Background: I'm using a NPN transistor (TIP-122) to switch power (digital, with PWM) to a ceramic resistor/heater. I'm trying to finalize the circuit and move it from breadboard to perfboard. I will want to periodically swap the heater. One heating element draws 0.58A (at 12V), the other will draw 5A (at 5V).
If I choose a resistor for the base such that the transistor is saturated and applying enough current for the high power heater, will it cause damage when used with a lower power output? Or is the base resistor only present to protect the signal input (which is only designed to supply a tiny current)?
Best Answer
It's a very normal operating mode to saturate your BJT transistor with more than necessary base current. The drawbacks are slightly wasted power in the base and slower than normal turn-off, but still faster than necessary for most applications.
In general, you are better off with MOSFETs in these types of applications since the gate could be considered load-independent, save for some very special occasions and cases you are unlikely to encounter. The drawback of MOSFETs compared to BJTs here is that your digital PWM with Aurdoinos and what not tend to be very low voltage, say 3.3 V, and most MOSFETs wants 10 or so V to stay happy where your BJT only needs ~0.7 V.