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10 amp NPN Transistor switch circuit

Discussion in 'General Electronics Discussion' started by VizBiz, Apr 29, 2013.

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  1. VizBiz

    VizBiz

    4
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    Apr 29, 2013
    Hi,

    I am looking at building the below circuit to switch 10 amp LED Driver from a small rotary switch Link. It is rated at 150ma.

    Here is the NPN Data Sheet I intend to link

    My supply voltage is a 10 cell lipo (Voltage is 44-30).

    I am trying to calculate what resistor to use at R1....
    I don't think I have been successful so far...

    Can I please have a value for R1?

    Thanks in advance

    C1.png
     
  2. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

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    Jan 21, 2010
    The switch is connected to the wrong side of the battery.

    R1 needs to be approximately: (Vbatt - 1)/((Ic/hfe)*2)

    for hfe use a value in the datasheet for the Ic desired (and if a range is given, use the lowest one). (edit: and for Vbatt, use the lowest value)

    In this case it is (30 - 1)/((10/8)*2) = 29/2.5 = 11.6 ohms.

    That might work, but it's going to waste a lot of power. You're also going to get close to a lot of maximum ratings (like Ib)

    I would recommend you find a high current darlington or a mosfet.

    If you're going for a darlington, I would recommend one of the following:

    BDW42G, BDW83C, BU941, 2STW100, BU931, 2SD2560

    In all these cases, the voltage drop will be up to 3V, so power dissipation will be up to 30W and hence a heatsink WILL be required.

    The gain at 10A is >300 for all of these (in some cases much greater), so the minimum value resistor will be: (30 - 1)/((10/300)*2) = 29/0.067 = 435 ohms.

    It might be even better to use a mosfet. An additional resistor is required (presuming you're not switching the load on and off many times per second), but the maximum dissipation in the mosfet would be far lower than in a transistor. It is quote likely that a heatsink would not be required.
     
  3. VizBiz

    VizBiz

    4
    0
    Apr 29, 2013
    OK - Thanks.

    I have had another look at my LED driver, and it is only pulling 5 amps (5.6 max).

    I have had a look in the "Parts Box" and I have found some of the below:

    BD649 NPN (LINK)

    and also

    BUZ73A 5.8A, 200V, 0.600 Ohm, N-Channel Power MOSFET (LINK)

    What would be better then?

    Heat sink is not an issue, but i don't want to waste to much power.....

    Cheers
     
  4. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

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    The BD649 is certainly an option. Using 750 as the hfe, you'll get something like 2k. The power dissipation is about 15W.

    The mosfet would run close to it's maximum rating, so I wouldn't advise that one, , and the power dissipation would be about 15W as well.

    A mosfet with a lower Rds(on) will dissipate less power. For example if it is 0.1 ohms, the dissipation would be 2.5W.

    power dissipation IS wasted power, so reducing it is a good thing :)
     
  5. BobK

    BobK

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    Jan 5, 2010
    And you can easily find MOSFETs with Rdson of more like 10mOhms, which would lower the power to well under 1 Watt and not need any heat sink.

    Bob
     
  6. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

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    Everything Bob says is true.

    Although you *may* need to be more careful about how you turn them on and off.

    There are mosfets in very tiny packages that can switch truly enormous currents.

    However, if you don't switch them quickly (and this refers to the method of driving the gate) they can be very rapidly destroyed.

    A mosfet that *can* be attached to a heatsink (even if it doesn't need to be) is likely to be able to handle a lot more abuse during switching.
     
  7. BobK

    BobK

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    Jan 5, 2010
    Steve,

    That is interesting. I thought the switching problems only showed up at high frequencies, no?

    Bob
     
  8. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

    25,497
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    Jan 21, 2010
    It depends how slowly you're switching the device :)

    If, as in this case, you have a voltage higher than Vgs(max), you need something to limit it.

    In this case, you might choose to use a voltage divider. This will limit the maximum Vgs, and also provide a path to discharge the gate and turn the mosfet off.

    However, there are going to be issues with switching speed, especially at lower battery voltages.

    A very small device may dissipate enough energy while performing a single switching transition to over temp the device and kill it.

    A larger device will have more thermal inertia and the temperature rise will be limited.

    This information is covered in the graphs for pulsed dissipation. Often there is a graph showing the limit of dissipation for a single pulse of varying duration. These graphs will have a much higher dissipation for a given pulse length in physically larger devices.

    In addition, smaller high current mosfets (often in SMT packages) have very large gate capacitances which makes this effect even more pronounced.
     
  9. duke37

    duke37

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    Jan 9, 2011
    44V on the gate of a mosfet may be too high. You would need a resistor and a protective zener to limit the voltage to, say, 15V.
     
  10. CDRIVE

    CDRIVE Hauling 10' pipe on a Trek Shift3

    4,960
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    May 8, 2012
    I guess he could also do this. C1 should prevent the FET's Cgs from slowing switching time. It doesn't compensate for contact bounce though. ;)

    Chris
     

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