Current Reduction

I'm working on a project that needs 18 V so I am using three 6V batteries in series. When I measure the current on my multimeter I get around 5.5 Amp. I need about half that current. So I used a 2 ohm 1/4 W resistor, but it burns out everytime. I don't understand why my other 1/4 W resistors say 470 ohm do not burn out. What type of resistor would I need for this to keep the resistor from burning out, maintaining ~18 V, and having around 2 Amp?

What exactly am I working on?

I'm building a geiger counter and I am attempting to increase the battery life by switching from 2- 9v's to 3 - 6 V batteries. The circuit involves a LM358 set up as a square wave generator that feeds a TIP31 to feed a transformer and voltage multiplier to produce 500 V. It works fine with the 2- 9V batteries in series (TIP31 does get a little hot) . When I connect three 6 Vs in series the TIP31 and LM358 fry. I think I need to step back the current... What watt rating resistor would be appropriate for this. The 2-9V batteries produce about 2.5 amp on my multimeter, so I would like to reduce the 3-6Vs to that. Using V * A = W wouldn't 18 * 2.5 = 45 W. I've never seen a 2 Ohm resistor in that watt rating... Any ideas?

The Power Dissipation in the resistor would be (Voltage Drop across the resistor) squared / Resistance in Ohms
This would give you power in W and then you can size your resistor accordingly

Maybe simply recalculate a pulse and transformer ?
I do not understand what exactly You doing but P=U8I is math... You can't change it.
So if You low down voltage You have to increase current ...
But if You say that LM and TIP blowes up it's mean that You have to recalculate device...
You give to long pulses or bad frequency...

Maybe use a different IC to generate pulse switching power supply ? Like MC34063 ? (it can work with external transistor and transformer ... see datasheet.

B regards
Mark

I think that having a circuit's performance being dependent on the maximum power output of the power supply is concerning. If bigger batteries mean more amperage at the same voltage, when what you're looking for is simply longer performance period, there's a fundamental problem with the design. Putting a large watt resistor may mask the issue but you're going to lose a large chunk of power as excess heat and may as well just stick with the 9V batteries.

It may be as Mark mentioned with some tweaking of the pulses but I'd definitely look into where that power is going and why your ICs can't handle the bigger battery. I'd bet there is a circuit design issue rather than a power supply problem. My two cents. Good luck!

If you give us diagrams of the circuit designs you have tried, and tell us what happened when you connected each of them, we will understand the problem and be able to give you better advice.

Maybe I'm misunderstanding something, but the circuit should basically draw the same current from any 18 volt source. Step down Hoover Dam to 18v DC and the circuit should still draw what it did with the two 9v batteries.

You do say the measured amps of both battery packs are unequal. What about the volts, measured under the circuit load? I don't THINK the internal resistance of the 3 6's should be all that different from the 2 9's ... but something strange is going on.

Did the 9v batteries tend to overheat? Maybe you weren't getting a full 18v out of them because the circuit was trying to draw more amps than they could really provide.