In trying to learn as much as possible about power supplies for use with a nichrome wire heating element, I think I've figured out about how much amperage I need, based on how many watts I need. Somewhere along the way I recall running into the notion that, in a straight power-supply circuit, with no resistors or divided voltages required to accomplish anything but watts, a deficiency in amps could be made up by supplying some excess voltage. In other words, volts can turn into amps to get a desired result. I've tried to find the passage that reveals that, but I can't, so I opted to open myself to ridicule and howls of derisive laughter by asking the question here. IS there any way that could work, have I been misinformed, or am I hopelessly out to lunch?

 

Power expressed as Watts = The Voltage * The Current. If we look at heating elements they normally call out a wattage and voltage. I have a strip heater laying here. It is labeled 240 Volts 500 Watts so we can assume 500 Watts / 240 Volts = 2.083 Amps will be the current draw at 240 Volts applied. However, there is more to it. This element and any like it have a hot working resistance and a cold resistance. The heating element has a resistance so what this means is I can't apply for example 480 Volts and expect my element to draw half the current if I double the voltage. I can't change the resistance, it is what it is. So if I look at things we can say 240 Volts / 2.083 Amps = 115.218 Ohms (hot). See where this is going?

Ron

 

a deficiency in amps could be made up by supplying some excess voltage. In other words, volts can turn into amps to get a desired result

Uh, no. Increasing the voltage will also increase the current into any given load.

The power a supply can provide is basically in Watts. If you increase the voltage, you decrease the amperage.

Bob

 

An analogy to help understand ohm's law and Watts law
Is a system of plumbing pipes. The voltage is equivalent to the water pressure, the current is equivalent to the flow rate, and the resistance is like the pipe size There is a basic equation in electrical engineering that states how the three terms relate. It says that the current is equal to the voltage divided by the resistance or I = V/R. This is known as Ohm's law.
Let's see how this relation applies to the plumbing system. Let's say you have a tank of pressurized water connected to a hose that you are using to water the garden.
What happens if you increase the pressure in the tank? You probably can guess that this makes more water come out of the hose. The same is true of an electrical system: Increasing the voltage will make more current flow.
Let's say you increase the diameter of the hose and all of the fittings to the tank. You probably guessed that this also makes more water come out of the hose. This is like decreasing the resistance in an electrical system, which increases the current flow.
Electrical power is measured in watts. In an electrical system power (P) is equal to the voltage multiplied by the current.
The water analogy still applies. Take a hose and point it at a waterwheel like the ones that were used to turn grinding stones in watermills. You can increase the power generated by the waterwheel in two ways. If you increase the pressure of the water coming out of the hose, it hits the waterwheel with a lot more force and the wheel turns faster, generating more power. If you increase the flow rate, the waterwheel turns faster because of the weight of the extra water hitting it.Electrical systems are more efficient when a higher voltage is used to reduce current
What are amps, watts, volts and ohms?
In an electrical system, increasing either the current or the voltage will result in higher power. Let's say you have a system with a 6-volt light bulb hooked up to a 6-volt battery. The power output of the light bulb is 100 watts. Using the equation I = P/V, we can calculate how much current in amps would be required to get 100 watts out of this 6-volt bulb.
You know that P = 100 W, and V = 6 V. So, you can rearrange the equation to solve for I and substitute in the numbers What are amps, watts, volts and ohms?
In an electrical system, increasing either the current or the voltage will result in higher power. Let's say you have a system with a 6-volt light bulb hooked up to a 6-volt battery. The power output of the light bulb is 100 watts. Using the equation I = P/V, we can calculate how much current in amps would be required to get 100 watts out of this 6-volt bulb.
You know that P = 100 W, and V = 6 V. So, you can rearrange the equation to solve for I and substitute in the numbers.
I = 100 W/6 V = 16.67 amps
What would happen if you use a 12-volt battery and a 12-volt light bulb to get 100 watts of power?
I = 100 W/12 V = 8.33 amps
So, this latter system produces the same power, but with half the current. There is an advantage that comes from using less current to make the same amount of power. The resistance in electrical wires consumes power, and the power consumed increases as the current going through the wires increases. You can see how this happens by doing a little rearranging of the two equations. What you need is an equation for power in terms of resistance and current. Let's rearrange the first equation:
I = V/R can be restated as V = I*R
Now you can substitute the equation for V into the other equation:
P = V*I substituting for V we get P = I*R*I, or P = I2*R
What this equation tells you is that the power consumed by the wires increases if the resistance of the wires increases (for instance, if the wires get smaller or are made of a less conductive material). But it increases dramatically if the current going through the wires increases. So, using a higher voltage to reduce the current can make electrical systems more efficient

 

Volts and Amps are not independent. When one is changed, so does the other.

You need two basic equations to work out power.

I = V/R
and
P = I * I * R

From the first equation we can substitute
V = I * R
R = V / I

Hence we can write the power formula in three ways

P = I * I * R
P = V * V / R
P = I * V

Hence if you know any two of I, V, R, and P, you can calculate the other two.

 

Dear all, taken together, your replies have set me straight and given me a rough idea how I got the initial wrong impression. I'll stick to working with Ohm's law and Watt's law, since most of the specs I can find on the heating-element subject offer watts as well as V and I. My apologies that it took so long to respond to this thread. I got into finishing the plastic bender, mocked it up with a power supply and a PWM controller, and Behold! it worked perfectly. There's more to be done, but I couldn't have got this far without bouncing my brute ignorance off this forum. When I get it adjusted and powered by a dedicated supply, I'll include a photo.

 

There are huge electrical circuits that use inductive heating like melting iron ore.
There is a world of difference in annual electrical cost when the frequency does not match.
It is understandable that a switched mode power supply used to drive a heating element would encounter complex explanations and people all over the world are using these power supplies wondering why it got so difficult. People cannot understand why their wall warts are so light. Bang good meant well product. Surviving is understanding the repair of the bangood mean well PS.

Glad you got past the switch mode drama and finished your plastic bender. Most just give up.

 

Bang good meant well product. Surviving is understanding the repair of the bangood mean well PS.

Surviving is throwing one away and replacing with another which you have on the shelf as a spare because life experience has taught you it's not a matter of if but when it fails.

Ron