What are the disadvantages of switching off a load with a shunt MOSFET?

Question

Normally, if I want to use a MOSFET to supply power to some device and switch it on/off with a logic signal I'd use this very typical MOSFET as a switch configuration:

simulation

When the switch closes, the MOSFET allows current flow, thing turns on. Nice. Here's another example with a lamp:

I've come across a design like this:

simulation

Here, when the switch closes, current gets pulled to ground, turning the thing (R_L) off. This is pretty much like an I2C data bus where we have a pull-up resistor and a transistor that pulls the bus low, only the goal here is power output, not logic.

I'm interested in why someone would want to switch off a device in this way, and also any drawbacks/flaws. Is this just a bad design in the first place?

One obvious drawback you can also see in the simulations is that R_L gets less voltage in the second configuration - just 8V. Assuming 8V is enough for R_L, what other problems could we have here?

Answer 1

the goal is to switch off the device as quickly as possible

Depends on the load:

• Resistive load: no difference
• Semiconductor load: no difference
• Capacitve load: the shunt circuit (second circuit) is faster, but you need a small resistor is series to limit the current
• Inductive load: the series circuit (fist circuit) is faster, but you need a high-voltage TVS diode across the transistor to absorb the high-voltage kick-back

Answer 2

The second example is inefficient:

• When on, the load only sees part of the 12 V.
• When off, current is wasted from the 12 V source

The only time the second circuit makes sense is at low power levels (for example, < 100 mW). In that case, the advantage is that you can control a ground-referenced load with a ground-referenced signal using a single transistor. The alternative (a "high-side switch") is more efficient but it uses two transistors.

Answer 3

You can always put a switch in series or parallel to a load, whichever makes more sense and does not cause a short circuit.

In your latter case, when switch FET is off, load gets current via the series resistor, and when switch FET is off, it simply short circuits the load so it shunts all the current through the switch.

Depending on if that is a good or bad way depends on what you want to achieve it and how. The real difference is that there will always be current flowing and the circuit consumes more power when switch is on and load is off.

In some cases it may be an advantage to do it like that.

Answer 4

If you want to deactivate the load very, very quickly then circuit #2 is what you need to go for. If you don't care about deactivation speed then circuit #1 is best because it wastes no power. Another disadvantage of circuit #2 is that the voltage appearing across the load is load resistance dependent. In other words, regulation may be an issue and you might end-up going for a high side switch and a low-side shunt.