You can use this 5V to 12V converter circuit to easily get 12 volts from a 5V similar source.
This is based of the theory that inductor holds current and passes in opposite direction. This is a DC to DC step up converter, also known as boost converter and it has efficiency of 60-80%.
As the efficiency of this circuit is not very high, so we can’t use it for large projects, which needs max power.
We can use it for low power consuming models. Like 12 Volt 3 watt model, which require 250mA current, and since we are giving 5 volt as input, we have to spend around 720mA current with 80% efficiency.
In this circuit we are going to put DC pulse through an inductor and take the boosted current from the inductor when it’s off. We are using a general purpose high current MOSFET, P55NF06 here which is mostly used in inverters.
And for making the inductor coil, we need to follow the below chart for some expected voltage range.
- 6v to 12v @ 1Amp : 80 turns of 24swg wire in a 0.5mm ferrite core
- 6v to 12v @ 500mA : 60 turns of 36swg wire in a 0.5mm ferrite core
- 12v to 18v : 120 turns of 24swg wire in a 0.5mm ferrite core, and so on…
If we are using it for circuit applications, we need to attach the zener diode at output, and if we are going to charge a battery, no need of zener diode.
The circuit below is shows 6 volt input, however you can use it with input voltage range of 4.5 volt up-to 9 volt.
One thing to remember, we can increase and decrease output voltage by altering the value of the zener diode, but increasing much voltage will produce less current and less efficiency, and also we have to attach a larger capacitor there in output.
We are driving this mosfet with high frequency square pulse generated by a 555 timer I.C. High frequency for higher efficiency, as current loss within inductor will be less. This is a voltage converter, not a constant current output converter, so we’ll get small ripples at output.
The above circuit diagram is of 6Volt to 12Volt converter. Change zener diode value to change voltage yourself.
Comments on this article before 07-12-2011, here jump to recent comments .
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Jay on May 8, 2011 - 3:38 pm
Very nice tutorial – keep it up!
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Rubel on June 7, 2011 - 3:44 am
Why we are using the 10K Pot ? What will be it’s value ? It is converting 6V to 12V – for USB 5V to 12V boosting do I need to change any thing ?
And any kind of ferrite core is not available in my market . In which familiar old circuit can I get this 0.5mm ferrite core ? And what will be it’s turns for 5V to 12V for two 12V 170mA Fans at a time ?
Thanks
Arup Reply:
June 7th, 2011 at 7:51 amFine tune the pot to minimize heat in the inductor coil. This will work in 5V also.
12Volt 170mA = 2.04Watts. And for these two fans you need power of 4.08Watts. If this circuit has 70% efficiency, then it require 5.83 Watts of power at input.
Since each USB port can deliver max 500mA current at 5V that means 2.5Watts, it isn’t enough to run these fans.Rubel Reply:
June 7th, 2011 at 10:12 pmThanks Arup . Why do you not add a FORUM in this site ? It will be really blessings for dummies like me . Your site is bookmarked .TahnksArup Reply:
June 7th, 2011 at 10:17 pmIf I start a forum, even if I make it public, users can only see posts and comments, but to open a thread, or for commenting they need to signup. This is a lengthy process, and not all are interested in these. That’s why I started it as a blog. -
Chinmoy Mitra on October 27, 2011 - 7:28 am
Dear Arup, It is nice to visit your website, and I appreciate the work you are doing here. I had posted comments a number of times before, I don’t know whether you have read them or not. Anyhow, I have a few suggestions for you, basically for the benefit of the readers who read your articles.
I suggest you read your articles thoroughly before posting, since I have noticed a number of critical mistakes in your articles. I am not here to criticize you, but this is for the newbies who want to take up electronics as a hobby. Electronics can be a very satisfying and creative hobby, but a very small mistake in a schematic can destroy a project and discourage the person trying to build it! I have seen many frustrated beginners trying to make circuits work which are basically doomed, because of silly mistakes (or maybe they are intentional?). To cite the latest problem with your circuit, you have added a zener diode straight across the output of this voltage booster circuit. This zener is going to blow the instant you switch on the circuit. Where is the current limiting resistor? Secondly, do you really suggest to use a Zener Diode to regulate the output voltage of a 500mA to 1 Amp power supply? You must be joking!! I think you should refresh your theory before posting any more circuits. I have nothing to comment about the numerous “very grateful for your help/suggestions/circuits etc.” posts and comments here. I feel sorry for those poor newbies!
To end this long comment, I would like to apologize if I sound harsh to you, but believe me, I don’t mean any harm. I have 35 years of experience in this field, and I also started as a hobbyst. Wish you all the best, and looking forward to more articles from you.Arup Reply:
October 27th, 2011 at 12:41 pmThanks for dropping a such valuable comment.
As per my thought, the inductor coil sends pulses of small currents through the IN4007 diode to charge the capacitor(output) at higher voltage, whose voltage is limited by the value of the zener diode.
I maybe wrong, as I haven’t done any degree in electronics, and whoever is expert in this field, goes for professional life rather than helping others. -
art on November 3, 2011 - 3:08 pm
just a question here…
why is so many parts to use?
why there is a timer?why not just use LM7812.. you can convert any dc supply to 12V
Arup Reply:
November 4th, 2011 at 3:34 pmLM7812 can regulate(not convert) higher voltages to 12V. The above circuit boosts 6volts to 12 volts.
kamal says
Hi. I need to boost low voltage(below 1v) into 3.7v at any expense of current output. Can u make changes in this ckt to do so?
Arup says
Boosting voltages below 1V is insane. You'll definitely end up with huge current loss if you do it. Near 80% or 90% loss which means 10% efficiency.
manoj says
your projects are excellent,arup
I am a 10 th class student interested in electronics.
Is there any way to learn its basics now.
Arup says
Head to http://www.bcae1.com/
In the right side browse from chapter 2 to the last.
olayemi oluwababamigbe says
arup am working on dc of output1.4 volt i want to get 5v kindly give me step by step compenet built togerther to produce an input of 5v
Arup says
AAsaver should be best option for you.
Chatta says
That Good DIY
sanjay kumar sharma says
dear brother Arup,
today i have seen ur effort to boost knowledge of electronics with ur DIY.it is realy praising.
I want to know the relationship between Inductence and swtiching frequency with illustration and formula of calculation with an example . what should be the Bmax while designing a transformer of ferrite core(iron dust core).
If you can help me in this i will be thankful to u.
with regards sanjay
Arup says
I'm afraid I can't help you in this matter as I have very limited theoretical knowledge about this technique.
Saurabh says
i am std of 10th
i want to boost 2-3 volt to 4.3 - 5 v.
because i want to charge a mobile phone.
-Saurabh (Dhule,Maharashtra)
Babin Basak says
ami ki kore 5v ke 12 v te boost korbo?
rakesh says
Please find the meaning of simple
sammao says
Thank you for post it,how to get 1a for this circuit,I know english a littlebit sorry.
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Babin Basak says
i need to boost 12v to 17v.
can you change the circuit to do so?????
Scott McKie says
Hello, Working from this scheme, I need an opinion. Does impedance matching enter into connecting the increased V+ voltage output from a capacitor in a series resonant circuit that is to be continuously connected to a separate tank circuit capacitor, with a duty cycle controlled MOSFET controlling the ground circuit between the two, thus momentarily completing the circuit. The scheme is for increasing the input voltage level to a resonant tank circuit. Presently, connecting just the V+ output output to the tank eliminates the V+ cap's output voltage. Thanks
Arnab says
Hi Scott, please clarify your question/opinion a little more, it's really hard to understand for me.
As this circuit's operating frequency is just few tens of kilohertz, may be 10 or 20 kHz, impedance matching is not a big deal.
Malik Shahid says
Hi arup,
I am working on a project in which I need about 48-50 volts out put from 12 volt or less in put. Clearly from the electric system of 1:00 Ltr. Car or 1000 cc car. Can you send me schematic diagram and parts list as I have almost zero knowledge of electronics. You can contact me on shahidshabbir68 at gmail.