Would these two circuits perform the same?

Discussion in 'The Projects Forum' started by imbaine13, Jan 4, 2014.

  1. imbaine13

    Thread Starter Member

    Oct 6, 2013
    62
    0
    Hello everyone.

    Below are two options for the driver circuit of a square wave inverter power stage.
    Would these two produce the exact same wave form?
    The only difference is one uses a PNP darlington pair while the other uses an NPN darlington pair.
    Please help, I need your opinions asap.
    Thanks guys.

    Driver 1.JPG

    Driver 2.JPG
     
  2. Athineos

    New Member

    Dec 18, 2013
    9
    1
    I think the only problem has to do with the different polarity of the produced pulses.
     
  3. tubeguy

    Well-Known Member

    Nov 3, 2012
    1,157
    197
    You can eliminate the emitter to base resistor between T1 and T2 in both circuits.
    Google Darlington transistor.
     
    imbaine13 likes this.
  4. Ramussons

    Active Member

    May 3, 2013
    557
    92
    If you do a deep core analysis, NPN transistors perform better than PNP ones.

    In a Generic circuit that you have shown, both will perform equally well.

    Ramesh
     
    imbaine13 likes this.
  5. ScottWang

    Moderator

    Aug 23, 2012
    4,853
    767
    Choosing the first circuit, but using two seperated transistors to replace the darlington pair, T1 change to NPN, T2 still using PNP.

    The specs of your circuit didn't clear, so you just try the values by yourself.
     
    imbaine13 likes this.
  6. MikeML

    AAC Fanatic!

    Oct 2, 2009
    5,450
    1,066
    Depends on the supply voltage?
    Depends on the amplitude of the driving signal?
     
  7. bountyhunter

    Well-Known Member

    Sep 7, 2009
    2,498
    507
    I don't think either circuit will work worth beans because the transistor T3 only has active pull up (turn on) at it's base and nothing to turn it off by actively pulling down.
     
    imbaine13 likes this.
  8. alexfreed

    Member

    Oct 8, 2012
    72
    10
    I disagree with bountyhunter: the BJT's emitter junction is a low resistance sink: on the order of 25 ohm.

    The first circuit is better in one respect: the output transistor can be saturated, i.e. collector can go down to say 0.3 volts or so. The second schematic is not good in this respect due to 3 times 0.7 volt drops. More power will heat up T3.

    Also I'm not sure 3 transistors are needed. The output power transistor will have a gain of say 30. Needs 100 ma to drive a 3 amp load. Easy with a single driving transistor. And last but not least power MOSFETS are cheap and better for switching applications.
     
    imbaine13 likes this.
  9. bountyhunter

    Well-Known Member

    Sep 7, 2009
    2,498
    507
    The OP said:

    He wants a square wave. He doesn't spec the frequency, but if it's some kind of square wave power inverter, it's probably going to have to be 20 kHz or higher. I guarantee you, the turn off edges of that output will look like crap because it lacks any turn off drive. It will not be a good square wave.
     
    imbaine13 likes this.
  10. imbaine13

    Thread Starter Member

    Oct 6, 2013
    62
    0
    The driving signal is a square wave with max voltage 12 volts and minimum voltage 0 volts. (From the bjt astable multivibrator)
     
  11. imbaine13

    Thread Starter Member

    Oct 6, 2013
    62
    0
    Thanks bountyhunter,
    Hadn't seen that! I'll add that, I'm thinking a 1000Ω resister would work.
     
  12. imbaine13

    Thread Starter Member

    Oct 6, 2013
    62
    0
    Thanks Alexfreed, I dont understand how the three 0.7 voltage drops would affect T3. Would you please elaborate on that? Actually, T3 has a gain of 20 minimum (2N3055) and in the schematics, I would have loved that it full turned on. I'm seriously considering MOSFETs as a replacement for T3 though.
     
  13. bountyhunter

    Well-Known Member

    Sep 7, 2009
    2,498
    507
    WHAT FREQUENCY?

    My point is that without some kind of strong active turn off on the power transistor's base, the turn off wave edge will not be sharp. This is the #1 cause of switching power losses in any switching power converter. If your frequency is 100 Hz, no worries, 10 kHz or above, sluggish turn off edge may be noticeable.
     
Loading...