Analag BJT Inspection Analysis

Thread Starter

sinatra39

Joined Oct 18, 2018
14
Hello folks,

Im watching currently a video lecture series on analog integrated circuits from the university of Berkeley and in video 4 the topic is inspection analysis.

At around 1:14 of ( http://www.infocobuild.com/education/audio-video-courses/electronics/EE140-Fall2013-Berkeley/lecture-04.html )the lecturer starts to derive the current through the circuit and comes to the conclusion that Icc is determined by -UBE-UEE/REE. This is what I dont understand because in theory at this point we should also take the transistor into account (and do the load plot etc). I mean in his analysis by choosing for instance REE -> 0 the current could go to infinity which it wont of course due to the transistor so it is obviously wrong, so I guess it is a good approximation and why is it for this example and when can I do this so easily?

Thank you very much in advance for the help!
 

Papabravo

Joined Feb 24, 2006
14,412
Ic (the collector current) will be zero if Vbe - Ve < 0.7 V (approx.)
When Vbe - Ve ≈ 0.7 V then Ic + Ib = I e ; The emitter current is the sum of the base current and the collector current, and Ib << Ic
The actual currents throught the transistor are mediated by external components, not any internal values of intrinsic resistance. You will seldom if ever see the terminals of a transistor connected across a voltage source.
 

Thread Starter

sinatra39

Joined Oct 18, 2018
14
Yes I'm aware of that but the problem of understanding is the following: In the video the lecturer wants to know the current in order to get the dc operation point. The only thing he knows about the transistor is that the base is grounded, so he calculates the emitter voltage to -0.7 V, assuming it runs. And with just this information and UEE and REE he calculates the current with simple ohms law. I mean in my bachelor I learned that calculating the op requires the transistor graph and the load line and the intersections represents the operating point. Why can I neglect this procedure here and just use ohms law and neglect any transistor effects other than the base-emitter voltage drop?
 

Papabravo

Joined Feb 24, 2006
14,412
Yes I'm aware of that but the problem of understanding is the following: In the video the lecturer wants to know the current in order to get the dc operation point. The only thing he knows about the transistor is that the base is grounded, so he calculates the emitter voltage to -0.7 V, assuming it runs. And with just this information and UEE and REE he calculates the current with simple ohms law. I mean in my bachelor I learned that calculating the op requires the transistor graph and the load line and the intersections represents the operating point. Why can I neglect this procedure here and just use ohms law and neglect any transistor effects other than the base-emitter voltage drop?
If you can do it correctly, then there is no problem.
 

Jony130

Joined Feb 17, 2009
5,183
The situation will look like this:


assd.PNG

And the KVL for Vee loop is :


Vee = Vbe + Iee*Ree therefore Iee = (Vee - Vbe)/Ree

No matter what.

Any additional questions?
 

LvW

Joined Jun 13, 2013
1,085
Vee = Vbe + Iee*Ree therefore Iee = (Vee - Vbe)/Ree
@Sinatra: And what does this relation (undoubtly true) tell us?
It clearly shows that the uncertainty of Vbe (0.65 or 0.7 volts ?) plays a minor role (if compared with Vee) for the resulting current Iee. This is one of the advantages of negative DC feedback caused by Ree.

As far as your example (Ree=0) is concerned: For Ree=0 we have Vbe=Vee and the transistor would be destroyed, unless you reduce Vee down to Vbe.
A similar effect applies also to a simple ohmic resistor: I=V/R. For R=0 the formula tells us that I becomed infinite. So - something must be wrong with this formula...?
 
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