Driver Saves Power In Energized Relay

March 22, 2011
Relay driver switches from full-wave to half-wave rectification after relay pulls in in order to save power.

The addition of a components and an unused NC contact to a dc supply provides a power-reducing feature to a relay driver circuit.

Electromagnetic relays need much less power for holding than they do to get energized. To take advantage of this fact by reducing the driving voltage of the energized relay to save power and reduce coil heating, many engineers have designed circuits to reduce the driving voltage of a relay once it’s energized.

Often, however, such circuits suffer from drawbacks such as excessive complexity or the waste of power saved in the relay coil in some other component, such as the relay driving transistor. The simple circuit described shown in the figure avoids those drawbacks. It needs only two diodes, a capacitor and a spare “normally closed” (NC) relay contact of the relay being driven. More importantly, it does not waste any power.

The relay driver design leverages resources available in the unregulated power circuit for a main system not shown here. A transformer with 9-0-9-V ac secondary voltages, diodes D1 and D2, and capacitor C1 form the power circuit, and their ratings depend on the main system circuit’s requirements.

The relay driver design adds diodes D3 and D4, filter capacitor C2, and a spare NC contact of the relay being driven. In this design, the relay has a 12-V dc specification and a coil resistance of about 150 Ω. The relay gets energized at around 8 V, and its holding voltage is about 3 V.

When the relay isn’t energized, the no-load voltage across C2 is about 12 V. The voltage across C2 will drop to about 8.7 V (due to load and chosen capacitance value) when driving transistor T1 receives the relay energizing command and begins conducting. The relay will readily energize and hold at this voltage, but energizing the relay will open the NC contact and remove D4 from the circuit.

Removing D4 converts the power to the relay driver from a full-wave to a half-wave source. This conversion further reduces the voltage across C2 to about 5.5 V, which is still sufficiently above the relay’s holding voltage to ensure reliable operation. This two-stage voltage reduction, from 12 to 8.7 to 5.5 V, significantly reduces the coil’s power consumption and heating without wasting power anywhere else in the circuit.

Tests on a prototype showed no chattering of the relay contacts at switch-over. Also, the main circuit power remained unaffected. For more information about this circuit, check out Anoop’s Analysis with the online version of this article at

Anoop’s Analysis

Chendvankar has an interesting idea to save power while driving a relay by switching from a full-wave rectified power supply to a half-wave rectified power supply. The circuit has the merit of simplicity and works fine with the set of components shown (a 9-0-9 transformer and a 12-V relay). Even if there is a spare contact, however, the design has several drawbacks.

First, the IFD requires a particular power-supply topology (i.e., a center tap transformer with a two-diode full-wave rectifier) and can’t be used with the more common, generic power schemes of an external dc power supply (like a power brick or battery) or an ac-dc switching power supply.

The design also needs an NC contact on the relay, which may not be available. Ensuring availability of the contact may mean using a double-pole double-throw (DPDT) relay instead of single-pole double-throw (SPDT), or double-pole single-throw (DPST) instead of single-pole single-throw (SPST), adding cost.

Second, if the main design does call for its own unregulated dc supply, using a center-tapped transformer in the supply also carries a cost penalty. Center-tapped transformers have twice as many windings as those for the same output voltage that have no center tap. As a result, they’re much more expensive—1.5 times to twice the price. Center-tapped transformers are also less commonly available, restricting design choices.

Third, because the relay driver uses unregulated dc, the transformer secondary voltage must be carefully matched to the relay being used, further restricting design choices. If using a relay rated at 9 V dc, for instance, the transformer will need to provide 8-0-8 V at rated load current, and 8-0-8-V transformers aren’t as common as 9-0-9-V transformers.

When matching the transformer and relay, there also needs to be consideration given to the range of load current that the “main circuit” (not part of the IFD) may draw. The peak voltage available to charge C2 depends on the main circuit’s load current draw. In the IFD’s given circuit, this voltage may be as high as 18 to 19 V and as low as 12.6 V (9 · √2).

Fourth, this circuit’s hold voltage depends not only on the main circuit’s load current draw but also on C2’s value and the relay coil’s resistance. Thus, the design approach does not guarantee a particular relay hold voltage, which may lead to unreliable operation.

And fifth, if the relay’s coil resistance is high, even when the relay is energized, the drive voltage across C2 may remain high even when the relay is energized. Thus, the IFD may not save much power. Voltage across C2 will still reach close to the transformer’s rated peak voltage. Even D4 is out of the circuit if the relay coil’s load current draw is small. Opening the relay’s NC contact, then, does not guarantee a significant voltage drop and corresponding power savings.

If saving power in the relay coil is very important, I would rather go with a relay that has very high coil resistance (a small cost premium) or use a solid-state relay (at a high cost premium) that requires only a few milliamps of drive current regardless of the load current it supports. If saving power were critical to my design, I could also use a latching relay (at a different cost premium) and get zero power dissipation in either an energized state or a non-energized state as the relay would only use power during state transitions.

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