Help for someone who is completely electrically illterate trying to design something

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Reduce the voltage that feeds the resistor. When the voltage is 1/2 then the dissipation in the resistor is 1/4.
 
audioguru said:
Reduce the voltage that feeds the resistor. When the voltage is 1/2 then the dissipation in the resistor is 1/4.
Click to expand...

Thanks for responding! Reducing the voltage is not an option, I am trying to power from a portable energy source and want full voltage... I want to limit the energy used on the resistor side, not on the power source side, if that makes any sense.
 
loblahblahblah447 said:
Thanks for responding! Reducing the voltage is not an option, I am trying to power from a portable energy source and want full voltage... I want to limit the energy used on the resistor side, not on the power source side, if that makes any sense.
Click to expand...

Makes sense, but not possible. The wattage power dissipated by the resistor is determined by it's resistance value in ohms and the applied voltage to the resistor. To lower the power dissipated by the resistor you either can lower the voltage or raise the resistance.
 

So the only way to get 15W output is to spec a 15W resistor?
 
loblahblahblah447 said:
So the only way to get 15W output is to spec a 15W resistor?
Click to expand...

I think you have to be a little more specific about what you working with and what you are trying to do with it.

When selecting a resistor there are two key specification that have to be determined, the resistance value needed in ohms and the maximum power dissipation required. One would not specify a 15 watt resistor that was going to actually dissipate 15 watts, that would be subjecting the resistor to damage as it would be too close to it's maximum allowed wattage. If a specific resistor was going to dissipate 15 watts of power one would specify a 25 watt or more power rating.

Lefty
 
Well basically what I want to do is run a heating element on a disposable battery. As I understand it, the heating element is essentially a resistor.

The spec I have for the heating element is a maximum wattage of 40W, but I want the power output to be 15W. Also, there is a spec for 120V and 240V, am I to understand that this is the minimum voltage required?
 
The power rating of a resistor has nothing to do with how much heat it makes. A low voltage across a resistor makes it heat less. A high voltage makes it heat more.
Ohm's Law determines how much a resistor heats.

A resistor with a high resistance heats less than one with a low resistance if they both have the same voltage.

Why are you wasting power by heating a resistor. What is the circuit?
 

As Audioguru mentioned in the first reply, the heat output is dependent on the voltage applied - to reduce the heat you need to reduce the voltage.

So what voltage is the element 40W at?.

Also, how long do you expect the battery to last, taking 15W from a disposible battery will drain it pretty rapidly - with heat output dropping all the time.
 

The heating element will have to be run at the voltage it is rated at for the wattage rating to apply. If you run a 120vac heater at 12vdc it will not generate it's rated heating power. A 40 watt heater designed to run at 120vac will only output .4 watts of heat if run at 12 volts.

Lefty
 
A thermostat controls the amount of heat from an electric heater by turning it off when it is warm enough then turning it on when it has cooled a little. If the heater is large then the on-off timing can be slow. You can turn it on and off quickly with an electronic circuit so the temperature varies only a small amount. The duty-cycle of the on and off determines the amount of heat.
 

Did you actually mean 4 watts or 10% of voltage = 1% of power? Thanks lefty.
 
loblahblahblah447 said:
Did you actually mean 4 watts or 10% of voltage = 1% of power? Thanks lefty.
Click to expand...

No, I meant .4 watts. It's a square function, watts = (voltage X voltage) divided by resistance. The resistor's ohms value is fixed, but the wattage consumed is based on the applied voltage.

Lefty
 
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Well let's say I have a D battery, 1.5V, with a good energy capacity. All I know about the heating element is a max power spec of 40W, is current a somehow fixed value in this situation? Without a resistance value for the heating element am I SOL to calculate anything? I don't know what kind of assumptions I can make.
 

hi,
I think you are misunderstanding the heater spec,
it says quote:a max power spec of 40W.

This is the max power/heat at which it can be used, NOT any other value less than this.
Lets assume the heater has a resistance of say, 12 ohms.
If I connect the heater across a 12V battery the Iamps= 12Volt/12ohm = 1amp.

Watts = V * I or in this case,
V^2/R = 12^2/12 = 12Watts.

Whats the voltage value you want to use and the resistance of the heater coil???
 
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An alkaline battery cell is 1.5V when brand new. Its voltage drops as it is used.
A brand new D cell will be 1.3V when it has a 1A load into 1.3 ohms which is only 1.3W of power in the resistor.
Its voltage will be 1.2V and the power will be 1.1W in half an hour.
Its voltage will be 1.0V and the power will be 0.77W in 6 hours.

If the heater produces 40W with a 120V supply then it will produce 15W with a 45V supply. It will produce 0.006W with a 1.5V supply.
 

It will produce 0.006W with a 1.5V supply.
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BUT last a very long time.
 

What do you mean when you say that it can't be used at a value less than 40W?

Basically, I need a resistance value for the heater to calculate the possible power output?

Also, thanks guys, this has all been extremely helpful so far!
 
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Look up Ohm's Law on Wikipedia, blahblah.

It's taught in high school.

It's the most basic, fundamental concept in electronics.

It's part of how the universe around you works.
 
loblahblahblah447 said:
What do you mean when you say that it can't be used at a value less than 40W?

Basically, I need a resistance value for the heater to calculate the possible power output?

Also, thanks guys, this has all been extremely helpful so far!
Click to expand...

It dosnt say it cant be used at less than 40Watts.
You keep quoting the maximum wattage as though its got to be run at that maximum wattage.

Measure the resistance of the heater coil, if you dont have a ohm meter, your local TV repair shop will measure its resistance for you.
 
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