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Question about MOSFET

Discussion in 'General Electronics Discussion' started by pizoman, Mar 18, 2013.

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  1. pizoman

    pizoman

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    Mar 10, 2012
    I'm using a n-channel MOSFET to control a DC motor (one of those small toy motors) where I'm using a photoresistor to switch the circuit on and off. When I directly connect the circuit to a 9 volt battery (the battery is actually only at ~6.2volts, it's a little drained), it works fine. But when I connect it to a voltage regulator set to 5 volts, the circuit doesn't work. I tried testing the regulator circuit with a simple two transistor LED to see if it works and it does.

    My question is: Do Mosfets have a minimal voltage requirement to operate properly? Or am I asking the wrong question? There's something I'm not getting and I'm trying to figure out why my circuit is not working.
     
  2. davenn

    davenn Moderator

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    two transistor LED ??


    remember a LED only requires ~ 25mA
    your motor probably much much more
    At 6.2 V the battery is basically flat and wont be able to supply any signif. current
    get a new battery

    Dave
     
  3. pizoman

    pizoman

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    Mar 10, 2012
    Ok, I found a fresh new 9 volt battery. Swapped it out and still doesn't seem to want to power my mosfet controlled DC motor circuit.

    What I meant by "two transistor" is just the two transistors that I used in my circuit.

     
  4. pizoman

    pizoman

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    Mar 10, 2012
    I'm wondering if my regulator circuit is not supplying enough current. Yet again, a battery doesn't provide current but only a potential unless I'm wrong there. With the new battery, I increased the voltage up to 8.5 volts. I'm using a LM1084 regulator.
     
  5. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

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    Perhaps you can show us the circuit you're using and tell us what mosfet you've got.

    It would also be useful to know if the motor works when powered directly from the battery and/or the power supply.

    Do you know that you have to be careful with mosfets? Handling them can destroy the gate.

    Batteries supply current too (that's what they do). Same as power supplies of any kind -- it's their purpose in life.
     
  6. pizoman

    pizoman

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    Mar 10, 2012
    I didn't know that handling a mosfet can damage them, but I'm using an IRL510. Yes, the motor does work when directly connected to the battery.

    I'm drawing a diagram right now. One sec.
     
  7. pizoman

    pizoman

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    Mar 10, 2012
    [​IMG]
     
  8. duke37

    duke37

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    Jan 9, 2011
    Two things

    1. What is the regulator for? What is its output voltage.

    2, I have not looked up the specification of the IRL510 but you may not be poviding sufficient voltage at the gate to turn it on.

    Where are your two transistors?
     
  9. KrisBlueNZ

    KrisBlueNZ Sadly passed away in 2015

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    Nov 28, 2011
    1. Your schematic has a curved line from the 9V supply rail coming down to the ADJ pin of the regulator. I don't know what it's supposed to mean. It shouldn't be there.

    2. Is your 9V battery the "PP3" type? That's the kind that's about the size of two AAs with a snap-on connector. These batteries can't supply much current. If you try to draw a lot of current from them, e.g. by trying to supply a motor that draws too much current, the battery voltage will droop.

    You can test for this by measuring the battery voltage with a multimeter before and after the motor is connected. A voltage drop of more than say 10% means you should be using a bigger battery, or a smaller motor.

    3. MOSFETs do have a minimum gate-source voltage required to bias them ON so they can pass current in their drain circuits. For the IRL510 this voltage is around 3~4V. Have a look at the data sheet. On page 3 the top right diagram shows how the drain current relates to the gate-source voltage for a typical device.

    4. In that circuit's current form, the MOSFET is not switched cleanly ON and OFF. There is a range of light levels for which the gate-source voltage will cause the MOSFET to operate within its linear region, where it is only partly conducting. This can cause significant power dissipation in the MOSFET, because there is simultaneously current through it, and voltage across it. (Static power dissipation is the product of voltage and current; if the MOSFET is switched cleanly, there is only ever one of these factors present, so the product, i.e. the power dissipation, is always low.)

    You can add this behaviour using a circuit called a Schmitt trigger. It's widely used in this type of application. Look it up on Wikipedia, Google it, or search other threads on this site.
     
  10. pizoman

    pizoman

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    Mar 10, 2012
    Kris,

    I think you're right, the MOSFET is not switching on. I'm trying to operate the whole circuit off of 6 Volts, so I regulated the voltage with a regulator to 6 volts. I did measure the voltage of the mosfet, negative probe on the source and positive probe on drain and got a reading of 3.1Volts. I assume this means that the mosfet is not fully turning on.

    Would the Schmitt trigger help with the situation?
     
  11. KrisBlueNZ

    KrisBlueNZ Sadly passed away in 2015

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    Yes, a Schmitt trigger is needed. This design should do what you want.

    [​IMG]

    This circuit will work with any supply voltage from 6V to 12V. R3 and D2 and the emitter-to-source connection create hysteresis, the necessary characteristic of a Schmitt trigger that makes it switch cleanly.

    The voltage across the motor will be about 0.8V less than the supply voltage, because of voltage drop across D2. If this is a problem, I can change the design to remove that source of voltage loss. Another transistor or MOSFET will be needed.
     

    Attached Files:

    Last edited: Mar 19, 2013
  12. pizoman

    pizoman

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    Mar 10, 2012
    R3 is that suppose to be 15 ohm? It says 15R. Also, would it be possible to design the circuit to run off a regulated voltage source of 5 volts? Would this circuit still work at 5 volts even if the motor runs slower?

    Does the current from a voltage regulator affect the operation of this circuit. I was reading some papers on mosfets and came across threshold values.
     
  13. KrisBlueNZ

    KrisBlueNZ Sadly passed away in 2015

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    Yes, 15R means 15 ohms. Yes that circuit would run from 5V, but not much less. You need to have enough voltage at the MOSFET's gate to fully turn it on. I don't know what you mean by current from a voltage regulator affecting the operation of the circuit.

    Here's a better design, without the diode drop.

    [​IMG]

    Here's a circuit description.

    Start assuming it's dark. The LDR's resistance will be high; perhaps 1 megohm. VR1 pulls Q1's base down to 0V and Q1 remains OFF.

    Q1's collector is pulled high by R1, providing the full power supply voltage to Q2's gate and turning Q2 fully ON. Current through Q2's drain-source path energises the motor with almost the full supply voltage. (There is a slight voltage drop across Q2 because of its RDS(on) resistance.)

    In this state, the voltage divider formed by R2 and R3 provides about 3/4 of the power supply voltage at the top end of the LDR.

    As the light falling on the LDR increases, its resistance drops. This causes Q1's base voltage to rise. At a certain brightness level, Q1's base voltage is high enough that Q1 starts to conduct. It pulls its collector voltage down, in the process turning off Q2. The voltage across the motor falls to zero as Q2's drain voltage rises to the positive supply rail.

    As this happens, the voltage on R3, and the voltage at the top of the LDR, increase. This causes Q1's base voltage to rise even more quickly. This design feature is called positive feedback and it creates an effect called hysteresis.

    As the light level increases, the circuit's threshold is a (relatively) high brightness level. Once the brightness reaches that threshold, the circuit activates and turns off the motor, and in the process, reduces its brightness threshold, so the brightness must drop below a lower threshold before the circuit will turn the motor back ON.

    This produces a "deadband" effect, prevents wittering around the trigger level due to small variations, and ensures that the threshold decision-making and switching action is clearly defined and sharp.

    For more information look up hysteresis, positive feedback, and Schmitt trigger on Wikipedia.
     

    Attached Files:

    Last edited: Mar 20, 2013
  14. pizoman

    pizoman

    51
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    Mar 10, 2012
    What I meant is, if I have a power source with a high current of say 1 Amp, running through a voltage regulator, wouldn't that current be directly translated to the dark sensor circuit? With a transistor like the 2N3904, would it be able to handle such current? Or does it not matter in my situation?
     
  15. pizoman

    pizoman

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    Mar 10, 2012
    Kris,

    What program are you using to display your circuit diagrams? I like the simplified look of your schematic.
     
  16. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

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    Kris is going to have to put "I use a very old commercial EDA program called OrCAD/SDT III to draw schematics" in his profile. :D
     
  17. KrisBlueNZ

    KrisBlueNZ Sadly passed away in 2015

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    No. I think the problem is that you don't really understand voltage and current, and how they interact. You should probably Google for basic electronics tutorials that explain these quantities.

    Re the schematic diagrams I post, see https://www.electronicspoint.com/220v-led-dimmer-t258159.html#post1536734
    It's the last question I reply to in that post.

    Steve: Except in these two threads, no one's ever asked me about it before!
     
    Last edited: Mar 20, 2013
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