Use TL494 that has a reference and feedback loop to stabilize the voltage.
http://www.ti.com/lit/an/slva001e/slva001e.pdf
If you can use microcontroller, use its PWM channel and ADC to achieve this.
Any MCU can be used. For example, AtTiny series from Atmel.
You need to program the MCU with what you want.
MCU can also help you achieve temperature compensation of the charging voltage and many other things e.g. over-voltage and over-current protection.
Contrary to what has been said above, 555 is not unreliable and not inefficient.
It can be used as a regulated booster than a simple booster (as it is in your case) but you need to modify your design.
In your current design add a fixed reference voltage source and an op-amp. Let op-amp compare the voltage at the output to the reference voltage. If reference voltage is 1V, you need to divide the output voltage by 14.4 (using voltage divider network) and then let op-amp compare the voltages. Output of the op-amp should manipulate (increase or decrease) the duty factor so that the output voltage is fixed at 14.4V regardless of load or input voltage variations.
It looks like a perfectly normal DCM (discontinuous conduction mode) waveform to me.
The lowest, straight segment is when the switch (to ground) inside the LM2588 is conducting, charging the inductor with current.
The next segment, the high straight one, is when the switch has cut off, and the inductor is now dumping its energy into the output capacitor(s) through the diode.
At the end of that segment, the energy in the coil is exhausted — i.e., its current drops to zero. There is now a reverse bias across the coil, and the current tries to flow in the other direction, but this is blocked by the diode.
Now you have a coil that is open-circuit at one end (neither the LM2588 nor the diode are conducting), but its internal distributed capacitance is charged to Vout - Vin. The next thing you see is the self-resonance of the coil dissipating that energy in the form of a sinewave (superimposed on a DC bias equal to Vin). This waveform decays at a rate determined by the Q (quality factor, or internal resistance) of the coil. This continues to the start of the next switching cycle.
The sinewave part of the cycle represents a relatively tiny amount of energy. However, since it occurs at a single specific frequency, it can be an annoying source of EMI. A damping resistor or an R-C snubber can be used to suppress it if necessary.
The self-resonant frequency seems to have a period of about 1.4 µs (714 kHz). Given this, and the inductance of 390 µH, you can infer that the effective distributed capacitance is about 130 pF.
Best Answer
Almost all boost converters have a feedback loop to control the output voltage. This means that there is no voltage gain, the output voltage is independent of the input voltage (as long as it is withing practical limits).
Perhaps you are thinking of this type of circuit:
simulate this circuit – Schematic created using CircuitLab
The voltage gain of this circuit without any load is infinite ! Why is that so ? When S1 is closed the electrical energy from the battery charges the inductor L1. Yes an inductor can be charged with current ! This build up a magnetic field in the inductor. When S1 opens the inductor wants to keep the current flowing. So the current flows through D1 into C1. Since L1 behaves as a current source when it is charged but S1 is open, the voltage at the anode of D1 will simply reach the value that is required to make the current flow. So with ideal components the voltage will reach an infinite value.
Switching frequency has nothing to do with this, a lower switching frequency means that S1 opens and closes faster meaning L1 is charged less per open-close cycle.