Datasheet

C
2 A
R = 10 Ω 1 = 3.33 Ω
1.5 A
æ ö
´ -
ç ÷
è ø
OFF-PK
C G
ON-PK
I
R = R 1
I
æ ö
´ -
ç ÷
è ø
CC2 EE-P
G
OFF-PK
V V
R =
I
-
CC2 EE-P
C G
ON-PK
V V
R = R
I
-
-
V
CC2
V
C
V
OUT
V
EE-P
ISO5500
15 V
R
G
C
G
15 V
I
on-pk
V
C
C
2
-
V
E
E
-
P
V
CC2
V
C
V
OUT
V
EE-P
ISO5500
15 V
R
G
C
G
15 V
I
off-pk
V
CC2
- V
EE-P
V
E
V
E
R
C
ISO5500
www.ti.com SLLSE64C –SEPTEMBER 2011–REVISED JUNE 2013
DETERMINING COLLECTOR RESISTOR, R
C
Despite equal charge and discharge currents, many power devices possess longer turn-off propagation and fall
times than turn-on propagation and rise times. In order to compensate for the difference in switching times, it
might be necessary to significantly reduce the charge current, I
ON-PK
, versus the discharge current, I
OFF-PK
.
Reducing I
ON-PK
is accomplished by inserting an external resistor, R
C
, between the V
C
- pin and the V
CC2
- pin of
the ISO5500.
Figure 71. Reducing I
ON-PK
by Inserting Resistor R
C
Figure 71 (right) shows that during the on-transition, the (V
CC2
– V
EE-P
) voltage drop occurs across the series
resistance of R
C
+ R
G
, thus reducing the peak charge current to: I
ON-PK
= (V
CC2
– V
EE-P
) /(R
C
+ R
G
). Solving for R
C
provides:
(10)
To stay below the maximum output power consumption, R
G
must be calculated first via:
(11)
and the necessary comparison of P
OL-WC
versus P
OL
must be completed.
Once R
G
is determined, calculate R
C
for a desired on-current using Equation 10.
Another method is to insert Equation 11 into Equation 10 and arriving at:
(12)
Example
Reducing the peak charge current from the previous example to I
ON-PK
= 1.5 A, requires a R
C
value of:
(13)
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