BRT21 Datasheet, PDF (4/9 Pages) Siemens Semiconductor Group – SITACO AC Switches Without Zero Voltage Switch With Zero Voltage Switch

BRT21/ 22/ 23
Vishay Semiconductors
Coupler
Parameter
Test condition
Symbol
Min
Typ.
Max
Critical rate of rise of coupled
input/output voltage
I_T = 0 A, V_RM = V_DM = V_D(RMS)
dv_IO/dt
10000
Common mode coupling
capacitance
C_CM
0.01
Capacitance (input-output)
f = 1.0 MHz, V_IO = 0 V
C_IO
0.8
Isolation resistance
V_IO = 500 V, T_amb = 25 °C
R_is
≥ 10¹²
V_IO = 500 V, T_amb = 100 °C
R_is
≥ 10¹¹
Trigger current
V_D = 5.0 V, F - Versions
I_FT
1.2
V_D = 5.0 V, H - Versions
I_FT
2.0
V_D = 5.0 V, M - Versions
I_FT
3.0
Unit
V/µs
pF
pF
Ω
Ω
mA
mA
mA
Power Factor Considerations
A snubber isn’t needed to eliminate false operation of
the TRIAC driver because of the high static and com-
mutating dv/dt with loads between 1.0 and 0.8 power
factors. When inductive loads with power factors less
than 0.8 are being driven, include a RC snubber or a
single capacitor directly across the device to damp
the peak commutating dv/ dt spike. Normally a com-
mutating dv/dt causes a turning-off device to stay on
due to the stored energy remaining in the turning-off
device.
But in the case of a zero voltage crossing optotriac,
the commutating dv/dt spikes can inhibit one half of
the TRIAC from turning on. If the spike potential
exceeds the inhibit voltage of the zero cross detection
circuit, half of the TRIAC will be heldoff and not turn-
on. This hold-off condition can be eliminated by using
a snubber or capacitor placed directly across the
optotriac as shown in Figure 1. Note that the value of
the capacitor increases as a function of the load cur-
rent.
The hold-off condition also can be eliminated by pro-
viding a higher level of LED drive current. The higher
LED drive provides a larger photocurrent which
causes the phototransistor to turn-on before the com-
mutating spike has activated the zero cross network.
Figure 2 shows the relationship of the LED drive for
power factors of less than 1.0. The curve shows that
if a device requires 1.5 mA for a resistive load, then
1.8 times 2.7 mA) that amount would be required to
control an inductive load whose power factor is less
than 0.3.
1
Cs(µF) = 0.0032 (µF)* 10^(0.0066IL (mA)
.1
Ta = 25°C, PF = 0.3
.01
IF = 2.0 mA
.001
0
iil410_01
50 100 150 200 250 300 350 400
IL - Load Current - mA(RMS)
Figure 1. Shunt Capacitance vs. Load Current
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Document Number 83690
Rev. 1.4, 10-Jan-05