transistor problem with pic.

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atntias

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hi, im using a pic to control pwm of a 12V motor, im using TIP41 npn transistor. which i connected like following: base to output emitter to ground collector to motors ground and motors + to 12V which i have in circuit (im converting it to 12V which 5v regulator.)

anyways im doing this twice once with leds and other with motor, with the leds everything is ok but with the motor - it works yet the transistor heats up alot -> how hot you ask, well i can put my finger on it without burning but still it much hotter then normal, i think.

is it the connection or shuld i just use a "bigger" transistor?
 
I don't think you need a larger transistor. How much current does your DC motor draw?
You need to limit the current from the output of the PIC to the base of the transistor.

*EDIT: Maybe you need TIP120, 121 or 122. They are not larger transistor. Their max. current is almost the same as TIP41, but they have higher current gain compared to TIP41.
 
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i measured the current and i got 1.02 amp it seems allot to me dose it make sense or am i measuring wrong?
 
Motor draws much current.
Let's asy 1 amp, and the current gain is 10. So you need the base current to be 100 mA. But the maximum current sourced by the output pin of the PIC is only 25 mA. You need darlington NPN transistor with a base resistor.
 
You can't use the transistor to the max.
When the motor has no load, the current is 1 A, but it will be higher when there is load. Besides, it needs higher current for the motor to start up.

You can't connect the base directly to the output pin. Just imagine a loop. From the output of the microcontroller, act as +ve supply, across a diode (be junction) then to ground. How much current in this loop? So a resistor is needed.
 
10K will only give you .0005ma's into the base. The current gain is 1000 min, so that will give you a 500ma current, that'll generate a lot of heat. A 1k resistor will give you 5ma's to the base which should saturate it fine.
 
Banana is right though, you're operating the transistor at it's maximum curernt rating and a motor is a dymanic load. You need to know the stall current of your motor as that's the worst case scenario. It's how much current the motor uses when it's output shaft is locked
 
The output of PIC is not exactly 5 V but Vdd - 0.7 V, according to the datasheet. There is voltage drop too across the be junction.
 
yea i found this MJD112 up to 2A, which should do the trick.
thx so much guys.
just to know when would you use Darlington transistor? i mean why here? beacuse the TIP41 also is up to 2A...
 
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You're really pushing that transistor then, you should probably be using one rated for 2 or 3As just to keep heat down otherwise it's not going to last very long. If you want a nice strong drive current just use the lowest value resistor that keeps the output within it's current limit.
 
Using a darlington transistor on large current circuit is never a good idea because of its large saturation voltage.

Take TIP120 and TIP41 for example. At one ampere collector current, the Vcesat of TIP120 is 0.8V while that of TIP41 is only 0.15V. It is worse by over five times, wasting energy as heat.

What one needs is not a darlington transistor but two transistors connect up to make a "darlington" configuration, with the collector of the driving transistor sourcing current from supply rail instead of from the collector of the power transistor.
 
Do you mean this?
The resistors values are from the datasheet of TIP120, not sure how are they defined.
 

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