Help needed with Dc-DC Converter

Discussion in 'General Electronics Chat' started by gusmas, Sep 20, 2013.

  1. gusmas

    Thread Starter Active Member

    Sep 27, 2008
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    0
    Hey everyone

    I designed a SEPIC converter, and in simulation everything worked perfectly. Once I went to PCB however I experienced massive issues.

    First the SEPIC parameters:

    Vin= 36V -68V
    Vout = 48V
    Iout= 10A
    Fsw = 20kHz

    Like i said in simulation it worked fine, however in practical I experience the following:

    The load in simulation is 2.5 ohms, in the first practical test I used a 2.8 power resistor. I also used scaled down input voltages for safety purposes....

    (1) Rload = 2.8 ohms
    At a 50% duty cycle, I am expecting Vout = Vin - Vdiode - V_otherlosses
    My practical results:
    Vin = 5V
    Total current drawn from source = 3A
    Voltage measured at the Load = 4.26V
    Duty Cycle = 50%.
    Now, when I apply ohms law to calculate the current going through my load resistor, I = V/R = 4.26V/2.8 ohms = 1.5A, half of what is being drawn from the power supply.

    For my 2nd test I used a 4.7 power resistor:
    At a 50% duty cycle, I am expecting Vout = Vin - Vdiode - V_otherlosses
    My practical results:
    Vin = 5V
    Total current drawn from source = 1.2A
    Voltage measured at the Load = 4.98V
    Duty Cycle = 50%.
    Now, when I apply ohms law to calculate the current going through my load resistor, I = V/R = 4.98V/4.7 ohms = 1.05A

    For my 3rd Test i used a 22ohm power resistor:
    At a 50% duty cycle, I am expecting Vout = Vin - Vdiode - V_otherlosses
    My practical results:
    Vin = 5V
    Total current drawn from source = 270mA
    Voltage measured at the Load = 5.72V
    Duty Cycle = 50%.
    Now, when I apply ohms law to calculate the current going through my load resistor, I = V/R = 5.72V/22 ohms = 260mA, this seem to be working

    For my 4th Test i used a 22ohm power resistor:
    At a 55% duty cycle, I am expecting Vout = Vin - Vdiode - V_otherlosses
    My practical results:
    Vin = 5V
    Total current drawn from source = 400mA
    Voltage measured at the Load = 6.84V
    Duty Cycle = 55%.
    Now, when I apply ohms law to calculate the current going through my load resistor, I = V/R = 6.84V/22 ohms = 310mA

    For my 4th Test i used a 22ohm power resistor:
    At a 60% duty cycle, I am expecting Vout = Vin - Vdiode - V_otherlosses
    My practical results:
    Vin = 5V
    Total current drawn from source = 600mA
    Voltage measured at the Load = 8.51V
    Duty Cycle = 60%.
    Now, when I apply ohms law to calculate the current going through my load resistor, I = V/R = 8.51V/22 ohms = 380mA

    For my 5th Test i used a 22ohm power resistor:
    At a 65% duty cycle, I am expecting Vout = Vin - Vdiode - V_otherlosses
    My practical results:
    Vin = 5V
    Total current drawn from source = 1A
    Voltage measured at the Load = 10.4V
    Duty Cycle = 65%.
    Now, when I apply ohms law to calculate the current going through my load resistor, I = V/R = 10.4V/22 ohms = 472mA

    For my 6th Test i used a 22ohm power resistor:
    At a 70% duty cycle, I am expecting Vout = Vin - Vdiode - V_otherlosses
    My practical results:
    Vin = 5V
    Total current drawn from source = 2A
    Voltage measured at the Load = 12.75V
    Duty Cycle = 70%.
    Now, when I apply ohms law to calculate the current going through my load resistor, I = V/R = 12.75V/22 ohms = 580mA

    For the final test I increased the duty cycle to 71% and then all of a sudden the current draw from my source jumped to 3A.
    The measured values were:
    Vin = 5V
    Total current drawn from source = 3A
    Voltage measured at the Load = 13.16V
    Duty Cycle = 71%.
    Now, when I apply ohms law to calculate the current going through my load resistor, I = V/R = 13.16V/22 ohms = 600mA

    I really don't know what is causing so much loss in power..
     
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    Last edited: Sep 20, 2013
  2. THE_RB

    AAC Fanatic!

    Feb 11, 2008
    5,435
    1,305
    Inductor saturation?

    Can you scope the inductor current waveform (across your 1.5 milliOhm resistor)?
     
  3. gusmas

    Thread Starter Active Member

    Sep 27, 2008
    239
    0
    the resistors in the circuit is just the representation for the esr of the inductors
     
  4. Papabravo

    Expert

    Feb 24, 2006
    10,149
    1,791
    The inductor is not properly sized and your supply is running inefficiently. You should shoot for at LEAST 80% efficiency. Remember that you need substantial current handling capability in the inductor to make things efficient and you might also want to consider your switching frequency. Too low and too high are both bad.
     
  5. gusmas

    Thread Starter Active Member

    Sep 27, 2008
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    The inductor wire is rated for 12 amps....
     
  6. THE_RB

    AAC Fanatic!

    Feb 11, 2008
    5,435
    1,305
    And the wire won't melt until 700 degrees? :)

    The limiting factor in a SMPS inductor current is usually the core. Do you have any experience with that?

    How closely does your real world inductor match the simulator inductor that worked well?
     
  7. gusmas

    Thread Starter Active Member

    Sep 27, 2008
    239
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    Well I got to around 50uH to each inductor. I simulated with the pratical measurement of the inductors and the simulation was still fine.

    And no I do not have experience with that.
     
  8. crutschow

    Expert

    Mar 14, 2008
    13,028
    3,238
    You need to determine what the saturation current is for the inductor. For that you need some way to measure the current through the inductor when it is operating using an oscilloscope. A small resistor in series with the source to ground of the MOSFET would allow you to do that. When saturation is reached there will be a sudden increase in the inductor current during the time the MOSFET is ON.

    Increasing the switching frequency will help that problem.
     
  9. gusmas

    Thread Starter Active Member

    Sep 27, 2008
    239
    0
    Ok, the first circuit is the one what you suggested.

    Reason for the 2nd one is, my converter is currently on PCB, to add the resistor at that place is going to be extremely hard. What if I add the resistor between my source and the converter, that should also work correct (2nd attachment)?

    Lastly, what do you suggest the resistance should be of that resistor?
     
    Last edited: Sep 22, 2013
  10. t06afre

    AAC Fanatic!

    May 11, 2009
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  11. crutschow

    Expert

    Mar 14, 2008
    13,028
    3,238
    Using the second circuit would require a differential measurement.

    What type of oscilloscope do you have?

    Try about 0.1 ohm resistor.

    You might try measuring the small voltage from the MOSFET drain to ground when its ON. It should show a noticeable rise in ON voltage due to the large increase in current when the inductor saturates.
     
  12. gusmas

    Thread Starter Active Member

    Sep 27, 2008
    239
    0
    Ill post my findings for the day soon. :)
     
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