Question about reducing current to a simple 12v 110w DC heating element

Thread Starter

Martoia

Joined Sep 5, 2021
4
I’m a beginner, you have taught me much already, this is my first time posting about a project so please be gentle :)

I have a 12v aprox-110w DC water heater element, which for excellent reasons I would like to use to boil 500mls of water. Wattage is aprox because the manufacturer gives contradictory wattage specs, and mine appears to operate between 106-110w in use.

The device does function as intended when connected to my battery system’s cigarette lighter output (12v, 10a max). However, its cable and terminals become very hot during operation. Other users of the device have noted this fault too. I fear that the insulation will burn or the casing melt.

I would like to modify and improve it so as to eliminate this risk whilst keeping current at or below 10a.

As it stands, the device draws 106-110w @12v from my battery system. By my calculations, this means its drawing 8.8-9.2a – just within the capacity of the battery system's 10a fuse. Some users of the same device have simply replaced the stock cable with thicker core, but my understanding (courtesy of Herr Doktor Ohm) is that this will increase the current drawn by the device, and therefore blow my battery system’s 10a fuse.

My idea is to add resistance to the circuit instead. I wondered if wiring a second identical heating element in series with the first would achieve this whilst still enabling me to boil water? Or, put directly: what would the effect on current and thermal output be of wiring two heating elements rated at 110w in series on a 12v circuit?

Thanks so much for your help :)
 

dendad

Joined Feb 20, 2016
3,921
If you have 2 in series, the current will halve. So will the total power, so each heater will now heat at 25% of the original.
I think better wire is the way to go, and if you want to reduce the power, use a Pulse Width modulator (PWM) controller so it does not just waste the unused power as heat, as a resistor world.
There are plenty of the PWM motor controllers on Ebay , and go for a 20A or 40A version.
 
Last edited:

KeithWalker

Joined Jul 10, 2017
1,933
I’m a beginner, you have taught me much already, this is my first time posting about a project so please be gentle :)

I have a 12v aprox-110w DC water heater element, which for excellent reasons I would like to use to boil 500mls of water. Wattage is aprox because the manufacturer gives contradictory wattage specs, and mine appears to operate between 106-110w in use.

The device does function as intended when connected to my battery system’s cigarette lighter output (12v, 10a max). However, its cable and terminals become very hot during operation. Other users of the device have noted this fault too. I fear that the insulation will burn or the casing melt.

I would like to modify and improve it so as to eliminate this risk whilst keeping current at or below 10a.

As it stands, the device draws 106-110w @12v from my battery system. By my calculations, this means its drawing 8.8-9.2a – just within the capacity of the battery system's 10a fuse. Some users of the same device have simply replaced the stock cable with thicker core, but my understanding (courtesy of Herr Doktor Ohm) is that this will increase the current drawn by the device, and therefore blow my battery system’s 10a fuse.

My idea is to add resistance to the circuit instead. I wondered if wiring a second identical heating element in series with the first would achieve this whilst still enabling me to boil water? Or, put directly: what would the effect on current and thermal output be of wiring two heating elements rated at 110w in series on a 12v circuit?

Thanks so much for your help :)
If you connect two heating elements in series, the resistance in circuit will be twice the original, so the current and power dissipated will be 1/2. Will a total of 60W be sufficient to heat the volume of water? If so, that would be an effective solution to your question.
 

Tonyr1084

Joined Sep 24, 2015
6,339
However, its cable and terminals become very hot during operation. Other users of the device have noted this fault too. I fear that the insulation will burn or the casing melt.

I would like to modify and improve it so as to eliminate this risk whilst keeping current at or below 10a.
Welcome to AAC.

Yes, its normal for the wires to get hot. That's because they have some resistance. The calculation of amperage is "Voltage divided by Ohms (resistance)". Smaller wire has higher resistance, therefore it gets hotter. Larger wire is more capable of conducting the higher current because it has lower resistance. The longer the wire the higher the resistance also applies.

You don't want to add resistance, you want to increase the current capability of the wire. At 110 watts you should be seeing (110W ÷ 12V = ) 9.2 amps. A 14 gauge wire should be sufficient for that kind of amperage. If you're running on 16 gauge wire - you're close to its max rating of 10 amps. AC wiring and DC wiring are a little different, so an on-line wire gauge calculator can yield reliable information on the subject. Here is a basic wire chart explaining a little about wire gauge.
 

Thread Starter

Martoia

Joined Sep 5, 2021
4
Thanks all for the really helpful replies and resources: question answered and my theory knowledge improved! :) Hope this thread'll be of use to others like me who are working through the basics.
 

Papabravo

Joined Feb 24, 2006
16,938
The title of this thread indicates a distorted view of the relationships between voltage, current, and power. The simple principle is this:
If two of the three quantities, voltage current, and power are fixed, then the third is determined and cannot be altered.

You can alter the situation as in post #3, but I'm pretty sure that is NOT what you had in mind.
 

Thread Starter

Martoia

Joined Sep 5, 2021
4
The title of this thread indicates a distorted view of the relationships between voltage, current, and power. The simple principle is this:
If two of the three quantities, voltage current, and power are fixed, then the third is determined and cannot be altered.

You can alter the situation as in post #3, but I'm pretty sure that is NOT what you had in mind.
Might it have been more accurate to title the thread “reducing the current to a 12v DC heating element rated by its manufacturer as 120w”? Point well-taken though: watts = current x voltage, so altering the current whilst holding voltage constant necessarily changes the wattage at which the device operates. I think my reasoning there is correct?

Anyway, thanks for the clear statement of the fundemental principles - I think they're sinking in. I’m also feeling confident that the practical problem I raised (hot wiring/fire hazard) can be solved.
 

Papabravo

Joined Feb 24, 2006
16,938
Might it have been more accurate to title the thread “reducing the current to a 12v DC heating element rated by its manufacturer as 120w”? Point well-taken though: watts = current x voltage, so altering the current whilst holding voltage constant necessarily changes the wattage at which the device operates. I think my reasoning there is correct?

Anyway, thanks for the clear statement of the fundemental principles - I think they're sinking in. I’m also feeling confident that the practical problem I raised (hot wiring/fire hazard) can be solved.
It is true that a smaller AWG wire gauge will allow you to carry current with a smaller rise in temperature. As a rule of thumb, I select wires with 700 circular mils per ampere of current flowing. You also have to look at the insulation with respect to temperature rise. Wire with insulation having a higher rating WILL be more expensive.

Circular mil - Wikipedia
 

Orson_Cart

Joined Jan 1, 2020
76
The usual way to gain good control here is a DC chopper ckt, which can be as simple as a TLC555 cmos timer, and grunty mosfet and a handful of parts including a potentiometer for control - this way you can vary the power to the load such that your ( apparently not big enough ) wires can handle the current
 

MisterBill2

Joined Jan 23, 2018
9,463
If you connect two identical heating elements in series, both the voltage and the current will be cut in half, and so the power will be cut to a quarter of the power. That may, or not, be adequate to boil the water. instead of 110 watts you will have 27.5 watts.
And your concern about the present wire overheating is quite valid.
Wire sizes for an application are determined by both voltage drop and conductor heating considerations. The electrical code sems to be aimed mostly at fire prevention, and so following that thinking, for a load of about ten amps the wire size should be number 16 (US wire size).
My guess is that the battery system is in a motor vehicle and so it will be a challenge to alter that part of the electrical system. You would need to, instead, add an additional connection direct from the battery, with an adequate fuse rated 15 amps, and using #14 wire, to power your 110 watt heater element.
 

Ian0

Joined Aug 7, 2020
3,463
If there is an engine running, or the 12V is on charge, then that 12V might be as high as 14.7V, in which case the power is 50% higher: 165W and the current is 11.2A.
Cigarette lighter connectors tend to be a cheap and nasty bit of folded-up tin, only designed for intermittent use (how long did the original cigarette lighter element take to heat up? Just a few seconds).
I suspect the heating in the cable might be heat that has conducted there from a poor contact in the connector. You need a really good plug and socket to work at 10A for any length of time. Some plugs have built-in fuses and that adds to the heat.
 

KeithWalker

Joined Jul 10, 2017
1,933
If you connect two identical heating elements in series, both the voltage and the current will be cut in half, and so the power will be cut to a quarter of the power. That may, or not, be adequate to boil the water. instead of 110 watts you will have 27.5 watts.
And your concern about the present wire overheating is quite valid.
That will be 27.5W per element, a total of 55W.
 
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