And while we’re at it, let’s compare a regular transistor (a.k.a. BJT) to yesterday’s MOSFET.
Again a smal test setup, but this time it also needs a 10 kΩ resistor between input signal and base:
The reason for that extra resistor, is that the base of an NPN BJT is essentially connected to ground via what looks like a forward-biased diode. So the voltage on the base doesn’t normally rise above 0.7V. Without current limiting resistor, the transistor would get damaged (and perhaps also the source circuit into it).
Compare this to yesterday’s screen shot and you’ll see that a BJT behaves like a MOSFET, sort of:
The main difference is that the switching point is much lower, around 0.7V – which happens to be just about the point where the base-to-emitter junction starts to conduct.
Here’s the same as X-Y graph (with again the X axis adjusted to 500 mV/div for full scale):
Compared to the MOSFET, the switch-over is steeper, i.e. more like a digital on-off switch. Note also that although the base-to-emitter voltage will be at 0.7V, the collector-to-emitter voltage is in fact below that, almost zero!
What might not be immediately apparent from the above plot, is that a transistor has a much more linear behavior (even if steeper, i.e. with more amplification). In that small range between about 0.65V and 0.75V, it’s in fact a great linear amplifier – which is what transistors were initially used for, and on a huge scale.
A simple way to describe them is that BJTs are current-driven, whereas MOSFETs are voltage-driven.
For a nice article about how to use BJT’s for signal amplification, see this page on the PCBheaven website.
The BJT was at the start of the semiconductor revolution, decades ago. The MOSFET added a new and very different component, perfect for switching enormous loads with amazingly little power loss.
For the dual-voltage supply of a few days ago, either a MOSFET or BJT will probably work. With the BJT, there will be a higher residual voltage – so a check is needed to make sure it switches properly with a feedback pin voltage of only 0.41V. The MOSFET has no such issues, it’s essentially a controllable resistor: no bipolar junctions or diode-like behavior in sight.
Actually, with n-MOSFETs the body (Usually tied to the source pin and channel) is P-type semiconductor, while the channels are N-type (Source + drain). This means that the P-type body and N-type drain channel form a PN diode junction, which becomes forward biased if the drain is more negative that the source. So don’t reverse bias your MOSFETs more than 0.5v or so :)