help on simple power supply using circuit wizard

Discussion in 'Homework Help' started by Johnson Andres Gangob, Jan 26, 2019.

  1. Johnson Andres Gangob

    Thread Starter New Member

    Jan 24, 2019
    4
    0
    good day

    i'm new in using circuit wizard and i'm having a hard time setting up a simple power supply, especially on the transformer windings. I can't get a 3 amps 24v during simulation.

    given:
    transformer:
    primary: 240v 1amps
    secondary:24v 3 amps
     
  2. ericgibbs

    Moderator

    Jan 29, 2010
    7,668
    1,546
    hi JAG,
    Welcome to AAC.
    I do not use Circuit Wizard, but where are you measuring the Vdc output from the bridge rectifier, also where is the smoothing capacitor.?
    Mark it on your diagram.
    E
     
  3. Johnson Andres Gangob

    Thread Starter New Member

    Jan 24, 2019
    4
    0
    thank you, here is the updated diagram. i measure the vdc and i got 23v, however, i only get 100mA using an altimeter. in addition, when i replace the voltmeter on the circuit wizard with an filter capacitor 1000micro farad 100v and put an fuse, all components explode
     
  4. jayanthd

    Active Member

    Jul 4, 2015
    863
    72
    There is no load resistor and ground connection in bridge rectifier circuit.
     
    Last edited: Jan 26, 2019
  5. Johnson Andres Gangob

    Thread Starter New Member

    Jan 24, 2019
    4
    0
    where should i put it and what should be the value.
    btw can you recomendme a free electronic offline simulator other than circuit wizard, it seems i cant change the transformers output current
     
  6. jayanthd

    Active Member

    Jul 4, 2015
    863
    72
    What load current you want ? See attached video.

    Extract the .rar file using WinRaR 5.x and view the .wmv video file in Windows Media Player or VLC Player.

    AC mains voltage = 240V rms 50 Hz
    Trf ratio is 10:1 step-down
    Secondary voltage is 24V AC rms
    DC output voltage across load = (24V * 1.4142) - 1.4 [2 X 0.7, (0.7 = 1 diode drop] = 33.9408 - 1.4 = 32.5408V DC.

    You cannot change transformer current without changing the load resistor.

    Vpri/Vsec = Npri/Nsec = Isec/Ipri

    Set RL in my circuit to get the required secondary and primary currents.

    I don;t know about any free offline electronic circuit simulator. I have licensed version of Proteus. I don't use Circuit Wizard much.





    CktWiz_2019-01-27_00-36-30.png
     
    Last edited: Jan 26, 2019
  7. jayanthd

    Active Member

    Jul 4, 2015
    863
    72
    Primary currents depends upon the secondary current and secondary current depends upon load RL here.

    Vpri/Vsec = Isec/Ipri = Npri/Nec

    240/24 = 10 is not equal to 3A/1A = 3 but 240/24 = 10 = 10:1 ratio that is Npri/Nsec = 10/1; 10 turns of primary = 1 turns of secondary.

    If you want Isec = 3A rms then

    Ipri = Isec / (Vpri/Vsec) = 3A/(240V/24V) = 3A/10V = 0.3A = 300mA.

    Edit:

    LaTeX

    {Ipri = Isec / (Vpri/Vsec) = 3A/(240V/24V) = 3A/10 = 0.3A = 300mA}
     
    Last edited: Jan 26, 2019
  8. jayanthd

    Active Member

    Jul 4, 2015
    863
    72
    Reply to PM (Private Message).
     
  9. bertus

    Administrator

    Apr 5, 2008
    19,507
    3,972
    Hello,

    You made a mistake in your explanation:

    Ipri = Isec / (Vpri/Vsec) = 3A/(240V/24V) = 3A/10V = 0.3A = 300mA.

    There can not be V at the given point.

    Ipri = Isec / (Vpri/Vsec) = 3A/(240V/24V) = 3A/10 = 0.3A = 300mA.

    Bertus
     
  10. jayanthd

    Active Member

    Jul 4, 2015
    863
    72
    @ bertus

    Good catch.
     
  11. Johnson Andres Gangob

    Thread Starter New Member

    Jan 24, 2019
    4
    0
    Thank you, i further understand from your explanation how my power supply should work.
     
  12. jayanthd

    Active Member

    Jul 4, 2015
    863
    72
    AC mains voltage = 240V rms 50 Hz
    Trf ratio is 10:1 step-down
    Secondary voltage is 24V AC rms
    DC output voltage across load = ((24V * 1.4142) - 1.4) [1.4 = 2 X 0.7, (0.7 = 1 diode drop] = (33.9408V - 1.4) = 32.5408V DC.

    Ipri = Isec / (Vpri/Vsec) = 3A/(240V/24V) = 3A/10 = 0.3A = 300mA.

    In image RL (Load Resistance) = 10k and IL (Load Current) = 3.22mA

    IL = Vdc/RL = 32.5408/10k = 3.25408mA (approx same as in simulation (image)).


    CktWiz_2019-01-27_00-36-30.png
     
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