Neamen D. Microelectronics: Circuit Analysis and Design

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68 Part 1 Semiconductor Devices and Basic Applications

–

Diode

rectifier

Voltage

regulator

Filter

voltage

source

Power

transformer

LOAD

Figure 2.1 Diagram of an electronic power supply. The circuits that characterize each block

diagram are considered in this chapter.

Ideally, the output voltage of a rectiﬁer circuit is a dc voltage. However, as we will see, there may be an

ac ripple voltage superimposed on a dc value. For this reason, we will use the notation

as the instanta-

neous value of output voltage.

2.1 RECTIFIER CIRCUITS

Objective: • Determine the operation and characteristics of diode

rectiﬁer circuits, which form the ﬁrst stage in the process of converting

an ac signal into a dc signal in the electronic power supply.

One application of diodes is in the design of rectiﬁer circuits. A diode rectiﬁer forms the

ﬁrst stage of a dc power supply. A dc voltage is required to power essentially every elec-

tronic device, including personal computers, televisions, and stereo systems. An electri-

cal cord that is plugged into a wall socket and attached to a television, for example, is

connected to a rectiﬁer circuit inside the TV. In addition, battery chargers for portable

electronic devices such as cell phones and laptop computers contain rectiﬁer circuits.

Figure 2.1 is a diagram of a dc power supply. The output voltage

is usually

in the range of 3 to 24 V depending on the particular electronics application.

Throughout the ﬁrst part of this chapter, we will analyze and design various stages in

the power supply circuit.

Rectiﬁcation is the process of converting an alternating (ac) voltage into one that

is limited to one polarity. The diode is useful for this function because of its nonlin-

ear characteristics, that is, current exists for one voltage polarity, but is essentially

zero for the opposite polarity. Rectiﬁcation is classiﬁed as half-wave or full-wave,

with half-wave being the simpler and full-wave being more efﬁcient.

Half-Wave Rectiﬁcation

Figure 2.2(a) shows a power transformer with a diode and resistor connected to the

secondary of the transformer. We will use the piecewise linear approach in analyzing

this circuit, assuming the diode forward resistance is

= 0

when the diode is “on.”

The input signal,

, is, in general, a 120 V (rms), 60 Hz ac signal. Recall that the

secondary voltage,

, and primary voltage,

, of an ideal transformer are related by

(2.1)

2.1.1

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Chapter 2 Diode Circuits 69

(a) (b)

Slope = 1

–

Figure 2.2 Half-wave rectiﬁer (a) circuit and (b) voltage transfer characteristics

where

and

are the number of primary and secondary turns, respectively. The

ratio

is called the transformer turns ratio. The transformer turns ratio will

be designed to provide a particular secondary voltage,

, which in turn will produce

a particular output voltage

Problem-Solving Technique: Diode Circuits

In using the piecewise linear model of the diode, the ﬁrst objective is to determine

the linear region (conducting or not conducting) in which the diode is operating.

To do this, we can:

1. Determine the input voltage condition such that a diode is conducting (on).

Then ﬁnd the output signal for this condition.

2. Determine the input voltage condition such that a diode is not conducting

(off). Then ﬁnd the output signal for this condition.

[Note: Item 2 can be performed before item 1 if desired.]

Figure 2.2(b) shows the voltage transfer characteristics,

versus

, for the

circuit. For

< 0

, the diode is reverse biased, which means that the current is zero

and the output voltage is zero. As long as

< V

, the diode will be nonconducting,

so the output voltage will remain zero. When

> V

the diode becomes forward

biased and a current is induced in the circuit. In this case, we can write

− V

(2.2(a))

and

= i

R = v

− V

(2.2(b))

For

> V

, the slope of the transfer curve is 1.

is a sinusoidal signal, as shown in Figure 2.3(a), the output voltage can be

found using the voltage transfer curve in Figure 2.2(b). For

≤ V

the output volt-

age is zero; for

> V

, the output is given by Equation (2.2(b)), or

= v

− V

and is shown in Figure 2.3(b). We can see that while the input signal

alternates

polarity and has a time-average value of zero, the output voltage

is unidirectional

and has an average value that is not zero. The input signal is therefore rectiﬁed. Also,

since the output voltage appears only during the positive cycle of the input signal, the

circuit is called a half-wave rectiﬁer.

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70 Part 1 Semiconductor Devices and Basic Applications

(a) (b)

(c)

Figure 2.3 Signals of the half-wave rectiﬁer circuit: (a) sinusoidal input voltage, (b) rectiﬁed

output voltage, and (c) diode voltage

–

(t) =

sin wt

(t)

+ V

–

(a) (b)

(t)

Figure 2.4 (a) Half-wave rectiﬁer used as a battery charger; (b) input voltage and diode

current waveforms

When the diode is cut off and nonconducting, no voltage drop occurs across

the resistor R; therefore, the entire input signal voltage appears across the diode

(Figure 2.3(c)). Consequently, the diode must be capable of handling the peak

current in the forward direction and sustaining the largest peak inverse voltage (PIV)

without breakdown. For the circuit shown in Figure 2.2(a), the value of PIV is equal

to the peak value of

We can use a half-wave rectiﬁer circuit to charge a battery as shown in Figure 2.4(a).

Charging current exists whenever the instantaneous ac source voltage is greater than

the battery voltage plus the diode cut-in voltage as shown in Figure 2.4(b). The resis-

tance R in the circuit is to limit the current. When the ac source voltage is less than

the current is zero. Thus current ﬂows only in the direction to charge the battery. One

disadvantage of the half-wave rectiﬁer is that we “waste” the negative half-cycles. The

current is zero during the negative half-cycles, so there is no energy dissipated, but at

the same time, we are not making use of any possible available energy.

EXAMPLE 2.1

Objective: Determine the currents and voltages in a half-wave rectiﬁer circuit.

Consider the circuit shown in Figure 2.4. Assume

= 12

R = 100 

, and

= 0.6

V. Also assume

(t) = 24 sin ωt.

Determine the peak diode current,

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Chapter 2 Diode Circuits 71

maximum reverse-bias diode voltage, and the fraction of the cycle over which the

diode is conducting.

Solution: Peak diode current:

(peak) =

− V

24 − 12 −0.6

0.10

= 114 mA

Maximum reverse-bias diode voltage:

(max) = V

+ V

= 24 +12 = 36 V

Diode conduction cycle:

= 24 sin ωt

= 12.6

ωt

= sin

−1



12.6



⇒ 31.7

◦

By symmetry,

ωt

= 180 −31.7 = 148.3

◦

Then

Percent time =

148.3 − 31.7

360

×100% = 32.4%

Comment: This example shows that the diode conducts only approximately one-

third of the time, which means that the efﬁciency of this battery charger is quite low.

EXERCISE PROBLEM

Ex 2.1: Repeat Example 2.1 if the input voltage is

(t) = 12 sin ωt

(V),

= 4.5

V, and

R = 250 

. (Ans.

(peak) = 27.6

mA,

(max) = 16.5

36.0 %)

Full-Wave Rectiﬁcation

The full-wave rectiﬁer inverts the negative portions of the sine wave so that a unipo-

lar output signal is generated during both halves of the input sinusoid. One example

of a full-wave rectiﬁer circuit appears in Figure 2.5(a). The input to the rectiﬁer con-

sists of a power transformer, in which the input is normally a 120 V (rms), 60 Hz ac

signal, and the two outputs are from a center-tapped secondary winding that provides

equal voltages

, with the polarities shown. When the input line voltage is positive,

both output signal voltages

are also positive.

The primary winding connected to the 120 V ac source has

windings, and

each half of the secondary winding has

windings. The value of the

output volt-

age is

120 (N

)

volts (rms). The turns ratio of the transformer, usually desig-

nated

)

can be designed to “step down” the input line voltage to a value that

will produce a particular dc output voltage from the rectiﬁer.

The input power transformer also provides electrical isolation between the pow-

erline circuit and the electronic circuits to be biased by the rectiﬁer circuit. This iso-

lation reduces the risk of electrical shock.

2.1.2

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72 Part 1 Semiconductor Devices and Basic Applications

During the positive half of the input voltage cycle, both output voltages

are

positive; therefore, diode

is forward biased and conducting and

is reverse

biased and cut off. The current through

and the output resistance produce a posi-

tive output voltage. During the negative half cycle,

is cut off and

is forward

biased, or “on,” and the current through the output resistance again produces a posi-

tive output voltage. If we assume that the forward diode resistance

of each diode

is small and negligible, we obtain the voltage transfer characteristics,

versus

shown in Figure 2.5(b).

For a sinusoidal input voltage, we can determine the output voltage versus time

by using the voltage transfer curve shown in Figure 2.5(b). When

> V

, D

is on

and the output voltage is

= v

− V

. When

is negative, then for

< −V

−v

> V

, D

is on and the output voltage is

=−v

− V

. The corresponding

input and output voltage signals are shown in Figure 2.5(c). Since a rectiﬁed output

voltage occurs during both the positive and negative cycles of the input signal, this

circuit is called a full-wave rectiﬁer.

Another example of a full-wave rectiﬁer circuit appears in Figure 2.6(a). This cir-

cuit is a bridge rectiﬁer, which still provides electrical isolation between the input ac

powerline and the rectiﬁer output, but does not require a center-tapped secondary wind-

ing. However, it does use four diodes, compared to only two in the previous circuit.

During the positive half of the input voltage cycle,

is positive,

and

are

forward biased,

and

are reverse biased, and the direction of the current is as

shown in Figure 2.6(a). During the negative half-cycle of the input voltage,

is neg-

ative, and

and

are forward biased. The direction of the current, shown in

Figure 2.6(b), produces the same output voltage polarity as before.

Figure 2.6(c) shows the sinusoidal voltage

and the rectiﬁed output voltage

. Because two diodes are in series in the conduction path, the magnitude of

two diode drops less than the magnitude of

One difference to be noted in the bridge rectiﬁer circuit in Figure 2.6(a) and the

rectiﬁer in Figure 2.5(a) is the ground connection. Whereas the center tap of the

secondary winding of the circuit in Figure 2.5(a) is at ground potential, the secondary

–

–V

Slope –

∼

–1 Slope –

∼

–v

on D

onD

T 2T

(b)(a)

(c)

Figure 2.5 Full-wave rectiﬁer: (a) circuit with center-tapped transformer, (b) voltage transfer

characteristics, and (c) input and output waveforms

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Chapter 2 Diode Circuits 73

winding of the bridge circuit (Figure 2.6(a)) is not directly grounded. One side of the

load R is grounded, but the secondary of the transformer is not.

EXAMPLE 2.2

Objective: Compare voltages and the transformer turns ratio in two full-wave recti-

ﬁer circuits.

Consider the rectiﬁer circuits shown in Figures 2.5(a) and 2.6(a). Assume the

input voltage is from a 120 V (rms), 60 Hz ac source. The desired peak output volt-

age

is 9 V, and the diode cut-in voltage is assumed to be

= 0.7

Solution: For the center-tapped transformer circuit shown in Figure 2.5(a), a peak

voltage of

(max) = 9

V means that the peak value of

(max) = v

(max) + V

= 9 +0.7 = 9.7V

For a sinusoidal signal, this produces an rms value of

S,rms

9.7

√

= 6.86 V

The turns ratio of the primary to each secondary winding must then be

120

6.86

∼

17.5

For the bridge circuit shown in Figure 2.6(a), a peak voltage of

(max) = 9

means that the peak value of

(max) = v

(max) + 2V

= 9 +2(0.7) = 10.4V

For a sinusoidal signal, this produces an rms value of

S,rms

10.4

√

= 7.35 V

–

: N

–

T 2T

(a) (b)

(c)

Figure 2.6 A full-wave bridge rectiﬁer: (a) circuit showing the current direction for a

positive input cycle, (b) current direction for a negative input cycle, and (c) input and output

voltage waveforms

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74 Part 1 Semiconductor Devices and Basic Applications

The turns ratio should then be

120

7.35

∼

16.3

For the center-tapped rectiﬁer, the peak inverse voltage (PIV) of a diode is

PIV = v

(max) = 2v

(max) − V

= 2(9.7) − 0.7 = 18.7V

For the bridge rectiﬁer, the peak inverse voltage of a diode is

PIV = v

(max) = v

(max) − V

= 10.4 −0.7 = 9.7V

Comment: 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 of the

secondary winding of the center-tapped transformer is utilized at any one time. Sec-

ond, for the bridge circuit, the peak inverse voltage that any diode must sustain with-

out breakdown is only half that of the center-tapped transformer circuit.

EXERCISE PROBLEM

Ex 2.2: Consider the bridge circuit shown in Figure 2.6(a) with an input voltage

= V

sin ωt

. Assume a diode cut-in voltage of

= 0.7

V. Determine the frac-

tion (percent) of time that the diode

is conducting for peak sinusoidal voltages

of (a)

= 12

V and (b)

= 4

V. (Ans. (a) 46.3% (b) 38.6%)

Because of the advantages demonstrated in Example 2.2 the bridge rectiﬁer cir-

cuit is used more often than the center-tapped transformer circuit.

Both full-wave rectiﬁer circuits discussed (Figures 2.5 and 2.6) produce a posi-

tive output voltage. As we will see in the next chapter discussing transistor circuits,

there are times when a negative dc voltage is also required. We can produce negative

rectiﬁcation by reversing the direction of the diodes in either circuit. Figure 2.7(a)

shows the bridge circuit with the diodes reversed compared to those in Figure 2.6.

The direction of current is shown during the positive half cycle of

. The output

voltage is now negative with respect to ground potential. During the negative half

cycle of

, the complementary diodes turn on and the direction of current through

the load is the same, producing a negative output voltage. The input and output volt-

ages are shown in Figure 2.7(b).

and D

(b)

–

: N

(a)

Figure 2.7 (a) Full-wave bridge rectiﬁer circuit to produce negative output voltages.

(b) Input and output waveforms.

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Chapter 2 Diode Circuits 75

Filters, Ripple Voltage, and Diode Current

If a capacitor is added in parallel with the load resistor of a half-wave rectiﬁer to form

a simple ﬁlter circuit (Figure 2.8(a)), we can begin to transform the half-wave sinu-

soidal output into a dc voltage. Figure 2.8(b) shows the positive half of the output

sine wave, and the beginning portion of the voltage across the capacitor, assuming

the capacitor is initially uncharged. If we assume that the diode forward resistance is

= 0

, which means that the

time constant is zero, the voltage across the ca-

pacitor 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 ca-

pacitor discharges exponentially with time (Figure 2.8(c)). During this time 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, then the output voltage will de-

crease exponentially with time, as previously indicated. An exaggerated sketch of

these two voltages is shown in Figure 2.8(d). The output voltage decreases at a faster

rate than the input voltage, which means that at time

, the difference between

and

2.1.3

Time

(e)

(b)

Time

–t/RC

(a)

–

w tw t

a sin w t

–t/RC

Figure 2.8 Simple ﬁlter circuit: (a) half-wave rectiﬁer with an RC ﬁlter, (b) positive input

voltage and initial portion of output voltage, (c) output voltage resulting from capacitor

discharge, (d) expanded view of input and output voltages assuming capacitor discharge

begins at

ωt = π/2

, and (e) steady-state input and output voltages

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76 Part 1 Semiconductor Devices and Basic Applications

Figure 2.9 Output voltage of a full-wave rectiﬁer with an RC ﬁlter showing the ripple voltage

, that is, the voltage across the diode, is greater than

. However, this condition

cannot exist. Therefore, the diode does not turn off immediately. If the RC time

constant is large, there is only a small difference between the time of the peak input

voltage and the time the diode turns off. In this situation, a computer analysis may

provide more accurate results than an approximate hand analysis.

During the 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 turns 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 ﬁl-

ter capacitor. The steady-state output voltage of the RC ﬁlter is shown in Figure 2.8(e).

The ripple effect in the output from a full-wave ﬁltered rectiﬁer circuit can be seen

in the output waveform in Figure 2.9. The capacitor charges to its peak voltage value

when the input signal is at its peak value. As the input decreases, the diode becomes

reverse biased and the capacitor discharges through the output resistance R. Deter-

mining the ripple voltage is necessary for the design of a circuit with an acceptable

amount of ripple.

To a good approximation, the output voltage, that is, the voltage across the ca-

pacitor or the RC circuit, can be written as

(t) = V

−t



/τ

= V

−t



/RC

(2.3)

where



is the time after the output has reached its peak value, and RC is the time

constant of the circuit.

The smallest output voltage is

= V

−T



/RC

(2.4)

where



is the discharge time, as indicated in Figure 2.9.

The ripple voltage

is deﬁned as the difference between

and

, and is

determined by

= V

− V

= V

(1 − e

−T



/RC

)

(2.5)

Normally, we will want the discharge time



to be small compared to the time

constant, or



 RC

. Expanding the exponential in a series and keeping only the

linear terms of that expansion, we have the approximation

−T



/RC

∼

1 −



(2.6)

We can show that the difference between the exponential function and the linear approximation given by

Equation (2.6) is less than 0.5 percent for

RC = 10T



. We need a relatively large RC time constant for this

application.

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Chapter 2 Diode Circuits 77

The ripple voltage can now be written as

∼







(2.7)

Since the discharge time



depends on the RC time constant, Equation (2.7) is

difﬁcult to solve. However, if the ripple effect is small, then as an approximation, we

can let



= T

, so that

∼





(2.8)

where

is the time between peak values of the output voltage. For a full-wave

rectiﬁer,

is one-half the signal period. Therefore, we can relate

to the signal

frequency,

f =

The ripple voltage is then

2 fRC

(2.9)

For a half-wave rectiﬁer, the time

corresponds to a full period (not a half pe-

riod) of the signal, so the factor 2 does not appear in Equation (2.9). The factor of 2

shows that the full-wave rectiﬁer has half the ripple voltage of the half-wave rectiﬁer.

Equation (2.9) can be used to determine the capacitor value required for a par-

ticular ripple voltage.

EXAMPLE 2.3

Objective: Determine the capacitance required to yield a particular ripple voltage.

Consider a full-wave rectiﬁer circuit with a 60 Hz input signal and a peak output

voltage of

= 10

V. Assume the output load resistance is

R = 10 k

and the rip-

ple voltage is to be limited to

= 0.2

Solution: From Equation (2.9), we can write

C =

2 fRV

2(60)(10 × 10

)(0.2)

⇒ 41.7 μF

Comment: If the ripple voltage is to be limited to a smaller value, a larger ﬁlter ca-

pacitor must be used. Note that the size of the ripple voltage and the size of ﬁlter

capacitor are related to the load resistance R.

EXERCISE PROBLEM

Ex 2.3: Assume the input signal to a rectiﬁer circuit has a peak value of

= 12

V and is at a frequency of 60 Hz. Assume the output load resistance is

R = 2k

and the ripple voltage is to be limited to

= 0.4

V. Determine the

capacitance required to yield this speciﬁcation for a (a) full-wave rectiﬁer and

(b) half-wave rectiﬁer. (Ans. (a)

C = 125 μ

F, (b)

C = 250 μ

F).

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