Circuit for motor/relay control

Discussion in 'Homework Help' started by AD633, Jun 24, 2013.

  1. AD633

    Thread Starter Member

    Jun 22, 2013
    96
    1
    The circuit following uses the outputdigital of microprocessor (0V or 5V) to drive a relay or motor.both the transistors work in regime ofcommutation (ON /OFF) and the circuit can be fed with any tension between 6Vand 18 V.

    Could someone please help me understand how this circuits works?

    In general terms i suppose this should be something like ,due to the fact that the npn transistor switch between cutting and saturation due to voltage from the microprocessor, the FET is on or off triggering the relay and motor, or not depending on its state.


    2)considering hFE = 240 VBE = 0.7V, VCEsat= 0.4V, determine the intensity of current that the output of microprocessor has of provide to the circuit work correctly.


    For this

    <br />
<br />
Voutput+R//R*Iouput+VBE=0V<br />
<br />
5V+(1*10^3)//100*10^3*Ioutput+0.7=0<br />
<br />
Ioutput=4,34 m A<br />

    Is this correct?

    3)Determine a value minimum possible to the tension of threshold VT of
    MOSFET channel N.Justify.

    I have no idea how to do this.What should i start to determine this?


    Thanks
     
  2. Jony130

    AAC Fanatic!

    Feb 17, 2009
    3,957
    1,097
    For the Vin = 5V we have this situation

    The base current is equal to:

    Et = 5V *100K/(1K + 100K) = 4.95V


    IB = (Et - 0.7V)/(100K||1K) = (4.95V - 0.7V)/990Ω = 4.29mA

    And Ic = 240*Ib = 1A but the maximum current that can flow through collector and Rc resistor is equal to

    Ic_max = Vcc/Rc ≈ 18mA ... 6mA depending on Vcc value.

    And because Ic > Ic_max the BJT is in saturation region.
    And the collector current is equal to:

    Ic = (Vcc - Vce(sat))/Rc = 17.6mA....5.6mA

    And Vgs voltage seen by the MOSFET gate is equal to 0.4V. So the MOSFET is OFF.

    But if we change input voltage to 0V. The BJT is cut-off. But now the MOSFET gate is pull-up to Vcc by Rc resistor. So voltage between gate and source is equal to Vcc. So if Vt < Vcc the MOSFET will be ON. And that will energize (trigger) the relay.
     
    Last edited: Jun 26, 2013
    AD633 and screen1988 like this.
  3. screen1988

    Member

    Mar 7, 2013
    310
    3
    Jony,
    We often don't know exactly the value of β, and how can we use this way to check if the transistor is in saturation or not.
    Does the β here, 240, is the typical value?
    I assume that β ranges from 50 to 600.
    With βmin = 50 then Ic = β*Ib = 50*4.29mA = 214.5 mA > Icmax => the transistor is always in saturation region.
    Is that right?
     
  4. Jony130

    AAC Fanatic!

    Feb 17, 2009
    3,957
    1,097
    We can use β_min from datasheet.

    No, in his first post AD633 give as Beta value.


    Yes, you got that right.
     
    screen1988 likes this.
  5. AD633

    Thread Starter Member

    Jun 22, 2013
    96
    1
    I understood it now.

    Then the minimum value for the voltage VT (Treshold) has to be VT>VCEsat (0.4 V) and hence VT at least has to be greater than 0.4 V, is that it?


    Thanks
     
  6. Jony130

    AAC Fanatic!

    Feb 17, 2009
    3,957
    1,097
    Yes, you right. But in real life you don't have to bother about this condition Vt > 0.4V. Because most of a MOSFET has Vt < 2V.
     
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