Diodes and Circuit Applications

Diodes and Circuit Applications

1- pn junctionWhen p-type and n-type semiconductors are brought in contact a pn-junction is formed which is the basis of the semiconductor diode, a widely used circuit element.

The charge separation causes a contact potential to exist at the junction. This potential is typically on the order of few tenths a volt and depends on the material (for Silicon about 0.6 to 0.7 V). The contact potential is called the offset voltage Vγ.

The existing of electric field across the junction helps the minority carriers (holes in the n-type and electron in the p-type) to drift across the depletion region creating a small reverse saturation current Isflows in the reverse direction when the diode is reverse biased.

Is is independent of the junction voltage and is determined by thermal carrier generation. At room temperature, Is is in the order of nanoamperes (10-9 A) for Silicon.

Another current caused by diffusion of majority carriers (holes in the p-type and electrons in the n-type) across the junction called diffusion current Id.

The diffusion current flows in the forward direction and is largely dependent on junction voltage, thus, it increases as the forward bias voltage increases.

SD IIi =−= 0

Reverse-biased diodeThe effect of the reverse bias is to increase the contact potential. The only current is the drift current

Forward-biased diodeThe forward voltage acts in opposition to the contact potential therefore, the diffusion current is aided by the applied voltage

KTqvd

DeII /0=

The net forward current is,

diode equation

where,vD is the voltage across the pn-junction, K=1.38 x 10-23

J/K is Boltizman’s constant, q is the charge of one electron, and T is the temperature in kelvins.Since I0 is very small, iD is approximated to,

The diode is essentially conducts current in only one direction that is when the junction is forward-biased

)1( /00 −=−= KTqv

dDDeIIIi

KTqvD

DeIi /0=

Diode i-v characteristics

We can summarize the diode behaviour as follows: vD ≥ Vγ the diode is forward-biased (on), conducts current, and acts as a short circuit. -VZ ≤ vD ≤ Vγ the diode is reverse-biased (off), acts as an open circuit, and allows a very small reverse current to flow I0 . vD ≤ -VZ the diode operates in the reverse breakdown region, conducts a current in the reverse direction.

Note that the reverse breakdown region represents the behaviour of the diode when a sufficiently high reverse bias voltage is applied. Aided by the electric field, sufficient energy is imparted to the charge carriers that reverse current can flow. The process of conduction is very like an avalanche breakdown in which a single electron can ionize others.

2-Diode circuit models2.1-Ideal diode model• Ideal diode is a theoretical model for the practical diode. The diode is modelled as a simple on/off device.

To determine the conduction state of an ideal diode:

1. assume a diode conduction state (on or off).2. substitute ideal circuit model into the circuit (short circuit if “on”, open circuit if “off”.

3. solve for diode voltage and current using linear circuit analysis techniques.

4. if the solution is consistent with the assumption, then the assumed state is correct. If not, the diode conduction state is opposite to the assumed one.

ExampleDetermine whether the ideal diode in the circuit below is conducting. Vs=12 V, VB=11 V, R1=5 Ω, R2= 10 Ω, R3=10 Ω

SolutionAssume initially the diode is off and replace it by an open circuit

Apply KVL to the right-hand-side mesh:

The results indicates that the diode is reverse-biased and confirm the initial assumption. Thus, the diode is offIf the initial assumption is that the diode is on, replace the diode with a short circuit

VV

RRR

v S 8105

10

21

21 =

+=

+=

VvorVvv DBD 31181 −=−=+=

i1

i

Apply KCL: i = i1 + i2

note that v1=v2

Since v1=v2 <VB =11 V, we must conclude that current flow in the reverse direction of the assumed one. This observation is inconsistent with the initial assumption. Thus, the diode is off.

3

1

2

1

1

1

RVv

Rv

RvV BS −

+=−

Vv

v

Rv

Rv

Rv

RV

RV BS

75.8

)101

101

51(

1011

1512

1

1

3

1

2

1

1

1

31

=

++=+

++=+

Example (P9.7)Determine whether the diode is conducting or not. Assume that the diode is an ideal diode.

SolutionAssume the diode is on, then the current is:

KVL:

The result contradict the assumption since vD is negative and current sign means its direction is opposite to the assumed one. Thus, the diode is off

A

VVi iB 1333.0

151210

105−=

−=

+−

=

VvVviV

D

iDB

997.3121333.0x15100)510(−=−−=

=++++−

2.2- Offset diode modelWhile the ideal model is useful in approximating

the large-scale characteristics of a physical diode, it does not account for the presence of an offset voltage, which is unavoidable component in semiconductor diode.

The offset diode model consists of an ideal diode in series with a battery of strength equal to the offset voltage, Vγ .

Vγ

for vD < Vγ the diode acts as an open circuit (off) .for vD ≥ Vγ the diode is on and acts as a battery of Vγ V

For Silicon diode:vD ≥ 0.6 V 0.6-V batteryvD ≤ 0.6 V open circuit

ExampleIf VB=2 V, use the offset model to determine the value of v1 for which the diode D1 first conduct.

SolutionAssume the diode is off and replace it with the offset modelKVL:

then the condition for the diode to conduct is v1-2.6≥0or v1≥2.6 V

6.226.00

111

11

−=−−=

=+++−

vvvVVvv

D

BD γ

Example (P9.23)Use the offset diode model to determine the output voltage of each of the following circuits:

Solutiona) D1 and D3 are reverse-biased, D2 and D4 are forward-biased.

Vout=-5+0.7= -4.3 V

b) D1 and D2 are reverse-biased, D3 is forward-biased

Vout=-10+0.7= -9.3 Vc) D1 and D2 are reverse-biased

Vout= -10 V

3- Diode operating pointDiode circuits analysis aims to determine the dc operating point referred to as Quiescent-point (Q-point) when the diode operates in the forward region. i.e., to find the dc operating current, IQ, and the dc operating voltage, VQ.To graphically determine the operating point of a diode, Reduce the circuit to Thevenin or Norton equivalent circuit with the diode as the load. Write the load-line equation. Solve graphically by finding the intersection of the diode curve with the load-line curve. The intersection of the two curves is the operating point.

Consider the Thevenin equivalent circuit of an arbitrary linear resistive circuit connected to a diode then:

load-line equation

diode equation

DTDT vRiv +=

)1( /0 −= KTqv

DDeIi

The diode equation gives rise to diode iD-vDcharacteristics curve while the load-line equation describes a line with slope -1/R and y intercept given by VT/RT and x intercept given by VT.

TT

DT

D VR

vR

i 11+−=

ExampleDetermine the operating point of the 1N914 diode in the circuit shown below and compute the total power output of the 12-V battery. R1 =50 Ω, R2 =10 Ω, R3 = R4 =20 Ω,

SolutionWe first compute Thevenin equivalent circuit

RT= R1 || R2 + R3 + R4

RT =20+20+(10||50)=48.33 Ω

VVRR

RV ST 212x6010

21

2 ==+

=

Using load line equation, y intercept is VT/RT=41 mAand x intercept is VT = 2 V.Curves intersection isthe Q-point, thus:VQ=1 VIQ=21 mA

The battery power is:PB=12 x IBIB is equal to the current through R1 which is equal to the current through R2 and the diode current.

WPARVI

VVRRIV

B

RR

QQR

46.2)184.0021.0(x12184.010/84.1/

84.1140x021.0)(

222

432

=+====

=+=++=

4- Rectifier circuitsRectification is one of the important applications of the diode. Rectification is the ability to convert the AC signal to a DC signal..4.1- Half-wave rectifierHalf-wave rectifier is a simple but inefficient rectifier. A single diode is used to build this rectifier. Consider an AC source voltage, vi, connected to a load via a series ideal diode.

Vm

)()()(0)()()( tvtvbiasedforwardonisdiodepositive

VtvbiasedreverseoffisdiodenegativeiiL

Ltv =

=

Thus, the appearance of the load voltage will be as shown below:

Note that one-half of the AC energy is not recovered that why the half-wave rectifier is inefficient.

When the diode is conducting, the unknown iD and vLcan be found as follows:

when vi ≥ 0And

The rectified wave-form has a nonzero DC (average) voltage, where as the average input waveform voltage is zero. The average value of the load voltage is obtained by integrating the load voltage over one period and dividing by the period.

L

iD R

vi =

LDL Riv =

∫ ∫ =+=2/

0 2/, ]0sin[1 T

mT

TmDCload

VdtdttVT

vπ

ω

ExampleCompute and plot the rectified load voltage vR in the circuit shown below using the offset diode model.

Solution

Assume the diode is off. Apply KVL to the circuit (a):vD = vS – 0.6 (i = 0)Then, to be consistent with the assumption that the diode is off:vS < 0.6 V (vD <0 diode off condition)Since i=0 then,vR =0 V (diode off)If vS ≥ 0.6 V , the diode conducts and the current i is obtained by applying KVL to circuit (b),

6.0

6.0

−==

−=

SR

S

viRvand

Rv

i

We summarize the result as follows:

Observe that the load voltage is shifted down by amount equal to the offset voltage, vγ

≥−

<=

Vvforv

Vvforv

SS

S

R6.06.0

6.00

4.2- Full-wave rectifierFull-wave rectifier offers a substantial improvement in efficiency over the half-wave rectifier.The first section of full-wave rectifier circuit includes an AC source and a center-tapped transformer with 1:2N turns ratio. The transformer is used to obtain the desired voltage amplitude prior to rectification and to isolate the rectifier circuit from the AC source since there is no direct electrical connection between input and output of the transformer.

During the positive half cycle, the top diode is forward-biased (on) and the bottom diode is reverse-biased (off). Thus,

The offset voltage is neglected here since in most cases the AC source is 110 V rms, 60 Hz.During the negative half cycle, the top diode is off and the bottom diode is on. Then,

Note that the direction of iL is the same in both cycles because of the manner of diodes connections.

01 ≥== S

L

SL v

RNv

ii

02 <−

== SL

SL v

RNv

ii

The load voltage appearance is as shown below:

The full-wave rectifier results in twofold improvement in efficiency over the half-wave rectifier.

4.3- The bridge rectifierBridge rectifiers employ four diode commonly available as a single integrated circuit.

Uncertified source voltage Ideal diodes Offset diodes

Note that the effect of the offset voltage is to shift the waveform down by twice the diode offset voltage.

5- Ripple voltageAlthough rectifiers circuits give a non-zero average

voltage, their output still an oscillating waveforms. Instead of a smooth, constant voltage, the rectifier generates a sequence of sinusoidal pulses at a frequency double that of the original AC signal frequency in case of full-wave rectifier and equal that of AC signal frequency in case of half-wave rectifier.

Fluctuations about the mean voltage of the rectified output is referred to as output ripple or ripple voltage. The ripple voltage is undesirable component in the rectified load voltage.

A simple and effective means to eliminate most of the ripple is to use a capacitor filter that works as a low-pass filter which preserves the DC component of the rectified voltage while filtering out components at frequency at or above twice the AC signal frequency.

The smoothing (filtering) operation can be explained as follows: During the conduction period, the capacitor will charge to the peak value of the rectified signal, Vm . The load voltage vL(t) = vC(t). During the off period, the load current is supplied by the capacitor which starts to discharge through the load RL

As the capacitor discharge, vL(t) decreases until the capacitor voltage has fallen to the value of the now rising next half-cycle of rectified voltage, then another conduction and capacitor charging cycle starts. If the time constant, CRL, is large in comparison to the period of the AC waveform, then a reasonable accurate approximation can be made by assuming that the capacitor voltage falls linearly.

vL

VLmax=VLmin 2

rV

t2t1

The peak-to-peak ripple , Vr, voltage can be calculated by considering the capacitor discharge equation as follows:

CRL = time constant, the capacitor needs to discharge through the load

Vm= the peak value of the rectified outputNote that:

Note: t in the above equations is substituted by t2 the time at which vL(t) reaches its minimum value

)1( //minmax

CRtm

CRtmmLLr

LL eVeVVVVV −− −=−=−=

CRtm

rdcL

LeVV

VV /min 2

−=−=

2maxr

dcmLV

VVV +==

ExampleA half-wave rectifier is used to provide a DC supply to a 50-Ω load. If the AC source voltage is 20 V (rms), find the peak and average current in the load. Assume an ideal diode.Solution

Average current=

ARv

Rv

iL

rms

L

peakpeak 567.0

5020x22

====

∫ ∫ ∫

+==

T T T

TL

peakDC dtdtt

Rv

Tdtti

Ti

0

2/

0 2/

0)sin(1)(1 ω

AR

v

L

peak 18.050x20x2

===ππ

ExampleIn the full-wave power supply shown, the diodes are 1N4001 with a rated peak inverse voltage (also called peak inverse voltage) of 25 V. If n (turn ratio)=0.05883, C=80 μF, RL=1 kΩ, Vline=170cos(377t) V.a- determine the actual peak reverse voltage across each diodeb- explain why these diodes are or are not suitable for the specifications given.

Solutiona) The peak voltage of Vs1 is:

at ωt=0: D1 is on , D2 is offKVL to L1:

-10+0.7+Vm=0 , Vm=9.3 V=peak value of vL (t)KVL to L2:

10+ vD2-reverse +9.3=0 , vD2-reverse =-10-9.3=-19.3 VThe same analysis can be conducted for D1-reverse.voltage but at ωt=π at which D1 off and D2 onb) Since the rated peak reverse voltage (25 V) > calculated actual peak reverse voltage by adequate margin, the diodes are suitable for the given specifications.

VnVV imSm 10170x05883.0 ===

0)()( 11 =++− − tvvtv LonDS

0)()( 22 =++ − tvvtv LreverseDS

ExampleIn the bridge-rectifier circuit shown, if IL=85 mA, VL=5.3 V, Vr=0.6 V, ω=377 rad/s, and vline=156cos(ωt).Determine the value of:a- the peak value of Vs

b- the capacitor C

Solution a-

KVL: at t=0,

VSm=peak value of VS VSm+0.7+0.7+Vm=0 ,

Note:

b- at t=t2,

t2=2.725/377=7.23 ms

0)()( 31 =+++− tvvvtv LDDS

VVSm 76.57.07.0 =++=

04487.0156

7===

im

Sm

VVn

radV

VvvtSm

LDD 725.2)(cos min4212 =

−−−= −ω

031 =+++− −− monDonDSm VvvV

0)()cos( 422 =+++ tvvvtV LDDSm ω

RL=VL/IL=5.3/0.085=62.353 Ω

Ce 353.62/00723.016.56.0 −−=

FC

e C

µ1023)893.0ln(00723.0

353.62

893.06.56.01353.62/00723.0

=−=

=−=−

)1( /2 CRtmr

LeVV −−=

6- Voltage regulationDC power supplies contains voltage regulators, that is, devices that can hold a DC load voltage relatively constant in spite of possible fluctuations in the DC supply.

The most common device employed in the voltage regulation is the Zener diode For voltage regulation, Zener diode must operate at the Zener breakdown region to hold the load voltage nearly constant at Zener voltage, VZ.

When the reverse voltage is lower than –VZ , the diode behaves as a voltage source providing a voltage reference that is nearly constant provided that the unregulated supply voltage remains above VZ

The slope is not infinite at the Zener breakdown region. The finite slope causes the Zener voltage to change slightly. For the purpose of analysis, VZ is assumed constant

Circuit symbol

Consider the circuit below where the unregulated DC source VS is to be regulated to the value of VZ. An ideal Zener diode is assumed here. Note that the diode must connected upside down to obtain a positive regulated voltage.

To analyze the regulator operation the following observation are considered:1) The load voltage must equal VZ, as long as the Zenerdiode is in the reverse breakdown region

(ideal diode, rZ=0 Ω, VL=VZ)

1) Any current in excess of that required to keep the load at the constant voltage VZ is “dumped” to ground through the diode.

2) The source current is given by

(ideal diode, VL=VZ)

L

Z

L

LL R

VRVi ==

ZSL iii −=

S

ZSS R

Vvi

−=

If practical Zener diode model is used then,

VL= VZ + rZiZ

Zener diode is usually rated in terms of its maximum allowable power dissipation.

( ideal model, rZ=0 Ω)

If rZ is considered then:

S

LSS R

Vvi −=

ZZZ ViP =

S

ZZZS

RriVv −−

=

ZZZZZ riViP 2+=

In practice, the range of load resistance RL is constrained to:

RLmax is typically limited by the power dissipation of the Zener diode and RLmin is limited by the maximum supply current

Note that when the load disconnected suddenly or when RL becomes greater than RLmax , larger currents will be dumped by the diode and power dissipated may exceed the diode rating

Also when RL becomes smaller than RLmin the unregulated supply may not be able to provide the current required to sustain the load voltage.

To avoid the above worst cases RL value must stay within the specified constrains

maxmin LLL RRR ≤≤

To ensure the Zener diode remains in the voltage-regulation region, a proper value of the current limiting resistor R must be used.

The minimum value of R is calculated at iZmax that caused by: maximum source voltage, minimum load current, and minimum Zener voltage.

The maximum value of R is calculated at iZmincaused by minimum source voltage, maximum load current, and maximum Zener voltage

ZL

ZZZS

ZL

LS

S

LS

iiriVv

iiVv

iVv

R+−−

=+−

=−

=

maxmin

maxminmaxmin

ZL

ZZZS

iiriVv

R+

−−=

minmax

minmaxminmax

ZL

ZZZS

iiriVv

R+

−−=

ExampleIf vs=24 V, VZ=12 V, RS=50 Ω, RL=250 Ω, determine the minimum acceptable power rating of the Zener diode. Assume an ideal Zener diode.

Solution A

RVv

iS

ZSS 24.0

501224

=−

=−

=

ARV

iL

ZL 048.0

25012

===

and the Zener current ,

Corresponds to a nominal power dissipation

=0.192 x 12=2.304 WHowever, if the load accidently removed, all the load current is diverted to flow through the diode. Thus, the worst-case Zener current is

and

Safe design would exceed the value of PZmax computed above. 3-W Zener diode is suitable in this case.

Aiii LSZ 192.0=−=

ZZZ ViP =

Aii SZ 24.0max ==

WViP ZZZ 88.2maxmax ==

ExampleDetermine the minimum and maximum value of the series resistor may have in a regulator circuit whose output voltage is to be 25 V. The input voltage varies from 35 to 40 V, the load current is 75 mA and the Zener diode used has a maximum current rating of 250 mA. Assume iZmin=0 A.

Solution

but rZ=0 (ideal diode) then:

iZ

maxmin

maxminmaxmin

ZL

ZZZS

iiriVv

R+

−−=

Ω=+−

=+−

= 15.46250752540

maxmin

maxmin

ZL

ZSS ii

VvR

ExampleWe desire to hold the load voltage constant at 14 V. Find the range of the load resistance for which regulation can be obtained if the Zener diode is rate at 14-V, 5 W. Use ideal model.

minmax

minmaxminmax

ZL

ZZZS

iiriVvR

+−−

=

Ω=+−

= 3.1330752530

maxSR

SolutionThe minimum load resistance is calculated by assuming all the source current goes to the load.

The maximum current through the Zener diode that does not exceed the diode power rating is,

Then, So the maximum load resistance is,

max2.120101450

Ls

Zs

s

LsS iA

RVV

RVVi ==

+−

=−

=−

=

Ω==== 7.112.1

14

maxmin

s

Z

L

LL i

ViVR

AVPi

Z

ZZ 357.0

145

max ===

Aiii ZSL 843.0357.02.1maxmin =−=−=

Ω=== 6.16843.014

minmax

L

LL i

VR

ExampleDetermine the minimum and maximum values of the series resistor R if the following regulation specifications are required.VZ=12 ± 10% , rZ=9 ΩiZmin=3.25 mA , iZmax=80 mAVS=25 ± 1.5 V , IL=31.5 ± 21.5 mA

Solution

7- Diode clipper (Limiter)Diode clipper employed to protect the load against excessive voltage, i.e., to keep

maxmax )( VtvV L ≤≤−

Assume ideal diodes:

D1 on

Condition for D1 to conduct:

vL(t)=Vmax

D2 analysis can be conducted by analogy with the preceding D1 analysis. The results is the load voltage clipped from both side as shown below, assume Vmax = 5 V

Condition for D1 to be off :

8- The diode clampDiode clamper is used to clamp the signal waveform to a fixed DC value.

VC= capacitor voltage=Vpeak

Clamp positive peaks to 0 V, i.e., Shifted down to 0 levelnecessary condition: RC >> T

VC= Peak -VDC

Clamp positive peaks to Vpeak-VDC V

Necessary condition: VDC <Peak

peaksCsout VtvVtvtv −=−= )()()(

DCpeaksout VVtvtv +−= )()(

clamping to 0 V (shifting down)The clamping also works with the diode in the

reverse direction. The capacitor will charge to –Vpeakand

In this case the negative peaks shifted up to 0 V. A battery VDC also can be connected in series with the diode to shift the waveform up by a voltage different from Vpeak V.

peaksout Vtvtv += )()(

DCpeaksout VVtvtv −+= )()(

ExampleDesign a biased diode clamp to shift the DC level of the signal vs (t)=5cosωt up by 3 V. Use ideal diode model.Solution

(shifting up)

To have a DC level of 3 V, the capacitor must be charged by 3 V or: . Since Vpeak =5 V then VDC should equal to 2 V

DCpeaksout VVtvtv +−= )()(

VVV DCpeak 3=−

9- Peak Detector

+

-vi(t) +

-

D

vo(t)C

0

Vm

-Vm

vi(t)

t

vo(t)

t

0

Vm

Peak detector circuit is similar to clamper circuit except that the diode and the capacitor swapped their places. The output, vo(t) , is taken across the capacitor

When vi(t) increases initially from 0 V to Vm (its peak value), the diode is forward biased and immediately charges the capacitor up to the peak value of the input voltage, Vm , thereafter, the diode remains reverse biased as the capacitor will keep its voltage constant at Vm

Peak detector circuits are used in peak-reading voltmeters and in smoothing pulsating waveforms

10- Photo diodesPhoto diodes are fabricated with a surface material that is transparent to light. When the light reaches the depletion region it generate electron-hole pairs by a process called photoionization. As a consequence, the reverse current depends on the light intensity.

the i-v curve is shifted to lower values according to the value of photoionization current IP.

L1 curve represents normal forward-biased diode operation. Q-point in the positive i and positive v. Power is positive and diode dissipates power.

L2 curve represents photodiode operates as solar cell. Q-point in the negative i and positive v. Power is negative and diode generate power.

L3 curve represents photodiode operates as light sensor. The diode is reverse-biased and the current determined by the incident light intensity.11- Light-emitting diodes (LEDs)Diode operating in this mode emits light when forward-biased. LEDs exhibit a forward voltage (offset) of 1 to 2 V. In principle, by forwarding the diode and causing a significant level of recombination, some of the energy released is converted to light energy.

Gallium arsenide (GaAS) is one of the more popular substrates for creating LEDs. LED operating current can range from 20 to 100 mA.

ExampleDetermine: a) the LED power consumption. b) the resistance RS. c) the power required by the voltage source. VS =5 V; ILED =40 mAVLED=1.7 V

Solutiona) PLED = VLED x ILED =1.7x0.04=68 mWb) VS= ILED RS + VLED

RS =82.5 Ωc) PS= VSx ILED =200 mW