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I have some of these (RGB LEDs with drivers). They connect to 12V, and they draw a max of 1.8A total (full white):

Load

I wanted to add a low-side switch to turn the load on/off via a PIC I/O pin output. After getting some advice in this thread, I created the circuit below:

switch expected

The values above are what I expected the switch to behave like. According to the TIP31C transistor datasheet, Ic=1.8A gives a Vbe(sat) of 1.0V, and a Vce(sat) of 280mV, which apparently shouldn't get very hot:

datasheet

Instead, I'm seeing this:

switch actual

After a few seconds of being on, the transistor starts smoking and is way too hot to touch.

Can somebody please explain why there is a difference between what I expect to see and what I am seeing?

Note: my electronics knowledge is pretty basic, so please assume I don't know much, I won't be offended :). If I've left off any necessary information, please comment. Thanks!

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    \$\begingroup\$ Your transistor is hot because it is dissipating 4v *2A = 8watts. You need to increase the base current to drive the transistor into saturation and thereby reduce Vce. Try reducing the 100 Ohm base resistor to 47ohms and see if Vce reduces. If the PIC is not able to supply more current, use a second transistor and make a Darlington pair. \$\endgroup\$ Commented Oct 2, 2015 at 21:02
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    \$\begingroup\$ Your load must be lower in impedance than you expected. You thought it would draw 1.8 A with 11 V across it, but instead it is drawing 2.1 A with only 8 V across it. Since your load is demanding more current then you designed for, it's not surprising that the driving circuit is overheating. \$\endgroup\$ Commented Oct 2, 2015 at 21:03
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    \$\begingroup\$ More base current! \$\endgroup\$ Commented Oct 2, 2015 at 21:08
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    \$\begingroup\$ Even though smoking transistors is very satisfying, it can be very bad for you in the long run if you develop a lifelong habit. \$\endgroup\$ Commented Oct 2, 2015 at 22:13
  • \$\begingroup\$ I'm sorry to hear you smoked a transistor. I did advise a small heatsink, just in case... I didn't get from your previous question enough details on what you intended to drive the transistor with (PIC). Also LEDs have highly variable resistance. It's not a constant with voltage! Your previous question didn't even say you wanted to drive LEDs. The LEDs you link to have no datasheet whatsoever. Measuring the current through LEDs at 12V is no guarantee for the current at 8V. See what @The Photon said. \$\endgroup\$ Commented Oct 3, 2015 at 5:13

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You'll notice that in Figure 2 it states "Ic/Ib = 10".

This means that to saturate the transistor you should drive it as if the Hfe is only 10, not the almost 100 in Figure 1. Note also that figure 1 is the typical figure not the guaranteed value. The guaranteed value in the data sheet is only 10 at 3A collector current.

You need to provide more base drive.

The PIC however will not be able to drive the required 180mA output. You need a darlington transistor such as a TIP120 or these days a better solution is to use a MOSFET. The TIP120 would have a saturated voltage drop of ~1V and so would require a heat sink. A mosfet would have a very low drop and would work without one.

Even there you have to be careful as many MOSFETs will require 10V drive, more than the PIC can provide. For this there are so called Logic Level FETs that switch with only 5V on the gate.

Something like FQP30N06L would work.

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    \$\begingroup\$ Curses. You beat me to it. \$\endgroup\$ Commented Oct 2, 2015 at 21:13

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