Re: Transistors
- From: John Larkin <jjlarkin@xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx>
- Date: Mon, 19 Nov 2007 21:23:50 -0800
On Tue, 20 Nov 2007 04:36:29 GMT, JosephKK
<joseph_barrett@xxxxxxxxxxxxx> wrote:
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
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c
+10v----1K------b
e
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5K
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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.
Well, it's stranger than it looks. Transistors that saturate well in
this setup have high reverse betas, namely beta with the emitter and
collector switched. Some do, some don't.
John
.
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