Arduino output to open transistor, not getting enough current.

I am attempting to use an arduino output pin to open a transistor that is connected to one of these: http://www.robotroom.com/Launch-Controller/Estes-model-rocket-igniter-before-and-after-launching.png
It is an electric match for launching model rockets. The requirement for igniting these should be about 1 amp for 50 milliseconds.
I have an NPN transistor with the arduino output connected to the transistors base. The unregulated battery(4xAA) voltage is connected to the transistors collector. The emitter is connected to the electric match witch is in turn connected to ground.
Now the problem: The transistor is turning on and allowing current to flow from collector to emitter, but won't provide enough current to ignite the match.
If I replace the match with an led the led will turn on.
I have used multiple transistors with a collector current up to 5 amps.
so how do I get the arduino to activate the transistor so that it will allow enough current to flow to ignite the match?
The transistor will light the match if I remove the arduino and connect the battery voltage to the transistor base instead. Also when I connect it like this the match will ignite much slower than if I connect it directly to the battery. Do some transistors react faster than others?

Hello,

look at Transistor - Wikipedia. There you see the basic connections. The emitter of the NPN transistor is connected to GND and the collector to one pin of your load. The other pin of the load is connected to plus of your supply voltage. To drive the transistor with your arduino you connect the base to your arduino output over a resistor (!!!). The voltage between base and GND is around 0.7V when the transistor is open. Your arduino gives you max. 20mA. Now you can calculate the value of the resistor.

About the transistor type:
Look at the current amplification factor (beta) of your transistor. This is the ratio of max. collector current and base current. When you need 1A collector current and you have 20mA base current then beta must be 50 minimum. Take a transistor that has beta of at least 100. On power transistors beta is much smaller than on small signal transistors. The datasheet of the transistor tells you how long it can operate at 1A.

Best regards

Elektrix

"Your arduino gives you max. 20mA. "

You can go up to 35mA safely.
Vout of Arduino will be >=4.2V.
B-E of transistor will be ~0.7V.

(4.2 - 0.7)/0.035 = 100 ohm
Use 120 in case Vout from Arduino is closer to 5V.

Do some measurements with the transistor you have, or read its spec sheet, and adjust accordingly.

Or ditch the transistor and try an opto-isolator , my 2c worth

Here is a datasheet for one of the transistors I have http://www.fairchildsemi.com/ds/PN/PN2222A.pdf
Where do I find the current amplification factor? Would it be the current gain bandwidth product which is listed at 300?

I also need to detect continuity across the igniter. That is why I have it connected the the transistors emitter. Here is a schematic:https://dl.dropboxusercontent.com/u/10327232/launcher.png
In order for the arduino to sense continuity across the igniter it needs to be on the same side of the transistor as the ground connection. Can you think of a way to detect continuity with the igniter on the collector side of the transistor?

Why not just connect the collector to an input pin with a 100K pull down resistor.
If you have battery connected to the igniter, you will see a 1. If not, you will see a 0.

2N2222A not a good part. Only rated for higher currents for very short times:

  • Pulse Test: Pulse Width ? 300?s, Duty Cycle ? 2.0%

You're gonna smoke it at 1A for >150 times that long.

I'd go with Logic Level N-Channel MOSFET with Low Rds (like in the 10s of milliOhms) in a TO-220 package.

Such as like this:https://dl.dropboxusercontent.com/u/10327232/launcher%20(2).png
With the igniter attached pin 2 will read high. But when it is not attached, won't pin 2 be floating?

Won't be floating if you have a pulldown resistor. High resistor value to pull it low, but not so much that there is any chance of lighting the ignitor.

Ok, you must mean like this: https://dl.dropboxusercontent.com/u/10327232/launcher%20(3).png
You said pull-down resistor and I imagined a pull up resistor.

So I will probably need two resistors there, one to connect to ground and another to protect the arduino input pin from the raw battery voltage.

But you said the battery was 6V (4xAA), so not much protection needed.