All About Circuits Forum  

Go Back   All About Circuits Forum > Electronics Forums > Homework Help

Notices

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.

Reply   Post New Thread
 
Thread Tools Display Modes
  #11  
Old 07-18-2013, 08:29 AM
AD633 AD633 is offline
Member
 
Join Date: Jun 2013
Location: Ciudad de México
Posts: 96
Default

Quote:
Originally Posted by Jony130 View Post
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 = ??
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
Reply With Quote
  #12  
Old 07-18-2013, 09:27 AM
t_n_k t_n_k is offline
Senior Member
 
Join Date: Mar 2009
Posts: 4,949
Default

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.
Reply With Quote
The Following User Says Thank You to t_n_k For This Useful Post:
AD633 (07-18-2013)
  #13  
Old 07-18-2013, 02:10 PM
AD633 AD633 is offline
Member
 
Join Date: Jun 2013
Location: Ciudad de México
Posts: 96
Default

Quote:
Originally Posted by t_n_k View Post
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.
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
Reply With Quote
  #14  
Old 07-18-2013, 11:13 PM
t_n_k t_n_k is offline
Senior Member
 
Join Date: Mar 2009
Posts: 4,949
Default

@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 by t_n_k; 07-19-2013 at 12:07 AM.
Reply With Quote
Reply   Post New Thread

Tags
, , ,


Thread Tools
Display Modes

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 Jump


All times are GMT. The time now is 06:15 AM.


User-posted content, unless source quoted, is licensed under a Creative Commons Public Domain License.
Powered by vBulletin
Copyright ©2000 - 2014, vBulletin Solutions, Inc.