MOSFET Question

Discussion in 'General Electronics Chat' started by TimCollins, Sep 20, 2011.

  1. TimCollins

    Thread Starter New Member

    Sep 19, 2011
    9
    0
    I have a simple question I think... I have a very simple test circuit using a 6v battery, light bulb and a MOSFET(IFRP06N4). I am just testing the MOSFET as a switch turning the light on and off. When I charge the gate by touching a wire between the source and gate the light comes on thats great. I can switch the light off by touching the drain and gate which is good also. Basically things work as I would expect... However, my question is this. When I turn the light on and measure the voltage I am only getting about 3V across the source and drain. I was expecting something closer to 6V? The reason I am asking is that if I take my battery and hook it to the bulb directly it is about twice as bright as with the MOSFET in the circuit. I thought a MOSFET acted basically as a switch so seeing the voltage drop basically in half does not seem right? Am I doing something wrong?
     
  2. SgtWookie

    Expert

    Jul 17, 2007
    22,182
    1,728
    Here is the datasheet for your MOSFET: http://www.irf.com/product-info/datasheets/data/irfp064n.pdf
    BTW, you mistyped it.

    You need to get the gate 10v more positive than the source terminal in order to get the MOSFET fully turned on. With a 6v battery, you're just over halfway there. As the gate voltage increases, the Rds(on) will decrease, up until the Vgs is around 10v.
     
  3. colinb

    Active Member

    Jun 15, 2011
    351
    35
    I think I'm not alone when I say that V_{GS(th)} ratings have caused me confusion before. It is not the fully-turned-on gate voltage, but rather the gate voltage at which the FET just begins to conduct. For the fully turned on gate voltage, look in the Electrical Characteristics section of the data sheet for R_{DS(on)} and then look at the Conditions column for the V_{GS(th)} value that provides this fully-on state. For this FET, it is 10 V.

    Also, hopefully you aren't trying to turn the FET on by connecting the gate to the drain. That is likely not what you want.
     
  4. Audioguru

    New Member

    Dec 20, 2007
    9,411
    896
    Without a schematic we are simply guessing.
    This is what the OP said is happening:
    1) The light bulb turns ON when the N-channel Mosfet gate and source are connected. But then the Mosfet is turned OFF.
    2) The light bulb turns OFF when the Mosfet drain and gate are connected. But then the Mosfet is partially turned ON.

    Is the Mosfet connected as a follower so the light bulb is from the source to ground? Then the gate must be +16V for the Mosfet to completely turn on.
     
  5. SgtWookie

    Expert

    Jul 17, 2007
    22,182
    1,728
    Thanks folks, I obviously didn't interpret Tim's statement correctly - and I'm sure that he has the source and drain terminals confused as well.

    Have a look at the attached schematics and simulations. The upper one shows your incorrect circuit where you connected the gate to the drain. What happens here is the gate winds up somewhere between the supplied voltage and the source terminal; somewhat higher than the threshold, but not fully turned on. This will result in high power dissipation in the MOSFET.

    A corrected schematic is below.
     
  6. TimCollins

    Thread Starter New Member

    Sep 19, 2011
    9
    0
    Thanks you for all of your responses. I think it is working now. When the light bulb is on and I measure the voltage between the source (pin3, right pin) and drain (pin2, center pin), I get 0 volts. Visually this appears correct because the bulb appears fully on. Then, when I ground the gate and the bulb goes off the voltage between source pin and drain pin comes very close to matching my battery voltage. I believe that is exactly what I should expect? or is there another way I should test that would be better?
    As I was testing, I did take your advice and I used a 12v battery rather than 6v. However, this was an interesting test also... After I got the circuit working as I think it should with the 12v battery (unless you shoot me down on my test above) I replaced it again the 6v battery just to see what would happen. I was expecting the light to stay off since I was not fully reaching the 10v Rds(on). However, the light still comes on. So it seems as though the 10v Rds(on) doesn’t really mean you have to have a full 10v before the circuit is closed. Rather I think what is happening with my 6v battery is that the circuit is probably closing just enough to allow 6v through so technically the circuit is not fully closed. Does that sound right or am I off somewhere?
    I think this was my problem before… I had the load in circuit between the source pin and the cathode. I reversed all of that. So now the load is between the drain and anode. The cathode is connected directly to the source now.
    Again thanks for sharing your knowledge in this area…
     
  7. Audioguru

    New Member

    Dec 20, 2007
    9,411
    896
    My guess about your wrong wiring was correct because it was connected as as follower.

    Each Mosfet is different. Some are very sensitive and turn on good with only a few volts from gate to source. Even if they have the same part number as the sensitive ones others need 10V. When you buy one you don't know its sensitivity so if you want to make a lot of circuits and want them all to work then you give them all 10V from gate to source. Or you can buy hundreds of Mosfets and test each one then pick the most sensitive one.

    When a Mosfet is not fully turned on and it has a heavy load then it gets very hot.
     
  8. SgtWookie

    Expert

    Jul 17, 2007
    22,182
    1,728
    That's correct; if the voltage across the source and drain is near zero, the MOSFET is conducting.

    That is what you should expect. You might consider measuring the current flow through the drain, but measuring current flow can easily wind up blowing fuses in your meter. Better to measure voltage instead.

    Are you sure you weren't connecting the gate to the drain instead of +6v before?

    Until Vgs is around 10v, the MOSFET won't be considered to be fully turned on; the Rds will be higher than specified for Rds(on).

    That doesn't make sense. There IS a body diode between the drain and source, cathode on the drain, but it's secondary to the MOSFET itself. You just call the terminals the gate, drain and source. But if you had the load connected between the drain and source, you would have been shorting the battery with the MOSFET when it was turned on!

    This also makes no sense.

    Your load should be connected between +V and the drain terminal.
    The source terminal is connected to -V.
    To power the load, connect the gate to +V.
    To turn the load off, connect the gate to the source terminal.
     
  9. Audioguru

    New Member

    Dec 20, 2007
    9,411
    896
    There is a translation problem here;
    1) Anode= positive supply.
    2) Cathode= 0V or ground.
     
Loading...