Circuit with BJT and JFET transistors

Discussion in 'Homework Help' started by AD633, Jul 13, 2013.

  1. AD633

    AD633 Thread Starter Member

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    Hi,

    For the following circuit determine the voltages VG,VC and VD

    VG

    <br />
<br />
VG=25 V *(18kOhm)/(18kOhm*91kOhm)=4,12 V<br />
<br />

    For determing Vc i need to know the value of the current Ic of the BJT.
    For determing Ic do i need to determine Id first?


    For determing Ib (not sure if this equation is correct)

    <br />
<br />
Vcc= 330kOhm*Ib+VBE<br />
 15V=330kOhm*Ib+0,7V<br />
  Ib=43,3uA<br />

    <br />
Ic=beta*Ib=160*(43,3uA=6,928mA<br />
<br />
Vc=1,1kOhm*6,928mA=7,6208 V<br />
<br />

    I need to know what the value of the current Id and Vgs in order to determine the value of VD right?

    Can i determine VD from this equation:

    <br />
Vcc=VC+VCE+VD+VDS<br />
Vcc=VC+(VC-VD)+VDS<br />

    Suposing that the BJT works in the ative region therefore IE is equal to Id

    <br />
<br />
Id=Idss(1-(Vgs)/(Vp))^2<br />
<br />
6,298mA=6mA(1-(Vgs)/(-6 V))^2<br />
<br />
Vgs=-1,788 V<br />
<br />

    How can i determine VD and VDS from what i already know..

    Thanks

    Attached Files:

  2. t_n_k

    t_n_k AAC Fanatic!

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    Firstly, check your incorrect equation for VG from which you (luckily) obtain the right value ....

    \text{V_{\small{G}}=\(\frac{18k}{91k+18k}\)25 =4.128 \[V\]}

    Next you need to set up some equations to solve for the FET drain current.

    The base equation is

    \text{I_{\small{D}}=I_{\small{DSS}}\(1- \frac{V_{\small{GS}} }{V_{\small{p}}}\)^2 \ --- equation(1)}

    You already know VG and VS is found from

    \text{V_{\small{S}}=I_{\small{D}}*R_{\small{S}}= I_{\small{D}}*1200}

    So

    \text{V_{\small{GS}}=V_{\small{G}} - V_{\small{S}}= 4.128-I_{\small{D}}*1200 \ --- equation(2)}

    You solve equations 1 & 2 to find FET drain current. The rest will follow from this.
  3. AD633

    AD633 Thread Starter Member

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    So i have to do a system of equations with:

    \text{I_{\small{D}}=I_{\small{DSS}}\(1- \frac{V_{\small{GS}} }{V_{\small{p}}}\)^2 \ --- equation(1)}


    \text{V_{\small{GS}}=V_{\small{G}} - V_{\small{S}}= 4.128-I_{\small{D}}*1200 \ --- equation(2)}

    I have found  Id=16,3mA and  VGS=-16 V

    Now suposing that the BJT is in the ative region.Id=Ie ,and Ie is aprox equal to Ic

    Therefore  Vc=25 V - (1,1 kOhm)*(16,3mA)=2,07 V <br />
<br />

    For determing Vd
    <br />
<br />
Vcc=Vc+Vd+Vds+1,2kOhm*Id<br />
<br />
25 V=(1,1kOhm)(16,3mA)+Vd+vds*1,2kOhm*16,3mA<br />
<br />

    The problem is that i don't know neither vd or vds.Do i need to find ib first for determing Vd?

    Thanks
    Last edited: Jul 16, 2013
  4. AD633

    AD633 Thread Starter Member

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    The problem is that i don't know neither vd or vds.Do i need to find ib first for determing Vd?
  5. Jony130

    Jony130 Senior Member

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    OK, but how can Id current be larger than Idss and and also Vp and Vgs?
    So you need solve this equation again.
  6. t_n_k

    t_n_k AAC Fanatic!

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    @ AD633
    .Remember that quadratic equations usually have two roots - not one. You chose the wrong root. The value must make sense physically as well as mathematically.
  7. AD633

    AD633 Thread Starter Member

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    \text{I_{\small{D}}=I_{\small{DSS}}\(1- \frac{V_{\small{GS}} }{V_{\small{p}}}\)^2 \ --- equation(1)}


    \text{V_{\small{GS}}=V_{\small{G}} - V_{\small{S}}= 4.128-I_{\small{D}}*1200 \ --- equation(2)}

    I have found  Id=4.6mA<IDSS and  VGS=-1.39 V

    Now suposing that the BJT is in the ative region.Id=Ie ,and Ie is aprox equal to Ic

    Therefore  Vc=25 V - (1,1 kOhm)*(4.6mA)=19.94 V <br />
<br />

    For determing Vd
    <br />
<br />
Vcc=Vc+Vd+Vds+1,2kOhm*Id<br />
<br />
25 V=(1,1kOhm)(4.6mA)+Vd+vds*1,2kOhm*4.6mA<br />
<br />

    The problem is that i don't know neither vd or vds.Do i need to find ib first for determing Vd?

    Thanks
  8. Jony130

    Jony130 Senior Member

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    Well the correct answer is
    Id = 4.23621mA and Vgs = -0.958448V

    And yes, you need to know Ib first to find Vd = Ve voltage.
  9. AD633

    AD633 Thread Starter Member

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    For finding Ib do i need to use a system of equations?Something like:


     Vcc=330kOhm*Ib*Vbe+VDG+VGS+1.2kOhm*ID <br />
<br />
        <br />
        Vcc=Vc+Vd+Vds+1.2kOhm*Id <br />
<br />


    Or assuming that the BJT is in the ative region  Ic=Beta*Ib and that Ie is approx equal to Ic

     Ic=160*(4.236mA)=677.76 mA

    Thanks
  10. Jony130

    Jony130 Senior Member

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    Yes, first assume that BJT is in active region and find Ib current.
    Then you need to check if your assumption about BJT active region was right.


    What?? Ic = ??
  11. AD633

    AD633 Thread Starter Member

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    I know that Id=4.236mA.Id=IE,and therefore if the BJT transistor is in the ative region Ic approx equal to Ie.

    I don't see another way to determine Ib.I could write:

     Vcc=330kOhm*Ib+VBE+VDG+VGS+1.2kOhm*Is  .... but there are many variables that i don't know.

    Thanks
  12. t_n_k

    t_n_k AAC Fanatic!

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    You know the emitter current for the BJT must equal the drain current of the FET.

    So \text{I_{\small{E}}=I_{\small{D}}=4.236mA}

    If the BJT is in the linear region you can simply apply

    \text{I_{\small{E}}=(1+\beta)*I_{\small{B}}}


    Hence

    \text{I_{\small{B}}=\frac{I_E}{\(1+\beta \)}}

    If you know IB then it's a simple matter to find the BJT base voltage and hence the emitter voltage.
    AD633 likes this.
  13. AD633

    AD633 Thread Starter Member

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    Ib can be found trough \text{I_{\small{E}}=(1+\beta)*I_{\small{B}}}<br />
IB=(4,236mA)/(1+160)=26,31uA


    Now i can find Vb

     Vb=Vcc*(330kOhm)/(330kOhm+1.1kOhm)=24,92 V<br />
<br />
<br />
Vd=Ve=Vb-Vbe=24.92 V-0.7 V=24.22 V<br />
<br />

    Is this correct?

    Thanks
  14. t_n_k

    t_n_k AAC Fanatic!

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    @AD633,

    You seem to have some worrying gaps in quite basic DC circuit analysis understanding.

    There's no logic in your assertion ...

     Vb=Vcc*(330kOhm)/(330kOhm+1.1kOhm)=24,92 V<br />
<br />

    How do you not see that the BJT base voltage is the supply voltage less the drop across the 330k base bias feed resistor?

    You clearly need to strengthen your basic skills before attempting problems like this one. Otherwise you'll end up with significant difficulties in the future.
    Last edited: Jul 18, 2013
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