calculating emitter stablizied biasing sat current

Hi all: I have a simple emitter stablized circuit. with Vcc = 12V, Rc = 2.2k, Re =

1.2k I did Vcc / (Rc + Re) to find the sat current = 3.53mA ( assuming Vsat = 0V ) but the answer on my textbook gave me 5.13 mA!!

Is it my problem or the textbook's problem? thanks for any help...

--
{ Kelvin@!!! }
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Telus News
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Your logic seems okay to me. But I'm not trained in electronics, so I'll listen to others on this, too.

I tend to think about BJTs as having Vce(sat) of about < 0.4V, very often in the near .05V-.15V range. But that's barely anything different to what you were saying. I can imagine how to increase the emitter current, of course, by simply jacking up the base voltage and dumping base current like it was going out of style. But the collector current would only then decline a little as the (Ic+Ib)*Re = Ie*Re voltage jacked up the collector voltage, which rides at Vce(sat) above it, and thus dropping the collector current somewhat. I assume they were talking about "sat current" as collector current, yes?

At 5.13mA, just through the collector resistor Rc alone, you've about used up all your 12V supply -- with only about .7V left over. This sounds a lot like someone imagining that the saturation current is the entire 12V supply going through the collector resistor and a diode to ground. If the wrong polarity BJT were wired up, it might act like that, I suppose.... But it really sounds more like you are assuming they were talking about a collector resistor when they were really talking about a base resistor. Or they assumed that and wrote something else.

Oh, well. I'm stumped.

Jon

Reply to
Jonathan Kirwan

Your arithmetic is correct, assuming Vcesat = 0. Isat will be even less with a non-zero Vcesat. It wouldn't be the first time that a textbook has a typographical error. (Or maybe the purpose is to find out if you're confident enough in your coursework to actually _find_ the error(s) in the textbook! ;-) )

Cheers! Rich

Reply to
Rich Grise

What's the Vee? -5.2V maybe?

....jerry

Reply to
Jerry R

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