Paul,

This forum may or may not be the first stop on the student's journey to solving the problems presented to them. They do not tell us nor do we ask. They rarely post the source of their uncertainty.

Couple that with "google translated" inquiries from non-Greco/roman languages, and it becomes a communication dance to get to the problem statement.

There are some assumptions made by all who answer the questions posed here. One of which is the students have already seen their problem worked out before and are struggling with replicating the effort with a similar problem. This is a failure in their associative learning process. These type failures demand the students to produce their work up to the present so the membership can "troubleshoot" and explain where the student went astray.

Sometimes it's like pulling teeth to get information from the OP, sometimes it's not.

Learning to differentiate between those wanting to learn and those wanting the answer is not easily seen by some who answer. Those who "answer" deserve an applause for participation while being cautioned about providing answers. It's a fine line.

A few years back I was asked how I discriminated against the two types of posters (those who want to learn and those who want only the answers). I told the person they would pick that up quick enough.

The Socratic method is the method used by the ebook and by association, this website.

We, the members, rarely get the Paul Harvey on the student's textbook or curriculum. Very few cite their source or provide a complete problem statement from the originator.

Everything considered, all the members do an outstanding job in the homework area.

I will agree with you, WBahn appears to be a very good teacher. Maybe he will "record" his lectures for his own vBlog.

 

Hi

Need some help... For the attached transistor circuit I want to determine:

• \(I_c\)
• \(U_{ce}\)

Stuck and need some help on how to approach this.

Thanks!

To calculate transistor currents we need to think a little and to work out. In any case of BJT(pnp or npn) the base-emitter junction is forward biased(like a diodes p-n junction) and the base-collector junction is reverse biased(diode with reverse polarity). The emitter of npn transistor is heavily doped with n-type carriers. As regards the collector region its less doped with n-type carriers and narrow in relative to emitter. So base is p-type and very thin. Electrons under the electrical field begin to move in p-type base and some of them recombine with the holes of base. But the rest of electrons are being attracted with the electrical field of base-collector junction, so creating the collector current. Because of recombination of some electrons in the base region collector current is less than emitter current (Ic=Ie - Ib; Ie=Ic+Ib). In the open position the voltage drop between base and grounded emitter is about 0.7v, so if the base signal is Vbb the Vrb(voltage on Rb) would be Vbb-Vbe=Vbb-0.7v(Vrb=Vbb-0.7v). The base current would be Ib=(Vbb-0.7v)/Rb=Vrb/Rb. Now we now the base current. To figure out collector current we multiply Ib and β. Ic=Ib * β. Vce is Vcc - Vrc. Vcc( supply voltage between collector and emitter), Vrc is voltage drop on the resistor in collector circuit. We know collector current Ic so voltage drop on Vrc will be Vrc=Ic*Rc. So Vce=Vcc-Vrc.

 

Thanks for all the help!

But do you understand how to do it now? (As opposed to just parroting an equation provided by someone else).

Thanks WBahn. Yes, I do understand how to do it now. The algebra behind it is basic. I was stuck and when you and shteii01 gave me the answer it was kind of a "duh" moment on my part...

The problem is that OP did not read the textbook.

shiteii01: I do read my textbook. I appreciate both WBahn and PRS's teaching methods. I will, to the bitter end, try to solve a problem presented to me. Sometimes, however, I get stuck and then it is nice with some help.... be it a slight push in the right direction or, as you did, a full presentation of the equation needed, thanks again. I did not sign up here and ask you all for help because I couldn't be arsed to do it myself.