Transcript Solution

CHAPTER TWO
POWER SEMICONDUCTOR DIODES
AND CIRCUITS
DESIGNED
BY
DR. SAMEER KHADER
PPU
“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
ID  Is (e VD n .VT  1)
where VD  diode volt age
Is  leakage current
VT  t hermalvolt age
VT  K.T
q;
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 pnjunction 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.
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 region
tb- 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.
Irr.ta Irr.tb Irr.trr
Qrr  2  2  2
tb  ta  ta  trr  trr 2  2Qrr
di / dt
2.Qrr
Irr 
 ta.di/dt ; ta.trr.di  2.Qrr.dt  trr  2Qrr
; Irr  2Qrr . di dt ;
trr
di / dt
trr  ta  tb , Irr  ta.di/dt
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:
SF  0.5 
 ta 
trr
1.5
tb
 trr  ta  tb
ta
5.106
6


3
.
334
10
s
1.5
di
 3.334106 * 80 / 106  266.72A
dt
Qrr  12 Irr.trr  12 266.72* 5.105  1333.6 C
Irr  ta.
III- (S.2.5) : Diodes Classification : Depending on the recovery characteristics, and manufacturing
1.
2.
techniques, there are three types:
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.
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:
Is1  Id 1  Is 2  Id 2
Vs  VD1  VD 2  VD 2  Vs  VD1
VD1
VD 2
VD1 VD 2
 Is 2 
;

 ( Is 2  Is1);
Rs1
Rs 2
Rs1
Rs 2
case1 : Rs1  Rs 2  R
VD1  VD 2  R.(Is 2  Is1)  VD1  Vs  VD1  R.( Is 2  Is1) 
Vs R
VD1 
 .(Is 2  Is1);
2
2
case2 : VD1  VD 2  Vs 2
VD1 VD 2
VD1. Rs 2


 ( Is 2  Is1)  Rs1 
Rs1
Rs 2
VD 2  Rs 2.(Is 2  Is1)
Is1 
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
Vs R
7000 80000
 .(Is 2  Is1) 

.(50  40).10 3  3900V
2
2
2
2
VD 2  Vs  VD1  3100V
VD1 
case2 : VD1  VD 2  Vs 2  3500V
3500.70000
 Let Rs 2  70k  Rs1 
 58.34k

3
3500 70000.10.10
 Rs1  58.34k; and Rs 2  70k
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
VD1  V 3  VD 2  V 4;
VD1  Rs 3.ID1  VD 2  Rs 4.ID 2
Let Rs1  Rs 2  R  R (ID 2  ID1)  VD1  VD 2
VD1  VD 2
R 
ID 2  ID1
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:
ID  ID1  ID 2  ID 2  ID  ID1  95  50  45A
VD1  VD 2
1.8  2
R 

 20m
ID 2  ID1
45  50
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:
Mode#1 D1 conducts: The current will flow from the source
to the load through D1, where the current i1:
di1
Vs
t
Vs  R.i1  L
i1(t ) 
(1  e  )
dt
R
VL (t )  Vs.e
t

;  L / R
Mode#2 Dm conducts: The current will continue to flow through the load and
the diode Dm., where the current i2:
di1
Vs
t
0  R.i1  L
i1( t ) 
(1  e  )
dt
R
Vs
At t  t1  I  I1 
;
R
i 2( t )  I1.e
t

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:
di
Vs
2 0 0.1 06
Vs  Ls
 Ls 

 4H
di
dt
50
dt
di
Vs
Irr  trr.
 trr.
dt
Ls
Vs
200
Vs
Io 

 2A  Ip  Io  trr .

R
100
Ls
200
 Ip  2  1 0.1 0 6
 5 0 2A
4.1 0 6
The excess energy due to current increasing is determined as:




1
1
2
2
WR 
Ip  Io .Ls 
5022  2 2 .4.106  0.504J
2
2
1
2.WR
WR  WC  Cs.Vc 2  Cs 

2
2
VC
2 * 0.504
 Cs 
 25.2 F
2
200
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.
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