Electronic – 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.

Best Answer

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:

enter image description here

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.