125
115
105
95
85
75
2015107.05.04.03.0
2.0
T
R
, REFERENCE TEMPERATURE (
°
C)
V
R
, DC REVERSE VOLTAGE (VOLTS)
Figure 1. Maximum Reference Temperature
1N5817
40 30 23
60
80
R
q
JA
(°C/W) = 110
125
115
105
95
85
75
2015107.05.0 304.03.0
40
30
23
R
q
JA
(°C/W) = 110
80
60
Figure 2. Maximum Reference Temperature
1N5818
125
115
105
95
85
75
2015107.05.0 304.0 40
R
q
JA
(°C/W) = 110
60
80
Figure 3. Maximum Reference Temperature
1N5819
40
30
23
T
R
, REFERENCE TEMPERATURE ( C)
°
V
R
, DC REVERSE VOLTAGE (VOLTS)
V
R
, DC REVERSE VOLTAGE (VOLTS)
T
R
, REFERENCE TEMPERATURE (
°
C)
1N5817, 1N5818, 1N5819
http://onsemi.com
3
NOTE 3. — DETERMINING MAXIMUM RATINGS
Reverse power dissipation and the possibility of thermal
runaway must be considered when operating this rectifier at
reverse voltages above 0.1 V
RWM
. Proper derating may be
accomplished by use of equation (1).
A(max)
where T
A(max)
=
T
J(max)
=
P
F(AV)
=
P
R(AV)
=
R
q
=
T
J(max)
− R
q
JA
P
F(AV)
− R
q
JA
P
R(AV)
Maximum allowable ambient temperature
Maximum allowable junction temperature
Average forward power dissipation
(125°C or the temperature at which thermal
runaway occurs, whichever is lowest)
Average reverse power dissipation
Junction−to−ambient thermal resistance
Figures 1, 2, and 3 permit easier use of equation (1) by
taking reverse power dissipation and thermal runaway into
consideration. The figures solve for a reference temperature
as determined by equation (2).
R
J(max)
q
JA
R(AV)
ubstituting equation (2) into equation (1) yields:
T
= T
− R
q
P
(
Inspection of equations (2) and (3) reveals that T
R
is the
ambient temperature at which thermal runaway occurs or
where T
J
= 125°C, when forward power is zero. The
transition from one boundary condition to the other is
evident on the curves of Figures 1, 2, and 3 as a difference
in the rate of change of the slope in the vicinity of 115°C. The
data of Figures 1, 2, and 3 is based upon dc conditions. For
use in common rectifier circuits, Table 1 indicates suggested
factors for an equivalent dc voltage to use for conservative
design, that is:
(4)
V
R(equiv)
= V
in(PK)
x F
The factor F is derived by considering the properties of the
various rectifier circuits and the reverse characteristics of
Schottky diodes.
EXAMPLE: Find T
A(max)
for 1N5818 operated in a
12−volt dc supply using a bridge circuit with capacitive filter
such that I
DC
= 0.4 A (I
F(AV)
= 0.5 A), I
(FM)
/I
(AV)
= 10, Input
Voltage = 10 V
(rms)
, R
qJA
= 80°C/W.
R(equiv)
. Read F = 0.65 from Table 1,
Step 1. Find ∴ V
R(equiv)
= (1.41)(10)(0.65) = 9.2 V.
Step 2. Find T
R
from Figure 2. Read T
R
= 109°C
Step 1. Find @ V
R
= 9.2 V and R
q
JA
= 80°C/W.
Step 3. Find P
F(AV)
from Figure 4. **Read P
F(AV)
= 0.5 W
@
I
(FM)
I
(AV)
= 10 and IF(AV) = 0.5 A.
Step 4. Find T
A(max)
from equation (3).
Step 4. Find T
A(max)
= 109 − (80) (0.5) = 69°C.
*Values given are for the 1N5818. Power is slightly lower for the
N5817 because of its lower forward voltage, and higher for the
N5819.
Circuit
Load
Half Wave
Resistive Capacitive*
Full Wave, Bridge
Resistive Capacitive
Full Wave, Center Tapped*†
Resistive Capacitive
Sine Wave
Square Wave
0.5
0.75
1.3
1.5
0.5
0.75
0.65
0.75
1.0
1.5
1.3
1.5
**Note that V
R(PK)
≈ 2.0 V
in(PK)
.
†Use line to center tap voltage for V
in
.
Table 1. Values for Factor F
