Is there any particular reason why I never see the MOSFET between the cathode of the LED and Vin? Assuming a Vin of 12V, and a 5V logic signal, should I expect the MOSFET to not conduct from drain to source? Is this why the switch is typically located between the anode and ground, after the load does its business?
Seems trivial, though I never thought the sequence of components in series mattered.
ablarry91:
Is there any particular reason why I never see the MOSFET between the cathode of the LED and Vin?
I think you mean "between Vin and the ANODE of the LED". Cathode is the negative side: ground.
Yes. It is easy to use a logic-level N-Channel MOSFET as a low-side switch. Put it between the load and Ground and connect the data pin to Gate through a current limiting resistor.
To use a MOSFET as a high-side switch you need a P-Channel MOSFET and you need to raise the gate to Vin to get it to turn off. You can do that with a pull-up on the gate and an NPN transistor to drag it to Ground when you want to turn it on but that's two more components.
Post a schematic or photo of a hand drawn schematic and then refer to it using the component labels.
Is there any particular reason why I never see the MOSFET between the cathode of the LED and Vin?
Maybe because it wouldn't turn on if it was reverse biased. (with the cathode connected to Vin and the anode connected to the Mosfet drain)
The mosfet sinks the led current so the led cathode is always going to be connected to the Mosfet Drain and the mosfet Source is always going to be connected to ground.
ablarry91:
Is there any particular reason why I never see the MOSFET between the cathode of the LED and Vin? Assuming a Vin of 12V, and a 5V logic signal, should I expect the MOSFET to not conduct from drain to source? Is this why the switch is typically located between the anode and ground, after the load does its business?
Seems trivial, though I never thought the sequence of components in series mattered.
To conduct the gate needs to be +5V w.r.t. source. When the MOSFET is on the drain
and source are at the same voltage (nearly). If the drain was connected to +12V you'd
need +17V on the gate, rather than +5V. This is called source-follower configuration
and is not normally used for switching (*) due to lack of a 17V supply!
(*) Actually it is commonly used for switching in most industrial applications, but this
requires use of bootstrapped floating gate-drivers that work at the same potential as
the source, which itself is pinging up and down as the device switches. Doing this
allows an H-bridge or 3-phase inverter circuit to use all n-channel devices which are
superior to p-channel devices (about 3 times better carrier mobility).
Yes exactly - I would expect at least 12A and around 10V to call this high current.
Is there any arduino working with 12A mosfets around? Possibly mosfets that are DIL form.
