Relays continue to be useful and popular, but in low-power, battery operated applications they tend to be power wasters. The following relay economy circuits reduce coil power significantly by simply adding a series resistor. As soon as the relay picks up, the coil voltage is reduced either by the normally closed contact opening or the charging of a capacitor that is connected in parallel with the economy resistor. Included is the standard relay economy circuit as well as two additional circuits that may be new to the world.
Note that these relay economy circuits may be used in conjunction with the relay driver circuits previously posted: www.electroschematics.com/7123/relay-driver-2/
Figures 1 & 2 show how high and low side drivers can be connected to provide relay economy. The unused normally open relay contact provides the output. If an isolated contact is required, a two pole relay must be used as in figure 3.
Figures 4 & 5 solve the problem by simply adding a capacitor in parallel with the resistor. This allows the use of a single pole relay. The required capacitance is high, but voltage is low so that the size is physically manageable.
Figure 6 shows how to reduce the capacitor size by a factor of 20 by use of a simple transistor integrator. The collector voltage of Q1 integrates to the maximum voltage allowed by R1 in about 25mS.
Operation of transistor integrator
Red arrows indicate capacitor charge path
Green arrows indicate capacitor discharge path
C1 integrates slowly due to the limited current flowing through R2. The voltage across R2 is limited by the Vbe of Q1 (0.7V) while integrating.
Relay operate time
I did an experiment on relay operate time so I could determine series capacitor RC size. I believe that most small relays have similar operating times.
Schematic of test setup
When switch is closed, coil voltage is conducted through 10K resistor.
When contacts close the above voltage is shorted.
Contact bounce is visible at the instant of contact closure—observe voltage spikes generated
Contact operate time = 4.5mS
For conservative initial power application time, make RC (time constant) equal to approximately 5 * operate time or 5 * 5mS = 25mS. Note that the resistance used in determining R*C is the parallel resistance of both the relay coil and added resistor.
Voltage reduction considerations
If the relay current is reduced to 67%, relay power drops to 45%, but since source voltage (e.g. 12V) remains the same, the total circuit power is reduced to only to 67%. (not bad!) For this, make the series resistor equal to half of the coil resistance.
If the relay current is reduced to 50%, relay power drops by a factor of (4), but again since source voltage is again unchanged, the total circuit power is reduced to 50%. (good savings!) For this, make the series resistance equal to the coil resistance.
Which factor should I use—67% or 50%? This is a function of relay pickup (and dropout) voltage performance. Some relays perform much better than published specifications while others barely make the 80% guaranteed voltage pick up spec. Some remain picked up down to 20% voltage or better while others drop out near 50%. Generally, use the 67% figure unless you actually evaluate the relay using a variable voltage power supply.
When gradually increasing voltage, some relays exhibit a stepped response, so when evaluating observe armature motion (if exposed) and/or listen very, very carefully. The 1st step is the initial moving of the armature and mechanical linkages and perhaps the closing of the contacts, while the 2nd step is the final magnetic circuit completion when the armature actually touches the coil pole. In the case of a stepped response, use the 67% factor. 50% voltage may not be reliable.
During evaluation, also experiment with the effect of vibration. At reduced voltage, a relay may easily drop out upon vibration—this may be aggravated if the relay has a stepped response.
A typical example
I have a relay circuit powered by a 9V battery. The best available relay selection is 6V, 60mA (100Ω) that draws 90mA @ 9V. What options do I have?
Add a 200Ω series resistor. Resulting voltage and current in economy mode is 3V and 30mA. Add capacitor across resistor. Parallel resistance = 100 * 200 /(100 + 200) = 67Ω. C = 25mS /67 = 373uf minimum (use 330uf, 16V aluminum electrolytic).
For the future
Voltage doubler relay driver — reduces total circuit power to 25%