Changing resistance on soldering iron

[heidesw]

So being curious, I opened up my cheapo radioshack battery powered soldering iron and to my surprise there was basically just a resistor and a contact switch which connected to the batteries and the soldering tip. Does this seem correct or am I mistaken.

My purpose for opening it is because the temp. on the iron is way too high and I am trying to lower it to an appropriate range (~200 C) I think right now it is around 500 C. Could I just increase the resistance in order to decrease the power that is going to the soldering tip? If so how would I go about calculating the amount of resistance necessary to get the required temp?

I am a little rusty on the physics of electricity but trying to get back into, so any help in the right direction would be much appreciated. Feel free to reference relevant equations, as I said I need to restudy some of these things.

I'm not sure if this is right but the resistor appears to be 51 ohm, and there are four AA batteries which means 6V.

 

[vk6kro]

That can't be right.

If there was a 51 ohm resistor in there in series with the heating element, the total power would be less than 0.7 watts. Not enough to get the sort of heating you are describing.

Somewhere near the tip, there will be a heating element which will be made of nichrome wire or similar. It will get very hot when power is applied.

You may be able to reduce the voltage to this element so that it doesn't get so hot, but you need to know what is already in there.

 

[berkeman]

So being curious, I opened up my cheapo radioshack battery powered soldering iron and to my surprise there was basically just a resistor and a contact switch which connected to the batteries and the soldering tip. Does this seem correct or am I mistaken.

My purpose for opening it is because the temp. on the iron is way too high and I am trying to lower it to an appropriate range (~200 C) I think right now it is around 500 C. Could I just increase the resistance in order to decrease the power that is going to the soldering tip? If so how would I go about calculating the amount of resistance necessary to get the required temp?

I am a little rusty on the physics of electricity but trying to get back into, so any help in the right direction would be much appreciated. Feel free to reference relevant equations, as I said I need to restudy some of these things.

I'm not sure if this is right but the resistor appears to be 51 ohm, and there are four AA batteries which means 6V.

How are you measuring the 500C? What adverse affects are you seeing in your solder joints? How much did the RS soldering iron cost? What kinds of things are you soldering? Wires? Through-hole components? SMT components? Other?

 

[Adjuster]

Just a guess, but does this contraption have a little light to show when it is on?

51 ohms might be about right to tap off a few milliamps for an LED.

 

[heidesw]

@Adjuster: Yeah, you're right it does have an LED, that part makes sense now, it's just drawing off the small bit of power to turn the led on when the iron is heating up. Great thanks for the clarification on that.

I work with a low melting point solder (tin, silver, coppper) and would like to use my wireless iron sometimes, that is why I am trying to lower the temperature on this iron.

@berkeman: The temperature is based off the specifications given by the manufacturer and from my own testing. I didn't spend much on the iron, but I just wanted to see if this wouldn't be a good project to ease my way back into studying physics, since the iron uses such a simple circuit. But my problem is that I can't think of how I would use the equations to figure out how to incorporate the temperature. Could I possibly reduce the number of batteries (there are currently four AA), or would it be easier to use a resistor to decrease the power? Wouldn't a resistor decrease the power too much since a 51 ohm resistor limits the power to .70W?

Please let me know if I am off track, thanks for the help...and the patience.

 

[Adjuster]

Haven't you just agreed that the 51 ohms is for a pilot light? If that is so, this resistor limits the power used by the light to less than 0.7W, but it does not affect the heating element. The element needs to pass a few amps at 6 Volts in order to give a reasonable heating power. You don't say what the iron is rated at, but at 6V an element of two ohms resistance would pass 3 Amps, giving 18W.

You could consider putting a suitably low resistance in series with the element, but note that resistor would consume a fair few watts and so would generate quite a bit of heat. Your idea of missing a battery out may be a better option, provided that the iron really does use 1.5V disposable cells, so that there is no question of a charger which might be upset by the lower voltage.

If you do want to lower the battery voltage with minimal alteration to the iron, consider making up a dummy cell with the two ends connected together. That way you can always go back to full power later by fitting all four cells.

 

[Adjuster]

Incidentally, if this thing really works off a four ordinary 1.5V AA cells, how long can it work for? Are you sure the batteries aren't some kind of rechargeable type?