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  1. Dec 7, 2008 #1
    Hi Guys,

    I need to design something that will isolate our DAQ (Data Acquisition Card) for a basic simulation we need to do. I was thinking about designing a mosfet switch, where the DAQ sends some voltage to the gate above the threshold voltage to turn the MOSFET on and 0 V to turn it off. The MOSFET needs to be able to drive a computer fan (resistance is around 7k) and it runs at 160 mA. I was thinking about using common source. Anyone have any suggestions on the type of MOSFET i should be using in order to achieve 160 mA? Any help would be appreciated!
  2. jcsd
  3. Dec 8, 2008 #2


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    You might be able to turn it on, but you may be operating in saturation rather than linear mode--you should apply as close to the rail voltage (operating voltage of the fan) as possible to ensure that it turns on in linear mode (i.e. acts as a switch, with very low drain-source voltage). To that end, you can use a BJT (in common-emitter mode, designed to operate in either saturation or cut-off) to bootstrap the TTL voltage (if this is what is being generated) to your rail voltage--a high side driver, basically.

    As for what FET to use, well, if you don't need really fast switching, a basic 2n7000 would probably meet your current requirements.
  4. Dec 8, 2008 #3
    Please spare a min to watch the MOSFET Switch circuit.
    I was wondering how a 30V, 2A voltage source at drain can drive a 12V load?
    am i missing something??
  5. Dec 8, 2008 #4


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    Without knowing the resistance of the light bulb, I'd assume that he's either causing the MOSFET to operate in linear mode (and applying approximately 30V to the lightbulb) or that the MOSFET is operating in saturation mode, and applying less than 30V to the lightbulb.

    If you didn't know, the 2A refers to the maximum current that the supply can output, not what it outputs all the time. Actually, since the lightbulb draws 2.5A (30W / 12V), he would exceeding the maximum output current of the voltage source, and thus is probably drawing a little over 2A, but causing the voltage source to sag (output at a lower voltage due to excessive current draw). Clever little trick.
  6. Dec 8, 2008 #5
    Yes I understand ratings!!
    well but how is he achieving 12V , 30W across light bulb?
    Any maths to support it?
  7. Dec 8, 2008 #6


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    As I implied above, he's probably not. But a light bulb is a pretty simplistic device that only requires a certain amount of current running across it to glow. I don't think I can (maybe someone else can, though) support this with an equation because he's probably relying on the voltage supply to limit the current going to the lightbulb.

    You can very crudely determine what the voltage source's output voltage is by guesstimating the internal resistance (R_int) of the voltage source (make a Thevenin equivalent that assumes the short-circuit current is 2A and the open circuit voltage is 30V). And if you assume that the MOSFET is in linear mode, you can assume that the MOSFET drain-source path acts as a wire (or, find the R_ds from the datasheet--which should be only a few ohms--negligible compared with the other elements).

    With the assumptions made above, you now just have a voltage divider consisting of R_int and the lightbulb, driven by the (ideal) voltage source.

    EDIT: My back-of-envelope calculations gives R_int as 15 ohms, and R_bulb as 4.8 ohms, which tells you something right there :-D
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