What is the difference between synchronous and conventional dc-dc boost converter? which one is best and why?
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\$\begingroup\$ As a hobbyist only, I believe that synchronous is more efficient since the gated conduction drop can be less. In practice, it's not so clear to me. So I think that unless you are a designer of such ICs, stop worrying about that detail. Just look at all your other criteria and pick what works better for you. But I may learn something from answers here, too. \$\endgroup\$– jonkCommented Aug 4, 2017 at 19:59
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\$\begingroup\$ @jonk At light loads a diode can be more efficient than sync rectification since you have the gate drive losses in driving the FET. So many sync rect controllers have a "diode emulation" mode where they stop switching the sync FET when the load gets below a certain threshold and let the body diode conduct. \$\endgroup\$– John DCommented Aug 4, 2017 at 20:11
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\$\begingroup\$ @JohnD That comports fully with my prior notions. (I had edited out a phrase "in theory" as part of "is more efficient", which now looking back I should have left in.) That's why I added "just look at your criteria" and told the OP to let things flow out of other design details and stop worrying about the specific mechanism. \$\endgroup\$– jonkCommented Aug 4, 2017 at 20:25
4 Answers
Synchronous rectification refers to the practice of using an active element such as a MOSFET switched at the appropriate instants in place of a diode. Here's a simplified example of a boost converter:
simulate this circuit – Schematic created using CircuitLab
Not shown here is the logic to switch the transistors. In the synchronous case it's important that M2 and M3 are never both on at the same time, as this would short C2 to ground, possibly damaging the transistors, and severely decreasing efficiency.
The advantage is the conduction losses through a MOSFET (M3) may be less than through a diode (D1). Consider the voltage across D1 will be about 0.6V, or perhaps 0.2V if current is low enough that a Schottky diode is feasible. However, a properly selected MOSFET in many circumstances may have a VDS even lower, and thus lower losses.
The disadvantage is higher complexity and cost.
The distinction is actually about rectification on the output/secondary side of the converter. Synchronous rectification means that a transistor (normally a MOSFET) is used to control the current flow for minimum loss.
Asynchronous rectification refers to the use of one or more passive diodes to control the current flow.
The latter is simpler, and quite effective at low power levels. At higher power levels (to a point), the lower VDS of a full-on MOSFET results in lower losses.
At very high power levels (high current levels), BJTs, IGBTs or other types of devices may have lower losses than MOSFETs.
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\$\begingroup\$ It all depends. On a buck converter the diode does not carry that much current so something like 200W buck converter could very well work just fine with async topology. Not so much with boost topology! \$\endgroup\$ Commented Aug 4, 2017 at 20:07
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2\$\begingroup\$ @Barleyman, I don't think that's correct. The diode or MOSFET on the low side in a buck converter has the same peak currents and the high side device. You cannot change the current in an inductor instantaneously so when the high side devices opens, the same current that was flowing through the high side flows through the low side. See current's Ia and Ib in the diagram on Figure 2 of this webpage: analog.com/en/analog-dialogue/articles/… \$\endgroup\$ Commented Aug 4, 2017 at 20:21
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1\$\begingroup\$ @Dave Tweed, good answer. Could use some diagrams to make it clear to OP. Technically though it's not "high power" that leads to higher losses for asynchronous design, it's having to handle high current. \$\endgroup\$ Commented Aug 4, 2017 at 20:23
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\$\begingroup\$ @VincePatron all right, let's say average current then. On a low step down ratio your mosfet is going to carry most of the grunt work. As I said it depends on your application. On a high power boost topology the rectifying diode is your biggest headache power dissipation-wise. \$\endgroup\$ Commented Aug 4, 2017 at 20:53
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\$\begingroup\$ With power semiconductor current carrying capacity usually far outlies the power dissipation capacity. You may be perfectly fine with 20A peak current but your part will die in fire with much less average current unless you have vigorous cooling solution in place. \$\endgroup\$ Commented Aug 4, 2017 at 21:01
The term synchronous rectification (abbreviated sync rect) implies the replacement of a diode by a controlled switch, usually a MOSFET. This transistor can be self-driven, meaning its \$V_{GS}\$ is naturally self-generated by the converter (an auxiliary winding for instance) or it requires an extra driving circuit to generate the appropriate signals. In continuous conduction mode (CCM) where the inductor current never returns to 0 within a switching cycle, the control of the sync rect can be complex to avoid shoot-through (both switches are on during a short period of time). This is even more complicated with isolated designs since the main switch lies in the primary side while the controlled switch is in the (isolated) secondary side: a second signal is usually needed (but not always) to ensure minimal shoot-through.
In discontinuous conduction mode (DCM), the diode naturally turns off (the inductor current cancels withing a switching cycle) and the control of the sync rect is greatly simplified. Some converters such as those operated in boundary mode (at the boundary between CCM and DCM), lends themselves well to sync rect control because they can't enter CCM.
Sync rect controllers observe the drain-source voltage of the sync rect MOSFET. As its body diode conducts first, the drain-source voltage spontaneously drops and later on, the MOSFET is turned on: turn-on losses are small considering the zero-volt switching (ZVS) action. Drive losses also benefit from this ZVS action as the Miller effect is greatly reduced. When the current decreases in the MOSFET, the voltage tends to go down as well and as it passes a given threshold, it signals the control chip to turn the MOSFET off. This is a difficult phase because you want to extend the conduction duration as long as you can (to benefit from the low \$r_{DS(on)}\$ but, on the other end, you need to avoid shoot-through losses. Finally, parasitic inductance corrupts the signals by extra offset and ringing implying rigorous PCB layout rules (shortest paths, smallest areas etc.) to avoid false turn-off.
The conduction losses of a diode depend on its dynamic resistance \$r_d\$ and its threshold voltage \$V_{T0}\$ (\$\approx\$0.6 V for Si and \$\approx\$0.4 V for a Schottky). The resistive term is sensitive to the squared rms current while the dc source is affected by a the average current: \$P_d=r_dI_{rms}^2+V_{T0}I_{avg}\approx V_fI_{avg}\$ where \$V_f\$ is the diode total drop at a given operating point (extracted from the I-V curves). You thus realize that the diode dissipation is related to the average current flowing in its junction (considering a low ripple current of course).
For a MOSFET, the conduction losses are classically given by \$P_d=r_{DS(on)}I_{rms}^2\$ where the term \$r_{DS(on)}\$ represents the on-resistance at the highest junction temperature (100 °C or so). This where you realize it is sometime hard to beat the classical diodes, especially considering a MOSFET contribution sensitive to the ac ripple of the current while a diode is less. It can be the reason while you often see paralleled MOSFETs to better perform than a simple diode: paralleling MOSFETs help reducing the heat and can avoid a heatsink while there is not much you can do for a diode beside selecting a bigger die or adding more aluminum around it.
Finally, the adoption of sync rect also depends on the diode voltage drop contribution to the overall output. A 0.4-V drop is large for a 3.3-V or 5-V output but does not affect efficiency the same way for a 24-V output for instance. Sync rect will make better sense in the first example unless you face size constraints which impose you to design without heatsink.
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\$\begingroup\$ Please, consider improving the format of your answer. As it stands it is almost a wall of text with no images or links to relevant sources. At least break the text into more paragraphs to aid readability, and thus aiding whoever will read it to scan quickly for relevant information. \$\endgroup\$ Commented Aug 6, 2017 at 10:13
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1\$\begingroup\$ Hello, the point is granted and I inserted some breakpoints. Thank you for pointing this out. \$\endgroup\$ Commented Aug 6, 2017 at 12:09
Instead of crafting my own wording in a broken English, a simple Google search for {difference between synchronous and conventional dc-dc boost converter} reveals the hit on top, from Rohm Tech Web,
it reads,
As shown in the figure, the difference lies in the fact that whereas in the nonsynchronous rectifying type, the low side switch is composed of a diode, in the synchronous rectifying type, as same as the S1, the switch is a transistor.