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ELECTRONIC CIRCUIT ANALYSIS AND DESIGN BY DONALD NEAMEN PDF

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Electronic Circuit Analysis And Design By Donald Neamen Pdf

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by Donald A Neamen · electronic circuit analysis. Preview Microelectronics Circuit Analysis and Design Donald Neamen 4th Solutions. Pages·· . Neamen, Donald A. Microelectronics: circuit analysis and design / Donald A. Neamen. and devices courses and electronic circuits courses. He is still a. Electronic Circuit Analysis and Design Second Edition. Donald A. Neamen. McGraw-Hill. Chapter One. Semiconductor Materials. and Diodes. McGraw- Hill.

The electric field may become large enough that covalent bonds are broken and electron-hole pairs are created. Electrons are swept to the n-region and boles to the p-region by the electric fieldgenerating II reverse-bias current. This breakdown mechanism is called the Zener effect. Another breakdown mechanism is caged avalanche breakdown.

The generated electron-hole pairs can themselves be in v olved in a. The reverse-bias current for each breakdown mechanism will be limited by the external circuit. The PIV of a diode must never be exceeded in circuit operation ir reverse breakdown is to be avoided.

Zener diodes are fabricated with a specifically designed breakdown voltage and are designed 0 operate in the breakdown region. These diodes are discussed later in this chapter. SWitching Transumt Since the pn junction diode can be used as an electrical switch, an important parameter is its transient response, that is, its speed and characteristics, as it is switched from one state to the other. Here, we neglect he change in the space charge region width, When a forward-bias voltage is applied.

This happens because the excess minority carrier electrons! The large reverse-bias current is initially limited by resistor RII to approximately 1. After this time. The fall time " is lypically defined as the lime required for the current to fall to 10 perc-ent of its initial value. The total hlrfl--off time is the sum of the storage time and the fall time.

These same transient effects impact [he switching of transistors. For example. The transient tdm-t 1l time is the time required to establish the forward-bias minority carrier distributions. During this time, the Chapter t Semiconductor Materials and Diodes voltage across the junction gradually increases toward its steady-state value.

Although the turn-on time for the pn junction diode is not zero. If Vp ;;;;; U. As we have seen, the diode is a two-terminal device with nonlinear i-v characteristics. The analysis of nonlinear electronic ci rcui ts is not 1: IS straightforward as the analysis of linen electric circuits.

Mathematical relationships, or models, that describe. An example is Ohm's law. When a reverse-bias voltage is applied, the current through the diode is zero Figure I. Assume that the input voltage "lis a sinusoidal signal. During the positive half-cycle of the sinusoidal input, a forward-bias current exists in the diode and the voltage across the diode is zero.

The equivalent circuit fOI this condition is shown in Figure 1. The output voltage Vo is then equal to the input voltage, During tae negative half-cycle of the sinusoidal input, the diode is reverse biased.

The equivalent circuit for this condition is shown in Figure L23 d , In this part of tile cycle. The output voltage of the circuit is shown in Figure 1.

Over the entire cycle, the input signal is sinusoidal and has a zero average value; however, the output signal contains only positive values and therefore has a positive average value- Conseqently, this circuit is said to rectify the input signal, which is tne first step in generating a de voltage from a sinusoidal ac voltage. A de voltage is required in virtually all electronic circuits.

As mentioned. Methods a and b are closely related and are therefore presented together. The graphical analysis technique involves plotting two simultaneous equations and locating their point of intersection, which is the solution to the two equations. We will use both techniques to solve the circuit equations. These equations are difficult 10 solve by hand because they contain both linear and exponential terms. Kirdaboff's voltage 1. In the remainder of this section in which de analysis is emphasized, the de variables are denoted by uppercase letters and uppercase subscripts.

Combining Equations 1. Equation 1. The use of iteration to find a solution to this equation is demonstrated in the following example. Determine the diode voltage and current for the circuit shown in Figure 1. We can write Equation '. This equation is referred to as the circuit load line, which can be plotted on a graph with Tn and VD as the axes. From Equation I. The load line can be drawn between these two points. Using the values given in Example 1. The seco..

The intersection of the load line and the device characteristics curve provides ile dc current 'D This point is referred to as the qliesce". However, the concept of the load line and the graphical approach are useful for "visualizing" the response of a circuit, and the load line is used extensively in the evaluation of electronic circuits. Test Your Understanding "1. Determine VD and lv, using the ideal diode equation and tbe iteration method. VY' we assume a straight-line approximation whose slope is 1 rf.

The equivalent circuit for this linear approximation is a constant-voltage source in series with a resistor Figure 1. In this case, the equivalent circuit is an open circuit Figure I. For V C! Determine the diode voltage and current in the circuit shown in Figure 1.

Assume piece-wise linear diode parameters of Vr The equivalent circuit is shown ill Figure 1. This solution, obtained using the piecewise linear model, is nearly equal to the solution obtained in Example 1.

In addition, if the cut-in voltage is 0. Which is not significantly different from the previous results. Therefore, the calculated diode current is not a strong function of the cui-in voltage, Consequently, we will often assume a cut-in voltage of 0. The Q-point changes for each load line. The load line concept is also useful when the diode is reverse biased.

The diode current lD and voltage V D shown are the usual forward-biased parameters. Applying Kirchhoff's voltage law, we can write 1. The diode characteristics and the load line are plotted in Figure 1.

Although the piecewise linear model may yield solutions that are less accurate than those obtained with the ideal diode equation, the analysis is much easier. The power dissipated ill the diode is to be no more than 1,05 mW. Determine the maximum diode current and the minimum value of R to meet che power specification. Such models can factor in many diverse conditions, such as the temperature dependence of various parameters.

One of the earliest, and now he most widely used. Determine the diode current and voltage characteristics of the circuit shown in Figure 1. The input voltage VI was varied de sweep from 0 to: Figure I ,30 b and c shows the diode voltage and diode current characteristics versus he input voltage.

I YI'I b Figure 1. Several observauons may be made from the results. The diode voltage increases at aunost: The piecewise linear model predicts quite accurate remits at the maximum mput voltage, However, these results show that there IS definitelya nonlinear rci: Jlillfl between the diode current and diode voltage.

We must keep in mind [hal the piecewise linear model IS an approximation technique that works very well in many applications. For the ideal diode equation.

For the piecewise linear model. In most cases, however, '1 is assumed to be zero unless otherwise given. When semiconductor devices with pn junctions are used in linear amplifier circuits. The rollowing sections examine these HC cha racteristics. To investigate this circuit. I J Q is the de quiescent diode current.

In addition, the diode voltage will contain a de value with an ac signal superimposed, as shown in Figure 1. The relationship between the diode current and voltage can be written as 1. We are neglecting the -I term in the diode equation. We see from these two equations that t', r. Circuit Analysis To analyze the circuit shown in Figure 1. Eumple 1.

Analyze tile circuit shown in Figure 1. Dl'iiLk the analysis into two parts: In other words. TDQ 0. The ac diode current is. JI IlA The ac component of the output voltage. Throughout the text. To do so, we will use separate equivalent circuit models for each analysis.

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Frequency Response In the previous analysis, we implicitly assumed that the frequency or tile uc signal was small enough thai capacitance effects in the circuit would be negligible. If the frequency of the ac input signal increases, the diffusion taplicitalH.. We must also add the junction capacitance, which is in parallel with the diffusion resistance and capacitance, and a series resistance.

What is the quiescent diode current? The solar cell, photodiode, light-emitting diode, and Zener diode are types of pn junction diodes with specific characteristics that make them useful in particular circuit applica tions. When Light hits the spacecharge region, electrons and holes are generated. They are quickly separated and swept out of the space-charge region by the electric field, thus creating a photocurrent. The generated photocurrent will produce a voltage across the load, which means that the solar cell has supplied power.

Solar cells are usually fabricated from silicon. Solar cells nave long been used to power he electronics in satellites and space vehicles, and also as the power supply to some calculators. Solar cells are also used to power race cars in a Sunrayce event.

Collegiate teams in the United Stales design, build and drive the race cars. The power from the solar array can be used either to power an electric motor or to charge a battery pack 1. An example is the pbotooMJde. Incident photons or light waves create excess electrons and holes in the space-charge region. As previously explained, when a forward-bias voltage is applied across a pn junction, electrons and holes Row across he space-charge region and become excess minority carriers.

These excess minority carriers diffuse into the neutral semiconductor regions, where: LEDs are fabricated from GaAs or other compound semiconductor materials, In an LED, the diode current is directly proportional to the recombination rate, which means that the output light intensity is also proportional to the diode current.

Monolithic arrays of LEDs are fabricated for numeric and alphanumeric displays. An LED may be integrated into In optical cavity 10 produce a. Such a device is a laser diode. The current-voltage characteristics of a Schottky diode ase very similar to those of a pn junction diode. The same ideal diode equation can be used for both devices. The current in a [ junction diode is controlled by the diffusion of minority carriers, The current in a Schottky diode results from the flow of majority carriers over the potential barrier at the metallurgical junction, This me-all, that there is no minority carrier storage in the Schottky diode.

S c -cond, the reverse-saturation current Is for a Schottky diode is larger than ,hat or a pn junction diode for comparable device areas.

Figure I.. U compares the characrerisucs of t he two diodes. Applying the piecewise linear model. In later chapters. Consider the circuit shown ill Figure J. The de calculations for a circuit containing a Schottky diode are [he same as those for a circuit containing a pn junction diode.

Another type of metal-semiconductor junction is also possible.

A metal applied to a heavily doped semiconductor forms, in most cases, an ohmic contact: Determine the forward-bias voltages required to produce I rnA in each diode. The difference in forward-bias voltages is 0.

Neamen - Electronic Circuit Analysis and Design 2e.pdf

Determine the reverse-saturation current of the Schottky diode. At some point, breakdown occurs and the current in the reverse-bias direction increases rapidly. The voltage at this point is called the breakdown voltage. The diode I-JI characteristics. Allhougll the breakdown voltage is on the negative voltage axis reverse-bias , its value is given as a positive quantity.

The large current that may exist at breakdown can cause heating effects and catastrophic failure or the diode due to the large power dissipation in the device. However, diodes can be operated in the breakdown region by limiting the current to a value within the capabilities of the device. Such a diode can be Figure 1. The diode breakdown voltage is essentially constant over a wide range of currents and temperatures, The slope of the I-V characteristics curve in.

Typically, r" is in the range of a few ohms or tens of ohms. The circuit symbol of the Zener diode is shown in Figure 1. Note the difference between this symbol and the Schottky diode symbol. TIle voltage V z 1S the Zener breakdown voltage, and the current Iz is the reverse-bias current when the diode is operating in the breakdown region.

Design Example 1. Consider a simple constant-voltage reference circuit and design the value of resistance required to limit the current in this circuit. Consider the circuit shown ill Figure 1. Assume that the Zeller diode breakdown voltage is Vz '" 5,6 V and the Zener resistance is t: As before, we can determine the current from the voltage difference across R divided by the resistance.

That is, The resistance is then Comment The resistance external to the Zener diode limits the current when the diode is operating In the breakdown region. In the circuit shown in the figure.

We will see further applications If the Zener diode in the next chapter. If the voltage across the Zener diode is 5. The current-voltage characteristics of the diode are nonlinear: The current IS an exponential function of voltage in the forward-bias condition, and is essentially zero in the reverse-bias condition.

A piecewise-linear model of the diode was developed so that approximate hand calcularion results can be easily obtained. The i-v characteristics If the diode arc broken into linear segments, wh. A srnull-ssgnal linear equivalent circuit wa, developed and is used Hl determine the rclanonship between the ac current and ac vohage. This Slime equivalent circuit will be applied extensively when the frequency response of rransistots is discussed. II' Understand the concept of intrinsic carrier concentrauon, the difference between ntype and p-type materials.

Section 1. Describe an Intrinsic semiconductor material. What jf meant by the intrinsic carrier concentration'! Dcscnbe the concept of an electron and it bole as charge carriers in the semiconductor material. Describe an extrinsic semiconductor material, What is the value of the electron conccntrauon ill an n-type material.

Describe the concepts of drift current and diffusion current in a semiconductor material. What is meant by a builr-in potential barrier. How is a junction capacitance created in a reverse-biased pn junction diode? Describe the meaning of Is and VT, 8. Under what conditions is the small-signal model of a diode used in the analysis of a diode circuit" Describe the operauon of a simple solar cell circuit.

Is the semiconductor n-type or p" type? Is the serniconductor n- or p-type? Should donor or acceptor impurity atoms he added to intrinsic s. Determine the maximum allowable temperature. The semiconductor conductivity is o "" 2. Determine the drift current in the semiconductor. Chapter I Semiconductor Materials and Diodes 43 1. Determine the conductivity of the semiconductor. As an approximation. An inductance or2.

Calculate the resonant frequencyI, or the circuit for reverse-bias voltages of: The diode current is I Determine the ratio or current at T to thai at C 1.

Determine the increase in forward-bias voltage thai will cause a factor of 10 increase in current. Analysis 1. Determine the diode voltage lind current.

Figure P1. The minimum diode current is to be iD min ;;;;;; 2 rnA. The maximum power dissipated in the diode is 10 be no more limn IOmW. Determine appropriate values of Rr and Rl' FlgweP1. A sinusoidal voltage is superimposed Oli V DQ such that the peak-to-peak sinusoidal current is 0.

J6 is biased with a constant current source I. A sinusoidal signal v. Assume-that C is large. Ul rnA. IO-u A and A, respectively. Determine the forward-bias voltages required to produce a current of f-1A ill each diode.

The difference in forward-bias voltages between the Iwo diodes is O. Determine the value of R such that the currents in the diodes are equal. R Figure P1. A at temperatures or: Plot the characteristics from a reverse- bias voltage of S V to a. Choose values of R to generate quiescent diode currents of approximately U , and IOmA.

From a computer simulation analysis. Do the computer simulation results compare favorably with the theoretical predictions? Each design should be verified by a computer simulation] ' We presented the ideal current-voltage retut ionship, and considered the piecewise linear model, which simplifies the de analysis of diode circuits. In this chapter, the techniques and concepts developed in Chapter I are used to analyze and design electronic circuits containing diodes A general goal of this chapter is to develop the ability to use the piecewise linear model and approximation techniques in the hand analysis and design of various diode circuits.

Circuits 10 be considered perform functions such as rectification, clipping. The conversion of an ac voltage to a de voltage, such as for a de power supply. Clipper diode circuits clip portions 01 a signal that are above or below some reference level. Clam per circuits shirt the entire signal by some de value. Zener diodes, which operate in the reverse-bias breakdown region, have the advantage that the voltage across the diode in this region is nearly constant over a wide range or currents.

Such diodes are used in voltage reference or voltage regulator circuits, Finally, we look. An LED circuit is used in visual displays, such as the seven-segment numerical display, The photodiode circuit is used to detect the presence or absence or light and convert this information into an electrical signal, Although diodes are useful electronic devices, we will begin to see the limi!

A diode rectifier forms the first stage of a dc power supply as shown in Figure 2. As we will see throughout the text, a de power supply is required to bias all electronic circuits. The de output voltage Vo will usually be in the range of 3 h 24 V depending on the particular electronics application.

Microelectronics Circuit Analysis and Design

Throughout [he firs; part of this chapter, we will analyze and design [he various stages in the power supply circuit, Voltage reg. The diode is useful for this function because of its nonlinear characteristics.

Rectification ill classified as balf-wa"e or full-wave. We will use the piecewise linear approach in analyzing Ihis circuit. VI, is, in general, a V rms , 60Hz ac signal. Recall that the secondary voltage, "s. Diode Circuits 51 where N! The ratio Nd N2 is called the transrormer tums ratio. Tile transformer turns ratio will be designed to provide a particular secondary voltage, rs. In using the piecewise linear model of the diode.

To do this. Determine the input voltage condition such that a diode is conducting on. Then lind the output signal for this condition. Determine the input VOltage condition such that a diode is not conducting off. Then find the output signal for this condition. Item 2 can be performed before item I if desired] Figure 2. V " the diode will be nonconducting. When I'. We can see that while Ihe input signal l's alternate, polarity and has a time-average value of zero. The input signal is therefore rectified, Also.

Consequently, the diode must be capable of handling the peak. For the circuit shown in Figure 2. Figure 2. Because the resistance R is It constant, the: As the load line sweeps across the diode I-V characteristics. We can use a half-wa ve rectifier circuit to charge a battery as shown in Figure 2.

Charging current exists whenever the mstantaneous ac source voltage is greater than the battery voltageplus the diode cut-in voltage as Chapter 2 Diodt Circllils wr- 8 b FIGure 2. The resistance Jl in the circuit is to limit the Current. When the ac source voltage is less than V 8. Th us current flows only in the direction to charge the battery.

One disadvantage of the halfwave rectifier is that we "waste" the negative half-cycles. The current is rem during the negative half-cycles, so there is no energy dissipated, but at the same time. Determine; ai Ihc peak diode current; b tbe maximum reverse-bias diode voltage; and c the fraction percent of the cycle over which tile diode conducts. One example of a full-wave rectifier circuit appears in Figure 2. The input to the rectifier consists of a power transformer, in which the input is normally a V rms , 60 Hi ac signal, and the two outputs are from a center-tapped secondary winding, that provides equal voltages v,.

When the input line voltage is positive. The primary winding connected to the V ac source has NI windings. The turns raCio of the transformer, usually designated Nd NJ. This isolation reduces the risk of electrical shock. During the pMit;"e hair of the illput v oltage cycle, bot!

The current through DI and the output resistance produce a positive output voltage. During the nega tive half cycle, D I is cut off and Di is forward biased, or "on," and the current through the output resisranee again produces a positive output voltage, If we assume that the forward diode resistance " or each diode is small and negligible, we obtain the voltage transfercharacteristics, 1'0 versus 's, shown in Figure 2. The corresponding input and output voltage signals are shown in Figure 2.

Since a rectified output voltage occurs during both the positive and negative cycles of the input signal. Another example of a full-wave rectifier circuit appears in Figure 2,7 a. This circuit is a bridge reedier. DI and D: WQ diodes arc in series in the conduction path, the magnitude of Vu is two diode drops less than the magnitude of I's.

One difference to be noted in the bridge rectifier circuit in Figure 2. Whereas the center tap of tile secondary winding of the circuit in figure 2. Ex;ample 2. Compare voltages and the transformer t urns ratio in two full-wave rectifier circuits. Consider the reenter circuits shown in figures 2-IXa and 2. The desired peak output voltage Vo is 9V. For the center-tapped transformer circuit shown in Figure 2. These calculations demonstrate the advantages of the bridge rectifier over the center-tapped transformer circuit.

First, only half as many turns are required for the secondary winding in the bridge rectifier. This is true because only half If tbe secondary windin, or the center-rapped transformer is utilized at anyone rime.

Utl, the peak inverse: Because of [he advantages demonstrated in Example 2. Ripple Voltage, and Diode Current lf a capacitor is added in parallel with Ihe load resistor of a half-wave rectifier to form a simple filter circuit Figure 2. If we assume that the diode forward resistance is 'f I , which means that the 'fC time constant is zero, the voltage across the capacitor follows this initial portion of the signal voltage.

When the signal voltage reaches its peak and begins to decrease, the voltage across the capacitor also starts to decrease, which means the capacitor starts to discharge. The only discharge current path is through the resistor. If the RC time constant is large, the voltage across the capacitor discharges exponentially with time Figure 2. During this lime period, the diode is cut off. A more detailed analysis of the circuit response when the input voltage is near its peak value indicates a subtle difference between actual circuit operation and the qualitative description.

If we assume that the diode turns off immediately when the input voltage starts to decrease from its peak value. An exaggerated sketch of these two voltages is shown in Figure 2. However, this condition cannot exist. In this situation, a computer analysis may provide more accurate results. During Ihe next positive cycle of the input voltage, there is a point at which the input voltage is greater than the capacitor voltage, and the diode urns back on.

The diode remains on until the input reaches its peak value and the capacitor voltage is completely recharged, Since the capacitor filters out a large portion of the sinusoidal signal, it is called a filter capacitor. The steady-state output voltage or the RC filter is shown in Figure 2.

The ripple effect in the output from a full-wave filtered rectifier circuit can be seen in the output waveform in Figure 2. The capacitor charges to its peak voltage value when the input signal is at its peak Determining the ripple voltage is necessary for the design of a circuit with an acceptable amount of ripple. However, if the ripple effect is small, then as all approximation, we can let T!

For a full-wave rectifier. Tp is one-halt the signal period. Assume the output load resistance- is R; IOk. From Equatio-n 2. If the ripple voltage is 10 he limited 10 a smaller value, a larger filler capacitor must be used.

The diode in a filtered rectifier circuit conduct. The diode current supplies the charge lost by the capacitor during the discharge time. We see that 2. Using Equation 2.

V,w lD. J5 The average diode current during the diode conduction time in a full-wave rectifier circuit is then 2. The average capacitor current iC. From Equauon 2. Comparing Equations 2.

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Design a full-wave rectifier til meet particular specifications. A full-wave rectifier is to be designed to produce a peak output voltage or 12 V, deliver mA to the load, and produce an output with a. All input line voltage of 12 lV rms , 60 Hzis available.

A full-wave bridge rectifier will be used, because of the advantages previously discussed. Dttslgn PoIoll1!

I A particular turns ratio was deterrained for the transformer. However, this particular transformer design is probably not commercially available. The PSpice circuit schematic and the steady-state output voltage are shown in Figure 2, We see that the peak output voltage is III the?

Spice simul! Test Vour Understanding 2. A filter capacitor is connected in parallel with Chapter 2! Assume the diode cut-in voltage is 7 V and the output resistance is 2,5 kn. Ans, C '" 2O, i",f 2. In the circuit in Figure 2. Terminal 2 on CI is positive with respect to terminal I. We assume that the time constant RL C2 is very long compared 10 the period of the input signal.

As the polarity of Vs changes to that shown In Figure 2. The same ri pple effect occurs as in [he output voltage of the rectifier circuits. There are also voltage tripler and voltage quadruplet circuits. These circuits provide a means by which multiple de voltages can be generated from a single ac source and power transformer. This makes the Zener diode useful in a voltage regulator, Or' a constant-voltage reference circuit. In this chapter, we will look at an ideal voltage reference circuit.

One reason is that a standard Zener diode with a particular desired breakdown voltage may not be available. However, this section will provide the basic concept of a voltage regulator.

For this ci rcuit, the output voltage should remain constant. The resistance R, limits the current through the Zener diode and drops the "excess" vohage between Vps and Vz.

Solving this equation lor the diode current, Iz, we get 2. The current in the diode is a minimum, Iz min. Equation 2. More stringent design requirements may require the minimum Zener current to be 20 to 30 percent of the maximum value. Then, combining Equation 2. The current in the radio win vary between 0 off to lOOmA full volume. The equivalent circuit is shown in Figure 2. R, Fig. The maximum Zener diode current Call he determined from Equa.

Tile minimum Zener diode current occurs for V,. We find ,, min:: I The variable input in this example was due to a variable battery voltage. We will again see later how more sophisticated designs can solve this problem. Assume a 5. Example 1. Consider a silicon p! Test Your Understanding 1. The applied voltage V R induces an applied electric field, EA, in the semiconductor.

The direction of this applied field is the same as that of the E-field in the space-charge region. Since the electric fields in the areas outside the space-charge region are essentially zero, the magnitude of the electric field in the space-charge region increases above the thermal equilibrium value.

This increased electric field holds back the holes in the p-region and the electrons in the n-region, so there is essentially no current across the pn junction. By definition, this applied voltage polarity is called reverse bias. When the electric field in the space-charge region increases, the number of positive and negative charges also increases.

If the doping concentrations are not changed, the increases in the charges can only occur if the width W of the J w l L.. Jure 1. Because of the additional positive and negative charges in rhe space-charge region, a capacitance is associated with the pn junction when a reverse-bias voltage ill ,I pplied.The same ideal diode equation can be used for both devices.

The ratio Nd N2 is called the transrormer tums ratio. During the nega tive half cycle, D I is cut off and Di is forward biased, or "on," and the current through the output resisranee again produces a positive output voltage, If we assume that the forward diode resistance " or each diode is small and negligible, we obtain the voltage transfercharacteristics, 1'0 versus 's, shown in Figure 2.

Terminal 2 on CI is positive with respect to terminal I. Consider the reenter circuits shown in figures 2-IXa and 2. Find the output of the parallel-based clipper in Figure 2. Analyzing muitidiode circuits requires determini ng if the individual devices are "on" or "off. For example, the de level of a TV signal may be lost during transmission.