Diff amp common mode model

Thread Starter

automagp68

Joined Nov 13, 2011
81
BJT diff amp in question here

In the diff mode it is true that ib1 = -ib2

For the common mode i was always under the impression that ib1 = ib2

Both inputs are the same v1 = v2 = vin

Hence the current out of the emitter is the sum of ib1 and ib2 going to an RE resistor.

Is this not true?
 

MikeML

Joined Oct 2, 2009
5,444
BJT diff amp in question here

In the diff mode it is true that ib1 = -ib2

For the common mode i was always under the impression that ib1 = ib2

Both inputs are the same v1 = v2 = vin

Hence the current out of the emitter is the sum of ib1 and ib2 going to an RE resistor.

Is this not true?
Not quite:
Ib1 = Ib2 only if the diff pair currents are balanced; otherwise not. Both Ib1 and Ib2 flow into the bases if NPN; out of the bases if PNP.
The emitter current of each half of the pair is β*Ib, so the current in the common emitter current sink (not necessarily a resistor) is the sum of β*Ib1 + β*IB2, which at balance might simplify to 2*β*Ib.
 

Thread Starter

automagp68

Joined Nov 13, 2011
81
I'm talking ideal cases

The flow into the base is the same

your putting two identical signals on each base in phase with each other.

So how is Ib1 not equal to Ib2

Hang on one second ill draw a picture
 
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Thread Starter

automagp68

Joined Nov 13, 2011
81
What i have here is the correct answer according to the book

However here is my problem

Because Ib1 = ib2 you get a negative common mode gain

If ib1 = -ib2 then you would get a positive common mode gain

I am 100% certain that ib1 = ib2
That is the definition of common mode

 

Thread Starter

automagp68

Joined Nov 13, 2011
81
I have no idea why the picture posted sideways sorry

The only way Ib1 = -ib2 is if you have the two signals out of phase

The definition of common mode is the SAME signal so how is it possible that ib1 does not equal ib2 is what I'm confused about

Like i said above i got the correct magnitude for the common mode gain but i get a negative
 

WBahn

Joined Mar 31, 2012
25,911
BJT diff amp in question here

In the diff mode it is true that ib1 = -ib2

For the common mode i was always under the impression that ib1 = ib2

Both inputs are the same v1 = v2 = vin

Hence the current out of the emitter is the sum of ib1 and ib2 going to an RE resistor.

Is this not true?
So if ib1 = -ib2 that means that the current is going in one base and out the other? Are you talking large signal or small signal?
 

WBahn

Joined Mar 31, 2012
25,911
What i have here is the correct answer according to the book

However here is my problem

Because Ib1 = ib2 you get a negative common mode gain

If ib1 = -ib2 then you would get a positive common mode gain

I am 100% certain that ib1 = ib2
That is the definition of common mode
You say that you are 100% certain that ib1 = ib2.

You say that if you have ib1 = ib2 that you get a negative common mode gain.

You got a negative common mode gain.

You say you got the correct answer according to the book.

So what's the problem?
 

Thread Starter

automagp68

Joined Nov 13, 2011
81
Standard reply from a rude moderator on a power trip. What else is new

No I'm not annoyed that i didn't get help in an hour. I simply stated for the last few months this forum has been very quite. You know why i can say that? BECAUSE OF BEEN A MEMBER ON THIS SITE LONGER THEN YOU! Therefor i have a longer history of how active this site TYPICALLY is.

But in typical moderator fashion you make it about something different.

The answer according to a solution i found by the author is +.22
That is the PROBLEM
As i stated above. You know the part you didnt bother reading

Just so happens I'm in Denver. I see your in larkspur. Since you are apparently an electronics genius why don't you come up here and explain it to me. Ill pay you for your time
 

Jony130

Joined Feb 17, 2009
5,176
Well for me the common mode gain is Acm ≈ Rc/(re +2Re) ≈ Rc/2Re ≈ 2kΩ/8.6kΩ ≈ 0.232V/V and usually we omit the minus sign because we now what this "minus" sign represent/means. This minus only informs us that the Vo voltage is 180 degree phase shift in respect to input voltage.
 

WBahn

Joined Mar 31, 2012
25,911
Standard reply from a rude moderator on a power trip. What else is new

No I'm not annoyed that i didn't get help in an hour. I simply stated for the last few months this forum has been very quite.
That's bull and we both know it.

You know why i can say that? BECAUSE OF BEEN A MEMBER ON THIS SITE LONGER THEN YOU! Therefor i have a longer history of how active this site TYPICALLY is.
Why not shout it out? What, do you want a prize? My response had nothing to do with being a moderator. But since you've been around so long and know so much about the boards, you already know this.

But in typical moderator fashion you make it about something different.
I suspect I was right on target. But that's neither here nor there.

The answer according to a solution i found by the author is +.22
That is the PROBLEM
As i stated above. You know the part you didnt bother reading
Where did you state this? Where is this part I didn't bother reading?

Did you state it in your first post?

BJT diff amp in question here

In the diff mode it is true that ib1 = -ib2

For the common mode i was always under the impression that ib1 = ib2

Both inputs are the same v1 = v2 = vin

Hence the current out of the emitter is the sum of ib1 and ib2 going to an RE resistor.

Is this not true?
Did you state it in your second post?

lots of crickets in here lately
Did you state it in your third post?

I'm talking ideal cases

The flow into the base is the same

your putting two identical signals on each base in phase with each other.

So how is Ib1 not equal to Ib2

Hang on one second ill draw a picture
Did you state it in your fourth post?

What i have here is the correct answer according to the book

However here is my problem

Because Ib1 = ib2 you get a negative common mode gain

If ib1 = -ib2 then you would get a positive common mode gain

I am 100% certain that ib1 = ib2
That is the definition of common mode

Here, not only do you NOT state that the author's answer was positive, you explicitly state that you DID get the same answer as the author (i.e., from the book) and the answer you show that you got is negative.

Apparently you think that everyone should be a mind reader and read what you wanted to state but couldn't be bothered doing so.

Just so happens I'm in Denver. I see your in larkspur. Since you are apparently an electronics genius why don't you come up here and explain it to me. Ill pay you for your time
Be happy to. My charge out is currently $125/hr, including travel time, plus $0.56/mile.
 

Thread Starter

automagp68

Joined Nov 13, 2011
81
O wait a second, i see the problem! You didn't read the entire post and made a bone head comment now want to blame it on me because you didn't bother to read the first 2 pages completely.
Man i hope wherever you work tolerates thats kind of stuff. Grow up you grumpy old man. Is this how you get your kicks now a days? 50 something year old man goes on a forum to birate people that know less then he does? What a big surprise, an engineer with no social skills

You know whats funny about you! In your quest to show everyone that the moderator is so superior you still can't answer the question! So not only are you on a power trip but you don't even know the answer to back it up!
 
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Thread Starter

automagp68

Joined Nov 13, 2011
81
Jony130

Thanks for your kind reply.

Can you please explain the justification for your answer
As you can see the from the picture i posted by my calculations the gain ACM is negative. I know what you mean by saying the negative sign just denotes the inversion. Thats fines of course. What I'm interested in is how your calculation yields a positive value

The Vout f the circuit is clearly -B*ib2*2k

Because both the input signals are the same and also in phase, ib1 = ib2

So Vout = -B * ib1 * 2k

Giving us a negative common mode gain
 

WBahn

Joined Mar 31, 2012
25,911
O wait a second, i see the problem! You didn't read the entire post and made a bone head comment now want to blame it on me because you didn't bother to read the first 2 pages completely.
Again -- what did I miss? Where did you state (in which post) that the author's answer was positive? What is so hard about pointing it out? Unless, of course, it isn't there to be pointed out.

Man i hope wherever you work tolerates thats kind of stuff. Grow up you grumpy old man. Is this how you get your kicks now a days? 50 something year old man goes on a forum to birate people that know less then he does? What a big surprise, an engineer with no social skills

You know whats funny about you! In your quest to show everyone that the moderator is so superior you still can't answer the question! So not only are you on a power trip but you don't even know the answer to back it up!
Pure childish games. Seen your tactics plenty of times. Grow up.

You should have been able to look at the original schematic and estimate, by inspection, that the common mode gain was approximately -(1/2)(2 kΩ / 4.3 kΩ) = 0.233. If you couldn't, that's fine. But you should take the opportunity to review it and see why you can make that quick estimate without doing a detailed analysis. Do you understand why the factor of 1/2 is there?

As for the missing minus sign, as Jony130 said it is frequently omitted. Part of the reason is that gain is often expressed in dB which can only deal with the magnitude of the gain (you have to specify phase to get the rest of the information). Personally, I think this is a bit sloppy unless you indicate that |Acm| = 0.22, but you have to get used to such things.
 

Thread Starter

automagp68

Joined Nov 13, 2011
81
Again -- what did I miss? Where did you state (in which post) that the author's answer was positive? What is so hard about pointing it out? Unless, of course, it isn't there to be pointed out.



Pure childish games. Seen your tactics plenty of times. Grow up.

You should have been able to look at the original schematic and estimate, by inspection, that the common mode gain was approximately -(1/2)(2 kΩ / 4.3 kΩ) = 0.233. If you couldn't, that's fine. But you should take the opportunity to review it and see why you can make that quick estimate without doing a detailed analysis. Do you understand why the factor of 1/2 is there?

As for the missing minus sign, as Jony130 said it is frequently omitted. Part of the reason is that gain is often expressed in dB which can only deal with the magnitude of the gain (you have to specify phase to get the rest of the information). Personally, I think this is a bit sloppy unless you indicate that |Acm| = 0.22, but you have to get used to such things.
Geeze it only took you 3 hours to say something useful

So your explanation is that yes ACM is typical expressed in DB
Yes i understand that a log rule only uses magnitude obviously because the function is not defined for a negative value

So please tell me how i should of seen this by inspection given the fact that I'm a controls person not to mention I'm just a student I'm comparison to your 30 years of work experience so please feel free to educate the next generation as to how i should of seen this by inspection
 

WBahn

Joined Mar 31, 2012
25,911
Geeze it only took you 3 hours to say something useful

So your explanation is that yes ACM is typical expressed in DB
Yes i understand that a log rule only uses magnitude obviously because the function is not defined for a negative value

So please tell me how i should of seen this by inspection given the fact that I'm a controls person not to mention I'm just a student I'm comparison to your 30 years of work experience so please feel free to educate the next generation as to how i should of seen this by inspection
Choosing to ignore your continued whining, and recognizing that it is likely hopeless to encourage you to try to see if you can figure it out for yourself (despite that being the better way to learn most things), imagine the voltage of both inputs going up by some amount, Δv_in. Because the base-emitter voltage drops are nearly constant, this means that the voltage across the emitter resistor will increase by this same Δv_in. This will increase the current in that resistor by Δi_e = Δv_in/Re. This increased current in the emitter resistor is shared equally between the two transistors, so only half of it shows up as increased collector current in the transistor driving the output. So Δi_c = Δv_in/(2Re). This increase in collector current will result in v_out going down by an amount Δi_c·Rc. Put it all together and you have:

Acm = Δv_out/Δv_in = -Rc/(2Re)

The actual common mode gain will be slightly less (in magnitude) than this for two reasons. Is it too much to ask you to ponder what those reasons might be? Hint: What two assumptions did I make about the behavior of the transistors in my analysis that are not quite true and what is the impact of the actual behavior on the analysis. There's actually a third assumption, but it is a bit more iffy.
 
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