How To Improve Boost DC/DC Converter?

Here is another application note from Vishay Siliconix for your reference.

Thanks again. However, now I face difficulties in my MOSFET. Mind to give some advice regarding this?

I tried to turn on the mosfet by doing the following. With the circuit as shown constructed, I measure the voltage drop across the Drain and Source and I obtain 5V. But I'm not sure if Im doing the right thing as even though I remove 5V to the Gate, I still obtain a reading of 5V if I were to tap my multimeter on the Drain and Source...

Mind to comment about this? Please? Is it trut that I'm suppose to read any values if I remove the 5V source for the gate?

By the way, the MOSFET used is a logic MOSFET, IRL2703.

The circuit is correct, you should measure 0V.

Since the MOSFET does not turn on, it means there could be a wiring problem or a defective MOSFET or gate resistor. Do you measure 5V between gate and source? If not, the resistor is open or you have a short from gate to source (or the MOSFET got shorted G-S, which can happen easily if you mishandle the part).

Removing the gate connection can cause the MOSFET to turn on or off (or partially on), depending on the charge that gets transferred to the gate due to leakage effects: dirt, flux on PCB, your touching the gate, etc. because the G-S impedance is extremely high.

Never leave the gate of a MOSFET open, since it can easily get damaged. The G-S capacitor can only withstand about 30V max, therefore it can easily get damaged from ESD or leakage from other parts of the circuit running at higher voltages (higher than 30V).

I just realized, looking at your other post, about the buck-boost, that you got the pins of the MOSFET wrong: the drain is in the middle, so you swapped the gate and the drain!
Make this modification and if your MOSFET is still intact, everything should be OK.

Hello, I am a university studying doing my final year project on Inductionless DC-DC Converter. I am trying to investigate the efficiencies of the DC-DC conversion. I understand that there are chips which offers efficiencies as high as 95%. But from a theory perspective it is impossible to gain such a high efficiency, is there anyone who understand this and could give me a hand to how companies achieve this efficiencies.

Thank you so much.

To comemon1010,
DC DC conversion --> you mean switching regulator or linear regulator.
For switching regulator, the theotical efficiency is 1 assuming no ESR (on cap/inductor/voltage source and power switches).
However, in reality, it is not the case.
Efficiency = Pout/Pin

It has some non ideal effects (Not all are listed)
1. switch voltage drop
there is a voltage drop across diode and there is current passing through when conducts --> power loss
2. ESR effect on ripple voltage --> often produce greater voltage ripple --> power loss
3. ESR on inductor --> Iavg pass through the inductor --> power loss.
4. Switching loss
Solutions to 1-4
1. ppls using synchronous MOS swich control to replace diode
2. use CAP with smaller ESR
3. use IND with smaller ESR
4. One way to reduce switching loss is to modify the circuit to make switching occur at zero voltage and/or zero current

With helps in 1 - 4, some company likes linear tech/ maxims can achieve DCDC switching converter up to 95% efficiency.