This is a solved example from the textbook, but I I don't understand how VBE1 and VBE2 were obtained. like why VBE1=VBE+Vid/2 ? I know about the voltage difference but why is it positive here and negative for VBE2 ?
Thanks
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\$\begingroup\$ vid is the differential input voltage. VBE is the same for Q1 and Q2, and is the DC operating point, or steady state voltage. All they are doing is assuming that vid is equally distributed between the two transistors. So Q1 sees vid/2, and Q2 sees -vid/2. The differential voltage, then, is vid/2 - (-vid/2) = vid. This is the standard way of teaching about the differential pair. Keep looking at it until it looks normal. ;-) \$\endgroup\$– user57037Sep 11, 2015 at 16:17
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\$\begingroup\$ my question is, where is Vid/2 located? is it the emitter voltage? so Q1 emitter voltage is Vid/2 and Q2 emitter voltage is -Vid/2 ? \$\endgroup\$– user65652Sep 11, 2015 at 16:23
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\$\begingroup\$ Yes - I understand your problems. To me it is NOT the standard way of teaching diff. amplifiers because it is not a circuit diagram that can work in reality. There is no dc path that allows DC base currents and it is not shown how Vbe biasing is established. Hence, there is no reference node for the input voltages. I never would use such an incomplete circuit for explaining the diff. amplifier. \$\endgroup\$– LvWSep 11, 2015 at 16:27
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\$\begingroup\$ I am just looking for an analytical way to drive the equations. so I can solve similar problems later on. \$\endgroup\$– user65652Sep 11, 2015 at 16:30
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\$\begingroup\$ What book is this from? I think the problem is that you are not fully up-to-speed on some of the conventions. If you are enrolled in a class, have a talk with the professor. If not, go re-read some earlier material. Note that the lower case v means a small signal voltage, so vid is a small signal voltage. VBE is the DC operating point. The circuit is incomplete, but that is understood. Incomplete circuits are often used in textbooks to help focus on whichever concept is currently being taught. \$\endgroup\$– user57037Sep 11, 2015 at 16:50
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3 Answers
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You can think of this as an assumption based on symmetry and linearity. If the differential input voltage is very small (<< 2Vt) then the input resistance of the two transistors will remain constant and the small differential voltage will divide equally.
The \$\alpha\$ factor is close to one for real transistors, so Ic ~= Ie.
They're really just saying that the collector currents are about I/2 in each case, and they increase/decrease by the transconductance gm multiplied by half the input voltage, and the collector voltages decrease/increase by the gm*Vid/2 times the collector load resistances.
Not that they have not separated the large signal bias from the small signal behavior in this example.
answered Sep 11, 2015 at 16:54
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This is a consequence of the assignment of v_id with the positive input connected to Q1. Subtracting the two voltages gives $$v_{id} (+) - v_{id} (-) = V_{BE1} - V_{BE2} = (V_{BE} + \frac{v_{id}}{2}) - (V_{BE} - \frac{v_{id}}{2}) = \frac{v_{id}}{2}+\frac{v_{id}}{2}) = v_{id}$$
answered Sep 11, 2015 at 16:02
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1\$\begingroup\$ No - VBE is a DC voltage and Vid seems to be the signal voltage. Please read my comment to your question above. The circuit is uncomplete - and there is no clear (and clean) method for solving the task. \$\endgroup\$– LvWSep 11, 2015 at 16:32
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\$\begingroup\$ @LvW - Yes, VBE is a DC voltage, and yes, Vid is the signal voltage. Nonetheless, given the conventions shown in the figure, my calculations are straightforward and correct. For instance, if VBE1 were lower than VBE2, the equations would not work out. VBE1 is higher than VBE2 (for positive Vid) simply because that's how Vid is defined. \$\endgroup\$ Sep 12, 2015 at 0:30
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\$\begingroup\$ WhatRoughBeast-I have to apologize because I have placed my comment at the wrong place. It was my intention to answer the second question/comment directly below the first contribution(problem description) from user65652 (he has asked if Vid/2 would be the emitter voltage). \$\endgroup\$– LvWSep 12, 2015 at 8:29
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All the answers are correct, but I think an alternative interpretation could be helpful.
Since both emitters are at the same voltage, you can define a new term that is the average of the two inputs: $$V_{BE}=\frac{v_{BE1}+v_{BE2}}{2}$$ Then you can define a new term that is the the difference between the two: $$v_{id}=v_{BE1}-v_{BE2}$$ This is helpful because we want to amplify the difference voltage \$ v_{id}\$ and essentially ignore/eliminate the average voltage \$V_{BE}\$. Now solve those two equations for \$v_{BE1}\$ and \$v_{BE2}\$, and you'll get the expressions you found.
