electronic principles chpt 2

CHAPTER 2

DIODE CIRCUITS AND THEIR APPLICATIONS

2.1 Diode Characteristics

2.2 Diode Circuit Analysis

2.3 Diode Application

2.4 Zener Diodes

1Prepared by AP Hj Amran Mohd Zaid

2.1 DIODE CHARACTERISTICS • Diode

- component that restricts the direction of flow of charge carriers. - Allow an electric current to flow in 1 direction but block it inthe opposite direction.

• Circuits that require current flow in 1 direction typically include 1 or more diodes in the circuit design.

• Today the most common diodes are made from semiconductor materials such as Si or Ge.

• Fig 2.1 - example of diode and symbol

Fig 2.1 Diode: (a) semiconductor crystal; (b) symbol of diode2Prepared by AP Hj Amran Mohd Zaid

• Complete IV characteristics - by combining fwd-biased and reverse-biased IV characteristic as shown in Fig 2.2.

2.1 DIODE CHARACTERISTICS (cont)

Fig 2.2 Complete IV Characteristic of pnjunction


Diode current-voltage relationship - Shockley equation:

vD = diode voltageIs = saturation current n = emission coefficient (1 < n < 2) VT = thermal voltage given as:

k = Boltzmann's constant (1.38 x 10-23 J/K) q = electron charge (1.6 x 10-19 C) T = temperature operation (K)



• To find the diode voltage given the diode current:

• By applying KVL to the diode circuit in Fig 2.3 , will obtain:


• Eqn 2.4 can be solved using additional eqnwhich can be obtained from Load Line Analysis of diode.

Fig 2.3 The diode circuit 5Prepared by AP Hj Amran Mohd Zaid

• The diode IV x-tic curve andeqn (2.4) is plotted in thesame graph as shown inFig 2.4.

• When ID = 0, yields VD = Vss Point A

• When VD = 0, yields ID = Vss/R point B.

• Connecting points, A and B, load line.

• The intersection betweenload line and diode x-tics iscalled the operating point.


Fig 2.4 Diode characteristic with load line


2.2 DIODE CIRCUIT ANALYSIS• In analysis, the diode is assumed to be ideal (either ON or OFF

state). • Fig 2.5(a) - Diode OFF (reverse-biased), Vcathode > Vanodeopen

circuit and Idiode = 0• Fig 2.5(b)- Diode ON (forward-biased), Vcathode < Vanode.

short circuit and Idiode ≠ 0.

Prepared by AP Hj Amran Mohd Zaid 7

Fig 2.5 Diode biased; (a) Reverse-biased; (b) Forward-biased

EXAMPLE 2.1 Referring to circuit below, determine ID and VD for the fwd-biased circuit (assuming the diode is Si).


Solution: The amount of current flows through the fwd-biased diode depends on source voltage E, the type of diode (Si, Ge) & resistor R. Considering the diode is Si, VD = VF = 0.7V. The current will be determined using the Ohm's Law;

hence

Solution: a. Since E < VBR the reverse-biased diode behaves like an

open circuit. Hence, the only current flow is the reverse saturation current Is, which is very small a few ηA for a Si diode and a few μA for a Ge diode.

ID = Is ≃ 0; VD = -E = -l0V b. Since E > VBR, reverse breakdown occurs, VBR drops

across the diode and the remainder of the source voltage (E - VBR) drops across the resistor R.


EXAMPLE 2.2Determine the diode current ID and diode voltage VD for the reverse biased circuit by assuming; (a) Diode breakdown voltage VBR = 90V, E = 10V and R = 1kΩ(b) Diode breakdown voltage VBR = 90V, E = 100V and R = 2kΩ

EXAMPLE 2.3

Determine the diode current ID and diode voltage VD for the circuit below.


Solution:

Convert the above circuit to a simple series circuit by using Thevenin' s Theorem.


• Thevenin's voltage is the voltage across R2, using voltage divider;

• Thevenin's resistance Rth is R1 //R2 ;

Inserting the diode into the Thevenin's equivalent circuit;


Since the diode is Ge,

EXAMPLE 2.4Determine the diode current ID and the output voltage Vofor the circuit below;


Solution: The voltage drop across Si diode is 0.7V and Ge is 0.3V. Hence, the voltage drop across both diodes is 1V.

2.3 DIODE APPLICATIONS

Widely used in electronic circuits applications such as rectifier, clipping and clamping circuits etc.

2.3.1 Half-wave Rectifier

In a half-wave rectifier, an ac source is connected to a diode and a load resistor, RL as shown in Fig 2.6 (a).


Fig 2.6 Half-wave Rectifier; (a) circuit; (b) output voltage(a) (b)

2.3.1 Half-wave Rectifier (cont)• Sinusoidal input voltage (Vin) goes +ve diode fwd biased &

conduct current through the load resistor (RL). The current produces output voltage across RL which has the same shape as the +ve input voltage as shown in Fig 2.6(b) for time t1 to t2.

• Input voltage goes –ve diode reverse biased no currentflow voltage across load resistor RL is zero. The output voltage for the 2nd cycle is shown in Fig 2.6(b) for time t2- t3.


• The net result of ½ -wave rectifier is only the +ve ½ cycle of the ac input voltage appear across the load resistor. The output for 2 cycles is shown in Fig 2.6(b).

Fig 2.6

2.3.1 Half-wave Rectifier (cont)Avg value for ½ wave rectified output voltage:

-value measured on a dc voltmeter.- Mathematically is area under the curve over a full cycle and divided by 2π, the number of radian in full cycle as shown in Fig 2.7.


- Eqn 2.5 shows the VAVGapproximately 31.8% of Vp

Fig 2.7 Average value of half-wave rectified signal

Vp = peak voltage.

2.3.1 Half-wave Rectifier (cont)The Barrier Potential (Si = 0.7V, Ge = 0.3V) will have an effect on the ½ wave rectifier. During the +ve ½ cycle, the input voltage must overcome the Barrier Potential before the diode become fwd-biased. The expression for the peak value is;


2.3.2 Full-wave Rectifier • Full-wave rectifier allows 1-way current through the load

during 3600 of the input cycle.• 2 types of full-wave rectifiers :

- Centre-tapped Rectifier- Bridge rectifier


Fig 2.8(a) Centre tapped full wave rectifier

2.3.2.1 Centre-tapped Rectifier • Is a full-wave rectifier that uses 2

diodes connected to the secondary centre-tapped x-former.

• The input voltage is coupled through the x-former to the centre-tapped secondary.

• Half of the total secondary voltage appears between the centre tap and each end of the secondary winding as shown in Fig 2.8(a).

2.3.2.1 Centre-tapped Rectifier (cont)

• For a +ve half-cycle input voltage, the polarities of the secondary voltages are shown in Figure 2.8(b).

• D1 - fwd-biased, D2 - reverse biased. • Current will flow to RL through D1. • Output will be the same as the half-cycle of the input

voltage.


Fig 2.8 (b) Centre-tapped full-wave rectifier - positive half-cycle

2.3.2.1 Centre-tapped Rectifier (cont) • For a -ve half-cycle input voltage, the voltage polarities on

the secondary are shown in Fig 2.8(c). • D1 - reverse-biased, D2 - fwd-biased. • Current will flow to RL through D2


Fig 2.8(c) Centre-tapped full-wave rectifier - Negative half-cycle

2.3.2.1 Centre-tapped Rectifier (cont)• X-former turn ratio have an effect to the output

voltage. • If x-former turn ratio =1, the peak value of rectified

output voltage is equal to half of the peak value of input voltage minus the potential barrier.

• To get output voltage with peak equal to input peak minus the potential barrier, a step-up x-former with turn ratio =2 must be used.


2.3.2.2 Bridge Rectifier • Uses 4 diodes shown in Fig 2.9. • When the input cycle is +ve,

- D1 and D2 - fwd-biased - D3 and D4 - reverse biased. - The current path is shown in Fig 2.9(a). - The output voltage obtained across RL will be the

same as the +ve half input cycle.


Fig 2.9(a) The full-wave bridge rectifier - +ve half-cycle

2.3.2.2 Bridge Rectifier (cont)• When the input cycle is -ve,

- D3 and D4 fwd-biased, D1 and D2 reverse-biased. - The current direction and the output voltage across RL

shown in fig. 2.9(b).


Fig 2.9 (b) The bridge full-wave rectifier - -ve half cycle

2.3.2.2 Bridge Rectifier (cont)• The current flowing through resistor RL is in the same

direction for both half-cycles. • During the +ve half cycle and -ve half cycle, 2 diodes

will always in series with the load resistor. If these diodes are taken into account, the output voltage is;


2.3.3 Clipping Circuit The diode clipper circuit divide into +ve Diode Clipper and -veDiode Clipper as shown in Fig 2.10. Positive Diode Clipper – Fig 2.10(a)- The circuit will clip (removed) the +ve part of the input

voltage. - As the input goes +ve, the diode becomes fwd biased and

conducts current. - Point A is limited to +0.7V when the input voltage exceeds

this value. - When the input voltage goes back to 0.7V, the diode is

reverse-biased and appears as an open circuit. - The output voltage look likes the -ve part of the input

voltage, with magnitude determined by the voltage dividerformed by R1 and the load resistor RL.



Fig 2.10 The clipper circuit; (a) +ve Diode Clipper ; (b) -ve Diode Clipper

2.3.3 Clipping Circuit (cont)

2.3.3 Clipping Circuit (cont)

Negative Diode Clipper – Fig 2.10(b)

• -ve half cycles of the input voltage will be removed. • When the diode is turned around as shown in Fig 2.10(b),

the -ve part of the input voltage is clipped off. • The diode is fwd-biased during the -ve part of the input

voltage. Point A is held at -0.7V by the diode drop. • When the input voltage goes above -0.7V, the diode is no

longer fwd-biased and the output voltage across load resistor RL is proportional to the input voltage.


Fig 2.10(b)

2.3.4 Clamping Circuit A clamper adds a dc voltage to the signal.Positive Clamper • Fig 2.11(a) - basic idea for a +ve clamper.• When a +ve clamper has a sinewave input, it adds a +ve dc

voltage to the sine wave shifts the ac reference level (normally zero) up to a dc level.

• The effect is to have an ac voltage centered on a dc level. This means that each point on the sine wave is shifted upward, as shown on the output wave.


Fig 2.11(a) Positive clamper shifts waveform upward;

Positive Clamper (cont)

• Fig 2.11(b) shows an equivalent way of visualizing the effect of a +ve clamper. An ac source drives the input side of the clamper. The Thevenin voltage of the clamper output is the superposition of a dc source and an ac source.

• The ac signal has a dc voltage of Vp added to it. This is why the entire sinewave of Fig. 2.11(a) has shifted upward so that it has a +ve peak of 2Vp and a -ve peak of zero.


Fig 2.11 (b) +ve clamper adds a dc component to signal.

2.3.4 Clamping Circuit (cont)

Positive Clamper (cont)• Fig 2.12(a) is a +ve clamper

- initially C uncharged. - 1st -ve half cycle of input voltage, the diode turns on -Fig. 2.12(b).

- At -ve peak of the ac source, the C has fully chargedand its voltage is Vp with the polarity shown in fig2.12(b).


Fig 2.12 (a) Ideal +ve clamper (b) at the +ve peak



• Slightly beyond the -ve peak, the diode shuts off (Fig. 2.12(c)). The RLC time constant is made much larger (at least 100 times) than the period T of the signal.

Stiff clamper: RLC > l00T ----(2.10)



• For this reason, the capacitor remains almost fully charged during the off time of the diode. To a 1st approximation, the capacitor acts like a battery of Vp volts. This is why the output voltage in Fig. 2.12(a) is a positively clamped signal. Any clamper that satisfies Eq. (2.10) is called a stiff clamper.


• The idea is similar to the way a ½ wave rectifier with a capacitor-input filter works. The 1st quarter cycle charges the capacitor fully. Then, the capacitor retains almost all of its charge during subsequent cycles. The small charge that is lost between cycles is replaced by diode conduction.




• Fig. 2.12(c) - charged C like a battery with a voltage Vp. This is the dc voltage that is being added to the signal. After the 1st quarter cycle, the output voltage is a +vely clamped sine wave with a reference level of 0; that is, it sits on a level of 0 V.



Positive Clamper (cont)• Fig 2.12(d) -Since the diode drops 0.7V when conducting,

the capacitor voltage does not quite reach Vp. For this reason, the clamping is not perfect, and the -ve peaks have a reference level of -0.7V.




Fig 2.12 +ve clamper (a) Ideal +ve clamper; (b) at the +ve peak; (c) beyond the +ve peak; (d) clamper is not quite perfect.




Negative Clamper• If diode in Fig. 2.12(d) turned around become -ve

Clamper (Fig. 2.13).• Capacitor voltage reverses, and the circuit becomes a

-ve clamper. • Clamping is less than perfect because the +ve peaks

have a reference level of 0.7 V instead of 0 V.

Fig 2.13 -ve clamper



2.3.4 Clamping Circuit (cont)Both +ve and -ve clampers are widely used. For instance, tv receivers use a clamper to change the reference level of video signals. Clampers are also used in radar and communication circuits .

2.4 ZENER DIODE • Use for voltage regulation.• Symbol - Fig 2.14. • p-n junction device which is designed for a specific reverse

breakdown voltage. • The reverse breakdown voltage for a zener diode << the

regular diode and referred as the nominal zener voltage Vz. • The breakdown voltage level Vz is set by carefully

controlling the doping level during manufacture.• Zener diode is designed to operate in reverse breakdown.


Figure 2.14 Zener Diode symbol

2 types of reverse breakdown in a zener diode are zener and avalanche.

Zener effect: high reverse voltages can provide e- enough energy to "jump" from valence band to conduction band, thus creating free e-. Hence, the diode conducts a high current under reverse bias.

Avalanche effect: Minority carriers in the depletion region are strongly accelerated by the electric field, thus creating e- hole pairs by impact ionization. The increase of free carriers increases the current, which provides more carriers to create impact ionization.


2.4 ZENER DIODE (cont)

From zener diode IV x-tic (fig. 2.15), when reverse voltage VR is increased, the reverse current IR remain extremely small up to the 'knee' of the curve. The reverse current is also called zener current Iz. At this point, the breakdown effect begins.



Fig 2.15 IV characteristic of zener diode



The internal zener resistance RZ begins to decrease as reverse current increases rapidly. From the bottom of the knee, the zener breakdown voltage Vz remains essentially constant as zener current Iz increases as shown in Fig. 2.15.

The ideal model of zener diode in reverse breakdown and the ideal x-tic curve is shown in Fig. 2.16. It has constant voltage drop equal to the nominal zener voltage. This constant voltage drop across the zener diode produced by reverse breakdown is represented by a dc voltage symbol even though the zener diode does not produce a voltage.



Fig 2.16 Ideal zener diode; (a) ideal model; (b) Ideal x-tic curve


Fig 2.16 Ideal zener diode; (a) ideal model; (b) Ideal x-tic curve


2.4.1 Application of Zener Diode

• Use as a voltage regulator for providing stable constant reference voltages.


• Figure 2.17 • zener diode is used to regulate a dc voltage.• As the input voltage varies (within limit), the zener

diode maintains a nearly constant output voltage across its terminal.

• However, as Vin changes, lz will change proportionally so that the limitation on the input voltage variation are set by the min and max current (IZK and IZM)

• R is the series current limiting resistor.

Fig 2.17 The zener diode as regulator

2.4.1 Application of Zener Diode (cont)

• Refer to Fig 2.17, the absolute lowest current that will maintain regulation is specified at IZK and for example zener diode 1N4740A is 0.25mA and represents no load current. The power specification is 1 W.


Thus

For min zener current, the voltage across 220Ω resistor is

Since

Then

For the max zener current, the voltage across the 220 Ω resistor is,

VR = IZMR = (l00mA)(220Ω) = 22VTherefore,

Vin(max) = 22V + 10V = 32VThis shows that this zener diode can ideally regulate an input voltage from 10.055V to 32V and maintain an approximate 10V output. The output will vary slightly because of the zener impedance, which has been neglected in this calculation.



EXAMPLE 2.5

Determine the min and max input voltages that can be regulated by the zener diode in fig below. Solution: From the datasheet for lN4733A: Vz = 5.1V at Iz= 49mA, IZK= 1mA and Zz = 7Ω at Iz. For simplicity, the equivalent circuit is show next.

Izk = Min Current

At IZK = 1 mA, the output voltage is, VOUT ≃ 5.1V - ΔVz = 5.1V - (IZ – IZK) ZZ

= 5.1V - (49mA -1mA)(7Ω) = 4.76V Therefore, Vin(min) = IZKR + VOUT = (1mA)(100Ω) + 4.76V = 4.86V

To find the max input voltage, 1st calculate the max zener current. Assume the temperature is 50oC or below and the power dissipation is 1 W.

At IZM, the output voltage is,

Therefore,


electronic principles chpt 2

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