All About Circuits Forum Circuit with BJT and JFET transistors
 Register Blogs FAQ Members List Today's Posts Search Today's Posts Mark Forums Read

 Homework Help Stuck on a textbook question or coursework? Cramming for a test and need help understanding something? Post your questions and attempts here and let others help.

#1
07-13-2013, 09:36 PM
 AD633 Member Join Date: Jun 2013 Location: Ciudad de México Posts: 95
Circuit with BJT and JFET transistors

Hi,

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

VG

$

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

$

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)

$

Vcc= 330kOhm*Ib+VBE
15V=330kOhm*Ib+0,7V
Ib=43,3uA
$

$
Ic=beta*Ib=160*(43,3uA=6,928mA

Vc=1,1kOhm*6,928mA=7,6208 V

$

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:

$
Vcc=VC+VCE+VD+VDS
Vcc=VC+(VC-VD)+VDS
$

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

$

Id=Idss(1-(Vgs)/(Vp))^2

6,298mA=6mA(1-(Vgs)/(-6 V))^2

Vgs=-1,788 V

$

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

Thanks
Attached Images
 circuit.png (38.2 KB, 25 views)
#2
07-14-2013, 01:44 AM
 t_n_k Senior Member Join Date: Mar 2009 Posts: 4,807

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
07-14-2013, 05:22 PM
 AD633 Member Join Date: Jun 2013 Location: Ciudad de México Posts: 95

Quote:
 Originally Posted by t_n_k 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.

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

$

For determing Vd
$

Vcc=Vc+Vd+Vds+1,2kOhm*Id

25 V=(1,1kOhm)(16,3mA)+Vd+vds*1,2kOhm*16,3mA

$

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 by AD633; 07-16-2013 at 03:05 PM.
#4
07-16-2013, 03:06 PM
 AD633 Member Join Date: Jun 2013 Location: Ciudad de México Posts: 95

Quote:
 Originally Posted by AD633 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 $ For determing Vd $ Vcc=Vc+Vd+Vds+1,2kOhm*Id 25 V=(1,1kOhm)(16,3mA)+Vd+vds*1,2kOhm*16,3mA $ Thanks
The problem is that i don't know neither vd or vds.Do i need to find ib first for determing Vd?
#5
07-16-2013, 04:05 PM
 Jony130 Senior Member Join Date: Feb 2009 Location: Poland/Wroclaw Posts: 2,286

OK, but how can Id current be larger than Idss and and also Vp and Vgs?
So you need solve this equation again.
#6
07-16-2013, 11:36 PM
 t_n_k Senior Member Join Date: Mar 2009 Posts: 4,807

.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
07-17-2013, 01:48 PM
 AD633 Member Join Date: Jun 2013 Location: Ciudad de México Posts: 95

Quote:
 Originally Posted by t_n_k @ 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.
$\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 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

$

For determing Vd
$

Vcc=Vc+Vd+Vds+1,2kOhm*Id

25 V=(1,1kOhm)(4.6mA)+Vd+vds*1,2kOhm*4.6mA

$

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

Thanks
#8
07-17-2013, 02:18 PM
 Jony130 Senior Member Join Date: Feb 2009 Location: Poland/Wroclaw Posts: 2,286

Id = 4.23621mA and Vgs = -0.958448V

And yes, you need to know Ib first to find Vd = Ve voltage.
#9
07-17-2013, 02:44 PM
 AD633 Member Join Date: Jun 2013 Location: Ciudad de México Posts: 95

Quote:
 Originally Posted by Jony130 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.
For finding Ib do i need to use a system of equations?Something like:

$Vcc=330kOhm*Ib*Vbe+VDG+VGS+1.2kOhm*ID

Vcc=Vc+Vd+Vds+1.2kOhm*Id

$

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
07-17-2013, 03:07 PM
 Jony130 Senior Member Join Date: Feb 2009 Location: Poland/Wroclaw Posts: 2,286

Quote:
 Originally Posted by AD633 Or assuming that the BJT is in the ative region $Ic=Beta*Ib$ and that Ie is approx equal to Ic
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.

Quote:
 Originally Posted by AD633 $Ic=160*(4.236mA)=677.76 mA$ Thanks
What?? Ic = ??

 Tags bjt, circuit, jfet, transistors

 Thread Tools Display Modes Linear Mode

 Posting Rules You may not post new threads You may not post replies You may not post attachments You may not edit your posts BB code is On Smilies are On [IMG] code is On HTML code is Off Forum Rules
 Forum Jump User Control Panel Private Messages Subscriptions Who's Online Search Forums Forums Home Electronics Forums     General Electronics Chat     The Projects Forum     Homework Help     Electronics Resources Software, Microcomputing, and Communications Forums     Programmer's Corner     Embedded Systems and Microcontrollers     Computing and Networks     Radio and Communications Circuits and Projects     The Completed Projects Collection Abstract Forums     Math     Physics     General Science All About Circuits Commmunity Forums     Off-Topic     The Flea Market     Feedback and Suggestions

All times are GMT. The time now is 06:53 AM.