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Document Number 83690

Rev. 1.4, 10-Jan-05

BRT21/ 22/ 23

Vishay Semiconductors

Coupler

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.

Parameter

Test condition

Symbol

Min

Typ.

Max

Unit

Critical rate of rise of coupled

input/output voltage

10000

V/

µs

Common mode coupling

capacitance

0.01

pF

Capacitance (input-output)

0.8

pF

Isolation resistance

Ω

Ω

Trigger current

1.2

mA

2.0

mA

3.0

mA

Figure 1. Shunt Capacitance vs. Load Current

iil410_01

400

350

300

250

200

150

100

50

0

.001

.01

.1

1

IL - Load Current - mA(RMS)

Cs(

µF) = 0.0032 (µF)* 10^(0.0066IL (mA)

Ta = 25°C, PF = 0.3

IF = 2.0 mA