Electronic – Power Dissipation of Schottky Diode

power-dissipationschottky

I am implementing a circuit as shown below :-

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Source is a 15V Solar panel & Load is the Battery. The intent is that if someone tries to connect a bigger battery on the output, Diode(Schottky is what coming to my mind) shall block reverse flow being reverse biased.

In normal scenario, all the current will pass through forward biased schottky diode. The rating of the system is 15V/20A. Upon looking into few schottky diodes available, which have a forward voltage drop of about 400mV, the power dissipation is coming out to be Vf*If = 0.4V*20A = 8W which is a lot of power to dissipate.

Am I using the right approach?? Any suggestions here.

Best Answer

Am I using the right approach?? Any suggestions here

Maybe you should consider the standard MOSFET circuit (but modified) that acts as a very low volt drop diode i.e. it protects a circuit from reverse polarity but turns the MOSFET on so that forward volt drop is barely a few milli volts. The basic circuit is this: -

enter image description here

+Vin is where you would connect the solar panel and +Vout is the battery connection. If you add a bipolar transistor and base resistor you get this: -

enter image description here

From a quick simulation, the BJT turns off the MOSFET when the battery voltage exceeds the SP voltage by about 730 mV (R3 is 10k not 100k). Between the voltages being equal and the battery being 730 mV higher there will be some hundreds of mA flowing back to the panel. However, if you are trying to protect the panels from the wrong battery being put in place (i.e. 24V instead of 12V) then it should do the trick.

It's just an idea and not proven other than by a quick sim. Caveat emptor!!