> -----Original Message-----
> From: James Tu [mailto:james@2-BIT-TOYS.COM]
> Sent: 05 December 2003 11:34
> To: PICLIST@MITVMA.MIT.EDU
> Subject: [EE]: 12 V light dimmer
>=20
> I tried the second method...using PWM from my microcontroller=20
> to control a=20
> MJE3055T.  12 V supply is connected to one lead of the bulb,=20
> the other lead=20
> of the bulb is connected to the collector.  the drain is=20
> connected to GND.

The MJE3055T does not have a drain or source as it's a bipolar transistor.

>=20
> The transistor got _really_ hot and the bulb was hardly lit...it just=20
> glowed a bit.  Do I just need a transistor with a higher amp rating?
> Here are the specs for it...
> VCEO Collector-Emitter Voltage (IB =3D 0) 60 V
> VCBO Collector-Base Voltage (IE =3D 0) 70 V
> VEBO Emitter-Base Voltage (IC =3D 0) 5 V
> IC Collector Current 10 A
> IB Base Current 6 A
> Ptot Total Power Dissipation at Tcase =A3 25 oC   75 W
>=20
> It seems like it should be fine.  60V and 10A.  What am I=20
> doing wrong such=20
> that the transistor is heating up so much?

Check out the hfe (gain) figure for this device.  It is very low (as are
most bipolar power transistors).  The PIC is likely unable to supply enough
current into the base to switch the transistor into saturation.  One answer
is to avoid bipolar's altogether and use a logic level power MOSFET.  These
can be switched on fully by the 5 volt PIC pin.

I would suggest sticking with a PWM approach if you don't wish to have to
dissipate all the wasted power that a linear method produces.

Regards

Mike




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