Electronics > Projects, Designs, and Technical Stuff

Transistors completely fine with base high current

(1/3) > >>

Please anyone expert enough to list or tell which transistors that is completely  fine to have base high current  i.e. no base resistor in order to use it / Q as switch?

Why not just use a MOSFET?

Kim Christensen:
Define "high"....


--- Quote from: abdulbadii on March 31, 2023, 01:11:11 am ---Please anyone expert enough to list or tell which transistors that is completely  fine to have base high current  i.e. no base resistor in order to use it / Q as switch?

--- End quote ---
The transistor datasheet will have a specification figure for the maximum allowable base current. Stick within that specification limit.

Having excessive base current can cause problems with the turn-off performance of the transistor being slower while excess charge is swept out of the base region. Also, remember to take account the power dissipated in the base emitter junction when operating at high base currents.

If you feed the base of a small signal transistor from a supply line, unless you're running on a coin cell, you'll smoke it up even from a couple AA cells. So in general the number of such transistors is zero. In general - there are exceptions.

If you feed the base from an MCU pin, or from a CMOS output - those inherently act as current sources. Except, you then need to worry not about the transistor, but about the dissipation in the GPIO driver in the MCU.

Assuming - I apologize if I'm wrong - that such questions typically originate from beginner MCU aficionados, I'd say you'll probably overstress the MCU first. Each transistor will be dissipating the same power - about 15-50mW just from the base. Pretty much any small-signal general purpose type will not even notice that too much.

But the MCU will be dissipating progressively more the more transistors you attach that way. Even with just one transistor the 3.3V (I/O) MCU will be dissipating almost 4x more power than the base junction of the transistor. With high-drive 3.3V GPIO, the GPIO output structure will be dissipating in the ballpark of 100mW, and there will be as many of those dissipating as there are transistors driven on. So, drive 10 transistors, and suddenly there's a tiny fraction of a mm2 on the MCU that will be dissipating 1W of power. That becomes somewhat problematic for longevity and such. Relatively huge localized heat loads on the die.

The transistors so driven will be saturated good and proper and they'll take an order of magnitude longer to turn off than they did to turn on. That may be of no consequence, of course, just something to pay attention to.

Abusing MCU pins as current sources is better suited to LEDs, although even then you should set the drive level to the lowest available in the GPIO configuration, and still make sure you don't exceed the maximum VCC(IO) current or else the bond wire(s) may be the fuse. At least the LED will be dissipating more power than the GPIO output stage.

I still question the question, though: if you think those base resistors are somehow inconvenient to either calculate or mess with... just buy transistors that have them built-in! Problem solved.

In any case, the stouter the transistor, the more unhappy whatever is driving it will be about standing between the positive supply rail and the diode to ground that the B-E junction is in such cases. Power transistors can take a couple amps through the base. If you connect *all* MCU output pins to the base, emitter to 0V, and turn all the outputs on - the MCU will not last very long. The transistor won't even get a chance to get properly warmed up.


[0] Message Index

[#] Next page

There was an error while thanking
Go to full version
Powered by SMFPacks Advanced Attachments Uploader Mod