.

  • \$\begingroup\$ it's starting to become more clear. I guess my power supply is a CC (I am using an old Ac adaptator for a computer). I am not sure if there is an over-current protection. \$\endgroup\$ – MagTun Dec 10 '14 at 12:08
  • \$\begingroup\$ A current limit would not be a common feature in an old AC adapter. My guess is that this would overheat and trip a thermal limit or get very hot and melt. \$\endgroup\$ – akellyirl Dec 10 '14 at 12:18
  • \$\begingroup\$ My AC adaptater is about 4-5 years old... but made in china (it was for a dell inspiron). But it seems that if I increase the length of the cable to more than 20m, then this current issue will be solved.. \$\endgroup\$ – MagTun Dec 10 '14 at 12:25
  • \$\begingroup\$ yes 20m would give you 6 Ohms resistance or 60W of power required which is just about within spec for your stated PSU. \$\endgroup\$ – akellyirl Dec 10 '14 at 12:54
  • \$\begingroup\$ Ok, I understand now. Just to be sure: rather than having 20 m, I suppose I can also have another cable with a higher resistance/m or I can add a resistor of 3 ohms and have a 10m cable. Are those two options correct? \$\endgroup\$ – MagTun Dec 10 '14 at 13:11

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.

  • \$\begingroup\$ Thanks Majenko! Does this means that I should use a 20 m cable (at least) to be sure to get enought resistance so that the current won't be over 3.16A? \$\endgroup\$ – MagTun Dec 10 '14 at 12:00
  • \$\begingroup\$ With that cable and that power supply you would need at least 20m, preferably slightly more. It's never good to run a PSU right on its limit. You should aim for something like 10% headroom. \$\endgroup\$ – Majenko Dec 10 '14 at 12:02
  • \$\begingroup\$ That sounds really strange to me but I get it on the theoritical level. Also I will use this switch dimmer to be able to regulate the heat. How will this switch influence the current/voltage/power? \$\endgroup\$ – MagTun Dec 10 '14 at 12:07
  • \$\begingroup\$ I don't know. I don't know what that "dimmer" does. If it creates a PWM waveform, then it will rapidly turn the power on and off, and that will basically reduce the average voltage. \$\endgroup\$ – Majenko Dec 10 '14 at 12:13

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.

  • \$\begingroup\$ I didn't know an AC adaptator can operaates outside its design limits. This changes everything then... \$\endgroup\$ – MagTun Dec 10 '14 at 12:13

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