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Erm, right. I should probably know this, but...

Discussion in 'Electronic Basics' started by Jamie, Aug 30, 2003.

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  1. Jamie

    Jamie Guest

    I apologise for the HTML, if that's not liked in this group.

    Mosfets...

    If is was to take say, this (http://us.st.com/stonline/books/pdf/docs/2950.pdf) mosfet, and connect like this (excuse the drawing, oh and it'll only work in Courier, or some other fixed-width font like that):

    4.5V 18R
    -----------------\/\/\/\------
    | |
    \ |Source
    Switch \ ____|____
    | Gate | |
    ------------------| Mosfet |
    |_________|
    |
    |Drain
    ____|____
    | |
    | Load | 200mA (bunch of LEDs in parallel)
    |_________|
    |
    |
    0V |
    ------------------------------

    Now as far as I know, say the switch is open (ie. gate is at 0V) then no current will flow between the source and the drain, and the LEDs won't be lit, right? And if the switch is closed (gate at 4.5V and higher than the source voltage) then the mosfet will be 'on' and current will flow through the mosfet and the LEDs will light up, yeah? Or not? 'Cause I soldered it all together and the LEDs just stay lit, no matter what I do with the gate voltage. It's all getting abit annoying.

    Anyways, any help would be appreciated,
    Jamie
     
  2. Not only is usenet a HTML free zone, binaries are only allowed on
    binaries newsgroups. The one used by the sci.electronics.* groups is
    where you can post them as a
    GIF, JPEG, or other common image file. Then, post a message here giving
    the name of the post. If you try to post a binary file here, most
    people will never see it.
     
  3. Oh yes! - good catch. I didn't check the PDF. Also, (ideally) he should put
    the source to ground and the load and current limit resistor on the drain
    side, and then to 4.5V. This will maximize Vgs and ensure that it is
    sufficiently greater than threshold voltage.

    Steve B

    4.5V
    ___________resistor_________LEDs________Drain__________Source________GND

    |

    |

    |
     
  4. Jamie

    Jamie Guest

    Right, hang on. Your gonna have to explain that one again. I'm confused.

    Jamie
    (http://us.st.com/stonline/books/pdf/docs/2950.pdf) mosfet, and connect like
    this (excuse the drawing, oh and it'll only work in Courier, or some other
    fixed-width font like that):current will flow between the source and the drain, and the LEDs won't be
    lit, right? And if the switch is closed (gate at 4.5V and higher than the
    source voltage) then the mosfet will be 'on' and current will flow through
    the mosfet and the LEDs will light up, yeah? Or not? 'Cause I soldered it
    all together and the LEDs just stay lit, no matter what I do with the gate
    voltage. It's all getting abit annoying.
     
  5. Jamie

    Jamie Guest

    The whole switch thing was just for simplicity, the gate is actually connected to the output from an OR gate.

    Jamie
    Jamie,

    Use a SPDT switch so that you can ground the gate when in the off state, or use a 1K pull down resistor on the gate with your SPST switch. MosFETS have very high input impedance into the gate, and if left open as you have it, they can float in the on state (especially after they are already on). See diagrams below.

    Good Luck,
    Steve B


    Option 1:

    4.5V 18R
    -----------------\/\/\/\------
    | |
    \ |Source
    Switch \ ____|____
    | | Gate | |
    | ----------------| Mosfet |
    | |_________|
    | |
    | |Drain
    | ____|____
    | | |
    | | Load | 200mA (bunch of LEDs in parallel)
    | |_________|
    | |
    | |
    0V | |
    ------------------------------

    Option 2:

    4.5V 18R
    -----------------\/\/\/\------
    | |
    \ |Source
    Switch \ ____|____
    | Gate | |
    ------------------| Mosfet |
    | |_________|
    | |
    / |Drain
    \ ____|____
    / 1K | |
    \ | Load | 200mA (bunch of LEDs in parallel)
    | |_________|
    | |
    | |
    0V | |
    ------------------------------
     
  6. John Larkin

    John Larkin Guest

    Take a look at the little schematic on the first page of the fet
    datasheet. There's a parasitic substrate diode from source to drain.
    Basicly, a n-fet is intended to work with the drain more positive than
    the source. In your circuit, it's "upside down" so the diode is
    foreward biased all the time, so you can't turn the puppy off.

    So, use a p-channel fet, or, to use the same fet, rearrange things,
    putting the load in the drain circuit like Steve suggests. (Nice
    linear ascii schematic!)

    John
     
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