 # powering two SMA wires on one power source

I'm using a tip120 transistor to control some flexinol SMA wires. I can easily get one to work, but seem to be having issues when I try to drive both the wires with one power source, I'm assuming it has to do with fact that the current is being split between the two wires? and if so, not sure how to increase the current between the two-- would I just increase the voltage? with SMA it seems that current is the most important part.

if I increase the current I am not sure if that is ok with the tip120-- just started using these.

here is a schematic of how I am wiring it: https://www.dropbox.com/s/2fp01vgoz050r7h/Screenshot%202013-12-13%2012.01.28.png

the battery is actually a 9v 30mA transformer though, R1 is 2.2k and other resistors are the SMA wires.

here is also a link to the breadboard that can be forked http://123d.circuits.io/circuits/70276/

9v 30mA

I thought you were 0.006 wire that needed 400mA?

9v 30mA

I thought you were 0.006 wire that needed 400mA?

this is where I get a bit confused on amps. going to rubber ducky it... so amp rating on a transformer is how much you can pull and volts is actually volts of what comes out of the transformer.

I need 400mA or .4A so....

I have two 6in wires, thats 7.8ohms *2 = 15.6ohms resistance in the wires. and I have another transformer that is 12v 1A:

(12v/.4A)-15.6ohms = 14.4 ohms

ok so I need to add a resistor to this circuit... so it would be more accurate to do this: https://www.dropbox.com/s/aap9xehi068knlp/Screenshot%202013-12-13%2015.34.10.png

so by using those resistor values I am limiting current to .4A? I am not sure if that works with how I have it wired. Also since its going through the transistor would I still need a high watt resistor like a 5w? when I connected directly to transformer, it seems to be an issue, but when I had it on the tip120 it wasn't an issue, so not sure if it was limiting current somehow?

edit: also looking and schematic it seems I could simplify the pin out with just one R1 resistor as well...

Are you connecting the wires in series or in parallel?
The transistor will also have a voltage drop across it, probably approaching 2V with 400 or 800mA. You need to adust your resistor to account for that:
(12V - 2V)/.4A = 25 ohm - 15.6 ohm = 10 ohm resistor.
Presistor = I^2*R = 0.4 * 0.4 * 10 = 1.6W, so at least a 2W rated resistor.

Are you connecting the wires in series or in parallel? The transistor will also have a voltage drop across it, probably approaching 2V with 400 or 800mA. You need to adust your resistor to account for that: (12V - 2V)/.4A = 25 ohm - 15.6 ohm = 10 ohm resistor. Presistor = I^2*R = 0.4 * 0.4 * 10 = 1.6W, so at least a 2W rated resistor.

ah, interesting. how do I know what the voltage drop is? I tried to look up info on transistors but they mostly consist in explaining how the E C B works... if you have any suggested reading I would definitely put the work in.

they are in parallel I believe. here is another upload, not sure why other link isn't working http://i.imgur.com/HjzM0qK.png

Use a logic level MOSFET to switch the SMA wire, then the voltage drop will be tiny.

Oh, that diagram is just wrong! If you are wired like that, you have to fix that before proceeding.

http://www.fairchildsemi.com/ds/TI/TIP122.pdf

Problem is TIP120 is not a straight NPN transistor - it is 2 NPN connected in Darlington fashion, where 1st NPN provides extra drive to the 2nd for higher Collector-Emitter current flow. The drawback is higher Collector-Emitter voltage drop, closer to 2V than say 0.5 or 0.7V.

That may work out okay here tho:
(5V - 2V)/(7.8 ohm) = 384mA.

If the 5V is something else, re-do the math and add the appropriate current limit resistor.

CrossRoads: Oh, that diagram is just wrong! If you are wired like that, you have to fix that before proceeding.

http://www.fairchildsemi.com/ds/TI/TIP122.pdf

Problem is TIP120 is not a straight NPN transistor - it is 2 NPN connected in Darlington fashion, where 1st NPN provides extra drive to the 2nd for higher Collector-Emitter current flow. The drawback is higher Collector-Emitter voltage drop, closer to 2V than say 0.5 or 0.7V.

That may work out okay here tho: (5V - 2V)/(7.8 ohm) = 384mA.

If the 5V is something else, re-do the math and add the appropriate current limit resistor.

ah, you know what, I am new to schematics so I didn't realize how wrong that looked. I used a generic transistor on my build tool, I guess the pins were flipped (schematic gets build almost automatically from breadboard tool). I do in fact have the digital pins from arduino going to the base, not the collector. For the most part I have exactly what you have in my actual breadboard.

yeah I have 12v on it so the resistor was 14.4ohms.