Is driver necessary to drive P-FET Buck Converter using Microcontroller?

Question

I'm building a buck converter to charge an ultracapacitor from Hydrogen Fuel Cell. I'm having some issue with the gate switching for the converter.

Currently I'm testing the buck converter with a power supply, function generator and one power resistor as load.The buck converter I built is as below,

There are two 2ohm resistors at output side to prevent short circuit current to the supercapacitor. The converter is behaving like regular buck converter below 14V. But once it passes the 14V zone, the current will increase drastically and a lot of heat can be felt from the P-FET and Charging Resistors.

Is there any mistakes in my designs? I have listed the elements I installed on the converter below. The function generator is generating 40kHz, 5Vpp,2.5V offset and 50ohm load settings

Coupling capacitor: 100V,680uF

Inductor: 100uH, Saturation Current, Saturation Current 20.6A

P-FET: ON-Semicondcutor FQPF27P06

Diode: 1.3V Vf

Answer 1

The FQPF27P06 MOSFET is off (high resistance from source to drain) when the gate voltage is >=0V (i.e. positive) relative to the source. The MOSFET is on (low resistance source to drain) when the gate voltage is sufficiently negative relative to the source. The MOSFET resistance from source to drain decreases when the source-gate voltage is more than 4V (i.e. source is positive relative to gate) (also see FQPF27P06 datasheet, gate threshold voltage). As the source-gate voltage increases beyond 4V (up to the specified max = 25V; do not exceed!) the source-drain resistance decreases, becoming the minimum possible at 25V. However, little is gained by driving the gate more than 10V negative relative to the source; i.e. no need to apply full 25V. You must arrange your circuit to provide the indicated voltages for off and on operation of the MOSFET. Since V1 can be as great as 30V, it is unlikely that your function generator can directly provide the required voltages: relative to V1-, MOSFET off when gate = +30V to +55V (30V+25V) and MOSFET on when gate = +20V (30V-10V) to +5V (30V-25V). Thus, additional circuitry is likely needed to achieve the required gate voltages.

EDIT:

Designing and then building a switching DC-DC buck regulator is not a trivial effort! I have decades of design & fabrication experience, yet I would still purchase an off-the-shelf I.C. for such a design and I strongly suggest that you follow my advice. Ultimately your cost would be less and success more likely. Most manufacturers of DC-DC regulator I.C.'s provide example circuits either in the datasheet or in application notes. Your first attempt should be to follow such a design. Also, please search StackExchange for many other discussions relevant to your effort. An approach far more likely to lead to success would be to buy a complete DC-DC converter module that meets your requirements.

If you wish to continue your present course as a learning experience, then you have much to learn. Before you achieve an acceptable design you will almost certainly destroy some components, raising your costs and you will waste much time that you could have better spent on other aspects of your design.

In my opinion, the MIC5021 is not a wise choice for your current circuit for many reasons that I will not attempt to describe here.

Good luck. I hope that you decide to direct your efforts toward a less difficult goal.

Answer 2

Without feedback, the current through L1 might be increasing from cycle to cycle until it saturates and everything gets hot.

A flyback convertor can be connected to a fixed signal generator, because the inductor fully discharges and current resets to zero every cycle.

Buck convertor current doesn't reset to zero, just starts climbing from where it left off, so without feedback, you see a sawtooth that rises higher every cycle.

This also means that software switched flyback converters are easy to implement, but buck convertors are not.