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參數(shù)資料
| 型號(hào)： | AN5370 |
| 英文描述： | Using avalanche diodes without sharing capacitors |
| 中文描述： | 使用不共享電容器雪崩二極管 |
| 文件頁數(shù)： | 4/5頁 |
| 文件大小： | 82K |
| 代理商： | AN5370 |
www.dynexsemi.com
AN5370 Application Note
4/5
Fig.7 Series diodes
1.82
μ
s
1.3
μ
s
0.46
μ
s
0.52
μ
s
0
2A
3.9A
5.46A
5.8kW
8.2kW
b) avalanche power in fast diodes
a) current
3A/
μ
s
APPENDIX 1
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 :
t
1
=
√
(2.Qmin/(di/dt))
t
2
=
√
(2.Qmax/(di/dt))
I
R1
= (di/dt) x t1
I
R2
= (di/dt) x t2
P1 = I
R1
x V
(AB)R
P2 = I
R2
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.
............................................(11)
............................................(12)
............................................(13)
............................................(14)
............................................(15)
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
.........(17)
0
where Z is the transient thermal impedance and T = T2
–
T1
The junction temperature calculated for the rectangular pulse is,
TA
TJ(TA) =
∫
P2.(dZ(TA
–
t)/dt).dt
.................(18)
0
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.
The two equations give :
TJ(T2) = b(T2-T1).(P1 + P2)
...
……………………
(19)
2
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)
APPENDIX 2
Worked Example
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/
μ
s. Fig 5 shows a maximum stored
charge of 5
μ
C. A general rule for stored charge distribution is that
Qmax = 2 x Qmin. so that in this case we can assume that Qmin
= 2.5
μ
C.
Using equations 11 to 16 we have
t
1
=
√
(2 x2.5/ 3) = 1.73
μ
s
t2 =
√
(2 x 5/3) = 1.82
μ
s
I
R1
= 3.87A
I
R2
= 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) = 0.46
μ
s
The current and power waveforms are shown in figure 7.
Therefore, no capacitors are required if the avalanche power of
8.2kW for 0.52
μ
s is within the diode rating. Reference to figure
2 indicates that the diode has a power rating of 9kW at 1
μ
s and
a T
j
of 150
o
C and therefore will be capable of handling 8.2kW for
0.5
μ
s.
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