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Driving AC relays with a zero-crossing opto-triac.

Contents


Introduction.

To start with the beginning - driving AC relays with a zero-crossing opto-triac is generally not a good idea, but sometimes you have no choice.
The low power factor of some AC relays can in some cases prevent one half of the triac to turn on so the relay is driven with a half-wave rectified AC voltage.
The relay will most likely pull in, but the contact pressure is too low, so if it is used near its rated load current it will probably not last long.
You can find a spread-sheet with the calculations here.


Warning.

This circuit is most likely connected directly to the mains voltage.
Unless you are qualified to work with mains powered circuits, do not build them. They can kill you.
All mains wiring must be done in accordance with your local safety standards.


AC relay power factor correction.

To calculate or simulate a relay, we need its DC resistance and self-induction with nominal coil voltage.
These are normally not specified in the data-sheet, but can be found from published data.

AC relay with power factor correction.
Fig.1: AC relay with power factor correction.

Table 1: List of variables used in this calculation.
ffrequency [ Hz ]
VACSupply voltage [ V ]
I1Relay current [ A ]
R1Relay DC resistance [ Ω ]
L1Relay inductance ( at nominal coil voltage ) [ H ]
Z1Relay impedance [ Ω ]
ZL1L1 impedance [ Ω ]
P1Relay power dissipation [ W ]
VA1Relay apparent power [ VA ]
PF1Relay power factor [ W / VA ]
Ph1Relay current phase with reference to voltage ( current is lagging voltage ) [ rad ]

When f, VAC, R1 and I1 are specified in the data-sheet:
Z1=VAC/I1.
ZL1=SQRT(Z1^2-R1^2).
L1=ZL1/(2*PI*f).

When f, VAC, P1 and VA1 are specified in the data-sheet:
I1=VA1/VAC.
R1=P1/I1^2.
Z1=VAC/I1.
ZL1=SQRT(Z1^2-R1^2).
L1=ZL1/(2*PI*f).

When f, VAC, R1 and VA1 are specified in the data-sheet:
I1=VA1/VAC.
Z1=VAC/I1.
ZL1=SQRT(Z1^2-R1^2).
L1=ZL1/(2*PI*f).

Additional equations:
P1=I1^2*R1.
VA1=VAC*I1.
PF1=P1/VA1.
PF1=R1/Z1.
Ph1=acos(PF1).

For unity power factor, the impedance of R2, C2 must be the same as R1, L1, so:
R2=R1.
C2=1/(2*PI*ZL1*f).
With these values R2 will dissipate the same amount of power as R1 and this is obviously a waste of power.
R2 can be reduced by a factor of 50..100 without reducing the power factor very much.
If you aim for a power factor around 0.95, R2 will have a reasonable value. Use the spread-sheet to calculate this.
The minimum value of R2 is set by the peak current the triac driving the relay will tolerate.
C2 must be an X-2 rated type.


Appendix A: Relay measurement.

Out of curiosity, I measured an AC relay.
The relay is a Goodsky EMI-SH-1115A:
Voltage: 115 VAC
Coil resistance: 8100 Ω
Coil current: 7.65 mA

Current vs. voltage plot for AC relay.
Fig.A1: Current vs. voltage plot for AC relay.

The lime plot is for the voltage going from 0 V to 145 V and the blue from 145 V to 0 V.
Both coil self-heating and hysteresis are visible.
In the voltage range from 15 V to 35 V the relay was buzzing audibly.
The following 3 plots are calculated from this measurement.
Coil heating was not included in the calculations ( coil resistance for calculations is 8100 Ω ).

Impedance vs. voltage plot for AC relay.
Fig.A2: Impedance vs. voltage plot for AC relay.

Self-induction vs. voltage plot for AC relay.
Fig.A3: Self-induction vs. voltage plot for AC relay.

Stored energy vs. voltage plot for AC relay.
Fig.A4: Stored energy vs. voltage plot for AC relay.


References.

[1] IL410 data-sheet.

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