Designing a Low-Side Switching Buck Converter for 100V Solar to 48V 8A Load with MIC5014

I am designing a buck converter to be used to power a load from a solar panel (100v OCV max). The load is 48v, 8A maximum. I have been trying to simulate this circuit to get an idea of the voltage and current ripple using NI MultiSim, but the transient analysis often fails or is total nonsense. I have found plenty of resources for calculating the values of my components, but often they are not for the low-side switching I am trying to do. I have the mosfet, schottky diodes, and inductor already, but I could get different ones if need be. I have selected the MIC5014 high/low side gate driver to be controlled by a 5v pwm signal. I am looking for design resources, as well as some direction on how I can determine if these parts will indeed work together. I would normally tinker with the parts, but I am afraid of damaging the load and/or myself with such high power. I also have little experience with generating high frequencies (I am using an arduino at the moment.) Any help is much appreciated!


Note: 12ohm load is shown, as well as a voltage sense circuit with a RC filter. I plan to be able to sense the voltage across the load this way in order to regulate it via a microcontroller. 

Hi Justin-That's kind of a strange circuit, with neither side of the load connected directly to either side of the voltage source.  There are many problems with that circuit, one of which being that it will be extremely difficult to measure the voltage across the load for feedback purposes.  I think there are much simpler ways to build a buck converter. Take a look at Webench Power Designer from Texas Instruments.  I plugged your requirements into their calculator and it came up with 10 designs.-Rick

Justin,
I agree with Rick that the circuit is a bit strange. However, I did a search and found this similar circuit.
https://electronics.stackexchange.com/questions/330471/low-side-n-mosfet-buck-converterThis circuit has the diode and inductor in different positions relative to the load and the switch. it might be worth trying this arrangement to see if it works better.

You need a diode D4 in series with V4, your 100 V solar panel, with D4-anode connected to V4 positive end, to prevent reverse current into the panel. Without D4, D2 cannot ever conduct.I would recommend not using this circuit. Having your switch between the ground end of the solar panel and the load, means that the uncontrolled capacitance between panel and load is across the switch. There are also EMI and safety hazards.
It's better and safer to build a low-power floating drive for MOSFET Q2, so you can tie the ground ends of panel and load together.

A single solar cell can be modeled as a current source (with the current proportional to the product of the cell area and the intensity of sunlight), in parallel with a forward-biassed diode. A solar panel can be modeled as a current source in parallel with a string of series-connected forward-biassed diodes (one diode for each solar cell in the panel's series string). The short-circuit current is only a little larger than the maximum-power current, so solar panels don't mind being shorted. This permits unique switching topologies, where the switching MOSFET is in parallel with the solar panel.
I have seen simple solar lead-acid battery chargers with the MOSFET in parallel with the solar panel. The battery is fed through a diode: When the MOSFET is off, the panel current goes to charging the battery. When the MOSFET is on, the diode prevents the battery from discharging through the MOSFET.
For the design here: Move the bottom end of R4 and its parallel 100 uF capacitor from the drain of Q2 to ground. Move the drain of Q2 to the positive end of V4, placing Q2 in parallel with V4. Remove D1 and D2, as they will never conduct. Put some resistance in series with the capacitor across V4 to limit transient current when Q2 turns on.
When Q2 is off, the current flowing from V4 through L1 to R4 increases over time, increasing the voltage across R4. When Q2 is on, the current flowing through L1 decreases, decreasing the voltage across R4.

Correction: Remove D1 because D1 would wrongly conduct when Q2 is on.

You may want a power zener diode across R4, with zener voltage a little more than the maximum voltage expected across R4, to take the load current in case R4 opens.