= Instantaneous forward voltage (pw = 300s, D = 2%)
I
R
= Instantaneous reverse current.
t
rr
= Reverse recovery time (See Figure 9), summation of t
a
+ t
b
.
t
a
= Time to reach peak reverse current (See Figure 9).
t
b
= Time from peak I
RM
to projected zero crossing of I
RM
based on a straight line from peak I
RM
through 25% of I
RM
(See Figure 9).
Q
RR
= Reverse recovery charge.
C
J
= Junction Capacitance.
R
JC
= Thermal resistance junction to case.
pw = pulse width.
D = Duty cycle.
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2
RHRP3060 — Hyperfast Diode
Typical Performance Characteristics
Figure 1. Forward Current vs Forward Voltage
300
Figure 2. Reverse Currnt vs Reverse Voltage
2000
175C
I
F
, FORWARD CURRENT (A)
100
100C
I
R
, REVERSE CURRENT (μA)
100
100C
10
10
175C
25
o
C
1
0.1
25C
1
0
1
2
V
F
, FORWARD VOLTAGE (V)
3
4
0.01
0
100
200
300
400
500
600
V
R
, REVERSE VOLTAGE (V)
Figure 3. t
rr
, t
a
and t
b
Curves vs
Forward Current
50
Figure 4. t
rr
, t
a
and t
b
Curves vs
Forward Current
100
T
C
= 25C, dl
F
/dt = 200A/s
t, RECOVERY TIMES (ns)
t, RECOVERY TIMES (ns)
T
C
= 100C, dl
F
/dt = 200A/s
40
t
rr
80
30
60
t
rr
20
t
a
t
b
40
t
a
t
b
10
20
0
1
10
I
F
, FORWARD CURRENT (A)
30
0
1
10
I
F
, FORWARD CURRENT (A)
30
Figure 5. t
rr
, t
a
and t
b
Curves vs
Forward Current
150
Figure 6. Current Derating Curve
I
F(AV)
, AVERAGE FORWARD CURRENT (A)
30
25
20
SQ.WAVE
T
C
= 175C, dl
F
/dt = 200A/s
125
t, RECOVERY TIMES (ns)
100
75
t
a
t
b
t
rr
DC
15
10
5
0
75
100
125
150
175
T
C
, CASE TEMPERATURE (
o
C)
50
25
0
1
10
I
F
, FORWARD CURRENT (A)
30
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3
RHRP3060 — Hyperfast Diode
Typical Performance Characteristics
Figure 7. Junction Capacitance vs
Reverse Voltage
150
C
J
, JUNCTION CAPACITANCE (pF)
125
100
75
50
25
0
0
50
100
150
(Continued)
200
V
R
, REVERSE VOLTAGE (V)
Test Circuit and Waveforms
Figure 8. t
rr
Test Circuit
V
GE
AMPLITUDE AND
R
G
CONTROL dI
F
/dt
t
1 AND
t
2
CONTROL I
F
Figure 9. t
rr
Waveforms and Definitions
L
DUT
R
G
V
GE
CURRENT
SENSE
+
I
F
0
dI
F
dt
ta
trr
tb
t
1
t
2
IGBT
-
V
DD
0.25 I
RM
I
RM
Figure 10. Avalanche Energy Test Circuit
I = 1A
L = 40mH
R < 0.1Ω
E
AVL
= 1/2LI
2
[V
R(AVL)
/(V
R(AVL)
- V
DD
)]
Q
1
= IGBT (BV
CES
> DUT V
R(AVL)
)
CURRENT
SENSE
Q
1
V
DD
DUT
L
R
+
V
DD
Figure 11. Avalanche Current and Voltage
Waveforms
V
AVL
I
L
I V
I
L
-
t
0
t
1
t
2
t
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4
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