# differential amplifier input stage of op-amp

Discussion in 'Homework Help' started by PG1995, Apr 12, 2012.

1. ### PG1995 Thread Starter Active Member

Apr 15, 2011
753
5
Hi

Please have a look on this attachment and help me with the queries, particularly with Q3. I need to understand this quickly and doesn't need broad understanding of the topic. Thanks a lot for your help.

Regards
PG

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2. ### steveb Senior Member

Jul 3, 2008
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Question #2: they say approximately equal to make it clear that the left and right side circuits are nearly idential. In other words, the transistors are well matched, and the temperatures are equal. Only well matched transistors, at the same temperature, would give equal currents.

Question #3: The base voltage on Q1 is increased slightly which increases Vbe on Q1, however, the emitter voltage goes up a little which decreases Vbe on Q2. The net result is that the total emitter current (sum of emitter current in Q1 and Q2) is about the same. The lower current in Q2 emitter means lower current in Q2 collector, which means that the voltage on the collector of Q2 must go up. Note that there is an inverse operation because the collector voltage is Vcc-IcRc. The negative sign make an inverse relation. Likewise, the Q1 emitter current goes up which means that the Q1 collector current goes up which means the Q1 collector voltage goes down. Again, the voltage is inverted.

As to why the emitter voltage does go up slightly, this is a nonlinear effect resulting from the way the transistors operate. You need to look at the exponential relation between Vbe and Ie to get the exact numbers. However, intuitively you can think of the fact that the base voltage of Q2 is ground and the base voltage of Q1 is above ground. Compare this to the case where Q1 and Q2 both have their bases grounded. Intuitively you would expect a slight increase.

There is a slight imperfection in the book's explanation though. How can the emitter voltage increase slightly without there being an increase in the total emitter currect (they say nearly the same, so they are not really wrong, but what they say is misleading)? Most text books will describe the case of a slight increase in base voltage on Q1 and a slight decrease in voltage on the Q2 base. Under these assumptions the emitter current does remain exactly the same, and the emitter voltage does not increase. Then the increase in Q1 emitter current is exactly equal to the decrease in Q2 emitter current. This is the whole idea of a differential mode input signal. The left and right side oppose each other and keep balance, almost like a see-saw.

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3. ### PG1995 Thread Starter Active Member

Apr 15, 2011
753
5
Thanks a lot for the help, Steve. Very kind of you.

Please also help me with the query in this attachment. I'm just curious although I'm sure the book is correct. Thank you.

Regards
PG

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4. ### jtrent New Member

Mar 11, 2012
26
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A good explanation of question #3 by steveb but I think you are a little off about question #2.

Question #2 is not referring to the difference between the right and left parts of the circuit but is simply referring to the operation of any BJT transistor, indicating that the emitter current is approximately equal to the collector current. The emitter current being the sum of the collector current and the base current. The base current is often ignored because it is typically very small compared to the collector current.

Just a small clarification.

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5. ### steveb Senior Member

Jul 3, 2008
2,433
469
Thank you for the correction. I agree with you. Rereading the text, I see that I misread it.

6. ### jtrent New Member

Mar 11, 2012
26
4
What you seem to be talking about is more of a geometry problem in a sense. You are saying the area of one rectangle does not look the same as the area of the other rectangle. I tell my students to never assume a graph is drawn to scale. In other words, dont go by what your eyes tell you but rather what your calculations tell you. The formula does work out.

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7. ### Ron H AAC Fanatic!

Apr 14, 2005
7,050
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The graph is plotted on a log-log scale.
If it were plotted on a lin-lin scale, the rectangles would have equal areas. A*B=C*D. On a log-log scale, log(A)+log(B)=log(C)+log(D).
This means that, on the log-log plot, the sum of the ordinate and abscissa of the 1st rectangle will equal the sum of the ordinate and abscissa of the 2nd rectangle.
To prove it to yourself, draw the log-log plot on graph paper and sum the vertical and horizontal dimensions of each rectangle.

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