How to use 2N7000 as low side switch

Discussion in 'General Electronics Chat' started by other_santa, Dec 23, 2015.

  1. other_santa

    Thread Starter New Member

    Dec 22, 2015
    4
    1
    Hi All,

    I am learning ABC of electronics.

    This time I am learning how to use 2N7000 mosfet as switch.
    On breadboard I made a circuit "A" as shown in attachment and it will make LED ON when Gate is made high.

    However, when I change the position of load, LED is not being ON as I expected.
    Can someone please explain why circuit "B" does not make LED ON when G is made high?

    Thank you very much for reading this post.
     
  2. #12

    Expert

    Nov 30, 2010
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    Same as when you use an NPN transistor. The LED in the collector (drain) circuit sees the transistor as a switch. When you place the LED on the emitter (source) side, the voltage to the LED is always lower than the gate (base) voltage.
     
  3. Bordodynov

    Active Member

    May 20, 2015
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    See
    2N7000+Led.png
     
  4. other_santa

    Thread Starter New Member

    Dec 22, 2015
    4
    1
    Hi #12 and Bordodynov

    Thank you very much for the post.
    I did not expected so fast reply : )

    All the tutorials I see show similar circuit to what Bordodynov has shown.
    They put load on high side.
    Is it possible to use N channel mosfet(2N7000) as switch and put load on low side?

    Thank you once again for your kind posts.
     
  5. #12

    Expert

    Nov 30, 2010
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    Yes, but you will need more voltage so as to allow for the Vgs of the mosfet plus the voltage required by the LED.
     
  6. ScottWang

    Moderator

    Aug 23, 2012
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    The rating Vgs=10V of 2N7000, if the Vgs too low may not works properly.
    You can using circuit B to try, and move the led to the Vd to V+(Vbat), Vs connecting to V-(Vbat).
    If the led still didn't light up, then you need to change to other mosfet has very low Vgs as FDV301N.
     
  7. other_santa

    Thread Starter New Member

    Dec 22, 2015
    4
    1
    I used 9V battery and can see that LED is going ON.
    May be it is time for me to buy a P channel MOSFET for using in circuit B.

    #12, Bordodynov and ScottWang Thank you very much. It was my first post here and you guys are really helpful.

    Thanks!!!
     
    KJ6EAD likes this.
  8. ScottWang

    Moderator

    Aug 23, 2012
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    Normally you will need a current limiting resistor for led.
    R = (Vcc - V_led) / (I_led*80%)

    For example:
    R = (9V - 3V) / (20mA * 80%)
    R = 6V/16mA
    R = 375 Ω.
    You can using a 360 Ω or 390 Ω
     
    other_santa likes this.
  9. atferrari

    AAC Fanatic!

    Jan 6, 2004
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    I know this is not an actual answer to your question but it is what I had to use to drive a LED with a signal from an existing circuit (comparator). The disadvantage is that it is always drawing power.

    Just in case, here you have it. . .

    Drive the LED.png
     
  10. Bordodynov

    Active Member

    May 20, 2015
    637
    188
    other_santa said:
    ......
    All the tutorials I see show similar circuit to what Bordodynov has shown.
    They put load on high side.
    Is it possible to use N channel mosfet(2N7000) as switch and put load on low side?
    Yes.
    Use p-cannal. Example FDV302P or FDV304P.

    FDV304P+LED.png
     
  11. other_santa

    Thread Starter New Member

    Dec 22, 2015
    4
    1
    @atferrari that is an interesting circuit. Thanks for sharing.

    @Bordodynov Thanks for circuit and part numbers. I will try to get these parts and experiment.
    Thanks!!!
     
  12. Bordodynov

    Active Member

    May 20, 2015
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    You can use a voltage multiplier:

    2N7000+Led2.png
     
    absf likes this.
  13. ScottWang

    Moderator

    Aug 23, 2012
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    Do you think when you adding positive voltage(V2) to the Vgs will the p mosfet works?
     
  14. GopherT

    AAC Fanatic!

    Nov 23, 2012
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    Why do you ask? He has V2 = 0V in post 10.
     
  15. ScottWang

    Moderator

    Aug 23, 2012
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    It just quite strange and why not just connected to gnd?

    It looks like there is a voltage at the Vgs.
     
    GopherT likes this.
  16. Bordodynov

    Active Member

    May 20, 2015
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    In post # 10 I used a control voltage source V2. If there is no control, then why transistor? Then, only one resistor.
     
  17. ScottWang

    Moderator

    Aug 23, 2012
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    That is just a testing circuit, and if you need a simple control then it should be adding a resistor connecting from Vg to +V (+Vbat), and the control switch will be connecting from Vgs to ground (-Vbat).

    The circuit you attached is like a pulse control, but here is no needed, and normally when you using a P mosfet then you will need a npn bjt to drive the p mosfet, and the input can be input the pulse from 0V to 5V and 5V to 0V ...
    ┌┐┌┐┌┐┌┐
    ..└┘└┘└┘└┘
     
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