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Topic: Redundant solar/charging/battery/supply (Read 1 time) previous topic - next topic


Dec 12, 2017, 04:56 am Last Edit: Dec 12, 2017, 05:08 am by MorganS
but why not use an N-channel?
Good question. It took me 3 attempts to come up with this answer and I'm not sure it's the complete answer...

The N MOSFET is switched on by a positive VGS. That's the voltage between the gate and the source. The voltage on the drain is not important unless it gets high enough to destroy the device (the damaging voltage may be as low as 20V.)

If used as a high-side switch, the N MOSFET will have its source connected to the load. (The names are somewhat backwards because the source sources negatively-charged electrons.) If you've got a fully-charged 3.7V battery, the source will also be about 3.7V. With a 5V control voltage on the gate, VGS is only 1.3V. That's not enough to properly turn it on, even with a logic MOSFET.

I have seen many datasheets for things that use an N MOSFET as a high-side switch. Those devices usually have a voltage-booster circuit that gives about 10V above the supply voltage, which is enough to fully turn on a regular MOSFET.
"The problem is in the code you didn't post."


Dec 12, 2017, 02:44 pm Last Edit: Dec 12, 2017, 03:05 pm by cgorton
jremington: I am showing an N-MOSFET in my schematic. I think I have it hooked up correctly. My symbol looks different than yours, but my understanding is that battery is hooked to the source, the drain is hooked to the booster. That was my intent. And this booster has two grounds, but they are connected on the board (I checked with my multi-meter) so I just show one grounded.
10K between gate and source, got it!

morganS: that makes sense. I'll have to re-think using the N MOSFET.

I'll re-work this. Thanks again for the input.

Oh- jremington, the stall current is 1.7A. I'll need to look at that again.


Dec 12, 2017, 04:41 pm Last Edit: Dec 12, 2017, 04:44 pm by jremington
An N-MOSFET won't work as a high side switch, for the reasons that MorganS gave.

The voltage difference between gate and source controls the current flow.


Dec 13, 2017, 04:06 am Last Edit: Dec 13, 2017, 02:40 pm by cgorton
Having watched this, I think I get it:

The resistor between the gate and source (battery V+) keeps the gate at the same voltage as the source, and therefore keeps the switch off. When I pull the gate to 0V with the Arduino pin, it switches on the power to the booster and driver.

And I see I need the P-channel because with an N-channel I would not have enough Vgs.

I'll re-draw one more time tonight. :)

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