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A step-up transformer turns a lower alternating voltage to a higher one, while stepping down the current. This is due to the conservation of energy (energy to into the first coil must equal to the power coming out of the second coil).

I'm using a 12v DC to 240v AC inverter to power my soldering iron from a 12 volt DC battery. From what I've understood, soldering irons work by letting the high voltage current pass through the metal tip of the iron. This current then heats the metal to the high temperature required. Somewhere inside the inverter, a transformer is used to step up the voltage.

But if the power provided by the 12 volt DC battery (which goes into the first coil of a transformer) is the same as the power provided by the 240 volt inverter output (coming out of the secondary coil), why do I need the inverter at all? Essentially I'm turning electrical energy into heat, so if I were to connect the DC voltage supply directly into the iron wouldn't the iron heat the same way since the amount of power is being turned into heat?

The same question applied also for incandescent light bulbs: Since they work by heating a filament (which then emits light), should't they also work with lower voltage supplies as long as the provided current is higher? Then the power would again the the same.

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    \$\begingroup\$ Similar wattage can be achieved at lower voltage. Your inverter will have to pull the necessary currents already, so it's certain that your 12 VDC supply is already up to the task. Which is good. The problem will be in finding an appropriate soldering iron. I've never seen one that operates off of 12 VDC. But that doesn't mean one doesn't already exist (nor that it can't be designed and created.) Perhaps you can find one or otherwise set out to design and build one? \$\endgroup\$ – jonk Apr 9 '17 at 18:05
  • \$\begingroup\$ walmart.com/ip/… \$\endgroup\$ – user3528438 Apr 9 '17 at 18:14
  • \$\begingroup\$ sound like : how to convert transformer to reactance ? My opinion is that you have no knowledge of transformer reflexes. A magnetic field or a period conflict is possible. \$\endgroup\$ – dsgdfg Jan 9 '18 at 6:09
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If your soldering iron heater is intended for 240v, then giving it 12v would result in a very low heat output indeed. Feeding it with 1/20th of the voltage would result in 1/400th of the rated power (\$power=\frac{v^2}{R}\$), you might feel the chill taken off it, but no heat to speak of.

Before you look at 24v irons made by Weller, they run off 24v AC. I doubt the thermostatic switch is rated for DC, so you could expect the switch to fail in short order if you tried to run on off 24v DC.

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  • \$\begingroup\$ 12V DC soldering irons are available at a range of prices. \$\endgroup\$ – nekomatic Sep 7 '18 at 14:19
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In general, electrical and electronic devices are designed to operate with a specific input power which is defined as to type (AC or DC), voltage (12 volts, 120 volts, etc.) and frequency (for AC this is typically in the range of 60 to 400 Hz). If your soldering iron is designed to run on 120 VAC then it will not operate properly if fed 12 VDC, even if the DC source can provide the correct amount of power. Similarly, an incandescent light bulb designed for 120 VAC will not operate properly ion 12 VDC.It is certainly possible to design a soldering iron and an incandescent bulb to run on 12 VDC but then neither would operate on 120 VAC. A 12 VDC incandescent bulb, for example, would need a much beefier filament to handle the must higher current that it would need to produce the same light output. A 100 watt incandescent bulb operating at `120 VAC draws less than 1 ampere; however, a 100 watt incandescent bulb built to operate at 12 VDC would have to handle more than 8 amperes.

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  • \$\begingroup\$ Why does the filament need to be beefier since even with the higher current, the voltage is lower and the total power remains the same? Since the total power is the same, are we not providing the same amount of energy to the filament thus heating it with the same rate? \$\endgroup\$ – S. Rotos Apr 10 '17 at 9:54
  • \$\begingroup\$ But in case of filament, you can't "force" more current through same filament resistor if you "decrease" voltage, ohm's law! (Not applicable to all loads e.g. DC motor, motor draws as much current as needed because of its inherent feedback action, however Ohm's law is still satisfied by action of back emf) \$\endgroup\$ – Deep Oct 1 '17 at 15:13
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Both soldering irons and incandescent lights are resistors.

The current through a resistor is given by I = V/R. So the higher the voltage, the more current flows.

The power dissipated in a resistor is P = VI. Combining the two equations gives P = VI = V²/R. So if you lower the voltage, the current will drop, and the power will drop even more.

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  • \$\begingroup\$ Or in other words, you could, but you'd need to swap out the iron/lightbulb for one with a much lower resistance. \$\endgroup\$ – user253751 Apr 9 '17 at 22:51

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