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Need insight how to use 4017 counter with MOSfet

Discussion in 'Electronic Basics' started by Vencislav, May 25, 2004.

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  1. Vencislav

    Vencislav Guest

    Do you have any insight on how to use the decade counter with
    MOSfets to pulse power through an array of coils?

    I've heard from some that a diode needs to be connected in reverse across
    the MOSfet to prevent damaged from the Lenz Law Effect (or high voltage self
    induction due to the varying magnetic field of the pulse). The back
    EMF could otherwise destroy the MOSfet.

    Thank you!
  2. Tim Wescott

    Tim Wescott Guest

    To drive it from the 4017 you just need "logic level" n-channel MOSFETs.
    You want one of two diode protection circuits. The most commonly used
    one is:

    | |
    C| |
    C| -
    C| ^
    | |
    created by Andy´s ASCII-Circuit v1.24.140803 Beta

    But your relay will close quicker if you use:

    | |
    ||-+ \-\
    ||<- ^
    ------||-+ |
    | |
    | |
    === ===
    created by Andy´s ASCII-Circuit v1.24.140803 Beta

    with the zener diode voltage set to prevent the MOSFET drain-source
    voltage from being exceeded.
  3. Since the gates of mosfets appear to the outside world much like small
    capacitors, they can be directly connected to CMOS outputs, as long as
    the frequency is not so high that the gate capacitance can not be
    charged up and down in the duration of the output states. If you use
    N channel mosfets with their sources connected to the negative supply
    rail used by the cmos, then a log ic high output will turn the mosfet
    on, and a load connected between any more positive supply rail
    (assuming that supply shares a negative side with the CMOS supply) and
    the drain will get connected when the mosfet turns on.

    The diode protection you mention involves inductive loads. An
    inductance produces voltage in proportion to the rate of change of
    current, in a direction that tends to resist such change. This makes
    it easy to switch a voltage into an inductor, because as the inductor
    current begins to rise, the inductor produces a voltage that cancels
    the applied voltage and slows that current rise. So there is no turn
    on current surge.

    But when you try ot open a switch that is carrying inductor current,
    the inductor produces voltage that adds to the power supply, to keep
    the current from instantaneously falling to zero. This extra voltage
    must be interrupted by the switch. If the switch is mechanical, this
    effect causes arching between the separating contacts. If the switch
    is a solid state device, like a mosfet, this effect can cause over
    voltage failure of the device.

    A common solution to this problem is to parallel the inductive load
    (not the switch) with a diode, connected so that the diode is reverse
    biased by the normal on state voltage. This prevents the inductor
    from producing more than a forward biased diode drop when the current
    to the inductor is interrupted and its voltage reverses. Instead of
    the current falling quickly (which is what causes the inductor to make
    all that high voltage), the current simply detours to the parallel
    diode till the resistance of the coil and the voltage drop of the
    diode slowly use up the stored magnetic energy. The most extra
    voltage the switch can be exposed to is 1 diode drop above the
    positive supply.

    Of course, this protective method has the draw back that it makes it
    impossible to turn the inductive device off very fast.
  4. Ken Smith

    Ken Smith Guest

    This is not true if this is a CD4017 which can be run from +15V and drive
    the gate high enough for non-logic level MOSFETs. The down side is that
    CD4XXX parts are slow and can't output much current.

    [.. deleted suggested good spike catchers ..]

    In some cases a series resistor + capacitor combination across the coil
    is a good way to go too. Depending on the relay, it may actually open
    faster if the current rings than if it stops suddenly.
  5. That depends on the supply voltage of the 4017. If you power it from 10V
    or more any old MOSFET will do. That's the idiocy with this marketing
    term "logic level": There are about 5 different definitions of logic
    levels in common use and umpteen more in not so common use. It just
    confuses people.
    Surely you meant to say: "will open quicker"?
  6. Tim Wescott

    Tim Wescott Guest

    OK, I forgot this one, and I've been tagged for it twice.

    The "logic level" MOSFETs seem to be specified for both 5V and 3.3V
    these days, which is nice. I haven't seen any with specs for 1.8V yet,
    though :(.
    What, you mean you use normally open relays? :)
  8. We're cruel, aren't we?
    I've seen 2.5V specs. That's a start, at least.
  9. Here's one:

    Best regards,
    Spehro Pefhany
  10. Spehro Pefhany wrote...
    Yes, there are a host of 1.8V rated power mosfets. Furthermore,
    NEC has 1.5V parts, like the uPA675T and 2sk2159, etc., as does
    Sanyo, like the mch6602, mch6615, etc. Toshiba's HN1K05FU is
    rated at 1.2V, and there are a few others rated even lower.

    And don't forget bipolar transistors, rated at under 0.65 volts!

    - Win

    (email: use hill_at_rowland-dot-org for now)
  11. Ken Smith

    Ken Smith Guest

    Try Supertex, if you don't need too much power.

    TN2501 -> 1.0V
    VN4012 -> 1.8V
  12. Ken Smith

    Ken Smith Guest

    Germanium is better.

    JFETs have a Vgsth less than zero.
  13. Ken Smith wrote...
    Actually, the TN2501 has its Ron spec'd at 1.2V, if anyone
    is counting. That's not 1.0V, but it's still a pretty low
    voltage. How low? It's so low this MOSFET's Ron suffers
    substantially. Whereas it's Ron rating is 3.5 ohms max for
    Vgs = 2.0V, which is moderately useful, it degrades by over
    7x at 1.2V, all the way to 25 ohms. Whew, it's nearly off!

    Even though the FDG327N that Spef suggested isn't spec'd at
    1.2V (it's rated 0.14 ohms max at 1.8V), its curves (fig 4)
    show it'll work quite well there, typically about 0.2 ohms.

    The VN4012 is spec'd at 4.5V, and it has a fairly high
    threshold voltage of 1.8V, so it's not a candidate.

    - Win

    (email: use hill_at_rowland-dot-org for now)
  14. Fred Bloggs

    Fred Bloggs Guest

    The zener clamp sometimes requires a high peak power rating making for a
    pricey component- the rectifier diode clamp can be assisted for very
    fast turn-off like so:
    View in a fixed-width font such as Courier.

    | |
    | |
    | -
    | ^
    | | | VD,max-VCC-Vdiode
    IC | C| / Rc,MAX= ------------------
    | C| Rc IC
    | C| /
    \ / | \ Rc=Rcoil gets VD,MAX ~ 2x VCC
    | |
    | |
    o-----o <-- VD
    | |
    / |
    10K |
    / |
    \ |
  15. Tim Wescott

    Tim Wescott Guest

    If you're only energizing the relay occasionally you can also leave out
    the diode and just use an R or RC snubber network, at the expense of
    more current in the transistor.
  16. Or put an LED in series with the resistor (with an antiparallel diode
    or a parallel resistor) as an indicator.

    Best regards,
    Spehro Pefhany
  17. In Tim Wescott typed:

    Now why would that de-energize faster? The stored current can still
    flow after the transistor turns off, except that it flows to ground.
  18. The current decays faster only if the zener voltage is higher than the
    supply voltage, implying that there is a fixed voltage across the coil
    as the current decays. If the zener voltage is less than the supply
    voltage, the relay current never goes all the way to zero.
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