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Oscillating power transistor??

Discussion in 'Electronic Basics' started by Don A. Gilmore, Sep 18, 2004.

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  1. Hi guys:

    I'm working with a circuit where I take a square wave signal from an op amp
    comparator and use it to drive an inductive load. Obviously, the op amp
    can't handle much current, so I connect this output to the base of a power
    transistor and drive the inductive load with it.

    The problem is that, for some reason, I get a spontaneous oscillation from
    the circuit when I'm not even applying any signal. It's a strong signal and
    shows up clearly on the scope. I would filter it out except that it is
    relatively close in frequency to the signal I plan to apply to the inductor
    (on the order of hundreds of hertz).

    What is causing this and what should I do to eliminate/prevent it? Thanks
    for all replies.

    Kansas City
  2. If your driver applied a voltage across the inductor, and holds this
    voltage till a current develops through the inductor, and then turns
    off, the inductor must ring at its self resonant frequency. to
    produce a controlled voltage across the inductor, you must apply
    voltage to it from a source that has a low impedance at all parts of
    the waveform. Without seeing a schematic of your circuit, I have no
    way to know if it does this.
  3. Ban

    Ban Guest

    It is probably that your opamp is not able to swing close enough to the
    ground rail and the power transistor cannot switch off.
    In the first place you need a 1k resistor between the opamp output and the
    base, and another 1k resistor from base to ground. Then you will need a
    diode across your inductive load to prevent the transistor from avalancing,
    when it switches off. the anode to the collector, cathode to +
  4. CFoley1064

    CFoley1064 Guest

    Subject: Oscillating power transistor??
    Hi, Don. Try this Kansas City Standard method (view in fixed font or M$

    + +
    | |
    | |
    - C|
    ^ C|
    | C|
    | |
    VCC | |
    + '---o
    | |
    |\| 2.2K |
    -|-\ ___ |/
    | >-->|--|___|-o -|
    -|+/ | |>
    |/| .-. |
    | 2.2K| | |
    === | | |
    GND '-' |
    | |
    === ===
    created by Andy´s ASCII-Circuit v1.24.140803 Beta

    The diode in parallel with the load will help absorb the inductive kick. The
    diode in series with the op amp output will help ensure complete turnoff if
    your op amp doesn't go down to the negative rail. The resistors are necessary
    because transistors are current-driven devices.

    Good luck
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