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Discussion in 'Electronic Basics' started by KBSoftware, Aug 12, 2007.

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  1. KBSoftware

    KBSoftware Guest

    I a bit new to electronics, about 1.

    Now I'm learning about coils, and have a few questions.

    I have some coils and I know their Henry rating, I've found a lot of
    information on coils.
    Now I'm wondering if I took a coil let's say a 55uH applied maybe 3v,
    what would be the voltage resulting. I've seen a coil used in a 3v to
    9v dc-dc converter and that's why I'm wondering. I think the coil
    increases the 3v to some other voltage (the circuit used a 47uH), then
    an ic was used to regulate the output from the coil to 9v about.
    Is there a formula that could help me here.

    Or could I just build that simple circuit and using a multimeter
    safely read the voltage and amps that is being produced. What I'm
    worried that the voltage being produced is high and might damage my
  2. Tom Biasi

    Tom Biasi Guest

    To DC the coil only has wire resistance. When the DC is removed the magnetic
    field will try to keep the current constant and the voltage will spike.
    At AC the reactance of the coil is frequency dependant.
    Please look into inductance some more.
    Look into inductive reactance and impedance.

  3. Nobody

    Nobody Guest

    In electronics, they're generally referred to as "inductors":
    A step-up ("boost") converter works by allowing a current to flow through
    the coil into a low impedance, then redirecting that current through a
    relatively higher impedance:

    An essential property of an inductor is that the current cannot change
    instantaneously, so any increase in the series impedance produces a
    corresponding increase in voltage across that impedance.

    The output voltage of a boost converter depends upon the input
    voltage, inductance (Henries), switching periods (open/closed) and the
    load resistance. Because of the variation of output voltage with load
    resistance, practical boost converters typically use negative feedback to
    vary the switching periods in order to maintain a constant output voltage.
  4. default

    default Guest

    You have a coil and have a source of DC voltage through the coil - you
    interrupt the flow and the collapsing magnetic field attempts to keep
    the current flowing - no matter how much voltage that may take.

    In theory.

    In practice, the speed of change in magnetic field determines how high
    a voltage you can achieve. Just like a generator - faster the field
    changes the higher the voltage. The coil form - wire thickness - and
    other construction influences determine field collapse time.

    I use a 3 millihenry choke in a 2 volt DC NPN collector circuit
    switching at 10 KHZ and get 80+ volts out, and if I don't tame it, it
    eats the 30 volt transistor doing the switching.

    You won't damage the multimeter. You may get a reading that makes no
    sense - that's HF noise impressed on the signal you want to read. Use
    an old fashioned moving coil meter if you have doubts or filter the
    voltage to the multimeter (not always easy - high impedance meters
    pick up noise from the environment - that's the nature of things).

    Show us the circuit ?

    I use mine to get 10 volts from a 2.5 volt source so I can switch a
    mosfet that requires 8+ volts.
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