# How to convert a wide voltage range (1 V-12 V) to 5 V?

I have an unusual application where the input voltage can vary from 1-12 V DC. This must be converted to 5 V @ ~200-250 mA.

Since there are no available buck-boost converters suitable for this wide input voltage range, I came up with this (simplified) circuit:

simulate this circuit – Schematic created using CircuitLab

The XC61CC5002MR-G is a voltage CMOS supervisory IC which:

• Outputs VCC when VCC > 5 V - this enables the LDO and protects the boost converter from the higher voltage.
• Outputs 0 V when VCC < 5 V - this disables the LDO and biases the PMOS transistor to allow current to the boost converter.

The boost converter's absolute maximum voltage rating is 6 V, so it needs to be protected at the higher range of input voltage, whilst the LDO is OK up to 13 V.

The circuit worked as intended from 3-12 V, however the PMOS transistor couldn't switch when the input voltage was lower than 3 V, which shouldn't have surprised me since the VGS threshold was about 2 V at 250 mA.

I have also looked at high-side load switches and over-voltage protection ICs, but I cannot find any that will operate over the whole range of the input voltages after browsing for hours on Mouser and DigiKey.

Lastly I have explored using a N-channel MOSFET in the circuit above with an open-drain output variant of the same supervisory IC and a charge pump to bias the NMOS transistor when voltages are low, but to my surprise I couldn't find any charge pumps that work in the 1-5 V range.

I am looking for any suggestions to either make my circuit work with the lower voltages or how else I can achieve this 1-12 V to 5 V conversion without dramatically increasing the PCB footprint or cost. Unfortunately the input voltage cannot be changed, but the power supply can provide more than enough current to run the circuit.

• Welcome to the site. Commented Nov 7, 2018 at 15:03
• What part is the boost converter? Commented Nov 7, 2018 at 15:07
• ti.com/lit/ds/symlink/tps61099.pdf Commented Nov 7, 2018 at 15:08
• @TimWescott How did you obtain 7.5A?
– dim
Commented Nov 7, 2018 at 16:12
• @TimWescott surely it would be 1.25A + head room for inefficiency? Commented Nov 7, 2018 at 16:34

Allow me to make this circuit suggestion:

simulate this circuit – Schematic created using CircuitLab

For 1 V < Vin < 6 V the boost converter converts to a 6 V internal voltage and the LDO regulates that 6 V down to 5 V

For 6 V < Vin < 12 V the voltage at the Boost converter's output will "follow" the input voltage with some voltage drop due to a (Schottky) diode, so Vin = 7 V => Vmid = 6.5 V and Vin = 12 V => Vmid = 11.5 V.

Remember that boost converters have this basic circuit:

So when Vin is higher than the configured (regulated) output voltage, the output will follow the input voltage with a voltage drop from the coil's resistance and the diode.

• Are you aware of a boost converter that works down to 1 volt and can live with voltages at the input as high as 12 volts? Or are you suggesting "build from scratch"? Commented Nov 7, 2018 at 15:15
• @Andyaka No, I'm not. I just provide the circuit idea :-) Closest I know: CE8301 it can work from 0.9 V up to 10 V. Commented Nov 7, 2018 at 15:16
• The boost converter I am using can only tolerate up to 6v according to its absolute maximum ratings and I have looked around and don't think there are any that can tolerate any more than 6v whilst boosting from 1v. Commented Nov 7, 2018 at 15:20
• LT1615 seems to be a good match for the boost part. Starts up from 1V, doesn't burn up to 15V.
– dim
Commented Nov 7, 2018 at 15:22
• @TimWescott I don't understand your 10 W, according to me it is less than: 12 V - 5 V = 7 V => x 250 mA = 1.75 W for the dissipation in the LDO. Commented Nov 7, 2018 at 16:51

## A SEPIC might be worth a shot here

This situation (wide voltage range input, fixed voltage output, low-ish current requirements) strikes me as a decent opportunity to try one of the lesser-used switching topologies: the SEPIC. In particular, I would use a LM2621 converter IC with a low-drop Schottky diode -- this gives the chip the best chance of starting up at or about 1.0V (the datasheet specifies nominal startup at 1.1V and maximum startup at 1.2V over temperature, but the startup voltage likely depends on the drop across the diode).

As to passives, I would use the coupled-inductor version of the SEPIC topology -- this gives better ripple performance while using less board space than two separate inductor cores.

• Funnily enough, I discovered the LM2621 yesterday and I have ordered one, will post update if it works out. Commented Nov 10, 2018 at 12:46
• Thank you for linking that SEPIC article it was very informative, I have a question with the coupled inductor SEPIC topology would the voltage at the diode be limited to 5V? Because if so I could use a MAX40200 ideal diode current switch: (datasheets.maximintegrated.com/en/ds/MAX40200.pdf) which has a very small forward voltage but is only rated at maximum 6V. Commented Nov 11, 2018 at 11:15
• @jm212121 the voltage across L1b is Vin+Vout, so you'll need a much higher voltage part than that -- switching speed would also be an issue for many/most "ideal diodes". What you really need is a synchronous rectification controller if you want to do what you're trying to do... Commented Nov 11, 2018 at 15:38

To avoid the EXTREME switch, you need a flyback stepup with zener post-regulator, to provide the FET's gate drive.

Set the zener at 10 or 12 volts. Many huge FETS want a low max gate voltage (from what I recall).

Now you can use a LARGE FET, capable of 10 or 20 amps or whatever you need, that FET gate being driven 0/10 volts for efficient switching.

simulate this circuit – Schematic created using CircuitLab

• Could you explain in more detail what you mean here - that diagram is confusing me. In the diagram you have a 1-12 volt boost regulator, is that a boost regulator that accepts 1-12 volts as input, if so I can't seem to find one that accepts that range while maintaining a few hundred milliAmps at 1v. Commented Nov 7, 2018 at 20:39
• The bottom switcher ---- the flyback ----has to operate on 1 volt, or higher. Simply having a circuit that oscillates at 1 volt requires some thinking. Then you want that oscillator to operate a transistor in boost mode, with the output regulated by a 10 volt zener. The 10 volts is used in the very-high-current upper switcher, perhaps solely to drive the gate of the 20Amp? MOSFET. By having 10 volts to control the gate, instead of 1 volt, you make much more efficient use of the big MOSFET. Commented Nov 8, 2018 at 3:44

There's a lot of ways to skin a cat, and this is a particularly vicious cat. Here's my thoughts:

Boost set to 5.5V or so. I'd have to go shopping for parts to be sure, but I suspect you'll need something with an external switch to handle the 8-9A input at 1V. Follow that with a buck to 5V, to handle the case when the input's above 5V. Don't mess with the linear supplies unless you need really clean power.

If you can't find a suitable boost chip or switch FET, use a lightweight boost that's just big enough to power the "real" boost controller (and by extension, the "big" switch gate). Maybe even switch the lightweight guy out of the circuit if the input voltage gets too high.