# What is the difference between U(Th) and U(BQ) ?

#### Tobias Hildebrandt

Joined Jan 1, 2016
44
Good evening!

Last question for today, it is already very late... So here it comes:

So, R(3) = 10kOhm, U(BE) = 0.7 V , B = 200

My question is, I have painted exactly the same equivalent circuit and therefore I assumed that U(BQ) = U(Th). I doubt that it is a typo, because U(Th) ≈ 1.870026525 V and he wrote down 1.771 V. So what exactly is the difference ?

I hope, I didnt bother you to much today! In case, I forgot to translate sth, please let me know, I will fix it right away.

Tobias

#### WBahn

Joined Mar 31, 2012
26,398
What is U(Th)? U(BQ) is probably the quiescent base voltage. If U(Th) is some Thevenin voltage, then you need to make clear what the terminals in question are.

#### WBahn

Joined Mar 31, 2012
26,398
If you look at the Thevenin equivalent circuit for the bias circuit you get a Thevenin voltage of 1.87 V. If you then look at the voltage at the base, it is 1.77 V.

#### Tobias Hildebrandt

Joined Jan 1, 2016
44

Two points:
- I figured out, how to solve this circuit. ( I(BQ) + I(CQ) ) * R4 = 1.071 V = U(EQ) and for U(BQ) you just add the voltage drop.
- This being said, there is something fundamentally wrong with my understanding of circuits.
Why does 15 V * R2 / ( R1 + R2) give me not the right result? Why can't I use the Voltage divider result?

Tobias

#### Jony130

Joined Feb 17, 2009
5,335
U(th) is not the voltage at transistor base. U(th) is a thevenin's voltage

Why does 15 V * R2 / ( R1 + R2) give me not the right result? Why can't I use the Voltage divider result?
Because this equation do not take into account the transistor base current. And this current (Ib) is a load for your divider.
The voltage at base is equal to
U(BQ) = U(TH) - Ib*Rth

#### Tobias Hildebrandt

Joined Jan 1, 2016
44
Wonderful! Thank you!

PS: Btw, on some German math forums you have to tick a box when the problem is solved. Do you have sth similar here? Because I don't want to keep a thread open, when it is already answered. Anyway, thanks again for your help! It is very much appreciated!

#### WBahn

Joined Mar 31, 2012
26,398
View attachment 98960
Two points:
- I figured out, how to solve this circuit. ( I(BQ) + I(CQ) ) * R4 = 1.071 V = U(EQ) and for U(BQ) you just add the voltage drop.
- This being said, there is something fundamentally wrong with my understanding of circuits.
Why does 15 V * R2 / ( R1 + R2) give me not the right result? Why can't I use the Voltage divider result?

Tobias
Because the voltage divider ignores the effect of the base current. It only gives you the Thevenin equivalent voltage for the circuit consisting only of the voltage divider.

#### WBahn

Joined Mar 31, 2012
26,398
Wonderful! Thank you!

PS: Btw, on some German math forums you have to tick a box when the problem is solved. Do you have sth similar here? Because I don't want to keep a thread open, when it is already answered. Anyway, thanks again for your help! It is very much appreciated!
No. Here there is no way for the member to close a thread. Others may want to continue discussing the problem to further there own understanding and that's a major point of a forum is for one person's problem to benefit lots of people. If you do not want to receive e-mail notifications of further posts to the thread, you can "unwatch" it.

#### Tobias Hildebrandt

Joined Jan 1, 2016
44
I got lectured once, that I didnt mark a thread as "solved" at Matheplanet. I didnt want to upset the people in this forum. That is why I asked.

#### WBahn

Joined Mar 31, 2012
26,398
Perfectly reasonable to ask. Each forum has it's own set of rules and expectations -- and it can be very hard to keep them straight.

#### MrAl

Joined Jun 17, 2014
8,996
Hi,

Thevenin and Norton equivalents attempt to solve circuits in a 'piecewise' manner, where we are only looking at one part of the circuit for a time, and it may be only part of the solution. That means it would be good to learn more about these two equivalent circuit techniques.

For example, if you have a voltage source in series with a resistance as part of the circuit, then without looking too much at the rest of the circuit you can replace that with a current source in parallel with a resistance. Alternately, if you have a current source in parallel with a resistance you can replace it with a voltage source in series with a resistance.

Using these two ideas however does not imply that any given node is solved for in any way as there may be more work to be done first.
For example, if you have a voltage source in series with two resistors and you replace it with a current source and only one resistor, you'd still have more work to do in order to calculate the node voltage after the second resistor, and that would entail knowing any other component also connected to that node.
We could look at some simpler examples if you like because you should have been given more simple examples before you got into transistor circuits anyway. The normal flow of learning would start with resistor circuits with regular sources not dependent sources as the transistor model is made from (current controlled current source in your problem example).

#### Tobias Hildebrandt

Joined Jan 1, 2016
44
In the first year of this BSc Power Engineering & Renewable Energy degree program, we already had calculus, multivariable calculus and series in math, mechanics and vibrations as well as thermodynamics and waves in physics, C and C++ programming and of course the single and polyphase systems, electrical and magnetic fields, and analysis of ac and dc circuits.... I am pretty sure we talked about that in this particular class, BUT BUT BUT it became very clear to me that I either forgot Norton and Thevian equivalents already or that I never really understood them. So, YES, please if you can give me some easy examples to work through, that would be very much appreciated!!!

#### WBahn

Joined Mar 31, 2012
26,398
Probably the best way is to make up your own problems and work them. You have many circuit analysis techniques available to you that enable you to work the problems multiple ways so that you can have confidence that you've got the right answer and that allows you to see what parts of what problems seem to be tripping you up and to make new problems that focus on those issues. It also lets you start out as simple as you need to and then advance the problem complexity at whatever pace challenges you but keeps it acceptably within reach of where your current comfort zone is.