P channel mosfet as high side switch


New to electronics.

Trying to make a arduino monitored solar setup.

Please review my schematic and help me improve it.
Also I am bit confused regarding proper connections of P-Mosfet thus if anyone can confirm it on the schematic.

Thanks a lot in advance.

Your MOSFET connection is incorrect.
See Q7 and Q8 circuit below:

Please review my schematic and help me improve it.

Kinda hard to improve something we know little about.

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Your MOSFET connection is incorrect.
See Q7 and Q8 circuit below:

by this you meant that the BJT should be connected the correct way with the gate and collector in parallel and emitter grounded.
As per your MOSFET diagram it seems i have connected the mosfet source and drain correctly.

Also what I understand (I may be wrong) that

  1. as long as gate to source voltage is sufficiently high/equal, the P-MOSFET will be "OFF" (and conduct from source to drain) and
  2. as soon as the gate voltage is grounded/reduced sufficiently the MOSFET will turn "ON" and stop conducting from source to drain.
  3. Also during the "OFF" stage there should be no practical resistance between Drain and Source.
  4. if the gate voltage is not reduced sufficiently there would be some conductance b/w D and S with lots of resistance and thus heating of the MOSFET.

Please correct me if I am wrong.

Kinda hard to improve something we know little about.

Thanks a lot for the reply.
I am trying to construct a solar power monitor with auto load cutoff at low battery level. the RGB LEDs are for the battery level indication and single LED for Solar IN indication. Will add another LED to indicate load off state.
The ACS modules are for calculating the current at solar/battery/load. the resistor divider is to measure the voltage of solar and battery.
Presently using LM2576-ADJ buck converter (2 in parallel) to regulate the solar panel voltage. will be switching to a more robust 200W 9A commercial buck converter.
I have 100W 12V solar panel, with open circuit voltage at 20v.
Have lead acid 12V battery.

Hope this should clear some of the queries.

Guys please help.

Guys please help.

At this point I'm not sure what you want help with. The glaring problem: not driving that High-Side P-MOS correctly, has already been dealt with, right?

I can offer this:

I would add an over-voltage protection diode on the A0 and A1 inputs. Something like a 1N5817, with the Cathode to the Nano's +5V pin, and the Anode on the Analog Input. That way, if the voltage ventures above the Nano's +5 voltage, the diode will clamp it below the VCC+0.3V MAX input requirement.

Also, I would adjust the voltage divider values to get a full conversion range (i.e. 0-VCC, instead of the 0-1.8V, for the PV input, and 0~1.4V on the BAT input). That way, you'll get better conversion resolution. For instance:

In both cases, the lowest "+5V" voltage, on a Nano, is 4.95V

Also, in both cases, the divider current, when there is 4.95V on the Analog Input, is:

** **I[sub]D[/sub] = 4.95V/10k = [u]495µA[/u]** **

The PV can go to 20V open circuit, so, instead of 100k, I would use:

[b]R[sub]HI[/sub] = (20-4.95)/495µA = [u]30.4k[/u][/b]

The next available 1% value up from 30.4k, is 30.9k

For the BAT R3 value, assuming the Battery voltage will never go above, say 15V, instead of 100k, I would use:

[b]R3 = (15-4.95)/495µA = 20.3k[/b]

And, the next available 1% value up from 20.3k, is 20.5k

Now, all of that, of course, assumes you will be using that input over-voltage protection diode.

But, looking at your schematic hurts my eyes :stuck_out_tongue: , so either post a better resolution image, or ask specific questions. [or, maybe someone else, with better eyes, will fill in ;)[/i]

Also, the BAT voltage divider will be consistently drawing half a milliamp from the battery(ies). I don't think you've divulged the capacity of the batteries you plan to use, so I can't be sure, but if half a milliamp is an excessive drain, consider bumping up the resistances. BUT, if you do that, add a 100nF ceramic capacitor across the Analog Input, so the MCU's internal Sample-Hold can still get a good sample. This applies to any resistance greater than 10k, on an Analog Input.

I know that last suggestion might sound silly. I mean, if your Solar Panel is 100W, that's potentially 8As of current to the battery, and even if you're only storing an hour's worth of energy, at 0.5mA, it would take 2 months to draw those batteries down by 10%, and the battery's internal leakage will probably do worse than that, anyway--unless we're talking a lithium battery, in which case the charge will hold up pretty good. I merely added that last bit as an exercise in considering all possible "use cases".