1. Welcome to our site! Electro Tech is an online community (with over 170,000 members) who enjoy talking about and building electronic circuits, projects and gadgets. To participate you need to register. Registration is free. Click here to register now.
    Dismiss Notice

Class problem help (Converter U-I)

Discussion in 'Homework Help' started by Razvan, Jun 5, 2017.

  1. Razvan

    Razvan New Member

    Joined:
    Jun 5, 2017
    Messages:
    5
    Likes:
    0
    Location:
    Piteşti, România
    Hello everyone, I have this problem where i need to propose the supply/input voltage and the output current (that I want to obtain) and then to calculate and choose the correct components so my circuit can work.
    asd.png

    So, I choose the supply to 5V and the output that i want to obtain to 20mA. From my calculations, I got R = 250ohms, and i choose it 270ohms (standard). But now, my teacher wants to calculate the values for the NPN transistor and choose one properly (I choosed bc107a, but i dont think it's good, because i didn't make any calculations). I also tried to do my best with the schematic in orcad.
    asd2.png

    Can i get a little help?
     
  2. ericgibbs

    ericgibbs Well-Known Member Most Helpful Member

    Joined:
    Jan 4, 2007
    Messages:
    21,220
    Likes:
    644
    Location:
    Ex Yorks' Hants UK
    ONLINE
    hi R,
    Using a 5V supply and a 1K collector resistor will limit the possible current to < 5mA.
    I would use a higher voltage say +12v and also lower the 1K in the collector.
    OK.?
    E
     
  3. MikeMl

    MikeMl Well-Known Member Most Helpful Member

    Joined:
    Mar 17, 2009
    Messages:
    11,099
    Likes:
    562
    Location:
    AZ 86334
    How do you get 5V at the emitter of a transistor when it's base can only go to 5V? (Hint: what is the Vbe of any NPN transistor?

    Where do you buy an opamp that can pull its output pin to 5.00V when its supply voltage is 5.00V?
     
  4. dave

    Dave New Member

    Joined:
    Jan 12, 1997
    Messages:
    -
    Likes:
    0


     
  5. Razvan

    Razvan New Member

    Joined:
    Jun 5, 2017
    Messages:
    5
    Likes:
    0
    Location:
    Piteşti, România

    I think it's the diode that messes up everything.. Any suggestions please? I just need to make this circuit functional.
    [​IMG]
    Just ignore my orcad simulation.
     
  6. ronsimpson

    ronsimpson Well-Known Member Most Helpful Member

    Joined:
    Apr 17, 2007
    Messages:
    7,308
    Likes:
    969
    Location:
    Loveland, CO USA
    Diode is not a problem.
    Problem: +V must be larger than Vi.
    The voltage across R = Vi. For the transistor to work the voltage at the transistor's collector must be higher than Vi.
    There is voltage across Rl.
    If +V=12V, and Vacross Rl =2 volts then the transistor collector voltage will be 10V. Now a Vi of 0 to 5V is good.
     
  7. Razvan

    Razvan New Member

    Joined:
    Jun 5, 2017
    Messages:
    5
    Likes:
    0
    Location:
    Piteşti, România
    So.. i changed a lot, but now, i'm more confused about the Vi: doesn't matter it's value, the opamp output is always the same 12v like it's supply.
    Screenshot_1.png
     
  8. ericgibbs

    ericgibbs Well-Known Member Most Helpful Member

    Joined:
    Jan 4, 2007
    Messages:
    21,220
    Likes:
    644
    Location:
    Ex Yorks' Hants UK
    ONLINE
    hi R,
    Check the supply polarities.??
    A002.gif
     
  9. audioguru

    audioguru Well-Known Member Most Helpful Member

    Joined:
    Mar 16, 2004
    Messages:
    32,521
    Likes:
    943
    Location:
    Canada, of course!
    It is simple to calculate the maximum input voltage so that the opamp and transistor work properly:
    I deleted the attachment because it was calculated wrong.
     
    Last edited: Jun 6, 2017
  10. audioguru

    audioguru Well-Known Member Most Helpful Member

    Joined:
    Mar 16, 2004
    Messages:
    32,521
    Likes:
    943
    Location:
    Canada, of course!
    Your new circuit has the output voltage of the old 741 opamp much higher than shown on its datasheet. With a +12V supply its maximum output voltage is typically about +10V. Then the transistor's emitter voltage will be about +9.3V and the emitter current will be 9.3V/270= 34.4mA.
    But 34.4mA in the 1k load resistor needs a voltage of 34.4V that is not available.

    The maximum current is 9.8V/1270 ohms= 7.7mA. Then the max emitter voltage is 2.1V and the max output of the opamp is +2.8V. The max input voltage is +2.1V.
     
  11. audioguru

    audioguru Well-Known Member Most Helpful Member

    Joined:
    Mar 16, 2004
    Messages:
    32,521
    Likes:
    943
    Location:
    Canada, of course!
    Here is your new circuit with simple corrections:
     

    Attached Files:

  12. Razvan

    Razvan New Member

    Joined:
    Jun 5, 2017
    Messages:
    5
    Likes:
    0
    Location:
    Piteşti, România
    Thanks a lot guys!!!
    I got it now, so: i choosed my Vi=5V and the output current (Ie) that i wanted to obtain (≈20mA), because this circuit it's an U-I Converter. AND i got the Ie to 18.52mA, which is very ok and i'm happy.
    Screenshot_1.png

    Now i have calculate the VCE on the transistor (with formulas and stuff) and choose the correct transistor from the catalog (i put BC107a because it's the only transistor i know :banghead:). Also, do you know what is the role of the diode in the whole circuit?
     
  13. audioguru

    audioguru Well-Known Member Most Helpful Member

    Joined:
    Mar 16, 2004
    Messages:
    32,521
    Likes:
    943
    Location:
    Canada, of course!
    I showed you that your +5V input voltage is too high so the transistor is not controlling the current anymore when the input to the opamp is higher than +2.1V.

    Your original circuit showed no power supply for the opamp so assuming it is the +5V and ground then the diode did nothing.
    Now the opamp is powered from +12V and -12V so the diode prevents the output of the opamp from going negative enough to destroy the emitter-base of the transistor (max allowed is only 6V).
     
  14. audioguru

    audioguru Well-Known Member Most Helpful Member

    Joined:
    Mar 16, 2004
    Messages:
    32,521
    Likes:
    943
    Location:
    Canada, of course!
    R2 is 1k and is the load. The emitter resistor is not the load. Ohm's law shows that 20mA in 1k needs a voltage across the 1k to be 20V. But the supply for the transistor is only +12V, far from 20V.
    You correctly show the emitter current at 18.52mA but then the 1k collector resistor must have the same current then it will have 18.52V across it which is impossible.

    Why don't you understand that when the transistor is turned on as hard as it can then increasing the input voltage does not allow the transistor to control the current anymore?
     

    Attached Files:

  15. Razvan

    Razvan New Member

    Joined:
    Jun 5, 2017
    Messages:
    5
    Likes:
    0
    Location:
    Piteşti, România
    I understand now. Tomorrow i'll go to the teacher with the homework and I'll come back with a reply. Thanks a lot!!
     
  16. ci139

    ci139 Active Member

    Joined:
    Apr 12, 2012
    Messages:
    469
    Likes:
    37
    here's what i came up from ccxp-test_ES00-0010.png newer built coz don't like the 10+W going smoking up to the air (the switched v. requires a pulse-TF - but the variyng voltages complicate the design ....)
     
    • Dislike Dislike x 1
  17. Cicero

    Cicero Active Member

    Joined:
    Nov 21, 2014
    Messages:
    335
    Likes:
    36
    Location:
    UK
    What the actual heck?!?
     
  18. ci139

    ci139 Active Member

    Joined:
    Apr 12, 2012
    Messages:
    469
    Likes:
    37
    is the src. website - a tiny text below - https://www.quora.com/Can-we-use-voltage-source-to-generate-constant-current - that poped out from my recent web search for high amperage cc circuit - i found that example too . . . NOT constant current -- so ended up with that
    that is a bit more constant -- as the OP had the same circuit here -- i thought it may give some insight to the subj.
    . . .
    i don't understand your point the circuit is self describing - if you can read schematics - there should be no fuzz - if you think my input is irrelevant - vote it down , simple
     
    Last edited: Jun 8, 2017

Share This Page