Following up on a previous question, I've implemented the following scheme:


simulate this circuit – Schematic created using CircuitLab

The normal buck converter in the dashed box is used to charge a Li-Ion battery (of course, I've omitted irrelevant details). Its required power, factoring converting losses, is less than 5W. In order to extend the input voltage range of the L6902 to include 2.5…8V, I've prepended it with a custom boost converter, so the complete system behaves like this:

  1. When Vin ≤ 2.5V, no power is drawn from the generator (duty cycle of the boost converter is 0, and U1 is in under-voltage lockout);
  2. When Vin is in [2.5…8.5V], the boost stage is active and U1 sees around 8.5V at its input. The battery charging current is ramped up as Vin increases in a way that the L1 peak current doesn't exceed ~0.9A; the output current of the boost stage is less than 0.6A.
  3. When Vin ≥ 8.5V, the boost converter is in a "pass through" state, duty cycle = 0%, and we have Vstepup = Vin - diode drop. Again the current thru L1 and D1 is less than 0.6A (and usually much less due to the constant power load).

Since the device may work in 3) for extended periods I'm wondering are there any caveats to this mode. Of course there's D1 losses, which I can live with. I also reckon that if the load current changes suddenly (e.g. battery disconnected) the inductor in the power path may introduce some ringing, but given C1's value these should be well damped.

Since I'm far from an expert in DC-DC converters, I wonder whether there may be any caveats I'm not currently aware of?

  • \$\begingroup\$ In PWM language there is no such thing as 'zero' duty cycle. You simply shut OFF any and all pulses to the MOSFET under the conditions you mentioned. With the MOSFET OFF there are no hidden issues. \$\endgroup\$
    – user105652
    Commented Feb 13, 2018 at 1:44
  • \$\begingroup\$ @Sparky256 Zero duty cycle is another way of saying always off. 100% duty cycle is another way of saying always on. There sure are such things! \$\endgroup\$ Commented Feb 13, 2018 at 1:52
  • \$\begingroup\$ @immibis. True, but can a specific PWM IC actually do 0% or 100%?. I assumed the OP maybe not aware of his minimum and maximum pulse duty cycle. OP did not mention what PWM IC they are using. \$\endgroup\$
    – user105652
    Commented Feb 13, 2018 at 1:56
  • \$\begingroup\$ @Sparky256 Indeed the OP didn't mention a PWM IC at all, who even says there is an IC that specifically generates PWM (as opposed to, say, a microcontroller pin)? Just because your favourite tool can't do a thing, doesn't mean it's not possible... \$\endgroup\$ Commented Feb 13, 2018 at 2:01
  • \$\begingroup\$ @Sparky256, the gate signal is coming off a PIC pin. The PWM module inside allows the relevant duty cycle register to be set to 0, in that case the pin is held at GND. This makes my control loop really simple (no corner cases) \$\endgroup\$
    – anrieff
    Commented Feb 13, 2018 at 2:18

2 Answers 2


This is fine as long as D1 is a diode.

If D1 is instead a MOSFET (synchronous rectification), 0% duty cycle is not acceptable because switching is required for the gate charge pump.

  • \$\begingroup\$ +1 if you can tie your answer into the OP's actual question. Be careful of short answers, even if they are correct. They could get flagged as a LQ answer by the servers. \$\endgroup\$
    – user105652
    Commented Feb 13, 2018 at 3:19
  • 2
    \$\begingroup\$ "switching is required for the gate charge pump" ... or some alternative higher voltage supply for the rectifier's gate drive. Or a PMOS for the rectifier. etc. But you're correct, it's something to be aware of. \$\endgroup\$
    – user16324
    Commented Feb 13, 2018 at 11:53

When Vin >= 8.5V, as long as you can live with the diode losses, you don't saturate the inductor AND the diode is happy with those losses, it's fine. I would suggest a bigger package for the diode, maybe an SMC. You need to be able to spread the heat!!

However, driving a power MOSFET at 100kHz from a PIC pin is NOT a good idea. You'll drive the gates quite slowly and possibly damage the PIC and the MOSFET. Speaking of the MOSFET, the one you seems undersized. Did you build up this circuit the way it is? Did it work?

  • \$\begingroup\$ The MOSFET is undersized, yes, we eventually upgraded it to a DPAK part with similar gate capacitance. Otherwise yes, we built it and the circuit works well. Typically Vin=4V, the stepup output is ~8.5V and the input current is around 600mA. I've measured efficiency to be above 92%. \$\endgroup\$
    – anrieff
    Commented Apr 23, 2020 at 9:30

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