EIT/FE Exam EE Review 2 nd Session Prof. Richard Spencer. Transformer

Transcription

1 EIT/FE Exam EE eiew 2 nd ession Prof. ichard pencer Transformer Assume two coils are wound on the same core and that it has low reluctance (high permeability) If a current flows in one of the windings, the flux will be mostly confined the core and will flow through the other windings, inducing a oltage across them The oltages on the B two windings will be proportional to the number I of turns since the flux is the same

2 Transformer The two windings are usually called the primary and the secondary (determined by use) Assume the secondary has N-times as many turns as the primary, then V 2 = NV If the transformer is ideal, no energy is lost, so P = P 2, therefore, I = NI V2 N V P2 = V2I2 = = 2 2 NV VI = P = P2= eff V = = 2 2 N Diode A diode is an element that ideally only allows current to flow in one direction A practical diode I-V characteristic is shown here along with the I D /I equation and schematic symbol qv ( ) ( ) D kt V D V T I = I e = I e D V D /V T 2

3 Diode Example A diode can be used to conert AC into DC Consider the circuit shown below this is called a half-wae rectifier the output oltage is not constant, but it is only positie We can filter V! pencer/ghausi, Introduction to Electronic Circuit Design, e, 2003, Pearson Education, Inc. Half-Wae ectifier Circuit To change the oltage on a capacitor requires us to moe charge on or off the capacitor Therefore, a large capacitor resists haing the oltage across it change I if C is remoed pencer/ghausi, Introduction to Electronic Circuit Design, e, 2003, Pearson Education, Inc. 3

4 Diode B Diode FB Using linear models is a common way to sole nonlinear circuit problems. In this example, the diode is modeled as a 0.7 V battery when forward biased (FB) and as an open circuit when reerse biased (B). Combine Diode and Transformer Using a center-tapped transformer and two diodes, we can make a full-wae rectifier, which uses both half cycles This is more efficient It is easier to filter out the ripple pencer/ghausi, Introduction to Electronic Circuit Design, e, 2003, Pearson Education, Inc. 4

5 A diode bridge also performs full-wae rectification, but without needing a center-tapped transformer ne more example pencer/ghausi, Introduction to Electronic Circuit Design, e, 2003, Pearson Education, Inc. p Amp Circuits eiew of basic op amp operation Negatie feedback & irtual short circuit Gain circuits 5

6 Basic p Amp eiew The power supply connections are not usually shown The simplest circuit model is shown aboe and the output oltage is gien by: = a( P N) = a ID ID = Differential input oltage! Non-Inerting Amplifier I = N 0 I 2 Assuming the op amp input currents are zero, and 2 hae the same current and are in series. Therefore, = b ( ) + ab = a N + 2 a = + ab a = + ab ( ) = a = a b ID 6

7 Non-Inerting Amplifier a = + ab Note negatie feedback - key element With an ideal op amp, the gain is infinite, so a + lim = = = + a + ab b 2 2 Virtual hort Circuit We had; ) N = b + 2 ( ) 2) = aid = a b 3) a = + ab Therefore; ab ID = N = b = = + ab + ab NTE: as a, ID 0 VITUAL HT CICUIT A irtual short requires negatie feedback and large loop gain. 7

8 Virtual hort Circuit We had; ) N = b + 2 Therefore; ab ID = N = b = = + ab + ab NTE: as a, ID 0 VITUAL HT CICUIT It is a irtual short because no current can flow through it. ( ) 2) = aid = a b 3) a = + ab Virtual hort Circuit The syllogism often gien for the irtual short circuit (or irtual ground) is: ) = aid 2) a 3) must be finite 4), 0 This syllogism is not alid! Note that reersing the signs on the opamp inputs would not affect the syllogism, but the result is clearly wrong in that case. The problem is equiocation; a assumes that the opamp operates in its normal region, while saying o is finite requires that the output be saturated, and a approaches zero there! ID 8

9 Virtual hort Circuit The problem is equiocation; a assumes that the opamp operates here, saying o is finite requires that the output be saturated, and a approaches zero there! o, there is equiocation about the meaning of a, it does not hae one single meaning throughout the syllogism. Non-Inerting Amplifier I = N 0 I 2 Using the irtual short circuit idea lets us find the gain far more easily; first N = + 2 But the irtual short circuit requires that = = N = =

10 Inerting Amplifier N P Note that the irtual ground allows us to think of the operation as a V-to-I conersion followed by an I-to-V conersion. Negatie feedback and large loop gain imply a irtual short (or irtual ground here), so N = P = 0 and i = No input current implies that i so we obtain, = i 2 = i = i = 2 N A umming Amplifier Using this alternate iew of the operation allows us to see how to make a summing amplifier: i = i + i + i = i22 I-to-V conersion umming ( ) = i + i + i V-to-I conersions = + i