CHAPTER TWO

POWER SEMICONDUCTOR DIODESAND CIRCUITS

DESIGNED BY

DR. SAMEER KHADERPPU

“E-learning Project”

Diode circuits

I- (S.2.3) : Diode Characteristics: Power diode is a two-terminal pn-junction device ……….);The equivalent circuit and i-v curve are displayed as follows, where the diode conducts when a battery is connected across its terminals .

Symbol

P-N junction P-N junction

;qT.KVT

voltagethermalV

currentleakageI

voltagediodeVwhere

)1e(II

T

s

D

V.nVsD

TD

II- (S.2.4) : Reverse Recovery Characteristics: When the current falls to zero , the diode continues to conduct under the action of minority carriers that remain stored in the pn-junction and Bulk resistance. These carriers require a certain time to recombine with opposite charges and to be neutralized. This time is called the reverse recovery time trr . The figure shown below displayed the current falling process and the diode reverse parameters.

.di/dttI , tt t

.dt .Q.di.t t; .di/dt t trr

2.Q I

.tI

.tI

.tI Q

arrbarr

rr2rraarr

rr

2

rrrr

2

brr

2

arrrr

1- The reverse recovery time trr consist

of two parameters ta, and tb: trr =ta+tb ; Irr=ta.di/dt ta- due to storage charge in the depletion regiontb- due to storage charge in the bulk resistance.The softness factor Sf=tb/ta.Trr- depends on the junction temperature, di/dt, and the diode forward current

2- The reverse recovery charge Qrr : this is the charge carriers across the diode flows in the reverse direction due to changeover of the conduction state.

;2;/2

/2

.

2

tirrrr

rrrr

rrr

ddQIdtdi

Qt

dtdiQrttrrtatatb

Example 2.1: Given a diode circuit (p.2.1) with reverse recovery time trr=5 µS, and the rate of fall of the diode current is di/dt= 80A/ µS with softness factor Sf=0.5. Determine : 1- the storage charge Qrr; 2- the peak reverse current Irr.

Solution:

C6.133310.5*72.266t.IQ

A72.26610/80*10334.3d

d.tI

s10334.310.5t

t

tttt

t5.0S

5

2

1rrrr

2

1rr

66

t

iarr

6

5.1

6

5.1

rra

barra

bF

III- (S.2.5) : Diodes Classification : Depending on the recovery characteristics, and manufacturing techniques, there are three types:

1. General - purpose diodes: they have high trr=25 µS and with frequency < 1kHZ, applied in AC to DC circuits. The current rating up to 1000A, and up to 5kV.

2. Fast recovery diodes: they have small trr=5 µS and with frequency < 5kHZ, applied in DC to DC and DC to AC circuits. The current rating up to 100A, and up to3kV.

3. Schottkey diode: they have approximately zero reverse recovery time with high frequency up to 10kHz, and applied in high current low voltage applications. The current rating up to 300A and 100V circuit voltage.

IV- (S.2.8) : Series –Connected Diodes : In high voltage dc applications (mainly), the diodes are connected in series with purpose to

increase the reverse blocking capabilities.The difference in the i-v curve in the reverse blocking condition occurs due to manufacturing

errors and tolerances, therefore each diode should carry different voltage , while the leakage current is the same.

The solution is to force equal sharing voltage across the diodes by connecting a sharing resistances as well shown below:

THERE ARE TWO APPROACHES: 1. Equal sharing resistances connected across the diodes 2. Equal voltage sharing while the sharing resistances may differs. Mathematical Modeling:

).(1)(

2:2

);.(22

).().(

:1

);(;

1222

2.112

2

2

1

1

21

121

12111221

21

122

2

1

1

2

22

1

11

1221

2211

sssD

sDss

s

D

s

D

DD

sss

D

ssDsDssDD

ss

sss

D

s

D

s

Ds

s

Ds

DsDDDs

dsds

IIRV

RVRsII

R

V

R

V

VsVVcase

IIRV

V

IIRVVVIIRVV

RRRcase

IIR

V

R

V

R

VI

R

VI

VVVVVV

IIII

In general second case is the most applicable when sharing resistances are used.

Example 2.2: Two diodes are connected in series as well shown on up mentioned circuit, where the circuit parameters are: 7kV source voltage ( DC) , leakage current of first diode Is1=40mA and of second diode Is2=50mA . 1- Find the diode voltages if the resistances are equals Rs1=Rs2=R=80 kΩ. 2- Find the sharing resistances if the diode voltages are distributed equally.

Solution: Two cases must be described as follow Case#1: Rs1=Rs2=R=80 kΩ

k702Rsand;k34.581Rs

k34.58310.10.700003500

70000.35001Rsk702RsLet

V35002VsVV:2case

V3100VVsV

V3900310).4050.(2

80000

2

7000)II.(

2

R

2

VV

2D1D

1D2D

1s2ss

1D

V- (S.2.9) : Parallel –Connected Diodes : In high power applications, diodes are connected in parallel with purpose to increase the

current carrying capability. Due to some differences in the Bulk resistances of both diodes, there is a different current will flow through the diodes. Therefore by connecting resistances in series with the diodes the diode voltage is shared equally as well shown below:

The function of both Ls1 and Ls2 is to equally sharing the current under dynamic behaviors.When ID1 rises, the inductor voltage Ls1.d(Id1)/dt increases, and a corresponding voltage of opposite polarity is induced across inductor Ls2. This resulting low impedance in the circuit of D2, therefore shifting the current to flow through D2 path.

Mathematical Modeling: The following equation are derived under steady-state conditions

1D2D

2D1D

2D1D1D2D2s1s

2D4s2D1D3s1D

;42D31D

II

VVR

VV)II(RRRRLet

I.RVI.RV

VVVV

Example 2.3: Find the value of R required for adjusting the voltage across the diodes D1 and D2, if ID1=50A, ID= 95A, VD1=1.8V, and VD2=2V.

Solution:

m205045

28.1

II

VVR

A455095IIIIII

1D2D

2D1D

1DD2D2D1DD

Mode#1 D1 conducts: The current will flow from the source to the load through D1, where the current i1:

IV- (S.2.12) : Freewheeling Diodes If switch S in the figure shown below is closed for time t1, a current I1 is established through

the load; and then if the switch is opened for time t2 the current continues to flow through the inductor and the diode” closed path’. If there in no closed path the inductive energy induces a very high voltage and this energy is dissipated in form of heat and spark. The diode realized closed path is called usually FREEWHEELING DIODE.

The operation of the proposed circuit is divided into two modes:

RLeVstV

eR

Vti

dt

diLiRV

tL

tss

/;.)(

)1()(. 11

1

Example 2.4: Determine the value of Ls and Cs for the diode circuit shown behind with L=5mH;

R=100Ω; Vs=200V; and di/dt = 50A/ µS . Solution: by applying Lenz law the circuit

inductance Ls is determined as follows:

Mode#2 Dm conducts: The current will continue to flow through the load and the diode Dm., where the current i2:

t12

11

ts1

11

e.I)t(i

;R

VsIIttAt

)e1(R

V)t(i

dt

diLi.R0

A50210.4

20010102I

L

VtIIA2

100

200

R

VI

L

V.t

dt

di.tI

H450

10.200

dtdiV

Lsdt

diLV

66

.p

s

s.rrop

so

s

srrrrrr

6s

ss

The excess energy due to current increasing is determined as:

FC

V

WCsVCWW

JLIIW

s

C

RcsCR

sopR

2.25200

504.0*2

.2.

2

1

504.010.4.25022

1.

2

1

2

22

62222

Simulation Results: By applying the simulation software “ POWERSYS” for the same circuit is applied as follows for load voltage Vp1, circuit current I1, load current I2, excess energy due to current increasing is determined as:

The source current I1 at first instant rises significantly, and I2 rises also. At steady state the current I1 and I2 are stabilized at a value of 2A.

THANK YOU FOR LISTENING

QUESTIONS…..?

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BY EMAIL ON

sameer@ppu.edu