I (finally) succeeded on H-Bridge! ...now where am I losing 1.3v?

Discussion in 'The Projects Forum' started by masked, Jul 21, 2010.

  1. masked

    Thread Starter Member

    Jul 1, 2010
    48
    1
    Hi all,

    I finally succeeded with my one-input H-bridge by using the low voltage rail to power a PNP. It took 9 transistors, but works with both 12V and 27V :)
    ...but I can't track down which transistors are making me lose the 1.4V on the high output cycle.

    Any ideas? Here it is. (you can simulate it at http://www.falstad.com/circuit/index.html by pasting the attached .txt)
    [​IMG]
     
  2. Wendy

    Moderator

    Mar 24, 2008
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    All transistors loose 0.6 to 0.7V on the BE drop. If you have a Darlington pair you loose twice that, since Darlington transistors have two BE drops. This is also a fundamental problem with 555s, which use a Darlington Pair as part of their design.

    Another similar configuration called a Sziklia Pair has some of the same advantages and disadvantages, except it only has one BE pair.

    Just looking at your diagram again it doesn't quite have a Darlington pair, but it does have the two BE junctions.
     
  3. masked

    Thread Starter Member

    Jul 1, 2010
    48
    1
    I adjusted the resistors for better efficiency, but this is bascially the same circuit with labels...
    The question came up because I was suprised to NOT be losing voltage in the low state, and I expected both halves of the bridge to behave the same.

    After more fiddling, apparently the "dropped" voltage from Q1-Q4 returns back to the main circuit at the bottom. So if I replace the central resistor with a plain wire, I get a full 27V in both polarities.

    I'm still confused why adding the resistor only lowers the voltage in low state. Apparently this is something to do with the 6k resistor by Q3, whereas the "high state side" of the bridge is missing the analogous resistor.
    [​IMG]
     
  4. masked

    Thread Starter Member

    Jul 1, 2010
    48
    1
    One other question:

    It looks as if my bridge drivers will draw the same amps as the motor. So presently the motor and the bridge each draw 2.5mA for a total of 5ma.
    (You can drop them to 1.25 by doubling the lower resistor to 2k).

    Anyway, is that an efficiency concern for the bridge circuit to be drawing the same amps as the driven device?

    thanks again,
    -masked
     
  5. Wendy

    Moderator

    Mar 24, 2008
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    You have a dead short Collector Base on Q1 and Q2, this is not good. You may have the schematic drawn wrong too, it looks like you're going for a Darlington Pair.

    About your PM, You're welcome.

    What is your power supply voltage, it's not shown. Neither is your zener voltages. I might try redrawing it in a bit more legible format once we have the gophers nailed.
     
  6. Wendy

    Moderator

    Mar 24, 2008
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    True H bridges double the effective voltage (and it looks like AC) to their loads. Look at post #2 on AC/DC Inverters
     
  7. masked

    Thread Starter Member

    Jul 1, 2010
    48
    1
    Thanks, that's a great thread :)

    Turns out I just built a DC to AC inverter ...with a frequency of .01667Hz!
     
  8. SgtWookie

    Expert

    Jul 17, 2007
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    You're using Darlington configurations.

    You'll get voltage drops (and high power dissipation) as long as you use them.
     
  9. masked

    Thread Starter Member

    Jul 1, 2010
    48
    1
    Sorry about the drawing - I've just been using that simulator a lot since I can see the current flowing and change parts faster than on a breadboard.

    I've tested the power supply from 12V to 27v and varied the top left Zener to 7 less than the source. The other zener is overkill at 12V, and should eventually be a regular diode.

    Im building the real one on a breadboard at 27v at the moment, and will find out if the simulation holds up. -after that I'll do a proper schematic in paint.

    -masked
     
  10. Wendy

    Moderator

    Mar 24, 2008
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    You working the same problem? Try this instead...

    [​IMG]

    I'm using the second 555 as a straight forward inverter.

    Be careful though, the voltage across the load will be 60V P-P, enough to cause injury. The outputs can only handle 100ma before thing go wrong.
    .
    .
     
  11. masked

    Thread Starter Member

    Jul 1, 2010
    48
    1
    It's basically the same, except I'm only alternating the current direction once per minute, so the fact that it was an inverter hadn't occurred to me. (The timer cap is actually 100uF and the resistors are 47k and 800k in real life)

    I only have one timer so I can't make yours...and I only have 6 transistors so I can't make mine. I'm gonna start a beginners collection of spare parts and I'll build them both. I just need something-anything to build as a goal or I'll never get around to learning.

    I'd shied away from 2 timers because I didn't want to synchronize them... now that I see your schematic, connecting the 2nd chip's trigger to the output is much simpler than the 2 timing circuits I had in mind.

    Thanks much for the help!
    -masked
     
    Last edited: Jul 22, 2010
  12. Wendy

    Moderator

    Mar 24, 2008
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    What part of the world are you? Most of the old hands have added their locations on the corner, this helps on several fronts, getting parts for example.
     
  13. SgtWookie

    Expert

    Jul 17, 2007
    22,182
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    Masked,
    Try Linear Technology's LTSpice.

    It's a free download; Google is your friend.

    Join Yahoo!'s LTSpice User Group. It's also free, and there is a lot of help and support available.
     
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