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Unformatted text preview: 55L ’7’306 ﬁrmsa/oee, FEED/346k) 31/
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I: . ', . ei  ie ed nalis '_ The ultimate goal of analysis of electronic circuits is understanding, not just numbers or equa—
tions. This is especially true if you hope to design circuits. Unfortunately, our educational system
tends to condition you to do analysis for its own sake. I hope to encourage you to use what has
been called DesignOriented Analysis. Let me summarize some general principles of Design 1
Oriented Analysis that will help not only for design but also help you perform better on quizzes. ' In fact, these principles will help you in your careers, because they are good engineering prac
tices.  For design, an analysis that merely quantiﬁes the performance of one instance of some circuit is
of negligible value. You can usually ﬁnd the same information much more accurately, for
example, using a circuit simulator like SPICE. An equation that expresses the performance of a
whole class of circuits could be much more useful. ° The ultimate enemy of understanding is complexity. To handle complexity, encapsulate your
understanding of each part of a circuit or system; hide the details. (However, you still have to
understand how the different parts of the system interact.) * Use circuit tricks or previously derived equations to reduce the amount of algebra you need to
do. However, you must understand when these tricks or equations can be used and when not.  Break complex circuits up into pieces and analyze the pieces separately: “Divide and conquer."
' Use the simplest model that is consistent with the level of accuracy you need. I * Begin with symbolic solutions, then plug in numbers, and throw away as many terms as you
think you can get away with. Then, go back into the symbolic analysis with the simpliﬁed
results. Approximate as much as possible.  Plug in the numbers to check the validity of your approximations. — Without the numbers, all of the possible terms in an equation must be included for it to be
valid. Only with the numbers can we see for sure which terms can be eliminated. For
example, how can we simplify the symbolic expression Ri = r“ + (1 + B)(RE1 llRE2)?
You can’t unless you know something about the numbers that the symbols represent.  Suppose I tell you that in this case, [3 >> 1, and R52 >> RE]? Then we know that
RiE r1, + BREI = [Hi/gm + RBI) a BREI. Ifyou can show that ngEl » 1 you can
reduce this to Ri 2 BREI 0 Avoid all unnecessary analysis. Often the result you need is a special case of a more complicated
analysis you have already done.  Use dimensional analysis to help you spot errors in an algebraic derivation. ' Draw schematics in whatever way that makes it easiest to understand.  Currents ﬂow from top to bottom, for example
 PSpice gives really hardto—read schematics unless you work to avoid it. 8;; EEL 4306 Electronic Circuits II Lecture #8
H mall ' 31 Si tr te  First consider whether you have solved this problem before. Maybe this is a special case of some
more complicated analysis. If so, you can simplify that result and skip this analysis.  Otherwise,  Set voltages and currents that do not change to zero (short voltage source, open current source)
 Apply test input source ~ Deﬁne output variable — Deﬁne controlling variables for all dependent sources that change in response to your test input
~ Otherwise, avoid creating new variables (or else you will have to eliminate them)  Writing equations on the schematic diagram often allows simpliﬁcations ~ Eliminate intermediate variables. (Write all equations in terms of input and output variables.)  Compute the desired transfer function as the ratio of the output variable to the input variable. ° Rearrange the algebraic form of the result in different ways to highlight different aspects of the
circuit behavior. ' As a check, compare to previous analyses of simpler circuits, which should be special cases of
this one. 0 Once you know the equations for the basic transistor conﬁgurations, you should draw your
schematic using transistor symbols rather than a smallsignal equivalent Circuit.  Use dimensional analysis to check your results:
 Both sides of each equation should have the same dimensions (units).  All the terms in every sum should have the same dimensions.  If one step in your derivatiOn has wrong dimensions but the dimensions in ’the previous step are
selfconsistent, then this step must have an error.  w
536 EEL4306 Electronic Circuits II _ Lecture # g Mummngamaup) .o The general rule: Break the loop (anywhere), and terminate it so that the same conditions exist in
the open loop as in the closed loop. Apply a test source Xx; ﬁnd the returned signal X3. (Kx and
KY are either both voltages or both currents.) Find T= AB = Xy/Xx. Closed—loop circuit “Same conditions” means: a.) You must maintain the same DC operating point as in the original circuit. (This may be hard in
SPICE or with real circuits if the circuit depends on do loop gain to stabilize its Q point. You
could leave a huge inductor in series with the signal path and ﬁnd the gain using ac analysis.) 4 b.) ThelooP must be terminated with an impedance Zr equal to the (smallsignal) impedance Zi
seen looking into the rest of the loop. 25 is the Mg input impedance. In principle, to ﬁnd Zi
you must terminate the 100p at the other end with... Zi! This deﬁnition is mile; in principle,
an inﬁnite series of identical circuits would have to be interconnected to ﬁnd the prOper termina
tion. In practice, however, the impedance you see depends only very weakly on the conditions at g
the other end of the loop. (It’s like a rapidly convergent series in mathematics.) Note that the loop gain is a smallsignal parameter. en circu t Rt WUJU
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Shunt vs. Series Feedback I Signals can be applied as inputs to a feedback loop or can be deﬁned as outputs from a feedback
loop using either shunt or series connections. E‘ G
In the circuit below, the feedback signal path includes the following nodes: A, C, B, and F, plus all the‘internal nodes along RF. It also includes the following branches: C1, B2, 02, 133, E3, RF and E1. Mé’s_b,aLMXj ' SM 5W CaramelN“! Whether feedback is series or shunt, and therefore, whether resistance is raised or lowered by
feedback, depends on howlwhere the port is deﬁned. A port has two terminals. If one terminal is
smallsignal ground, and the other is a node that is part of the feedback signal path, then the port
has a shunt connection. To create a shunt input, the easiest way is to connect an input current
source from ground to the shunt node. Negative feedback will lower the resistance looking into
that port. If the loop gain were inﬁnite, we could consider the node a smallsignal virtual ground.
To form a shunt output port we can deﬁne the node voltage as the output variable. For inﬁnite loop
gain, the port would act as an ideal controlled voltage source (with zero output resistance). A few variations are possible. While we u5ually deﬁne the node voltage as the output, we could also deﬁne the current through a resistor across the port as the Output variable. Similarly, we usu
ally apply an input current source from ground to the shunt node. However, if there is a resistor in
parallel with the port in question we could put a voltage source in‘series with that resistor, and the
connection would still be shunt, as can be seen using the Nortonfl‘hevenin circuit transformations. ...
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 Spring '08
 Eisenstadt

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