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?

score 4 · Accepted Answer · answered Dec 17 '22 at 15:10

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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

answered Dec 17 '22 at 15:10

Davide Andrea

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If the load is a circuit board consisting of a variety of components, how would we determine whether this load is resistive or capacitive? – BobaJFET Dec 19 '22 at 15:03
1. Tells what the load is. 2. We tell you whether it's capacitive, inductive, resistive, a combination, or other. – Davide Andrea Dec 19 '22 at 15:57
What I can tell you is that the load has a capacitor at it's Vin pin. Perhaps the second configuration will discharge this capacitor faster? – BobaJFET Dec 19 '22 at 16:04
"What I can tell you ..." is not enough for me to answer your question, sorry. – Davide Andrea Dec 19 '22 at 22:22
Can you maybe give an example of a capacitive load that would benefit from your answer? – BobaJFET Dec 19 '22 at 22:36
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A large DC-DC converter. (Like, 100+ W) – Davide Andrea Dec 19 '22 at 23:03

Davide Andrea · Answer 2 · 2022-12-17T22:25:48.877

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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.

edited Dec 17 '22 at 22:25

answered Dec 17 '22 at 15:01

Davide Andrea

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The other advantages are faster switching (as Andy's answer says) and inverted logic; with the MOSFET turned on, the device is turned off and vice versa, without the need for an additional transistor. – Hearth Dec 17 '22 at 18:01
What constitutes 'low power levels' is subjective and depends on the application. So it weakens your otherwise good answer to put an actual value (100 mW) on it. – TonyM Dec 17 '22 at 19:36
@TonyM If you feel so strongly about it, feel free to edit out that part. I don't mind. – Davide Andrea Dec 17 '22 at 22:01
Oh, I don't feel anything about it, let alone strongly, am very chilled... :-) It's merely a technical observation, I'll leave it with you. – TonyM Dec 17 '22 at 22:09

Andy aka · Answer 3 · 2022-12-17T15:13:39.483

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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.

edited Dec 17 '22 at 15:13

answered Dec 17 '22 at 15:02

Andy aka

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Not really. It depends on the load. Circuit #1 turns off an inductive load faster that circuit #2. – Davide Andrea Dec 17 '22 at 15:21

score 2 · Answer 4 · answered Dec 17 '22 at 15:03

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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.

answered Dec 17 '22 at 15:03

Justme

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To add more context, lets say the goal is to switch off the device as quickly as possible. – BobaJFET Dec 17 '22 at 15:06
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@BobaJFET It might allow fast turn off but it really depends on what the device is, if it can even be turned off any faster. We don't know what the device is. It might have a diode and capacitor which prevents it from turning it off externally, and it might still work for a week. – Justme Dec 17 '22 at 15:12

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

4 Answers4