Are all buckboost converters bi-directional?
A non-synchronous switching converter uses a diode for one of the switches.
Such a converter typically transfers power in only one direction.
(Is there a better term for this than "non-synchronous switching converter"?)
A synchronous switching converter replaces each of those diodes with an actively controlled FET.
Such converters are inherently bi-directional and usually more efficient than
non-synchronous converters.
Do i still need transistors for the switches? Or will the chip do the switching for me.
If you only need 1 amp or so of input or output current,
there are several switching regulator ICs available
that have internal transistors that do all the switching for you.
For example,
you might look at the
LM2587T-12 datasheet figure 13,
which shows how to use that chip to produce 12 V regulated output
from input anywhere in the range 8 to 16 V.
(Alas, it is non-synchronous, transferring energy only in one direction).
If you need higher currents, you're pretty much forced to use external discrete FET transistors for the switches.
The typical circuit is shown on the first page of the
ADP1873 datasheet, which shows a
synchronous (and therefore bidirectional) 10 amp switching converter.
recommend good topologies
Some tips have been collected at http://opencircuits.com/Switching_regulator .
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.
Best Answer
In your schematic, if S1 and S2 are closed the power supply still cannot be shorted. However, the capacitor C would short. The peak current might be high enough to damage your switching mechanism. So dead time is probably still a good idea.