AN5370 Application Note
AN5370
Using Avalanche Diodes Without Sharing Capacitors
Application Note
Replaces July 2000, version AN5370-1.1
AN5370-1.2 July 2002
1. INTRODUCTION
The important property of the avalanche diode is its ability to
safely handle, without damage, relatively large reverse powers
levels. Normally diodes are limited to a reverse current of the
order of mA at room temperature and mA at elevated temperatures
but with avalanche diodes reverse currents of greater than 1 A
are possible.
The reverse characteristic of an avalanche diode is compared
in figure 1 with the characteristic of a normal diode. The
avalanche diode has low leakage current up to the avalanche
voltage and does not show the same increase in leakage current
as the normal diode. The silicon of avalanche diodes is regular
across the cross section therefore the majority of the reverse
current is conducted within the bulk of the silicon. In a normal
diode the majority of the reverse current is conducted close to
the edge of the silicon where the electric field strength is highest.
As a consequence of this, the reverse power handling capability
of the avalanche diode is expressed in kW for rectangular shaped
pulses of defined width. Normal diodes may conduct the reverse
current in a very small localised area so the reverse current has
to be limited.
The avalanche rating curve is shown in fig 2 for the MZ0409W,
a 2A, 1400V Vrrm (repetitive rating) and 1500V V(AB)R (avalanche
voltage) diode. The curve is valid for a case temperature of 25oC
and 150oC. The permissible avalanche power is 9kW for a
rectangular pulse of 10us and 0.9kW at 1 millisecond width.
Provided the diode is operated within the limits specified on the
data sheet, the excursion into the avalanche region is non
destructive.
2. DESIGN VOLTAGE
Avalanche diodes are used in series to make high voltage
strings. Due to their avalanche capability they can be used
without static sharing resistors or dynamic sharing capacitors.
For these strings, it is customary to allow a voltage safety margin
of 2.4 to 3 between the rated repetitive voltage of the string and
100
Duty Cycle
<0.001% Square Wave Pulse
Avalanche voltage
Instantaneous reverse power - (kW)
Reverse voltage
10
Normal diode
Reverse current
Avalanche diode
1
Tj = 25˚C
Tj = 150˚C
0.1
1
Fig.1 Reverse characteristics
10
100
1000
Surge duration - (µs)
10000
Fig.2 Non-repetitive peak reverse power
1/5
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AN5370 Application Note
the peak reverse applied voltage (PIV) . If N is the number of
series diodes and VRRM is the reverse repetitive voltage of the
diode then
N x VRRM = 2.4 x PIV (2.4 safety margin)....................(1)
In the case of bridge rectifiers
PIV = K x 1.57 x UDC ( single phase).........................(2)
PIV = K x 1.05 x UDC (Three phase)..........................(3)
Where UDC is the DC voltage and K is the mains overvoltage
factor typically 5%.
Using the above formula with K equal to 1.05 the number in
series is
N = 3.96 x UDC
This is inevitable given that any production contains diodes
having a spread of characteristics and that diodes do not
normally have to be selected. The only criteria for avalanche
diodes used in a string is that they should have the same
repetitive voltage rating. This means that the leakage current at
the repetitive voltage will be less than the value given in the data
sheet.
When the diodes are connected in series without sharing
components there is a unique leakage current such that the sum
of the voltages across the individual diodes will be equal to the
applied voltage. According to the design rules given above the
design voltage per diode, assuming they share equally could be
Vdiode = VRRM
…………….......................................(6)
2.4
VRRM
(Single phase)..........................(4)
N = 2.56 x UDC
VRRM
(Three phase)...........................(5)
3. VOLTAGE SHARING
a) Static
A series string of avalanche diodes will contain diodes with
different reverse leakage and differing avalanche voltage levels.
That is 416V for a 1000V diode and 583V for a 1400V diode.
The distribution of voltage between the diodes will depend upon
the applied voltage, the number of diodes and the leakage
characteristics ( these are temperature dependent)
The design voltage is considerably less than the avalanche
voltage however, it is conceivable that diodes of such extreme
characteristics could be in series causing one or more diodes to
avalanche during the reverse blocking of normal operation. This
is not serious because the current level will be of the order of
milliamps. If the equipment has been designed according to the
V(AB) min
V(AB) min
Reverse voltage
IF
Imax
Spread of
leakage currents
Commutation dI/dt
Spread of
avalanche voltages
Reverse current
Stored charge
Reverse applied
voltage
Imax = Maximum reverse current in a series string
V(AB) min = Minimum avalanche voltage
Fig.3 Spread of reverse characteristics
Fig.4 Commutation waveforms
2/5
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AN5370 Application Note
30
Tcase = 150˚C
25
Stored charge Qrr - (µC)
IF
IF = 1A
IF = 2A
IF = 3A
IF = 5A
IF =10A
IF = 20A
20
T2
T1
0
IR1
Fast diode
recovers
15
Slow diode
recovers
IR2
a) current
P2
10
P1
TA
5
(T2–T1)
b) avalanche power in fast diodes
0
1
10
Rate of decay of forward current - (A/µs)
100
Fig.5 Stored charge (max)
Fig.6 Series diodes
criteria above it is not possible for all diodes to avalanche,
therefore the current in the series string will always be less than
the current at which the diode with the highest leakage current
enters the avalanche region. ( see fig. 3)
current is assumed. A possible condition exists where the
applied voltage is sufficiently high for the fast diodes to be
subjected to the avalanche voltage at snap-off (time T1). The fast
diode starts the avalanche regime at time T1 with a current IR1
and finishes the regime at time T2 with a current IR2. The
instantaneous avalanche power at IR1 and IR2 is P1 and P2
respectively.
The fact that avalanche diodes in a series string can be allowed
to operate with voltage sharing determined by the diodes, makes
it unnecessary to use static sharing resistors.
b) Dynamic
The dynamic sharing is concerned with the reverse voltage
appearing across the series string when the current in the diodes
is being commutated. The diodes conduct a small current in the
reverse direction (fig. 4) at the end of the commutation and when
the current snaps off a voltage is induced in the commutating
circuit imposing a reverse voltage across the diodes.
The integral of the reverse current is called the stored charge and
is expresses in micro-coulombs. The limit case curve for the
MZ04xxW diode is shown in figure 5. The stored charge is
dependent upon the commutating current, the rate of commutation
and the diode junction temperature. Diodes taken from production
will show a spread of stored charge, the fast diode having a low
stored charge and the slow diode having a high stored charge.
The reverse currents of the fast and slow diodes could be as
shown in fig.6. In order to simplify the analysis, a triangular
current with an instantaneous snap-off at the peak reverse
P1 = IR1 x V(AB)R + ∆1)
P2 = IR2 x (V(AB)R + ∆2)
..............................................(7)
...............................................(8)
The ∆ values represent the increase in voltage in the avalanche
region. From the analysis given in the appendix it follows that the
avalanche rating of the diodes in a series string, if no dynamic
sharing capacitors are used, must be :
Avalanche Power = IR2 x (V(AB)R + ∆2),
Watts
...........(9)
for a time
TA = √(1/(2.di/dt) x (√Qmax - Qmin / √Qmax), µsec. .....(10)
Where Qmax & Qmin are the maximum and minimum diode
stored charge in µC and di/dt is the rate of change of commutation
current in Amps/µs.
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AN5370 Application Note
where Z is the transient thermal impedance and T = T2 – T1
2A
3A/µs
The junction temperature calculated for the rectangular pulse is,
1.82µs
1.3µs
TA
TJ(TA) = ∫P2.(dZ(TA–t)/dt).dt
0
.................(18)
0
3.9A
The transient thermal resistance for the short times encountered
( i.e. microseconds) in this analysis can be assumed to have a
derivative equal to a constant b.
5.46A
The two equations give :
a) current
8.2kW
TJ(T2) = b(T2-T1).(P1 + P2)
5.8kW
...……………………(19)
2
0.46µs
0.52µs
b) avalanche power in fast diodes
TJ(TA) = bTAP2
........………….........(20)
Equating equations 18 and 19 to solve for TA and using equations
11 to 16 gives,
TA =√(I.(di/dt)).[(√(Qmax ) - (Qmin / √Qmax)]
.........(21)
Fig.7 Series diodes
APPENDIX 2
APPENDIX 1
Worked Example
Avalanche Power during reverse recovery
The analysis considers the waveform discussed in fig. 6 and
assumes that the voltage drop in the avalanche region is negligible
compared with the avalanche voltage.
From the stored charge waveform we have :
t1 = √(2.Qmin/(di/dt))
............................................(11)
t2 = √(2.Qmax/(di/dt))
............................................(12)
IR1 = (di/dt) x t1
............................................(13)
IR2 = (di/dt) x t2
............................................(14)
P1 = IR1 x V(AB)R
............................................(15)
P2 = IR2 x V(AB)R ..........................................................(16)
The avalanche rating in the data sheets is normally for a
rectangular current pulse so it is convenient to convert the
avalanche power waveform into a rectangular shape. This can
be done by considering a rectangular pulse of power P2 and
width TA. The width TA is calculated so that the junction
temperature after time TA is the same as that occurring during
the actual pulse.
If it is assumed that the avalanche power varies linearly between
P1 and P2 then the temperature at time T2 is,
T
TJ(T2) = ∫{P1 +( (P2-P1)/T)t}.(dZ(T–t)/dt).dt
0
.........(17)
The example considers the diode MZ0414W whose avalanche
power rating and stored charge are given in figs 2 and 5
respectively. The avalanche power and pulse width TA, which
are required if no sharing capacitors are to be used, will be
calculated using the equations of Appendix 1.
It will be assumed that the commutation current is 2A and the rate
of the commutation is 3A/µs. Fig 5 shows a maximum stored
charge of 5µC. A general rule for stored charge distribution is that
Qmax = 2 x Qmin. so that in this case we can assume that Qmin
= 2.5µC.
Using equations 11 to 16 we have
t1 = √(2 x2.5/ 3) = 1.73µs
t2 = √(2 x 5/3) = 1.82µs
IR1 = 3.87A
IR2 = 5.46A
P1 = 3.87A x 1500V = 5.8kW
P2 = 5.46A x 1500V = 8.2kW
TA = (√1/(3 x 2)) x (√5 - 2.5 / √5) = 0.46µs
The current and power waveforms are shown in figure 7.
Therefore, no capacitors are required if the avalanche power of
8.2kW for 0.52µs is within the diode rating. Reference to figure
2 indicates that the diode has a power rating of 9kW at 1µs and
a Tj of 150 oC and therefore will be capable of handling 8.2kW for
0.5µs.
4/5
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