Re: Transistors

John Devereux jdREMOVE@xxxxxxxxxxxxxxxxxx posted to
sci.electronics.design:

JosephKK <joseph_barrett@xxxxxxxxxxxxx> writes:

Fred Abse excretatauris@xxxxxxxxxxxxxxxxx posted to
sci.electronics.design:

On Fri, 16 Nov 2007 16:32:35 -0700, Jim Thompson wrote:

On Fri, 16 Nov 2007 15:15:45 -0800, John Larkin
<jjlarkin@xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx> wrote:


<snip>


+5V
|
|
|
c
+10v----1K------b
e
|
|
|
5K
|
|
gnd



What is Vb? Ve?

John

John! You've created the universal employment test vehicle!

The amusing/distressing part of this is that no answers have been
forthcoming.

How about posting an E-mail address to send answers?

Then you can grade the respondents.

I'll lay you some side bets on who won't (be able to) answer ;-)

Assuming NPN, and 0.7V junction Vf:

Vb approx. 5.7V
Ve approx 5V
Ic is negative


I figure ib ~= ie ~=1.6 mA & 0.6 v Vbe.

Vb then is about 8.4 V and Ve 7.8 V.
Ic is well below uA.

How does that work?

There is a b-c diode junction, so Vb is pinned to 5V+0.7 = +5.7V.
b-e is forward biased so Ve is 5.7 - 0.7 = 5V. Ie = 5/5k = 1mA.

In other words, I agree with Fred.


Yep. i blew it, and Ic is about 3.3 mA.

.

Relevant Pages