Document Number: 83628

For technical questions, contact: optocoupler.answers@vishay.com

www.vishay.com

Rev. 1.6, 09-Jan-08

5

IL4116/IL4117/IL4118

Optocoupler, Phototriac Output,

Zero Crossing, High dV/dt, Very Low

Input Current

Vishay Semiconductors

Power Factor Considerations

A snubber isn’t needed to eliminate false operation of the

TRIAC driver because of the IL4116/IL4117/IL4118 high

static and commutating dV/dt with loads between 1 and 0.8

power factors. When inductive loads with power factors less

than 0.8 are being driven, include an RC snubber or a single

capacitor directly across the device to damp the peak

commutating dV/dt spike. Normally a commutating dV/dt

causes a turning-off device to stay on due to the stored

energy remaining in the turn-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 held-off 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 7. Note that the value

of the capacitor increases as a function of the load current.

The hold-off condition also can be eliminated by providing a

higher level of LED drive current. The higher LED drive

provides

a

larger

photocurrent

which

causes.

The

phototransistor to turn-on before the commutating spike has

activated the zero cross network. Figure 8 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.

Fig. 7 - Shunt Capacitance vs. Load Current vs. Power Factor

Fig. 8 - Normalized LED Trigger Current

iil4116_07

400

350

300

250

200

150

100

50

0

I - Load Current (mA)

L

0.001

0.01

0.1

1

C (µF) = 0.0032 (µF) x 10 ^ (0.0066 I (mA))

S

L

PF = 0.3

iil4116_08

1.2

1.0

0.8

0.6

0.4

0.2

0

PF - Power Factor

0.8

1.2

1.6

2.0

1.0

1.4

1.8

I

Normalized to I

at PF = 1.0

Fth

Fth