BJt current mirror with a twisted resistor

Thread Starter

KevinEamon

Joined Apr 9, 2017
269
1578269435184.png
Hi guys
I've an exam in a few days - the lecturer said there might be something in the test - similar to this above, but that he was going to mess with it.
Later he indicated one of the things he might do is add a resistor or two.
This above is easy peazy

But now I'm thinking to myself that he's going to slap that resistor - right in there where I've said "1" in the figure above.
Now I'm having a wee think to myself about this... but I've ended up in a madness.
Like look at this below for example... Have I created some type of logical crazy circuit here or what?

Please someone beat the sense back into me. I've been studying for days and I might have driven myself stupid.

1578270576964.png
 
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WBahn

Joined Mar 31, 2012
25,064
You are, as usual, completely ignoring units and just tacking on whatever units you want the answer to have.

Why are you doing 0 V / 3 kΩ? What is your basis for claiming that the 3 kΩ resistor has 0 V across it?

The assumption that Vbe is always a fixed voltage regardless of collector current is incompatible with how these circuits work -- they DEPEND on the collector current being a strong function of Vbe. For the original circuit, that assumption is sufficiently valid in order to get a value for Ro that is close to the value that is actually needed, but as soon as the two sides of the differential pair are no longer operating under identical conditions, you can no longer make that assumption for them.
 

Thread Starter

KevinEamon

Joined Apr 9, 2017
269
At least I'm getting slightly better... ;) And maybe I'm just fishing for your help good sir.
Seriously though I do usually use the unit values most of the time now. It gets a little tricky with fermi level stuff mind you.

K anyways if you look there at position 3: -
Well VB is grounded, and if they're set to drop 0.7V, then that must mean VE is -0.7V right? And then if you step up 0.7V on BJT 2 well VB there must be 0V... I think...

The assumption that Vbe is always a fixed voltage regardless of collector current is incompatible with how these circuits work -- they DEPEND on the collector current being a strong function of Vbe. For the original circuit, that assumption is sufficiently valid in order to get a value for Ro that is close to the value that is actually needed, but as soon as the two sides of the differential pair are no longer operating under identical conditions, you can no longer make that assumption for them.
So I've just created a non functional system... well that's nice.
Hmmm so do you think I'm wrong about where he's going to add this resistor? and if so... where might a naughty professor such as yourself, get up to a bit of badness?
 
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Thread Starter

KevinEamon

Joined Apr 9, 2017
269
Or maybe would he put resistors on both 2 and 3...?

Or a big nasty one on that current splitting wire, now that would be bad.
Sigh...
I hate not knowing what's coming...
 

WBahn

Joined Mar 31, 2012
25,064
Since 2 and 3 are written between the collector and emitter, I don't know what you mean by adding a resistor on both of them.

He could add a resistor in both of the base legs (where you have the 1 and the corresponding location on the other side), but that doesn't change anything given the assumptions in the problem.

As soon as you start talking about current mirrors, you need to pretty much stop using the assumption that Vbe is 0.7 V. Yes, it will still be about that and that can still be useful for finding the approximate quiescent operating point, but even that obscures the very basis behind which current mirrors operate -- they rely on that very strong relationship between Vbe and Ic in order to transfer the collector current in one transistor to another transistor by means of establishing the value of Vbe in the source transistor and then applying that value of Vbe to the image transistor.
 

Delta prime

Joined Nov 15, 2019
99
Please someone beat the sense back into me. I've been studying for days and I might have driven myself stupid.
Heed the warning and exceptional advice of the second thread. Relax stop studying you got this. The change the professor is going to add will be so simple you will embarrass yourself by answering it. I bet!
 

Thread Starter

KevinEamon

Joined Apr 9, 2017
269
KK breath... thx guys.
Right... But just to be sure, I mean if we take just the base circuit, in diagram 1 up there.
I can assume VE = -0.7V right?
 

Analog Ground

Joined Apr 24, 2019
269
Ask yourself what happens if you add two resistors. One from the emitter of transistor "2" and the collector of the current mirror transistor and one from the emitter of "3" to the collector of the current mirror. Or more simply, add two emitter resistors to the input transistors. Why might someone want to do this? Should the resistors be equal or different? Why? Good luck on your exam!
 

panic mode

Joined Oct 10, 2011
1,803
Please.....before you do ANYTHING.....annotate ALL circuit components, all nodes, all currents....
Then start writing equations using your schematics and annotation as a reference.

Only THEN, this will begin to make sense or earn you any credits. Do it right from getgo...

Btw, you not just need to use units but make sure they are correct. I1 is not 0.5A, and I2 is not 50mA. Both are off by several orders of magnitude.
 
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ci139

Joined Jul 11, 2016
1,086
in #1 the lower img. -- why R3 . . . it's just a current divider as shown on the upper img.
-- what goes down from Ro in principle goes up from other resistors (that , when /!\ neglecting /!\ the relatively low ß)
. . . if there is voltage drop on collector resistor and the emitter voltage is known ? whats the voltage drop on CE junction (that , if we 1-st neglect the base currents + later combining them - the voltage mirror is not exactly a voltage mirror ...)
 
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