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Questions about DC heating circuit

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Hey MrAl,
Okay then wire wound resitor it is! :3
I think I'll go for 30 watt, and arrange the resistamces under an aluminum plate as you memtioned. And of course, I'll keep you updated and show you the results when I am done.
I can't thank you enough MrAl, I appreciate your help and all the useful information you provided me with.
 
Hello MrAl,
I am trying to draw the final diagram now and write down all the components I will need. However, I am having a problem with the calculations. If my goal is 30 watt, would it be okay if I used 9Volt battery and a total of 3ohm resistances?
Because P=R*I^2 which mean 30watt=3ohm*3amp^2
The calculation looks accurate, but I feel that it is so wrong. I feel that it isn't possible to generate heat with only 3ohm! Is it possible?

I am a beginner thats why it is so confusing for me, I apologize if my question is too silly.
 
Hello MrAl,
I am trying to draw the final diagram now and write down all the components I will need. However, I am having a problem with the calculations. If my goal is 30 watt, would it be okay if I used 9Volt battery and a total of 3ohm resistances?
Because P=R*I^2 which mean 30watt=3ohm*3amp^2
The calculation looks accurate, but I feel that it is so wrong. I feel that it isn't possible to generate heat with only 3ohm! Is it possible?

I am a beginner thats why it is so confusing for me, I apologize if my question is too silly.

Of course I'll double the resistances so they last as you mentioned before. So, if my calculations were correct, I have to double the total resistances to 6 ohm instead of 3 ohm. Am I correct?
 
Hi again Sara,


Well Sara, we dont double the resistance we double the power rating of the resistor(s). Let me provide an example.

With a 9v source and a 3 ohm resistor, the power is:
P=9^2/3=27 watts (power equals V^2/R)

Now if we want to run the resistor at half power, that means we need to double the POWER rating of the resistor. Since we found we have 27 watts, double that is 54 watts, so we could get by with a 50 watt resistor.

However, 27 watts in one resistor means the heat may not spread out too well, so we might want to use four resistors instead of just one. Wiring the resistors in parallel, this means the resistance we need will be four times as high (because we are using four resistors now) so that's 4 times 3 which gives us 12 ohms. So each of the four resistors is 12 ohms.
Now since there are four resistors sharing the power, that means each resistor dissipates 27/4 watts, which is 6.75 watts. To verify, we use V^2/R again and we get:
P=9^2/12=6.75 watts.

Now we still have to double the power for each resistor, so we use resistors rated for at least 13.5 watts (6.75 times 2) so we find something close to that, maybe a 15 watt resistor, and we use four of them connected in parallel.

Make sense now?

Just one other thing: where are you going to get the 9v, 3 amp power source from?
 
Hey MrAl,
Yah it makes so much sense! I get it now. It's the first time for me to know that resistences have power! I thought they have only the resistance measure in ohms. I didn't know i can find resitinces with different watts.

About the source, doesn't a normal 9V battery work? I said 3 amp because i applied the formula :|
I=V/R .. I guess there is something wrong with that calculation, right?
 
I searched the battery types just now and found out that a 9V batter has only 500mA :/
Now all the calculations goes wrong..
How can I reach around 30 watt then with such a bettery? I don't mind adding another battery if that will solve it
 
Two or three little 9V batteries are nowhere near capable of delivering 30W (at least, for the time to heat your cup). The 500mAHr rating means 3A (even if you could squeeze that out of your 9V battery) would flatten the battery in 0.5/3 = 1/6 Hr = 10 mins. You need a BIG battery, such as in the above links. Note that you will also need a suitable battery charger.
 
An ordinary little 9V alkaline battery is too small to produce more than about 0.5A. It is rated at only 25mA (0.025A).
Energizer's datasheet shows that at 0.5A a new one lasts for only 36 minutes at 0.5A but its voltage has dropped to only 4.8V.
It might not be able to produce 1A but if it does then the battery might become as hot as its load.

Why not plug in a mug heater that is available? It uses ordinary electricity.
 
I searched the battery types just now and found out that a 9V batter has only 500mA :/
Now all the calculations goes wrong..
How can I reach around 30 watt then with such a bettery? I don't mind adding another battery if that will solve it

Hello again Sara,


Well as you probably gathered by reading the other posts before this one, using a regular little 9v battery isnt going to work very well. That's because the energy stored inside the battery depletes fast because there just is not much in there to begin with. You need something bigger like Clyde posted, a lead acid battery. But even that may not be enough depending on how long you want to run your coffee warmer.

For example, a 12v, 7 AH battery (that's 12 volts, 7 Ampere Hours) can supply less than 1 amp for 7 hours. At 3 amps it goes down three times faster, so divide 7 by 3 and we get 2.33 hours, which isnt that long really. And because we are draining it with such a high current of 3 amps it wont even last that long. So maybe we'd get 1.5 hours run time and then the voltage would go down fast so the heater/warmer would not work anymore.

What you really need here is a good plug in power supply or wall wart that can handle the 3 amp load without a problem. An old PC computer power supply could easily handle this, but it would be kind of big and bulky. So you may want to look around for a power supply or wall wart.

If you absolutely must run this as a portable device, then you should consider the size of the battery and the run time you get. Here are a few examples:
12v, 9AH, about 3 hours run time (battery not too heavy, about 6 pounds),
12v, 18AH, about 6 hours run time (battery might be a little heavy),
12v, 36AH, about 12 hours run time (quite a heavy battery though).

These are based on around 5 ohms load which would cause a 2.4 amp current draw on the battery, not 3 amps.

Note that when we go to 12v we increase the total resistance to around 5 ohms, but if we go down to 6v we'd have to decrease the resistance to 1.2 ohm which would draw 5 amps not 3 so it would run down even faster. So 12v is the recommended battery voltage because batteries 6v and 12v of the same ampere hour are often priced about the same.

So the question now is do you still want to use a battery or use a power supply of some type.
 
The datasheet for a little 9V battery shows how weak it is.
It can supply 0.5A (4.5W) to warm your coffee for only a few minutes then its voltage drops to 4.8V in 38 minutes. Then it is finished. The little battery will be as warm as your coffee.

You need about 3.3A at 9V (30W) to heat your coffee then it must use about 1A (9W) continuously to keep it hot.
 

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alec_t and audioguru,
yah as you mentioned, that battery unfortunately won't work :/
I don't wanna plug it into a mug heater, the idea is to make myself a portable mug thats able to heat. Thanks.

ClydeCrashKop,
Thank you for the batteries! they are bigger than what I hoped for but still it works, really thanks!

MrAl,
yah I figured that from the pervious posts.
So, if I used the 12V,7AH battery that clyde told me about, my circuit will work effectively but only for 1.5 hours?
If that's the case then no problem! It will work again after recharging the battery, right?
I don't mind having it working for only 1.5 hours each time, it works for me.
So, if I don't mind the time, can I use this battery?
And about the plug, I don't wanna use this, I really want it to be portable. Thank you MrAl.
 
The datasheet for a little 9V battery shows how weak it is.
It can supply 0.5A (4.5W) to warm your coffee for only a few minutes then its voltage drops to 4.8V in 38 minutes. Then it is finished. The little battery will be as warm as your coffee.

You need about 3.3A at 9V (30W) to heat your coffee then it must use about 1A (9W) continuously to keep it hot.
audioguru,
thank you so much for making that data sheet! it explains clearly why it won't work. I appreciate it very much.
Yah I get it know that I need 3A at 9V, but it is so hard to find a battery that has these features.
 
alec_t and audioguru,
yah as you mentioned, that battery unfortunately won't work :/
I don't wanna plug it into a mug heater, the idea is to make myself a portable mug thats able to heat. Thanks.

ClydeCrashKop,
Thank you for the batteries! they are bigger than what I hoped for but still it works, really thanks!

MrAl,
yah I figured that from the pervious posts.
So, if I used the 12V,7AH battery that clyde told me about, my circuit will work effectively but only for 1.5 hours?
If that's the case then no problem! It will work again after recharging the battery, right?
I don't mind having it working for only 1.5 hours each time, it works for me.
So, if I don't mind the time, can I use this battery?
And about the plug, I don't wanna use this, I really want it to be portable. Thank you MrAl.

Hello again Sara,


Yes that's about right. For a 12v battery though we can increase the total resistance to 5 ohms, which puts the drain current at 2.4 amps instead of 3 amps, so we get a little longer rum time. 7 ampere hours divided by 2.4 amps is close to 3 hours. The fact that these batteries are often rated in terms of a 20 hour drain current however means we wont see 3 hours but maybe 2.5 hours hopefully. At worst i would think at least 2 hours. But remember now we have to increase the resistance to 5 ohms to retain the same power (close to 30 watts) and get that extra run time too.

There are 12v, 7.2 Ampere Hour batteries for sale on Amazon for around $17 USD. Dont have any idea what the quality is though.
 
Hello again Sara,


Yes that's about right. For a 12v battery though we can increase the total resistance to 5 ohms, which puts the drain current at 2.4 amps instead of 3 amps, so we get a little longer rum time. 7 ampere hours divided by 2.4 amps is close to 3 hours. The fact that these batteries are often rated in terms of a 20 hour drain current however means we wont see 3 hours but maybe 2.5 hours hopefully. At worst i would think at least 2 hours. But remember now we have to increase the resistance to 5 ohms to retain the same power (close to 30 watts) and get that extra run time too.

There are 12v, 7.2 Ampere Hour batteries for sale on Amazon for around $17 USD. Dont have any idea what the quality is though.

Hello again MrAl,
that's awesome! Almost everything is solved now. So, 12V battery,7AHm will make it work for 30 watt for 2 hours.
If I don't mind having it working for only one hour, can I find a smaller size battery that will make it work? Because this one is good but so huge in size!
 
Hello Sara,

Yes. A 4.5 AH battery for example will allow it to run for around 75 minutes.

You can drain and recharge the battery about 200 times. It may loose capacity as time goes on however, so the run time will be shortened as you charge and discharge many times. If you dont drain the battery all the way you can get more recharge cycles.

If you dont use the battery for a period of 6 months recharge it anyway.
Try to keep the terminal voltage above 10.8 volts for longer life.
 
Last edited:
Hello Sara,

Yes. A 4.5 AH battery for example will allow it to run for around 75 minutes.

You can drain and recharge the battery about 200 times. It may loose capacity as time goes on however, so the run time will be shortened as you charge and discharge many times. If you dont drain the battery all the way you can get more recharge cycles.

Okay perfect! I found one in Amazon that is 12 V and 4.5 Ah, this is really good thank you!
So now shall I use this same total resistances for this battery, total of 5 ohms? which means four 0.8 ohm for each resistance in parallel, right?
 
Okay perfect! I found one in Amazon that is 12 V and 4.5 Ah, this is really good thank you!
...

Did you check its physical size? I used a 12v 4.5Ah battery in a motorcycle, it was about 6" x 3" x 4" high and weighed about 3 pounds. It's hardly goign to "fit into the base" of your portable coffee heater.

Also I think you might be still underestimating the amount of energy needed to heat the coffee, and underestimating the heat losses.

30W will keep the coffee hot ok, if applied well to the cup and the cup is in some insulating (styrofoam?) holder that surrounds at least the bottom half of the cup.

But if you still want to heat the coffee up from cold to drinking temperature, 30W is still too little power especially if your styrofoam insulating system etc is less than perfect. We worked out earlier in the thread that would take hundreds of watts, depending how quickly you want to heat the coffee and how it the final temp is.
 
Okay perfect! I found one in Amazon that is 12 V and 4.5 Ah, this is really good thank you!
So now shall I use this same total resistances for this battery, total of 5 ohms? which means four 0.8 ohm for each resistance in parallel, right?

Hi Sara,


Well when you connect resistors in parallel the total resistance decreases. So if you need 5 ohms and you are going to use three resistors, then 3 times 5 is 15 so you need 15 ohms for each of the three resistors.

The type of wire wound resistor only needs to be able to handle the power, and because of the way we are doing it each resistor needs to handle 1/N times the total power multipled by 2, so for three resistors that would mean 30*2/3=20 watts each resistor. The bolt down resistors are easy to use when you can bolt them down to some metal like aluminum. But in any case you may need to insulate the bottom so that most of the heat goes to the top surface to heat the liquid.
 
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