BJT question

Rusty? Indeed! There is too much to explain here. Go back to your textbook and _read_ carefully!

...Jim Thompson

--
|  James E.Thompson, P.E.                           |    mens     |
|  Analog Innovations, Inc.                         |     et      |
|  Analog/Mixed-Signal ASIC's and Discrete Systems  |    manus    |
|  Phoenix, Arizona            Voice:(480)460-2350  |             |
|  E-mail Address at Website     Fax:(480)460-2142  |  Brass Rat  |
|       http://www.analog-innovations.com           |    1962     |
             
         America: Land of the Free, Because of the Brave

You need to know the following equations:

Ie = Ic + Ib Ic = beta*Ib Vb = Ve + 0.7V

Now if you do not use any biasing for the transistor(a bad idea) and want Ic=50mA with beta = 50 then Ib = 1mA and Ie = 51mA.

So just sending in 1mA will get you about 50mA "out"(or in at the collector).

To do that you can ground Ve so that Vb = 0.7V(you can approximate this as

0V if you want but is essentially a diode drop).

Vcc ---- R ---- Vb (Base of transistor) "diode" ---- Emitter --- (ground)

R can then be calculated to get 1mA to the base, i.e.

(Vcc - Vb)/R = 1mA

If, say Vcc is 5.7V then

R = 5.7/1mA = 5.7kOhm

So using a 5.7k resistor will get you about 1mA base current which will get you about 50mA collector current.

This is not a good design though because its very unstable(beta will change with temperator and other factors).

You should read a book or look on the net for various "biasing" methods that help stabilize the transistor so that it is "independent" of beta and depends only on the components around it.

In the above case obviously Vce depends on the load we attach to the collector or emitter. (we have 51mA going through the emitter)

Jon,

Thanks for the help. I think I got it now...

Bo

a sheet

Isn't the data sheet hfe value a minimum? It's not '=3D0.83 mA', really, but rather '

Jim; you should have at least directed him to work on operation point / DC solution / quiescent point (Q-point), and the definition of edge of saturation.

[snip]

I would have, but the recitation of his problem sounded to me like he was reading from a homework sheet rather than expressing his true confusion.

If he'd expressed _any_ comprehension at all I would have jumped in to fill the gaps.

We already have enough bad engineers and technicians ;-)

...Jim Thompson

--
|  James E.Thompson, P.E.                           |    mens     |
|  Analog Innovations, Inc.                         |     et      |
|  Analog/Mixed-Signal ASIC\'s and Discrete Systems  |    manus    |
|  Phoenix, Arizona            Voice:(480)460-2350  |             |
|  E-mail Address at Website     Fax:(480)460-2142  |  Brass Rat  |
|       http://www.analog-innovations.com           |    1962     |
             
         America: Land of the Free, Because of the Brave

In your case, you just forget about power until you compute all the voltages and currents. Just stick to Ohm's law and the current loops in the transistor circuit. In the case of Vce you simply have Vcc=IcRc+Vce+IcRe+Vee, assuming an emitter bias resistor and power supply (Vee may be 0V), making Vce=(Vcc-Vee)/(Rc+Re). That formula neglects the Ib component of Ie but in most cases it is so small it is justified, less than 2% in your case. As for transistor saturation, recall that this is the condition of having both junctions forward biased. "On the brink of saturation" is taken to mean the BC junction is neither forward or reverse biased but has 0V across it. Then using the fact that Vce=Vcb+Vbe and Vcb=0, makes Vce=Vbe which will be around

0.75V for most small signal stuff at Ic=50mA. The transistor saturation curves show Vce vs Ic in deep saturation where the base drive is much much greater than the base drive in the active region, usually at Ib=Ic/10, so you don't want to use those curves.

whoops..that should be Vce=(Vcc-Vee)-Ic(Rc+Re).

On Thu, 13 Mar 2008 08:57:31 -0700 in sci.electronics.design, Jim Thompson wrote,

I read the original post and figured "here is exactly the kind of question Jim Thompson can answer if he wants to" so I marked the thread for retrieval, hoping to learn some electronics. Silly me.

If I posted one of my past tutorials, would you read it? Or ask disconnected, unreferenced questions like the OP?

I have a ton of stuff, some of which is already lurking there on my website, if you're interested.

...Jim Thompson

--
|  James E.Thompson, P.E.                           |    mens     |
|  Analog Innovations, Inc.                         |     et      |
|  Analog/Mixed-Signal ASIC\'s and Discrete Systems  |    manus    |
|  Phoenix, Arizona            Voice:(480)460-2350  |             |
|  E-mail Address at Website     Fax:(480)460-2142  |  Brass Rat  |
|       http://www.analog-innovations.com           |    1962     |
             
         America: Land of the Free, Because of the Brave

Legitimate viewpoint. I had perceived it as being so far down in the mud as to no longer be able to state a coherent question. Nor did i give OP anymore information than what i had stated.

On Sun, 16 Mar 2008 14:54:41 -0700 in sci.electronics.design, Jim Thompson wrote,

I always have plenty to read, so don't put yourself out on my account. I also read more than I ask questions. If you post an answer to any of my questions, I will certainly read it.