A problem about electronic load

Discussion in 'The Projects Forum' started by hejiang666666, Aug 12, 2013.

  1. hejiang666666

    Thread Starter New Member

    Jun 10, 2013
    9
    0
    Dear all
    I'm building an electronic load which can generate constant current to the power supply. In simulation, the current can be up to 10A(1V through resistor R1). However, when I build it on a bread board, the voltage which can up to 0.5V keeps decreasing, and MOSFET is overheating. So far, I think it may because of the power rating , but I don't know how to go further.
    Thanks in advance.
     
  2. Shayan22h

    Member

    Oct 24, 2012
    30
    3
    well i can show u another approach last sem we did some active load project where we were supposed to build an active load test device, testing a solar panel up to 10Amp and 40V, power dissipation of 400W, in this circuit below u can see both constant current mode and constant voltage mode,,, and those diodes are there for simulation of the solar panel :)
    cheers hope it could help

    i forgot to add make sure to use linear mosfets, they have been made for these purposes the one that u have used in ur simulation and the one i have used are not appropriate for this application , we used IXTH15N50L2 in our project worked quite nice
     
    Last edited: Aug 12, 2013
    hejiang666666 likes this.
  3. RamaD

    Active Member

    Dec 4, 2009
    254
    33
    Cannot understand this. Can you make it a little more clear please?

    If the voltage is 5V to the MOSFET, R1 drops 1V at 10A. Power dissipiation on the MOSFET will be 40W and R1 will be 10W. You need sufficient heatsinking for the MOSFET, preferably with a fan.
     
  4. hejiang666666

    Thread Starter New Member

    Jun 10, 2013
    9
    0
    hi, can you tell me what U2 R5 and R6 are used for? Many thanks.
     
  5. hejiang666666

    Thread Starter New Member

    Jun 10, 2013
    9
    0
    hi, can you tell me what U2 R5 and R6 are used for? Many thanks.
     
  6. Shayan22h

    Member

    Oct 24, 2012
    30
    3
    yes certainly i can explain well first of all this circuit has been designed to work both in voltage and current mode, if the S3 is closed S1 would be open and S2 would be connected to the - pin then we are in the voltage mode and we can change the voltage of the solar panel by sweeping our input control voltage,, the Op-amp U1 and its feed backs are used for both modes as the controller, the Mosfet is our active load but U2 is used only in the current mode where we control the current of the solar pannel and its main task is to amplify the signal 400 times the reason is (knowing the input voltage 0-5v) in order to maintain the feedback loop the voltage at both pins at U1 should be equal,, IR4=V is extremely small since the maximum current is 1.25Amp ( we use 8 of these circuits mosfets in parallel to distribute the power dissipation on the mosfets ) Vmax=ImaxR4=1.25Amp*0.01=0.0125V so we need to amplify this signal 400 times to make it equal to V+=5
    I hope it could help :)
     
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