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555 timer using a transistor to drive a 12 v. 30 amp relay

Please somebody help!
I'm not really an electronics guy, but I'm trying to learn. I'm working on a project making heated clothing for use in the cold Ohio weather. The heater will be powered with the motorcycle battery. So far, I've had success using a Velleman MK111 timer, modified with the supplied relay driving a 12 volt 30 amp automotive type relay. The heater circuit needs too much current to use the Velleman relay by itself. All is well and good except for one thing: The package is too big. My idea is to eliminate the Velleman relay and use a transistor in its place, thereby reducing the size of the package. I got a TIP3055 NPN transistor from Radio Shack thinking I could use it. So far, I've studied diagrams and made calculations on Excel for a week or so, but I'm still confused. I just don't know enough. I plan on wiring the circuit in common emitter mode (www.kpsec.com). The relay I want to drive has a resistance of 41.6 ohms. The input to the Velleman circuit is directly from the motorcycle battery. I have gloves hooked up drawing .5 ohm resistance. The wire in the gloves is 10 feet total of 30 AWG teflon coated copper. As a side note, 30 feet of the wire works fine connected directly to the battery, with no timer or other controls. I'm using the timer for just the gloves, or any future circuit with resistance too low to dissipate the heat. My questions - 1)Can I use the TIP3055? 2)What should the Base resistor be? 3)What am I overlooking? Thanks much.
 
You need 300ma for the relay coil. The TIP3055 has a gain of 20, so you need 15ma of base current. Actually calculate for 20ma to make sure the transistor is saturated.

So the base resistors is calculate to drop the voltage from Vin (the voltage driving to the base) to 0.7V (the typical Vbe of a transistor) at 20ma.

Ohms law:

V = I * R
Vin - 0.7 = 0.02 * R

R = (Vin - 0.7) / 0.02.

If the Vin is 12V, you would need a resitor of (12-0.7) / 0.02 = 565 Ohms.

BTW: The tip3055 is way overkill for this application.

Bob
 
You may be able to use a fet instead of the relay, you do not say what the resistance or the current drain of your heater is. Big fets can have a very low on resistance.
 
You may be able to use a fet instead of the relay, you do not say what the resistance or the current drain of your heater is. Big fets can have a very low on resistance.

I agree.
Another thing is that there is always a voltage drop due to the diode in a bipolar transistor. The voltage drop is more significant than it looks, because the power warming your hands is proportional to the square of the supplied voltage.
I wish I could suggest a decent power FET, but maybe someone else will have a suggestion.
Mark
 
Thanks all. Now I have a place to start. I hadn't considered an FET. I'll assemble the circuit and run some tests. Thanks again.
 
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