Batteries - Voltage/Amps/Ohms... I need help!?

Question

Hi! I'm building a model rocket launch controller and I am confused about something that has been giving me a headache for over a month now.
The device basically completes a circuit when the button is pressed, which heats up the resistance wire in the ignitor which then ignites the engine. My current system is running off of two 9v batteries wired in series and the problem is that when I compare it to the stock system which is just running off of 4 AAs, my system is barely getting the wire to ignite.

What could the problem be? Clearly it's not a matter of the voltage being too low. Do I need more resistance or something? Please help. I'd like to get this worked out so I can use it for my student's rocket launches.

Thanks!

This is helping me understand already...so the 9v doesn't have enough internal resistance to cause the nichrome wire to heat up as well as all those 1.5v AAs? I guess in this case resistance is a good thing.

I hadn't really understood the distinction between voltage and current capacity. Electronics is hard to understand without having been taught any of it.

Thank you both for your help. Any more input would be appreciated!

Answer 1

Switching to two 9 V batteries in series was certainly a bad move. I'll try to explain...

The ignitor is simply a short piece of nichrome wire. Electrically, its just a low-value resistor. Although it is possible to calculate the resistance of any piece of wire knowing its length, diameter, and the material it is made of, this isn't really necessary. Hardly could it exceed ½ Ω.

When batteries are connected in series, their voltages add; thus, two 9 V batteries in series turn out a voltage of 18 V. This is three times the voltage of 4, 1 ½ V AA batteries in series (4 × 1 ½ V = 6 V).

How voltage, current, and resistance relate to each other? Ohm's law states that I = V/R, where I stands for current (intensity), V for voltage, and R for resistance. Now, THEORETICALLY, according to Ohm's law, 18 V, being 3 times 6 V, should force a current 3 times stronger through a given resistor, whatever its value may be. Therefore, a 18 V battery should outperform a 6 V battery, at any rate.

Direct experiment has proved otherwise, however. How can this be? Well, batteries have some internal resistance. This resistance varies inversely with size. A large, lantern type 6 V battery has much smaller internal resistance than a AA cell, even a D size cell. 9 V batteries are actually made stacking 6 very small 1 ½ V ordinary cells in the package. Put another way, a 1 ½ V battery is a single cell; 9 V batteries are actually made from 6 cells, connected (internally) in series. The internal resistance of such batteries, understandably, is much higher than that of a AA cell.

By now, you have probably guessed that internal resistance is not really a good thing. Internal resistance renders a battery inefficient. As this resistance is internal, it is always in series with the "load", which is the name given to what you connect to the battery. To put things in perspective, if a battery, having 1 Ω internal resistance, is connected to a 1 Ω load, one half of the power generated by the battery is wasted inside the battery, as heat; only ½ of the power is delivered to the load to do useful work.

Thus, it is desirable, from efficiency standpoint, to operate a load with a battery whose internal resistance is considerably smaller than that of the load. Internal resistance, however, increases as the battery is being used. A "fresh" battery has much lower internal resistance than a spent battery of the same size. In fact, electrochemical reactions inside the battery do not change substantially as the battery is drained. A battery becomes inoperative because the rise in internal resistance, much more than any other reason.

It should became clear that the problem is due to very little power reaching the ignitor. With such two-9 V contrivance, most of the power never gets out of the battery! Switch back to a standard pack; or better still, get one of those square, large 6 V "lantern" type batteries. A deluxe solution would be a 6 V, 1.2 Ah unspillable, lead-acid storage battery. These are scaled-down storage batteries similar to automobile batteries. Their internal resistance is in the order of mΩ, and will fire quite a load of rockets on a single charge. Once it gets discharged, you just charge it again, instead of trashing it out.

Answer 2

I think that the problem is that the 9v batteries cannot supply sufficient current to heat the filament. The current capacity (limited by internal resistance) of 9v batteries is much less than AAs, and putting them in series doesn't help. Try connecting the 9v batteries in parallel. You don't need more voltage, you need more current capacity. Why not use the 4 AAs?

EDIT: On your additional information: the 9v batteries have TOO MUCH internal resistance. The internal resistance limits the amount of current that can be drawn from the battery.

Answer 3

If the stock system uses only 6 volts (4 AAs) i suggest you use the same, instead of using 18 volts...If there is some IC in the wiring control, you could have burnt it by using such a high voltage.I suggest you try your controller using a 6 volt source, and if it doesn't work, try making a new controller and use 6 volt source on that.

Answer 4

Can you send a picture of the system you made and the circuit diagram.

There is a lot of way to heat up a wire. Your requirement is very simple that's why I'm concern why it gives you so much headache.

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Answer 5

Number 10 hearing aid batteries because the cages are so tiny, there is barely enough room for the hens.