designing current amplification using darlington pair

Discussion in 'Homework Help' started by mahmoudsamneh, Jun 26, 2017.

  1. mahmoudsamneh

    Thread Starter New Member

    Jun 25, 2017
    8
    0
    hello
    looking for a help i need to design a circuit that amplify the current
    the input for it will be 0.6213 amp and its input voltage is about 24 volt
    so i need to get an output to about 3 amp and 24volt approx
    from where should i start calculation ?
    thanks in advance
     
  2. absf

    Senior Member

    Dec 29, 2010
    1,850
    505
    First using P=I x V

    Input power = 0.6213 x 24 = 14.9112W
    Output power = 3 x 24 = 72W

    No electronic components or circuits is able to do that. It's like asking a baby weighting 8 Kg to lift a 40 Kg stone.

    Allen
     
  3. mahmoudsamneh

    Thread Starter New Member

    Jun 25, 2017
    8
    0
    am designing a 48-24 v battery charger so i connected an ac source with 220v/50hz values with a transformer to step down the voltage to 48 v
    after that i used full wave rectifier circuit(bridge) to convert the signal to dc measring the current yields to about 15 amp but after filtering process it drops down to 0.6213 at 24 v
    that's why i'm thinking of using a darlington pair
    any hints that could help me
     
  4. MrAl

    AAC Fanatic!

    Jun 17, 2014
    5,131
    1,111
    Hi,

    What made you think you need a darlington pair?

    I ask because the inverse ratio of 0.6 amps to 3 amps is 30/6=5, so you only need a current gain of 5.
    What is unfortunate however is that if you really need 24 volts out with only 24 volts in and that extra current, you might need a boost circuit not a transistor. If you can live with maybe a 22v output then a transistor might work ok.

    If after filtering it drops down by so much though you should probably post the entire circuit because something else is happening also.
     
    absf and JoeJester like this.
  5. mahmoudsamneh

    Thread Starter New Member

    Jun 25, 2017
    8
    0
    i made a simulation to the circuit using matlab
    are you familiar with it ?
     
  6. JoeJester

    AAC Fanatic!

    Apr 26, 2005
    3,898
    1,731
    Post a screenshot of the schematic.
     
  7. MrAl

    AAC Fanatic!

    Jun 17, 2014
    5,131
    1,111
    Hi,

    There are a lot of simulators and not everyone uses Matlab, so if you post your circuit as Joe suggested we can then all take a look at it. That would help us understand your problem better.
     
  8. crutschow

    Expert

    Mar 14, 2008
    19,549
    5,449
    What you mean by that.
    What is the "filtering process"?
    What is the load at the output of the filter?
    Is 0.6213 the amperes into the load?
     
  9. mahmoudsamneh

    Thread Starter New Member

    Jun 25, 2017
    8
    0
    here is the full spec. used
    ac voltage source 220v/50hz
    transformer step down 220 to 48 v / nominal power 660 w / r1=r2=0 l1=l2=0 / rm=3 ohm / lm=2 h current at the primary side 74.7826 amps
    at the secondary part 11.7134amps
    4 diodes have 0.001 resistance and vf=0.7 v each
    the capacitor = 500e-5 f
    the current after capacitor 0.3442amps
    the resistance 50 ohm
    output voltage 29 volt aprox
    battery 24 volt and 0.32 ohm
    the circuit output should be 24 volt and 2 amps battery.png
     
  10. mahmoudsamneh

    Thread Starter New Member

    Jun 25, 2017
    8
    0
    1 using capacitor to purify the dc signal
    2. resistance 50 ohm and a battery to be charged 24 volt and 0.32 ohm
    3.yes but some how it becomes 0.3442 things are getting messy
    please take a look at the diagram above
     
  11. crutschow

    Expert

    Mar 14, 2008
    19,549
    5,449
    Things are getting messy since you don't appear to understand Ohm's law.
    How do you expect to get 3A through 50 ohms?
    That would require 150V. :eek:
    If you are trying to limit the battery charge current, you need a much smaller resistor.
     
    absf likes this.
  12. mahmoudsamneh

    Thread Starter New Member

    Jun 25, 2017
    8
    0
    i decreased resistance to 0.05 the current increased to 5.8 amps but the voltage equals to about 48 volt
    why i used 50 ohm in order to make the drop voltage i know the current would be in e-3amps so i decided to amplify the current using darlington pair
    or other solution that you guys might help me with
     
  13. crutschow

    Expert

    Mar 14, 2008
    19,549
    5,449
    Why not use a resistor value that gives you the voltage desired?
    Why did you use 50Ω?
    Do you know Ohm's law relating voltage, current, and resistance?
    If not, you need to learn it.
     
  14. mahmoudsamneh

    Thread Starter New Member

    Jun 25, 2017
    8
    0
    did you see the circuit diagram?
    am not trying to break ohm's law
    am saying i have a 48 volt at the secondary part of the transformer
    then i need at the output which in the circuit diagram is the battery to be 24 volt
    so i did use a 50 ohm to make a drop voltage that leads to let the battery have a 24 volt
    but the current become so small because of the resistance value
    so my main topic was asking about darlington pair and how to start designing and doing the math in order to amplify the current
     
  15. crutschow

    Expert

    Mar 14, 2008
    19,549
    5,449
    48Vac rectified will give about 65Vdc.
    If you want 2A @ 24V to the battery, then the needed series resistance (from Ohm's law) is (65-24) / 2 = 20.5Ω, so you need about a 20Ω, 100W resistor.
    So how did you arrive at 50Ω? :confused:

    If you want to use transistors to generate a regulated output voltage, that's a different thing.
     
  16. JoeJester

    AAC Fanatic!

    Apr 26, 2005
    3,898
    1,731
    Can you provide the exact assignment as written by the school or book?

    What kind of battery are you charging? do you have a particular model in mind?

    Have you visited Battery University?
     
  17. crutschow

    Expert

    Mar 14, 2008
    19,549
    5,449
    The difficulty is, you want us to implement your solution (which may not be the correct one) rather than help you to solve your problem (which is still not clear).
     
Loading...