HCNW3120 26. HCPL-3120 typical application circuit with negative IGBT gate drive. Application Circuit

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HCNW3120 circuit
Selecting  the  Gate  Resistor  (Rg)  to  Minimize  IGBT 
Switching  Losses.  (Discussion  applies  to  HCPL-3120
HCPL-J312 and HCNW3120
Step 1: Calculate Rg Minimum from the I
OL
 Peak Specifica-
tion. The IGBT and Rg in Figure 26 can be analyzed as a 
simple  RC  circuit  with  a  voltage  supplied  by  the  HCPL-
3120.
  (V
CC
 – V
EE
 - V
OL
) 
Rg  ≥  ———————  
      I
OLPEAK 
(V
CC
 – V
EE
 - 2 V) 
=  ———————  
   
        I
OLPEAK 
  (15 V + 5 V - 2 V) 
=  ———————  
             2.5 A 
=  7.2 Ω @ 8 Ω
The V
OL
 value of 2 V in the previous equation is a con-
servative value of V
OL
 at the peak current of 2.5A (see 
Figure 6). At lower Rg values the voltage supplied by 
the HCPL-3120 is not an ideal voltage step. This results 
in lower peak currents (more margin) than predicted by 
this analysis. When negative gate drive is not used V
EE
 in 
the previous equation is equal to zero volts.
Step 2: Check the HCPL-3120 Power Dissipation and 
Increase Rg if Necessary. The HCPL-3120 total power 
dissipation (P
T
) is equal to the sum of the emitter power 
(P
E
) and the output power (P
O
):
P
T
 = P
E
 + P
O
PE = I
F 
• V
F 
· Duty Cycle
P
O
 = P
O(BIAS)
 + P
O (SWITCHING)
    = I
CC
 • (V
CC
 - V
EE
)+ E
SW
(R
G
, Q
G
) • f
For  the  circuit  in  Figure  26  with  I
F
  (worst  case)  = 
16 mA,  Rg  =  8 Ω,  Max  Duty  Cycle  =  80%,  Qg  =  500  nC, 
f = 20 kHz and T
A
 max = 85 °C:
P
E
 = 16 mA • 1.8 V • 0.8 = 23 mW
P
O
 = 4.25 mA • 20 V + 5.2 µ J • 20 kHz
   = 85 mW + 104 mW
   = 189 mW > 178 mW (P
O(MAX)
 @ 85°C
   = 250 mW
-
15C*4.8 mW/C)
The value of 4.25 mA for I
CC
 in the previous equation was 
obtained by derating the I
CC
 max of 5 mA (which occurs 
at -40°C) to I
CC
 max at 85C (see Figure 7).
Since P
O
 for this case is greater than P
O(MAX)
, Rg must be 
increased to reduce the HCPL-3120 power dissipation.
P
O(SWITCHING MAX)
   
  = P
O(MAX)
 - P
O(BIAS)
   = 178 mW - 85 mW
  = 93 mW
   
E
SW(MAX)
   
   
   
   
   P
O(SWITCHINGMAX)
=  ——————— 
         f 
  93 mW 
=  ———— = 4.65 µW 
   20 kHz
For Qg = 500 nC, from Figure 27, a value of E
SW
 = 4.65 µW 
gives a   Rg = 10.3 Ω.
+5 V
1
270
2
CONTROL
INPUT
74XXX
OPEN
COLLECTOR
8
0.1 µF
7
+
V
CC
= 15 V
+ HVDC
Rg
Q1
+
V
EE
= -5 V
3
6
AC
4
5
Q2
- HVDC
Figure 26. HCPL-3120 typical application circuit with negative IGBT Gate drive.
HCPL-3120 fig 26
19