Skip to main content

0.6V to 6V/12V Boost Converter

In this post we learn how to make a 0.6V to 6V or 12V boost converter circuit using a single chip MC74VHC1G14, which uses under 1V to operate.

About the IC MC74VHC1G14


Normally, we all know that a silicon transistor would find it difficult functioning below 0.7V, unlike germanium counterparts which are capable of doing it with ease, however nowadays we don't often hear about these devices which have become quite obsolete with time.

The circuit discussed here uses an inexpensive Schmitt trigger NOT gate MC74VHC1G14 from the 74XX TTL  family which are designed to work with voltages well below 0.6V, to be precise even with as low as 0.45V. The device we employ is manufactured by Motorola.

The presented 0.6V to 6V boost converter circuit can be even modified to achieve upto 12V from a 0.6V source.

Referring to the figure below, we see a rather straightforward set up consisting of an oscillator stage using a single NOT gate inverter module as discussed above.

Simulation and Working


This NOT gate is very special since it's able to oscillate even at voltage as low as 0.5V which makes it very suitable for the present 0.6V to 6V or 12V boost converter application.

The oscillation frequency here is determined by R1 and C1 which is calculated to be around 100kHz.

The above frequency is fed to the base of an NPN transistor for the required amplification.

C2 makes sure the two IC and the BJT stages are kept isolated from direct contact in order to avoid the low input voltage from dropping below 0.5V

R2 and thee schottky diodes D1 keeps the BJT sufficiently biased for helping an optimal oscillatory response for the transistor.

D2 is another schottky diode which is introduced to keep the charge from C3 disconnected during the switch OFF periods of Q1 otherwise the stored charge inside C3 could get discharged or shorted via Q1.

The IC 7806 at the output is to maintain a fixed 6V irrespective of the boost level created by L1 and the associated converter stages.

L1 must be wound strictly over a ferrite core. The dimension and data of the coil is a matter of some trial and error or it may be procured as a ready made unit for the same.

Circuit Diagram




Need Help? Please leave a comment, I'll get back soon with a reply!




Comments

  1. HI bro,
    If i use 7805 inserted of 7806..than it use as a smartphone charger..??

    ReplyDelete
  2. Hi bro, both will work and can be used according to me.

    ReplyDelete
  3. may i know the code of the diode use here, i am going to try also

    ReplyDelete
  4. you can try BAT41 for the diodes

    ReplyDelete
  5. Hello,
    do you have any recommends for a similar IC ?
    I dont can find and get this IC on a cheap way. Do you have any recommends, does it need to be a Very High Speed CMOS?
    Thanks

    ReplyDelete
  6. hello, I am not sure about an equivalent IC, but a joule thief idea is one option which can be tried for achieving the same results.

    ReplyDelete
  7. Does it depend on the NOT gate or the TTL variant ?
    I really want use it, but I can not get this right one.

    ReplyDelete
  8. thanks neha!

    No it can't be used since a lemon would not produce sufficient current required for the boosting action....

    ReplyDelete
  9. How to connect the circuit of 0.6V to 6V/12V Boost Converter
    my mail is: camaradiop001@gmail.com

    ReplyDelete
  10. How to connect the circuit of 0.6V to 6V/12V Boost Converter
    my mail is: camaradiop001@gmail.com

    ReplyDelete
  11. the scheme of 0.6V to 6V/12V Boost Converter Circuit is complete?

    ReplyDelete
  12. how the connection to have 6 v at the output of the system.

    ReplyDelete
  13. the circuit complete, and you must build it exactly as its shown....click on the diagram to enlarge it.

    ReplyDelete
  14. Hello. Thanks for the reply. I would like to have a precision compared to al voltage input and sortie.de more I see + 0.5 v at the top of the schema.le less will be connected where.

    ReplyDelete
  15. Hello, the circuit can be used for converting as low as 0.5V into 6V provided the input current is high.

    ReplyDelete
  16. What is the number of turns it takes to the coil around the ferrite. If I use 6 v to the entered, what would be the output voltage of the circuit.

    ReplyDelete
  17. for 6V you can use any 555 based boost converter, this circuit is recommended only voltages lower than 2V

    you can also try a joule thief circuit for the same.

    ReplyDelete
  18. Hi Swagatham,

    I want a 24VDC to 48VDC boost converter.
    I got a 24DC battery as the voltage source.
    How can I build one. Please suggest.

    ReplyDelete
  19. Hi Uday,

    you can use any IC 555 based boost converter circuit from this website.

    https://homemade-circuits.com/?s=555+boost

    ReplyDelete
  20. Thank You Swagatham,

    I am need of a circuit where the voltage should decrease gradually over a specified range. As an example it should decrease from 15VDC to 5VDC continuously, but not as in one step, so that I can have control on the rate at which it is decreasing.
    Can you recommend something for this?

    Thanks a lot
    Uday

    ReplyDelete
  21. Uday, you can probably implement it using a PNP emitter follower BJT circuit, where a base/emitter capacitor would decide the slowly declining voltage across the emitter and positive supply of the BJT

    ReplyDelete
  22. Thanks Swagtham it worked.
    Thanks a lot!

    ReplyDelete
  23. You are welcome Uday, happy to help!

    ReplyDelete
  24. Sir can u recommend a circuit tha boost li ion bat 3.7 v to 5 v for android phone charger, tnx Sir!

    ReplyDelete
  25. Jindro, you can use the Joule Thief Circuit as discussed in the following article

    https://homemade-circuits.com/make-this-power-bank-circuit-using-37v/

    ReplyDelete
  26. Hi swagatam
    which battery do you recommend to be use in the Vin? can we use wrist watch battery 1.5V input if size is of utmost important in a given project? thanks alot.

    ReplyDelete
  27. Hi Abba, you can try any type of battery, size is not important but the output current will be as per the AH rating of the battery and proportionately reduced depending on boosted value of the output voltage

    ReplyDelete
  28. Hi,
    I would like a circuit to optimise the power of NiCad cells to drive Ultra Bright Leds.
    I have 3 cell Nicad packs rated at 3.6V which I'd like to use to drive the Ultra Bright Leds.
    To optimise the power available in the NiCad cells, I'd like the cells to continue to drive the Leds with a constant current as the cells voltage drops below the Ultra Bright Leds rated voltage.
    I was considering a Joule Thief Circuit or Voltage Doubler feeding a Constant Current circuit.

    Any suggestions/circuits please ?

    ReplyDelete
  29. Hi,
    you can try the first circuit from this post:

    https://www.homemade-circuits.com/how-to-make-simple-boost-converter-circuits/

    However current may not remains constant and may also drop as the voltage drops.

    ReplyDelete

Post a Comment