I need to convert a very low input voltage (around 0.5V) to a steady 3.3V output to deliver 50-80mA current with relatively high efficiency (>= 80%). After some research, it appears that some DC-DC boost converters from TI and Linear Technology are quite suitable for this application. However, they are somewhat expensive to purchase in small quantity. I'm wondering if it's possible to construct such a converter by myself, using components that are widely available. I don't have any prior experience in designing similar circuits, so I have no idea where to start. Any suggestions and/or pointers would be greatly appreciated.

PS. What I have in mind is a DC-DC boost converter with an inductor, but is it also possible to design an inductorless charge pump that meets the requirements that I mentioned above?

  • \$\begingroup\$ Good design costs a lot, a lot more than buying one of these chips and bad design costs even more. The question is how much do you "value" your time in trying to redesign the wheel? A typical inductorless boost circuit relies on charging capacitors in parallel and then reconfiguring them in series but because of the low input voltage, the capacitor switches used by this type of circuit are nearly useless. Kudos to anyone who finds one but no kudos to redesigning what TI and LT offer unless you plan to sell thousands of items per annum to plough back development cost. \$\endgroup\$
    – Andy aka
    Aug 26, 2013 at 17:30
  • \$\begingroup\$ @Andyaka It could be a hobbyist project where time is pretty much "infinite" and true monetary cost isn't. Though even then I doubt a discrete implementation would be cheaper than a single TI/LT part solution. The cost of many small parts adds up pretty quickly. \$\endgroup\$ Aug 26, 2013 at 17:37
  • \$\begingroup\$ @helloworld922 - I did ask the op how much they value their own time!! I'm thinking, buy the chip, get it working and then nip down the pub for a few beers with money you'd save building one from scratch LOL. \$\endgroup\$
    – Andy aka
    Aug 26, 2013 at 17:47
  • \$\begingroup\$ ... opportunity cost ... \$\endgroup\$
    – vicatcu
    Aug 26, 2013 at 17:56
  • 1
    \$\begingroup\$ Theres a Ti chip it works from 0.3 v - TPS61202 \$\endgroup\$
    – user54261
    Sep 29, 2014 at 23:00

2 Answers 2


A capacitor-based charge pump can itself not have better than 50% energy efficiency. There is a simple mathematical proof for this if you're interested.

Googling "boost converter" gives the basic schematic, which is an inductor, a switch node grounding transistor driven through PWM, a catch diode, and a filtering capacitor/output sense.


simulate this circuit – Schematic created using CircuitLab

The part numbers (Opamp, MOSFET) are whatever CircuitLab had available, not actually well-chosen parts for this circuit!

In this case, one challenging question is where the PWM would get its VCC from -- you need to bootstrap the circuit; make it run at a fixed PWM frequency before the error amplifier feedback kicks in.

Also, there will be two main losses: resistance in the inductor, and voltage drop in the diode. (There are some losses in the switch transistor and ESR of the output capacitor, too.) The faster your switch frequency is, the smaller you can make L1 and C1, and thus get smaller losses and smaller size there, but the more switching losses you will get in M1, and the harder the circuit board layout becomes.

You can also use a synchronous rectifier setup (second MOSFET instead of D1) to reduce D1 voltage drop losses, which will be huge in this circuit (almost 20% on its own!)

Given that the input is 0.5V, and you want output 3.3V/60mA, and the D1 drop is 0.7V, then you need a 8:1 voltage gain, so at least 560 mA through L1. L1 will have a resistive voltage drop, so you're going to want to look for 12:1 or better gain. (3.3V+0.7V)/12 = 0.333V, so the allowable drop over L1 is 0.167V, so Ohms law says maximum resistance of L1 is 0.298 Ohms. Yet another source of loss -- getting to 80% efficiency is going to be really hard!

  • \$\begingroup\$ Some of TI's charge pump products claim to have up to 90% efficiency: ti.com/lsds/ti/power-management/boost-charge-pump-products.page \$\endgroup\$
    – Aufheben
    Aug 27, 2013 at 2:59
  • \$\begingroup\$ I can see it's going to be quite hard in implementing such a converter from scratch, but I would still like to have a try. Can a simple 555 timer serve as the PWM? Can you give some more info on the bootstrapping part? Thank you very much. \$\endgroup\$
    – Aufheben
    Aug 27, 2013 at 3:04
  • \$\begingroup\$ That sounds like an entirely new question -- please ask something like "trying to bootstrap 555 PWM from 0.5V for boost converter." Although here's a hint: ti.com/lit/ds/symlink/lm555.pdf Datasheet, page 3, says supply voltage minimum for a 555 is 4.5V. \$\endgroup\$
    – Jon Watte
    Aug 27, 2013 at 3:17
  • \$\begingroup\$ Bootstrapping from 0.5V is a serious problem, as that's below the threshold voltage for lots of kinds of transistor. e.g. TI's "ultra low voltage" logic works from 0.8V upwards. ti.com/lit/an/scea033/scea033.pdf \$\endgroup\$
    – pjc50
    Aug 27, 2013 at 9:03
  • \$\begingroup\$ "A capacitor-based charge pump can itself not have better than 50% energy efficiency. There is a simple mathematical proof for this if you're interested." - That is incorrect. While they are limited in their voltage ratios there is no theoretical limit to their efficiency and commonly types such as the MAX7660 have efficiencies in the high 90%. \$\endgroup\$ Aug 12, 2015 at 22:59

Perhaps, you may want to use a simple push-pull generator on Germanium transistors and a transformer with 5-7 V output with following buck regulator.

A traditional boost converter (with MOSFET and inductor) requires relatively high voltage to drive FET's gate, and low drop BJTs will provide low efficiency. (And this is apart from problem of powering the PWM module!)

The transformer-based schematics implies two crucial choices: of a good core (you probably will have to wind the transformer yourself, taking care of winding resistance) and of a good transistor pair (or a single transistor if using more one winding).

Anyway a simulation with PSPICE is essential (I use OrCAD 16.5, strongly suggest involving 'Magnetic parts editor' for the transformer).


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.