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Old 7th July 2012, 07:50 PM   #21
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But the resistor is not connected to ground....
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Old 7th July 2012, 07:55 PM   #22
tsiros is offline tsiros  Greece
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through the transistor, of course. not sure how much current will go around and through the base, i'd hazard a guess it would be very little, negligible maybe

is the transistor operating in saturation now?
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Old 7th July 2012, 08:00 PM   #23
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does that imply that the Ie=β*Ib not only works right-to-left, but the transistor will try to balance Ie and Ib to satisfy that equation in other cases too?

is the transistor operating in saturation now?
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Old 7th July 2012, 08:01 PM   #24
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Ahh yes, but we know EXACTLY how much goes where. We know that the base current is 1/beta of the collector current. Or, vice versa, Ic = beta* Ib. Assume beta = 100, then Ib = (1/100)*14.4mA and Ic = (99/100)*14.4mA.

Now we connect the second transistor, base to base, emitter to gnd. It has the same Vbe, so the same Ib, so the same Ic...
So whatever you hang off that 2nd collector, the 2nd transistor will try to pull the same Ic through it - there's your current mirror!

jan
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Old 7th July 2012, 08:05 PM   #25
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Quote:
Originally Posted by tsiros View Post
does that imply that the Ie=β*Ib not only works right-to-left, but the transistor will try to balance Ie and Ib to satisfy that equation in other cases too?
I don't think a transistor has any concept of left or right - it just 'sees' what's connected to its three terminals. It cannot work different than the way it is supposed to.

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is the transistor operating in saturation now?
I don't think so. The Vce is equal to Vbe which is 0.6V. I think the definition of saturation is that Vbc is only a few 10's of mV. But others may know that better.

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Old 7th July 2012, 08:09 PM   #26
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A final one now: suppose I connect the 2nd collector of that mirror to the same +15V through a 500 ohms resistor, what is the Vce of the 2nd transistor?

jan
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Old 7th July 2012, 08:14 PM   #27
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Quote:
Originally Posted by janneman View Post
I don't think a transistor has any concept of left or right - it just 'sees' what's connected to its three terminals. It cannot work different than the way it is supposed to.
yes, well... i have two objections. you can ignore them if you deem them inappropriate or just moronic. I need to study more...

it can operate differently (depending on how you define "operate", i might be getting a bit boring or infuriating, please don't ban me) if it gets damaged...

second, and this is my main point, i do not know if i am conveying my thoughts properly:

i know if you stick a Ib, it will try to draw the appropriate Ie, but does the transistor try to balance the equation even if you stick a Ie and let Ib be variable?

maybe i have taken the "water valve" parallel a bit too seriously... because a valve sets the water flow, but the water flow can't set the valve position, which the transistor can.

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I don't think so. The Vce is equal to Vbe which is 0.6V. I think the definition of saturation is that Vbc is only a few 10's of mV. But others may know that better.

jan
okay. still, you and the rest of the gang have been of great help.
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Old 7th July 2012, 08:17 PM   #28
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Quote:
Originally Posted by janneman View Post
A final one now: suppose I connect the 2nd collector of that mirror to the same +15V through a 500 ohms resistor, what is the Vce of the 2nd transistor?

jan
Ic = 14.4 mA,

Vce = 15 V - 14.4 mA * .5e3 Ω = 7.8 V ?
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Old 7th July 2012, 08:23 PM   #29
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Perfect! Of course, if you're picky you may want to take account of the small differences caused by the 1/100 of Ic that goes through the base, but I never do that. It's close enough as it is.
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Old 7th July 2012, 08:29 PM   #30
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[snip]i know if you stick a Ib, it will try to draw the appropriate Ie, but does the transistor try to balance the equation even if you stick a Ie and let Ib be variable?
Well whatever you 'stick' as Ie needs an Ib to flow, right? For instance, if you ground the base and pull the emitter negative, the emittor will become negative until Vbe becomes 0.6V, then Ie will start to flow. In fact, Ie will increase so much that the current will develop enough voltage across whatever is pulling it negative so that the emitter will not drop any further. You can break the transistor that way, or it can go into heavy saturation, but the behaviour is exactly predictable by the way a transistor 'normally' works.

Now as you hinted, if the transistor gets into saturation, the nice Ic/Ib=beta breaks down. For instance, if you send 1A into the base there's no way that Ic becomes 100A (unless you have a really big one). But for 'normal' use the beta-concept works pretty well.

jan
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