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Operation of BJT in the circuit

Discussion in 'General Electronics Discussion' started by vivek20055, Nov 5, 2012.

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  1. vivek20055

    vivek20055

    7
    0
    Nov 5, 2012
    Hello,


    I like to know exactly the operation of BJT transistor in the circuit. Actually we can give directly pulse to MOSFET for switching purpose. what is the need of using BJT infront of MOSFET and giving the collector current to gate driver which drives the MOSFET.


    Thanks & regards,
    Vivek Alaparthi.
     

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  2. penfold

    penfold

    12
    0
    Oct 24, 2012
    If you look at what happens to the voltage at the collector on one side of the circuit when the MOSFET on the other side of the circuit is switched on it might give you a bit of a clue.

    The presence of the transistor will be to translate a voltage input to a current, this current affects the voltage at the collector of the BJT.

    In my opinion it doesn't look like a particularly well designed circuit, I can image a fair amount of noise spikes at the drain's of those FETs, and feed directly (albeit through a diode) into the input of an op-amp.
     
  3. vivek20055

    vivek20055

    7
    0
    Nov 5, 2012
    Hello DickCappels,


    Thank you for your reply,

    This circuit is a external synchronous self oscillator ZVS converter.
    As long as voltage at point Vg1 don't go below V1 * R5/(R5+R8) = 2.44V the MOSFET will be OFF. So we can force this by switching ON the BJT or circuit can do it
    "himself" via D3 or D2 if the BJT is OFF.

    Can you suggest me, how to replace the BJT with a AND gate.

    Thanks & regards,
    Vivek Alaparthi
     
  4. vivek20055

    vivek20055

    7
    0
    Nov 5, 2012
    operation of the circuit

    Hello,

    What is the operation R14 and R15 resistors in the circuit. I also need to know what is the operation of R6 and R7 resistors in the circuit. How the values of the
    resistors are specified in the circuit?



    Thanks & regards,
    Vivek Alaparthi
     

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  5. vivek20055

    vivek20055

    7
    0
    Nov 5, 2012
    How to calculate resistor values in the induction heating circuits?

    Hello,


    I am new to electronic circuits?. I need to calculate resistor values in the following circuit. Can anyone suggest me something about this?


    Thanks & regards,
    Vivek Alaparthi
     

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  6. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

    25,418
    2,788
    Jan 21, 2010
    Sure.

    You find the person that designed it and you ask them.

    or

    You find the circuit that you reverse engineered this from and you measure them

    or

    You go back to the web sire that published this and you look for them.

    But what you don't do is start a new thread when you have a perfectly good one already.
     
  7. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

    25,418
    2,788
    Jan 21, 2010
    Come clean vivek20055.

    You seem to have a circuit that you know nothing about, that you're making random changes to, and then asking us the function of parts involved in your own modifications. And then you ask what values are used in the original circuit!

    We can't give you any sensible advice.

    What are you trying to do? (And what is this circuit anyway?)

    edit: Oh, and you keep starting more and more threads about it to fully confuse us.
     
  8. CDRIVE

    CDRIVE Hauling 10' pipe on a Trek Shift3

    4,960
    648
    May 8, 2012
    This is an odd thread. :rolleyes:

    Chris
     
  9. CocaCola

    CocaCola

    3,635
    5
    Apr 7, 2012
    He/she is cross posting replies across multiple forums...

    Conjoined Sister Post
     
  10. CDRIVE

    CDRIVE Hauling 10' pipe on a Trek Shift3

    4,960
    648
    May 8, 2012
    Ugh! How irritating! We need an animated smiley with two guys. One has a bat!

    Chris
     
  11. CocaCola

    CocaCola

    3,635
    5
    Apr 7, 2012
    A little less violent...
     

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    Last edited: Nov 10, 2012
  12. CDRIVE

    CDRIVE Hauling 10' pipe on a Trek Shift3

    4,960
    648
    May 8, 2012
    That works for me. ;)

    Chris
     
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