Trying to lower the heat generated by a simple DC heating element circuit

Thread Starter

Blairege

Joined Feb 25, 2018
5
Hi all

I have a real simple DC circuit- just a Tenergy Li-on 7.4 Volt, 2600 mAh battery operating a heating element. The heating element has about 9 ohms of resistance. It is heating a little too hot so I am trying to come up with a fairly cheap inexpensive solution to not only lower the heat generated but also possibly have the battery run longer. For example, I assume if I just add resistors into this circuit to increase resistance, it may lower the temperature of the heating element but I assume it would also discharge the battery quicker. Is there a small, compact, inexpensive way to lower the battery output voltage?
 

wayneh

Joined Sep 9, 2010
16,162
For example, I assume if I just add resistors into this circuit to increase resistance, it may lower the temperature of the heating element but I assume it would also discharge the battery quicker.
Not so. Adding resistance in series reduces the load current and would extend battery life. Adding resistors in parallel would increase the load, make more heat, and deplete the battery faster.

Right now you have 7.4V across 9Ω. By Ohm's law V=IR, you've got a current of 7.4 = I • 9 or I = 7.4/9 = 0.822A. The power dissipated in resistive devices by ohmic heating is watts = I^2•R = 0.822^2 • 9 = 6W

Suppose you add 1Ω of addition resistance, to a total of 10Ω. Current goes down to 0.74A Power dissipation goes to 0.74^2 • 10 = 5.5W The resistor would be dissipating 0.74W and the heating element would handle the rest, about 4.7W. You'd want a resistor rated to at least about 2W or more for safety, if you choose this route. Another choice would be a 1W light bulb. I like using light bulbs because they provide visual feedback.

The diodes are a perfectly good solution as well but they will also dissipate heat just like a resistor would.
 

MrAl

Joined Jun 17, 2014
6,968
Hello there,

What you really want is a buck converter not simply a PWM or diodes in series.
It depends on how much you want to lower the voltage, but a high efficiency buck converter will do best as long as the output voltage is not too close to the battery voltage.

For example, say you have a 9v battery and you want to lower that to 7.6 volts.
If you do that with 2 diodes, the current will be 0.844 amps from the battery..
If you do it with just PWM the average current will be 0.844 amps from the battery.
If you do it with a 100 percent efficiency buck converter the current will be 0.713 amps from the battery.

Now obviously you cant get a 100 percent efficient buck converter, so what you have to do is see just how much you need to lower the output voltage. The lower you go, the more effective it is to use a buck converter with at least 90 percent efficiency. If you only have to drop a small voltage though, then resistance or diodes in series is good enough.
This assumes you want to get the maximum run time from the battery before having to recharge.
 

MrAl

Joined Jun 17, 2014
6,968
PWM also eliminates the extra heat that an added diode or a resistor would generate.
Hi,

I dont think that is true because PWM alone does not offer true power conversion. There has to be at least one energy storage element to get that effect. So the heat wasted should be the same.
We can run some numbers if you like but i've done this before although admittedly it was with LEDs not resistive load which could make the difference, but we can run some numbers.

Ok it does look better with resistive load. The problem comes in with LED drivers.
 
Last edited:

BobTPH

Joined Jun 5, 2013
2,204
For example, say you have a 9v battery and you want to lower that to 7.6 volts.
If you do that with 2 diodes, the current will be 0.844 amps from the battery..
If you do it with just PWM the average current will be 0.844 amps from the battery.
If you do it with a 100 percent efficiency buck converter the current will be 0.713 amps from the battery.
This makes no sense. PWM does not lower voltage, it lowers the percentage of time the voltage is applied. And it does so with nearrly 100% efficiency. A buck converter consists of a PWM controller plus an inductor and diode which are in the the path of the current going to the load. They both insert a loss that is not there with straight PWM. Using the same components for the oscillator and semiconductor switch, the PWM controller is always more efficient.

Bob
 

MrAl

Joined Jun 17, 2014
6,968
This makes no sense. PWM does not lower voltage, it lowers the percentage of time the voltage is applied. And it does so with nearrly 100% efficiency. A buck converter consists of a PWM controller plus an inductor and diode which are in the the path of the current going to the load. They both insert a loss that is not there with straight PWM. Using the same components for the oscillator and semiconductor switch, the PWM controller is always more efficient.

Bob

Hi there Bob,

Thanks for the inquiry.

The reason it does not make any sense to you is because you obviously have not done the math with different kinds of loads.
Try to power a high power LED with pure PWM and see what the numbers spit out.
Also keep in mind that the switch can not have zero resistance unless there is a small resistor in series with it like maybe 1 ohm.
This is where a buck converter comes in very nicely because it converts one form of energy to another form.

If you like i can show you some numbers.

With resistive load it is a different story because there is already some resistance in the circuit so no need to add more.
 

crutschow

Joined Mar 14, 2008
24,109
With a resistive load, PWM does lower the power at high efficiency, which is the case here.
Powering a LED is a different story, and is not pertinent to the TS's problem.

You can buy a PWM circuit as suggested, or you can build a fairly simple circuit using a 555 timer driving a MOSFET, configured as an astable oscillator with a pot adjustable duty-cycle -- your choice.
The amount of heat generated is then proportional to the PWM duty-cycle.

The MOSFET's dissipation equals (I² * Ron * PWM-dutycycle) where I is the load current.
Typically you want the MOSFET's Ron low enough so that the MOSFET dissipation is no more than a watt or so, so it doesn't need a heatsink.
 

MrAl

Joined Jun 17, 2014
6,968
Hello,

Yes i said that.
Also i need to clear up one more little detail.
When analyzing either circuit (LED, bulb, heater) we need to compare to the passive voltage drop method.
The passive voltage drop method uses either a resistor or diode(s) to drop the voltage rather than PWM or Buck.
So it is not just an absolute calculation it has to be comparative.
 

BobTPH

Joined Jun 5, 2013
2,204
Hi there Bob,

Thanks for the inquiry.

The reason it does not make any sense to you is because you obviously have not done the math with different kinds of loads.
How do you know what math I have and have not done?
Try to power a high power LED with pure PWM and see what the numbers spit out.
We are talking about a heating element, i.e. a resistor, not an LED.
Also keep in mind that the switch can not have zero resistance unless there is a small resistor in series with it like maybe 1 ohm.
This is where a buck converter comes in very nicely because it converts one form of energy to another form.
And, as I pointed out in my previous post, the buck converter starts with a PWM controller which also has a switch element, so switch lossses are a wash here.

But, we can do a lot better than one Ohm.

Try 400uOhms (yes, that is micro, not milli)

https://www.mouser.com/ProductDetail/Infineon-Technologies/IPT004N03L?qs=sGAEpiMZZMshyDBzk1/Wi7r8jvGh7mrdiB3BDvKkvig=

If you like i can show you some numbers.
Please do. For a heating element, which is what this thread is about.

Since you did not understand my previous post, maybe you will understand this.

Here are circuits for a PWM controller and a buck converter providing 50% power to a resistor. Please show me the math of how the second circuit is more efficient than the first.

upload_2019-4-28_17-11-48.png
 

oz93666

Joined Sep 7, 2010
731
Beyond dispute the most effective solution is to get inside the battery , it's made of 2 x 18650 3.6V cells ... rearrange them to run in parallel and deliver 3.6V into your heater....

Running the heater at half voltage will reduce the heaters temperature , and also it's resistance , so it may output perhaps a half or a third the power it did at 7.4 V ....

Solving this problem with circuitry only wastes energy in heating where heat is not required ..
 

wayneh

Joined Sep 9, 2010
16,162
Beyond dispute the most effective solution is to get inside the battery , it's made of 2 x 18650 3.6V cells ... rearrange them to run in parallel and deliver 3.6V into your heater....

Running the heater at half voltage will reduce the heaters temperature , and also it's resistance , so it may output perhaps a half or a third the power it did at 7.4 V ....

Solving this problem with circuitry only wastes energy in heating where heat is not required ..
Actually, halving the voltage reduces the power to 1/4. I^2•R and all. That's probably a bigger drop than the TS wants.
 

Sensacell

Joined Jun 19, 2012
2,544
PWM is by far the simplest and most efficient way to handle lowering the power delivered to a resistive load, given a fixed input voltage.
The pass element losses can be made vanishingly small, much smaller than any bucking voltage conversion circuit.

The only reason to change the voltage is if PWM is unfeasible for some other technical reason.
 

oz93666

Joined Sep 7, 2010
731
Actually, halving the voltage reduces the power to 1/4. I^2•R and all.
It does IF the heater resistance stays the same ... but fed 3.7V the heater will be at a lower temperature and so it's resistance will be lower than when fed 7.4Volts resulting in more than half the current it sees when fed 7.4V.

In addition the internal resistance of the battery be a quarter it was when configured for 7.4V .... so still more current , and the %power wasted in heating the cells (internal resistance losses) , is reduced.
 

Thread Starter

Blairege

Joined Feb 25, 2018
5
Hey guys

Thanks for all of the input. I probably need to lower the voltage from 7.4 volts to about 6 volts or so to get the desired heat out of the heating element. I am happy to do it any of the ways suggested by all of you, but I would like the battery to have as long of a run time as possible so I would prefer the method that uses the least amount of battery energy. Based on that, what do you guys think is the best solution?
 

dendad

Joined Feb 20, 2016
3,259
I would try the PWM method if it was my project.
With an Arduino, FET and a thermistor, you could control the temperature too.
 
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