Buck Converter Problem

Discussion in 'The Projects Forum' started by Phaisit, Feb 23, 2015.

  1. Phaisit

    Thread Starter New Member

    Feb 23, 2015
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    Hi. I'm facing a problem with a DC-DC Buck Converter here

    I'm on the project about to build a Solar Charging to battery

    I plan to use a 7.2V 2800mAh for the battery and 20V 1.12A Solar cell here

    and to use the buck converter I using Arduino as Microcontroller to control it on "Constant Current and Constant Voltage Mode" a

    And I used the Orcad to simulate the buck converter that I want the input 20V 50kHz & output around 0.6mA

    After this I tried to build it with MOSFET / Diode and use only resistance as load (Not using the Battery yet)

    but the voltage don't even go through the MOSFET even if I add the gate voltage to 15V (I use STB80NF55)

    here is the picture of the circuit of buck converter I simulated

    http://upic.me/show/54738967

    Any Suggestion on MOSFET and Diode to use & how to calculate the L,C of Buck ? Thx

    Edit : As if I know I have to use the Gate Driver here since the Vgs have to be over the voltage of battery, but atleast I want to know the output to see the different here)
     
    Last edited: Feb 23, 2015
  2. crutschow

    Expert

    Mar 14, 2008
    12,995
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    I can't see that picture.
    Can you attach it directly to your post?
    (Go to more options and click "Upload a File".
     
  3. Phaisit

    Thread Starter New Member

    Feb 23, 2015
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    Here's picture
     
  4. Phaisit

    Thread Starter New Member

    Feb 23, 2015
    26
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    I don't know how to calculate it when have load as battery

    Should I just calculate as Power like 7.2V from battery (Ni-Cd I use) and 0.6mA I want to become the power and then calculate the inductor or something . . .
     
  5. crutschow

    Expert

    Mar 14, 2008
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    Did the simulation work?
    Post the circuit diagram of the actual circuit that didn't work.
     
  6. MrAl

    Well-Known Member

    Jun 17, 2014
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    490
    Hi,

    Calculate what, the power in the load?
    The power in the load is the current times the voltage. The buck supplies the current to the battery.

    L and C would be similar to a buck with resistive load with the same output voltage as the battery and same charging current.
     
  7. Phaisit

    Thread Starter New Member

    Feb 23, 2015
    26
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    I simulate it with using MOSFET and Diode and it kinda work
    but when I try to build it for real it seem that the voltage doesn't get through the MOSFET

    I only use DCGen as source and to drive gate, is it have to have some resistance in gate pin ? and I kinda confuse about the ground of the control from arduino and the power from DCGen

    So it mean that I can use the 7.2V time 0.6mA equal to Power and calculate the L and C right ?
     
  8. crutschow

    Expert

    Mar 14, 2008
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    That is an N-MOSFET and requires a Vgs of 10V to fully turn on, thus you need to apply 30V to the gate (you are only applying 12V) with a 20V supply voltage.
    Typically this gate voltage is generated by a bootstrap driver circuit so you don't need a gate voltage source larger than the supply voltage.

    All grounds must be tied together.
     
  9. Phaisit

    Thread Starter New Member

    Feb 23, 2015
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    Can you suggest any way to do it ? Since I cant have a 30V Vgs by using only 20V Solar Cell
    Should I change to P-MOSFET or some way ?
     
  10. crutschow

    Expert

    Mar 14, 2008
    12,995
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    You can use a P-MOSFET instead with the source to the supply and the drain to the load.
    The P-MOSFET will be off when the gate voltage is equal to the source voltage and on when the gate is grounded.
    But you must be careful not to exceed the maximum Vgs rating of the MOSFET, which is typically 20V.

    Why do you have two threads on the same subject?
    That's bad form.
     
  11. Phaisit

    Thread Starter New Member

    Feb 23, 2015
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    0
    Really sorry about the new thread I thought I might have to show another result with another thread so I will leave this thread and you another one now
     
  12. MrAl

    Well-Known Member

    Jun 17, 2014
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    490
    Yes you can use the 7.2 volts times the 0.6 amps and get the power, and calculate L and C.
    You may also want to consider cases where the battery voltage is less too, as when it is charging. So if you calculate some L and C make sure it works the way you want it to work at a lower voltage too, like 5 volts for example. You should also be aware that if the voltage is too low it must be charged very slowly at first. If this is an Li-ion type battery pack you should read up on all the cautions of using these batteries first. There are a lot of precautions that go with charging these things which can lead to some serious injury and property damage.
     
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