Readers are advised to proceed with the construction of the presented circuits only after understanding the concepts from the core. Not adhering to this can lead to failures and frustrations.

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Friday, March 8, 2013

Simple DC to DC High Current Voltage Doubler Circuit - 3 to 10 Amps


The post explains a voltage doubler circuit which will almost double the voltage that's been applied at the input (up to 15V max), and also it becomes specifically useful since it allows higher current loads to be used at the output.


Since the voltage doubler circuit explained here is able to handle high current loads, the design becomes ideally applicable for raising solar panel voltages when there's no adequate amount of sun light incident on the panels.

Circuit Description

Looking at the given circuit diagram, let's assume we apply a 12V at the input of the circuit, the output would generate a potential of around 22V.

The circuit initiates its functioning when IC1a, R2 and C2 starts generating rectangular waves.

This signal also reaches at the output of IC1d, albeit in an inverted mode.

The presence of R2, C2 delays the output of IC1a which causes the output of IC1b to attain less than 0.5 duty factor, resulting in a waveform where the negative half may be shorter than the positive half).

The above also becomes true at the output of IC1c, were the input data is delayed with the help of C7, R5.

The output from IC1c which is in an inverted form is further buffered thrice via IC3f, IC3a and the gates in parallel IC3b-----IC3c.

The output from the above is finally used for driving the power mosfets.

Te transistor T1 is driven from the output of IC1b..... when T1 is ON, the point between R6, R7 attains a 2V potential, however since IC2a requires a 11 to 22V input, the negative potential for this chip is plucked from the positive of the input voltage, because the supply voltage and the collector of T1 is already subjected with the doubled voltage.

D1 is introduced to guarantee that the input to IC2a never drops below 10.5 V.

During the conduction periods of T1, T2 and T3 conduct alternately.

When T2 is switched ON, C10 gets charged with voltage equal to the input supply voltage through T3 and D3.

When T2 is turned OFF, and T3 gets ON, C9 goes through identical process as C10 above. However C10 holds the charge due to the presence of D3 which stops it from discharging.

Because the two capacitors are in series, the net voltage now attains a level that's almost twice that of the applied input voltage.

One interesting thing here is, since the circuit involves many inverting stages and also a few delay networks, the output mosfets can NEVER conduct together which makes the circuit extremely safe with the operations.

C1 buffers the input applied voltage in order to load the input with constant power irrespective of the varying current parameters across the output.

The components which are marked with dashed circles need to be appropriately cooled by adding large heatsinks to them.












22 comments:

  1. hi, this circuit can give constant voltage and current?

    ReplyDelete
    Replies
    1. It will depend on the input voltage, if it changes then the output would also change.

      Delete
  2. Hi swagatam i want to drive 25 3.4 20 ma leds in series with 6 v sla please post the circuit

    ReplyDelete
    Replies
    1. Hi Arun,

      For driving 25 nos leds would require a boost converter which would be unnecessarily complicated. instead you may connect all of them in parallel with 120 ohm resistor with each led.

      Delete
  3. Hi
    Can you give give me a circuit diagram of a booster circuit with DC i/p voltage varying from 1.5v to 3.3 volts 0.5amps to DC o/p of 9v 0.5 amps

    ReplyDelete
    Replies
    1. Hi, boosted voltage with same current output is never possible...

      Delete
    2. ok!!can you tell me the variation in current when we have the same i/p as mentioned above??

      Delete
    3. you may calculate the output current by dividing the input wattage by output voltage...it would be a little less than the calculated result.

      Delete
    4. thanks for the reply!!But can i have a basic circuit diagram of the booster circuit???it would be really helpful...

      Delete
    5. you may refer to the following circuit, you may have to modify the output winding for getting the specified output:

      http://homemadecircuitsandschematics.blogspot.in/2012/10/1-watt-led-driver-using-joule-thief.html

      Delete
  4. Sorry for the length.
    Hello Swagatam, I want to make a circuit in which I have a solar panel, a 12v rechargeable battery, and LED lights. LEDs should light up when there is no light in the room. Before you read my idea try to make a circuit that gives this service.

    My idea(very long! :D. Sorry I can't add an image)

    I tried to make a circuit where the positive end of the solar panel is connected to a diode then to the positive end of the battery but the negative end of it is connected directly to the battery, after I had connected the positive end of the battery to some sort of voltage divider where the second resistor would be a photodiode and then I connected to the wire that joins the first resistor (1 kohm) with the photodiode the LEDs. The negative node of the LEDs was connected to the negative end of the battery and the same thing was to the the phototdiode. Then I realised that I am wasting energy even when the LEDs are off.

    Can you help me! Please!

    ReplyDelete
    Replies
    1. Thanks Jad!

      I think the following circuit would fit your requirement:

      http://homemadecircuitsandschematics.blogspot.in/2013/03/simplest-automatic-led-solar-light.html

      Delete
    2. Thanks a lot mister Swagatam. :)

      Delete
    3. Here the transistor should be a field effect transistor. Right?

      Delete
  5. You should give credit to the original author. I saw that circuit published in a Elektor electronics book. Even worst you added your watermark !

    ReplyDelete
    Replies
    1. Nowhere I have mentioned that's it's my creation, in fact there are quite a few diagrams which I have borrowed from Elektor electronics and I have expressed it in those particular articles.

      All explanations are completely mine though, so may be while writing the post I missed the credit completely, because for me making the new hobbyists learn about any concept is too important and folks here are appreciating me a lot for that.

      I have added the water mark so that others do not pick it from here easily, because already hundreds of my original creations have been scraped online by different spam sites.

      If you have the name of the author you can provide it to me I will make sure it gets credited appropriately.

      Delete
  6. thanks for your time and effort that you have put so far
    i want to know if the circuit can actually increase the power output from the panel or just simply increase voltage while the current drops.
    If the power remains the same, don't you think that the charging rate of the panel will remain the same irrespective of the voltage that is boosted because charging is done by both current and voltage and not just voltage. which i think is power

    ReplyDelete
    Replies
    1. power will always remien same for input and output, so here the current is dropped for boosting voltage.

      If the current level after the drop matches the battery charging rate, then it will charge the battery optimally.

      Delete
  7. Thank you for the great circuit. I have a 12v 10A power supply that I need to get a little more voltage out of it. I would like to use the entire 10A however. Would this circuit be possible to modify to make 15 or 16v output from 12v? Thanks!

    ReplyDelete
    Replies
    1. Sorry that isn't feasible with any circuit. A higher voltage from a lower voltage source can be acquired only by sacrificing a proportionate amount of current such that the V x I value always remains equal.

      Delete

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