Wire gauge sizing equations for a 12 V system

Hi all,

I plan to control temperature in an enclosed system at 12 V using:

-A 60W Peltier TEC1-12706
-A 70 W Alpine 64 Ventilator with heatsink
-A 95 W AMD Ventilator AM2+/AM3 with heatsink

The system would be connected to a Mosfet and through a PWM pin, a PID function would control to which % duty the whole system would work.

That means that 1 cable should provide all the current for all the three devices.

What should be the right diameter/cross section area for this wire? 2,5 mm? I thought of using a copper wire.

I have stumbled upon many low voltage calculators but I want to see and understand the equations used.

https://www.tlc-direct.co.uk/Technical/Lighting/VoltageDrop.html

I have read about several parameters but I am unsure about the "voltage drop" I should have for example. Also, these websites say the results give no guarantees.

Total length of cable (m): 1
Max environmental temperature (°C): 30
Voltage drop (V/%): ?

For the tests and initial prototype, I will be using a power adapter (Model: JDT-001) with 3 A y 240 V, so it should suffice (3*240 = 720 W).

Thanks in advance!

The total voltage drop along a wire is easily calculated from the maximum current (I), the total length of the wire carrying the current (L) and the resistance per unit length (rho), which you can look up in the wire table.

Vdrop=ILrho

You decide on how much voltage drop can be tolerated, and then buy wire of at least the minimum diameter. From a 12V source, a drop of 1V means that 8% of the available power is wasted just heating up the wire.

Thermal limits to cable size are exactly the same for DC and rms AC - only the current matters
for thermal considerations. If the cabling is enclosed in trunking or similar the current
rating will be less. The length of cable is not relevant to this limit - this is about the cabling
not overheating and melting.

However for low voltage the voltage drop along the cable has to be considered too - this may,
or may not, be the limiting factor depending on the cable total resistance and current. At 12V
you might decide a total voltage loss of 1V is the maximum you can afford, so the IR product
for the cable (round-trip) then needs to be less than 1.
The cable length is obviously important to this calculation.

If this JDT-001 is the power adapter you plan to use, it won't work. 3A @ 12V = 36 Watts.

In the OP, you state that the devices draw a total of about 230 Watts, or around 19 Amperes from a nominal 12V source. I would suggest to use a 12V, 25A adapter instead.

Voltage drop calculation is easy:

You need the resistance of the wire, look here wire chart

You can estimate the length, remember your length must include both the positive and negative wires.

Your current:

60W + 70W+ 95W= 225Watts

Current = watts / volts so 225 / 12 = 18.7 Amps.

Your current will be slightly larger ~ 22 amps because a car is not 12V but 13 - 14 volts (when running).

Voltage drop will be 22 a * wire resistance = ______

I think you might be able to get but with 2 mm wire but 2.5 mm is much safer.

John

Thank you everyone for your answer!

jremington:
The total voltage drop along a wire is easily calculated from the maximum current (I), the total length of the wire carrying the current (L) and the resistance per unit length (rho), which you can look up in the wire table.

Vdrop=ILrho

You decide on how much voltage drop can be tolerated, and then buy wire of at least the minimum diameter. From a 12V source, a drop of 1V means that 8% of the available power is wasted just heating up the wire.

How much is normally acceptable?

As I said, it will be running only occasionally. Most of the cable will be on the cold side of the Peltier (less than 1 m3 in volume). That means that, since the peltier will be cooling, it will only reduce the output of the Peltier. My primary objective is that the cable does not burn.

I guess I could calculate the exposed cable heat emission and work out how much it will heat the environmnent.

The other part of the cable will be very small and will be enclosed in an insulated box, together with the rest of the electronics.

jremington:
The total voltage drop along a wire is easily calculated from the maximum current (I), the total length of the wire carrying the current (L) and the resistance per unit length (rho), which you can look up in the wire table.

Vdrop=ILrho

You decide on how much voltage drop can be tolerated, and then buy wire of at least the minimum diameter. From a 12V source, a drop of 1V means that 8% of the available power is wasted just heating up the wire.

Excellent, I will check the maximum temperature tolerated by the cable, together with the mentioned calculation of the heat emitted.

jremington:
If this JDT-001 is the power adapter you plan to use, it won't work. 3A @ 12V = 36 Watts.

In the OP, you state that the devices draw a total of about 230 Watts, or around 19 Amperes from a nominal 12V source. I would suggest to use a 12V, 25A adapter instead.

You are right, it is wrong. It is 72 W max (24 V * 3 A). I guess that if the system draws full power, my arduino (also connected to this power source), will not get enough power. I guess the best idea would be either to limit the System output or as you mentioned, use a more powerful adapter.

JohnRob:
Voltage drop calculation is easy:

You need the resistance of the wire, look here wire chart

You can estimate the length, remember your length must include both the positive and negative wires.

Your current:

60W + 70W+ 95W= 225Watts

Current = watts / volts so 225 / 12 = 18.7 Amps.

Your current will be slightly larger ~ 22 amps because a car is not 12V but 13 - 14 volts (when running).

Voltage drop will be 22 a * wire resistance = ______

I think you might be able to get but with 2 mm wire but 2.5 mm is much safer.

John

The regulator can be set exactly to the required voltage (it is not a car but a plug adapter what supplies energy).

Intensity = 225 / 12 = 18.7 A

Taking the Gauge 12 wire from your table: Wire resistance = 5.21 Ohm/km = 0.00521 Ohm/m * 1 m = 0.00521 Ohm

Voltage drop = I * wire resistance = 18.7 * 0.00521 = 0.097 V --> 0,81 % Voltage drop

I think I could go for a lower cable gauge, I will let you know!

I think you'll find you're at the thermal limit. Thinner wire will overheat, the voltage drop is not the issue
for a 1m run.

I think I could go for a lower cable gauge, I will let you know!

Unless there are some other restrictions we aren't aware I think you are overthinking the design.

While it is good to understand the concepts, you will find there is no "ideal" choice. There is however a wrong choice.

If you duty cycle is low (and short) you can get away with #12, else if a higher dutycycle or on for an extended period of time you would be best with #10.

You might be able to sneak by with a #14 but I personally don't see what the benefit of dropping to #14 and taking a risk. Also keep in mind, in most installations the supply and return wires are tied together making the thermal rise even worse (than one wire in free air).

I don't see any more than 5.5Amp mentioned in OP's original post.
The peltier using 5A@12volt, and the other items are computer fans (~250mA).

I think OP wants to move the heat of a 90watt TDP CPU with a 60watt peltier (with ~10% efficiency).
Not going to happen.
Leo..

@Wawa

Your right, I failed to take the time to understand the details of the OP's parts.

My bad :frowning:

Hi all,

sorry for the delay answering. I measured the intensity going through the ventilators and indeed, the Wattage I gave was the one they can dissipate. Their actual Wattage is much lower.

I did not want to bother anyone with the details of the project but it now seems appropriate. I am planning on controlling the temperature of a box, probably of less of than 1 m3 in volume. This module will take care of cooling if needed to a minimum of 10 °C and, maintain this temperature.

Another module will take care of the heating, as the peltier is not very efficient and I will trigger its use as much as possible.

My question now is, I have a “Geekcreit® 5M 1.27mm 20P Jumper Cable DuPont Wire” at home:
https://eu.banggood.com/Wholesale-Warehouse-5M-1_27mm-20P-Jumper-Cable-DuPont-Wire-Rainbow-Flat-Wire-Support-Wire-Soldered-wp-Eu-959792.html?akmClientCountry=DE&

The pitch is 1,27 mm, but I could not find anywhere in the specifications its cable diameter. Is it almost the same or related? That would mean I possess a gauge 16 cable.

I did all the calculations and data corrections and put them in a table, so anyone learning this could benefit (find attached).

FWIW, those jumper wires are approximately 34 or 36 gauge, suitable for logic level currents only. Quality stuff is good for about 100 millamps max with short runs. There is far more plastic insulation than there is copper in ribbon cable.

There are wire size specs for cable used with insulation displacement connectors. I would not expect Chinese wire to follow any specific requirements other than “made as cheaply as is possible”. Heck, some stuff from China wire isn’t even copper, aluminum and iron have been found with plastic coated being sold as wire.

Never buy wire that doesn't have a specification of strands and strand diameter, you will get
rubbish, probably CCA thinner than a human hair - unless its just for logic signals and not power.

I would use #16 (or metric eq) and at the smallest #18 (would be marginal IMHO)

On another note:

I've not used Peltier devices more that playing with one on the bench. I have read that they do not like PWM. Perhaps some of the other folks can provide better guidance for controlling the Peltier.

Some confusion seems evident here, some by the OP, some by those trying to answer.

-A 60W Peltier TEC1-12706
-A 70 W Alpine 64 Ventilator with heatsink
-A 95 W AMD Ventilator AM2+/AM3 with heatsink

The only power here that is important is the 60 W Peltier. The other devices are just a fan on a heatsink, rated to work on a CPU of that power level. As someone pointed out, those probably draw around 250mA.

As pointed out, that power supply is rated at 3A at 12V. Totally insufficient for this Peltier device. 60W/12V = 5A.

You should be using a power supply rated at least at 6A @ 12V. Considering how fast-and-loose they play it with specs, I'd prefer an 8A @ 12V supply.

Power your Arduino circuits from a separate supply. Preferably, bring up the Arduino supply, wait a few seconds, then power up the 12V supply. I'd put the Arduino outside of the cabinet. Use a single point ground connection between the Arduino and the two power supplies.

Don't use a fast PWM. The thermal mass of everything is pretty high, so a cycle time on the order of seconds is enough to control temperature.

A thermoelectric cooler has a built-in limited temperature differential of about 50C, with zero heat transfer. So don't expect much better than about 30C cooling versus ambient. Because that differential is between the hot side and the cold side, not internal box versus ambient.

If you decide to use one power supply for everything, be prepared to add a LOT of decoupling between the 12V line and the Arduino power line. Watch out for ground bounce. It is SO much simpler to just use a separate supply for the electronics.

polymorph:
Some confusion seems evident here, some by the OP, some by those trying to answer.

The only power here that is important is the 60 W Peltier. The other devices are just a fan on a heatsink, rated to work on a CPU of that power level. As someone pointed out, those probably draw around 250mA.

My apologies, I already updated the info with real measured values from the fans (attached on my previous post), which is, as you mentioned, much lower than expected.

polymorph:
As pointed out, that power supply is rated at 3A at 12V. Totally insufficient for this Peltier device. 60W/12V = 5A.

This was also updated on my last post. The power adaptor has 72 W, enough for testing, although insufficient for the whole of the project.

polymorph:
Power your Arduino circuits from a separate supply. Preferably, bring up the Arduino supply, wait a few seconds, then power up the 12V supply. I’d put the Arduino outside of the cabinet. Use a single point ground connection between the Arduino and the two power supplies.

If you decide to use one power supply for everything, be prepared to add a LOT of decoupling between the 12V line and the Arduino power line. Watch out for ground bounce. It is SO much simpler to just use a separate supply for the electronics.

I see. I was not aware of ground bounce possibility and the need of decoupling. I have been reading on this and the importance of calculating their capacitance:

https://www.arduinostartups.com/decoupling-capacitors-what-are-they-and-why-are-they-used/

What I showed was the most energy consuming module from all the ones I planned to connect to my arduino (module 1). Other modules (5 in total, 1 of them not fully defined) will also be part of the system. Please find a sketch attached. I hope it clarifies my final intentions for the system I am building.

It seems I will need to add a lot of decoupling because of all the modules. Would you still leave the Arduino separated from the other modules as you previously suggested? Would the point I marked with an arrow be the best to add as a single point ground connection?

My first idea was to connect all the ground ends from the MOSFETS on the Protoboard where I am connecting all the sensors that the Arduino has (not drawn) for commodity, but I think the marked point is the safest one for the system.

Please let me know what you think. I am very grateful to all of you for all the comments and ideas given