Efficiency about boost mode regulator connected to a supercapacitor

I am doing the math modeling of the regulator's efficiency and hoping someone can help me.

What I am studying is a boost mode regulator by Linear. Its datasheet gives an efficiency chart as shown. I guess the efficiency should be a function of V_in, V_out, I_out, Namely, E = f(V_in, V_out, I_out).

1. Suppose we connect a constant voltage source providing V_in to the regulator's input, and connect a super capacitor (whose initial voltage is within regulator's rated output voltage [V_out_min, V_out_max]), so V_out grows as the charging process goes on. Then what will be I_out? Will this always be the regulator's maximum rated output current?

2. Suppose the constant voltage source providing V_in to the regulator's input is replaced by a PV cell which has its I-V characteristics I_in = g(V_in), and the super capacitor configuration is the same as above. How can we determine I_out over time?

What I am thinking is, we have the following relations hold:

I_in = g(V_in);
E = f(V_in, V_out, I_out);
E = (V_out*I_out)/(V_in*I_in)
V_out = Integral(I_out)dt/C + V_out_0;

There are 5 variables (E, V_in, V_out, I_in, I_out), and only 4 equations, I feel that if I_out is fixed, we can solve all variable.

If you're concerned about efficiency from a solar cell, you should probably be trying to use a MPPT controller. Using a simple voltage regulator is going to make it though to get good efficiency, especially when the capacitor voltage starts out very low. In that case the boost converter is going to hit its current limit, which may be far more than the MPP of the solar cell, causing the cell voltage to collapse and getting unstable behavior.

Also if you want to maximize efficiency for a low voltage boost converter, you might look at getting a synchronous converter.

The chart that you have attached presents efficiency dependance on output current while all other variables you have mentioned are fixed (nominal values).

1. If you just connect the supercap on the output it is equivalent to not having a load at all, I_out will depend on many things and it will not be constant. Approximately It will be zero most of time, and it will follow inductor current while the synchrinous rectifier is ON inside the converter. You should add a load and measure the power delivered to the load. Be carefull that your load is adequate or to also include the voltage change on capacitor (it is also energy) in efficiency calculation.

2. mtweig is right. Matching internal resistance of the energy harvester is more important than the actual efficiency of your converter. Try to use some of the chips developed for such applications, for instance LTC3108.

In a boost converter, the current bursts do not need to go through the regulating IC.

During mosfet switch-On, power supply current goes straight through the mosfet and coil only.

Generally speaking, the current bursts are always the same amperage, for a given frequency and duty cycle.

During switch-Off, the coil generates current. This current must go somewhere. It goes through the diode to the output stage. The current bursts are pretty much the same amperage. They get shorter as the output voltage gets higher.

The supply voltage is added to the coil's emf. The coil will develop whatever level of emf is necessary, in order to overcome the charge level on the capacitor.

Theoretically your capacitor can eventually charge to hundreds of volts, if there is no load on it.

These are my observations from simulating boost converters.