Electronic – If I step up 9V to 12V with a boost converter, what happens when the input goes to 12V

When working with a boost regulator, the first thing you want to know is the critical inductance and critical current. Critical current defines the boundary between continuous conduction (CCM) and discontinuous conduction (DCM) in the inductor. Circuit dynamics are very different between the two modes of operation, and you want to be in one or the other mode.

Critical current for the boost is approximately:

\$i_{\text{crit}}\$ = \$\frac{V_o T_s}{16 L_{\text{crit}}}\$

In this case with L = 1.8uH, \$V_o\$ = 10.5V, \$T_s\$ = 1uSec; \$i_{\text{crit}}\$ would be about 0.37 Amps. Normally the load current is 0.2 Amps, but pulses to 1.2 Amps. That's bad.

During the pulse the regulator goes from DCM to CCM adding a pole the the modulator response.

• If the regulator is compensated for DCM, the move to CCM will make it unstable and it will oscillate.
• If the regulator is compensated for CCM, operation in DCM will likely be stable, but the transient response will be very poor.

To keep the regulator in CCM mode at a current of 0.15 Amps an inductor of about 4.7uH would be needed.

Another thing to keep in mind is that switching regulators have negative input impedance. This means that if the impedance of the source voltage is equal or greater than the input impedance of the regulator, the system will oscillate until it runs out of regulation range. In this case with about 15W of input power from 6.5V, the input source impedance needs to be less than about 1.4 Ohms everywhere below the loop crossover frequency (including any LC resonances). Looking at the input voltage variations in the pictures, it's not clear that the supply source is up to that.

The main power losses in a boost converter can be summarized as follows:

1. Power switch switching losses (e.g. MOSFET, BJT. Hereafter I will refer to the Power switch as the MOSFET)
2. MOSFET conduction losses.
3. DIODE switching losses.
4. DIODE conduction losses.
5. Other conduction losses (e.g. inductor resistance)

The efficiency of a converter is given by: eff = Po/Pin = (Pin - Plosses)/Pin.

As the losses change the efficiency therefore changes.

One can not make a blanket statement as to why the efficiency reduces or losses increase as the duty cycle increases because then one would need to know all five loss parameters as a function of current, voltage and switching frequency.

However, a simplified explanation of this phenomena is that the MOSFET conduction losses are unequal to the diode conduction losses. As the duty cycle increases, the MOSFET will conduct for a longer period and the diode for a shorter period. This in turn alters the power losses in the circuit. If the DIODE happens to have higher conduction losses than the MOSFET for example, then as the duty cycle changes, causing the DIODE to conduct for a relatively longer period than the MOSFET, then the efficiency will decrease. This is a simplified explanation, but the main principle is that as you change the duty cycle, the operating conditions for each element in the circuit change. Since the losses for each device depend on it's specific operating point, then changing the duty cycle changes the losses.