Electrical – Best way to pair a DC-DC boost converter and a MOSFET to compensate for MOSFET voltage drop

Using a large smoothing capacitor like C1, isolated from the boost output as with R1, may be the only way to use the LMR62014 for this. The LMR62014 has a switch current limit of about \$2A\$, which means that for \$L_o\$ of \$10 \mu H\$, \$V_{\text{in}}\$ of \$3.3 V\$, and \$f_s\$ of \$1.6 {\text{MHz}}\$, the most current delivered to a \$9 V\$ output will be ~ \$0.65A\$. So, it can't support high current pulses on its own. This may be a blessing, as there would be no need to try and get much bandwidth out of the control loop. Just let the bulk capacitance do the job.

Of course, C1 of \$470 \mu F\$ may not be enough depending on how much droop you can stand during the pulse. Using the equation provided by ThePhoton, shows that maximum expected droop for a \$300 \mu sec\$, \$2 A\$ pulse would be about \$1.3V\$, for a part with negligible esr.

It will be required to have some kind of load at the output of the boost at all times, otherwise \$V_{\text{out}}\$ will rise until something limits it by breaking down. A load of \$40 {\text{mA}}\$ (\$R_o\$ of \$225 \Omega\$) would keep operation in CCM if \$L_o\$ = \$10 \mu H\$ and \$V_{\text{in}}\$ < \$4.7V\$.

For best dynamic performance it's undesirable to change conduction modes, which will mean in this case stay in CCM. The CCM/DCM boundary is defined as:

\$L_{\text{crit}}\$ = \$\frac{(\text{dc}-1)^2 \text{dc } R_o}{2 f_s}\$ ; Where \$\text{dc}\$ = \$\frac{V_{\text{out}}-V_{\text{in}}}{V_{\text{out}}}\$

A good reference is "Understanding Boost Power Stages". If you're going to be designing a boost, that's a good place to start.

Given what you have provided, there is simply no way to tell.

On the one hand, your replacement FET has a nominal Rds(on) about 30% lower than the original, which means it should (all things being equal) dissipate about 30% less power. But, of course, not all things are equal. As Tony Stewart commented, the replacement FET input capacitance is nearly 3 times greater. This means that it's entirely possible that the gate driver will not provide a fast enough turn-on and turn-off, and dynamic heating may be much greater. Or maybe not. Without details of the gate drive it's all a big mystery.

And just as a passing comment, beefing up traces with solder isn't that effective, since the bulk resistivity of solder is about 10 times that of copper. You'll get better results if you lay down bare copper wire on the trace and solder that in place.