A colleague asked me the other day about a circuit I did with a transistor that turned LEDs on and off. This circuit was a tiny part of the board, just a way to save energy and prevent stray light interfering with optical detectors on other parts of the product. 
He asked why we usually tend to use FETs in common source configuration to switch power to LEDs or other loads. Why not connect the load to the source? Why use an p-channel transistor on the high-side? Is there some reason to use a FET instead of a bipolar?
“You can use FET or bipolar,” I told him, “You want the load on the drain or collector. If it doesn’t matter which side is switched, you usually switch the low-side.”
“I know all that,” he said, “but why?”
I have a solid understanding of transistors, but I cannot recall ever learning or working out which transistor type is best for switching power. So for this post I looked up some devices to get to a better answer than just heuristic-based intuition.
Why common-source / common-emitter?
The reason for putting the load to the drain instead of the source is straightforward. Putting the load on the source side of the transistor creates a common drain amplifier, also known as a source follower. This is good for buffering a high-impedance voltage. We want a circuit that approximates a switch, not a buffer.
Is a FET preferable to bipolar?
FETs are preferable because the voltage across them can go lower than a BJT’s. If a BJT’s collector-to-emitter voltage begins falling below 0.3V, the base-collector junction begins to forward bias. This fact means the voltage across the transistor “switch” will always be at least 0.3V, even if the load draws little current. A FET in triode mode (large gate voltage and a small drain-to-source voltage) operates like a resistor (value=R[DS,on]) right down to zero volts. To the extent this voltage drop is kept lower than the V[CE] on a BJT, a FET dissipates less power as heat.
A few minor differences pertinent to switching circuits:
- FETs cost a few cents more. - At 3k piece quantities, an inexpensive SMT FET costs $0.08. A similar BJT costs $0.03. If increased power dissipation necessitates a larger package BJT or a heatsink, BJTs' cost advantage disappears.
- FETs require no bias current at the drain.
Should the transistor go on the low side or high side of the load?
If transistor is being used to control a positive voltage, e.g. to provide sequencing for a part that needs a core voltage to be stable before the I/O voltage comes up, there is no choice. If the transistor is turning on/off an LED or other load not connected to I/O or other supplies, there are some advantages to switching the low side.
- If the transistor is on the high side, the output controlling the switch must be able to go high enough to turn the transistor completely off. This means its voltage must go almost as high as the positive side being switched.
- P-channel MOSFETs are slightly more expensive than N-channel. I used to believe PNP transistors were more expensive than NPN, but a comparison of some low-cost transistors suggests that NPN and PNP transistors provide the same performance at the same price point.
This leads me to the same answer that analog engineers know heuristically: In most circumstances use a FET in common source configuration, and use an n-channel unless there is some reason to switch the high side.