Lecture8_092005 - 55L ’7’306 firmsa/oee, FEED/346k)...

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Unformatted text preview: 55L ’7’306 firmsa/oee, FEED/346k) 31/ Dug. M/ 0/1/13 week— Fgao Bird: PERT/ct: EXAMPLES Fae/baylé FQV Bias; Es), Haoé 3,1 ‘fllése bl‘as Cursz ‘Fall J) Vac. I} 5 1—5:? (wcmg/M) (“WM/M) ' Vans f5 0 .— A‘p/VW/Mflc/fi MOMS.) #MSKJ'QV VMI' (m3 VMI‘st-fl) ‘— Pom‘L-v-‘c view/WIGA-re u-Pflwear 27.; Reiko—k =~> MLMQ funkawcy— fiurn [A79 "L’MflSA/L ' M bras Carbonic; +00 _ -w _. Wmew 3, a EEL 4306 Electronic Circuits II Lecture #8 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 Design-Oriented 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 ll-RE2)? 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 hard-to—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 small-signal 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 self-consistent, 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 (small-signal) 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 small-signal parameter. en- circu t Rt WUJU Ell-I'- UJLLI UJ'IJJIJJ 1323 www $88 v-C'xl' r-N-r- ‘(V'K' '7'?“ (NINE? NNN 51 En \1. ‘53}! m— EEL4306 Electronic Circuits II Lecture #8 H 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,aLMX-j- ' SM 5W Caramel-N“! 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 small-signal 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 small-signal 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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Lecture8_092005 - 55L ’7’306 firmsa/oee, FEED/346k)...

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