How does this work?

Discussion in 'General Electronics Chat' started by roadey_carl, Jul 14, 2010.

  1. roadey_carl

    Thread Starter Active Member

    Jun 5, 2009
    116
    5
    Hello again folks, just a quick one...

    I want to make this inverter circuit but I don't understand how it works connecting the Base of a PNP transistor to a positive output? Does it conduct on the flat line wave and the NPN conducts on the square wave?? :confused:
    http://www.sentex.ca/~mec1995/circ/555dcac.html

    And another thing, R4, I read it as..... 100kΩ to pin 1 of the pot, pin 2 of the pot connected to pin 2 and 6 of the 555, and pin 3 of the pot connected to ground via 0.1uf cap?


    Thanks alot for your help!:)

     
  2. bertus

    Administrator

    Apr 5, 2008
    15,638
    2,343
    Hello,

    This thread from last week is about almost the same circuit:
    555 timer inverter

    Bertus
     
  3. nomurphy

    AAC Fanatic!

    Aug 8, 2005
    567
    12
    What you refer to as pin-3 of the pot is also connected to pins 2, 6 on the IC along with the 0.1uF cap. Combined with R2, this will provide from 100K to 150K of resistance.

    When the 555 output pin-3 is low it will turn on Q2, when it is high it turns on Q1.
     
  4. roadey_carl

    Thread Starter Active Member

    Jun 5, 2009
    116
    5
    Ahhh I see!! Thanks alot, that makes sence :D
     
  5. Wendy

    Moderator

    Mar 24, 2008
    20,764
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    The transistors are a modified Common Collector configuration. The output (emitter) of the transistors follows the input voltage minus (or plus) the BE drop (0.6VDC). Since both transistors are trying to follow the input voltage you have a significant drive.

    There are much better ways of doing this. For such a crappy circuit this one has made the rounds. We keep getting threads asking why it doesn't work as advertised. This is because it can't, I doubt it was ever built, and no thought was given to the various Base Emitter drops. The 555, for example, has 1.3 VDC internal base emitter drop, since it's output uses a Darlington pair to Vcc. This means the positive output of a 555 can only be within 1.3V (give or take a little) of Vcc. Together all these losses add up in a big way.
     
    John Berry likes this.
  6. Audioguru

    New Member

    Dec 20, 2007
    9,411
    896
    No.
    When a power transistor like a 2N3055 has a load current then its max base-emitter voltage is as high as 1.5V when its current is only 4A and is higher with more current. The PNP transistor has the same voltage loss or more.
     
    John Berry likes this.
  7. sage.radachowsky

    Member

    May 11, 2010
    241
    38
    Can anyone here recommend a good inverter circuit that is relatively simple? I'm not the original poster, but I am interested in this resource. Thanks.
     
  8. Audioguru

    New Member

    Dec 20, 2007
    9,411
    896
    A simple inverter is too simple:
    1) Its output voltage is not regulated so is too high when the battery is freshly charged and when the load is low. Its output voltage is too low when the battery charge is running down and when the load is heavy.

    2) Its output is a simple square-wave instead of a sine-wave like mains electricity so electronic products that depend on the much higher peak voltage of a sine-wave and motor speed controls will not work.

    3) It has no protection circuits so it blows a fuse or blows-up if overloaded.

    4) It has no low voltage shut-down so it kills the battery by discharging it too low.
     
    John Berry likes this.
  9. Wendy

    Moderator

    Mar 24, 2008
    20,764
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    What are you trying to power? Square waves are not good for most applications.

    Also, what are your power requirements, 50Hz 220AC?

    You might consider starting another thread. Hijacking (which is what this is) is actively discouraged.
     
    John Berry likes this.
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