Designing an Electronic Load

Discussion in 'General Electronics Chat' started by rizwan.bashir, Aug 14, 2013.

  1. rizwan.bashir

    Thread Starter New Member

    Aug 14, 2013
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    Hi

    I actually want to have an electronic load of 15 ohms for maximum power transfer. Can I do it with a voltage controlled constant current source? If yes, how?

    I am not sure if I made myself clear or not, so here is the block diagram of what I am trying to achieve. Please see the image.

    Thanks
     
  2. crutschow

    Expert

    Mar 14, 2008
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    If you want maximum power transfer just use a 15 ohm resistor. Why do you want it to be controllable?
     
  3. THE_RB

    AAC Fanatic!

    Feb 11, 2008
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    Well IF the load is the microcontroller circuit on a regulated 5v, the max power transfer will be from a voltage regulator with low dropout and low quescent current. That transfers the max current available to the micro (at a fixed voltage of 5v).

    So the problem becomes "how to get max output current at regulated 5v from this circuit"?

    The O.P really needs to provide more info. :)
     
  4. wmodavis

    Well-Known Member

    Oct 23, 2010
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    I believe the essence of an electronic load is a controllable, variable impedance that will withstand up to some specified maximum power and is used for the purpose of testing the power source under controlled load. If you want what you said use a 15 Ohm resistor as cruts says. Sounds like an electronic load would be way overkill.
     
  5. rizwan.bashir

    Thread Starter New Member

    Aug 14, 2013
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    Yes, I think I should have provided more info.
    Actually I have a power source and I want to scavenge max power out of it to charge my capacitor bank for the regulator (please see attached image).
    The resistor will dissipate power, I want to make use of the available power. I have attached a more clear image now.
    Thanks
     
  6. rizwan.bashir

    Thread Starter New Member

    Aug 14, 2013
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    It is just that I want to have a fixed resistance of 15 ohm. Using R=V/I; if I can make V and I constant, I can have fixed value of R. So any suggestions on how to do it? If I need to provide any other info, please let me know.
    Thanks
     
  7. blueroomelectronics

    AAC Fanatic!

    Jul 22, 2007
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    A 15Ω resistor is 15Ω

    Why not tell us what you're trying to build?
     
  8. rizwan.bashir

    Thread Starter New Member

    Aug 14, 2013
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    I am trying to design a pilot over power line stuff. Have to transfer power over this pilot line and have to take care about the intrinsic safety criteria too.
    IS limits my power that I can send to my module at the receiving end. So I need maximum power to drive my module and power line communication modem.
     
  9. rizwan.bashir

    Thread Starter New Member

    Aug 14, 2013
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    I should be using the term electronic load instead of resistance. I need to design an electronic load that provides me an impedance of 15 ohms but it does not dissipate power like a resistor. That is why I had an idea of a voltage controlled constant current source. So any help now? Or am I still missing something?
     
  10. THE_RB

    AAC Fanatic!

    Feb 11, 2008
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    Yep, you are missing the best way to do it. ;)

    What voltage is coming from your power source? How much does it fluctuate? Does it go negative? Any 'scope shots of the power waveform?
     
  11. bountyhunter

    Well-Known Member

    Sep 7, 2009
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    Regardless of what magical approach you take..... any device that has a voltage drop across it and a current flowing through it will dissipate power EXACTLY like a resistor. The only advantage an electronic load has is that it's variable and easily adjusted.

    A controlled current source will dissipate exactly the same amount of power as a resistor. That's why elect loads have massive heatsinks and fans in them.
     
  12. ian field

    Distinguished Member

    Oct 27, 2012
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    Sounds like you need to charge your capacitor bank with whatever it can get and regulate for the load with a buck/boost SMPSU, as the voltage drops the switcher draws more current trying to keep its output up - so its sudden death if you don't keep the capacitor topped up.
     
  13. rizwan.bashir

    Thread Starter New Member

    Aug 14, 2013
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    Yes Ian, you are right, I need to charge up my capacitor bank with whatever I can get.

    The voltage waveform is attached now. The current in the line is around 110 mAmp.
    The voltage wave consists of three levels of pulses. +12V, -12V and +30V.
    +12V and -12V for 3 msec each and 30V for 4 msec. I need to charge my capacitor bank in +12V and +30V periods.

    So what kind of a circuit can I put within the block I marked as 15 ohm electronic load in my previous attachment?
     
  14. THE_RB

    AAC Fanatic!

    Feb 11, 2008
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    Thanks for providing the critical info. It would have been good if your opening post said something like; "This is the waveform my power source has, how do I get power from this to run my 5v DC microcontroller?". :)

    It looks like your power source makes >=12v DC for about 70% of the total time (ie 70% duty ).

    I would charge the cap bank through the diode, it will charge to approx 29v when there is no load. Then a buck converter to make +5v DC would be best (most possible power at 5v DC).

    How much current does your micro circuit need at 5v DC? If your micro only needs 10mA or 20mA like many circuits then you can probably just use a 5v DC linear regulator (ie 7805) after the cap bank. :)
     
  15. rizwan.bashir

    Thread Starter New Member

    Aug 14, 2013
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    Thank you for your reply. I need approximately 111mA.

    Please have a look at the attached image. I do not want my regulator to draw more than 110mA from the line. The zener diode is just used as a voltage regulator. So I have two questions here.
    1- Is there a better circuit to use instead of the zener diode to have a sharp cut off voltage?
    2- I want input current limit on my regulator so that it does not load the line (draw more than 110mA). The input voltage is 11V and the required output voltage is 3.3V. The output current limit is not required.

    Thanks again
    -Rizwan
     
  16. THE_RB

    AAC Fanatic!

    Feb 11, 2008
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    You're still not providing the required information!

    How much current will your 3.3v circuit require?

    Forget trying to solve the problem yourself by providing a circuit, what we need are the input supply specs (which you gave a voltage waveform and said would supply 110mA) and the OUTPUT specs, which is 3.3v out but how many mA are needed at 3.3v?
     
  17. rizwan.bashir

    Thread Starter New Member

    Aug 14, 2013
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    The required current is 110mA @ 3.3V.
     
  18. THE_RB

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    OK thank you. :) I wasn't sure if that is what you were saying before as you stated the 110mA was what the power source can supply.

    Since your power source can only supply power >=12v for 70% of the time, and can only supply 110mA at any time, the maximum average current you can get out will be 110mA x 70% = 3.3v at 77mA, and that is only IF everything is perfect.

    To get more than about 70mA output, you need a SMPS converter to convert a higher source voltage at low current to a lower output voltage and higher current.

    I'm not sure about your skill level, but one good way to do that is to use the power source to directly charge the cap bank (through the diode) so the cap bank will charge to close to 30v.

    That cap voltage will be close to 30v at low loads, and drop some amount of voltage when under load. Let's say it will be between 20-30v.

    Then add a SMPS buck converter (which you can buy from ebay) to convert the 20-30v to 3.3v.

    The buck converter will run about 80% efficiency. 3.3v 110mA out will be 0.363 W out (3.3v * 0.110A) so at 80% eff it will require an input of 0.454W. (0.363W / 0.80)

    0.454W input at average 20v would draw a current from your power source of about 23mA average. (0.454 / 20v)

    I have tried to keep the math part very simple there, hopefully that should be clear enough.
     
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