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I made a simple circuit to alternate its output between two inputs: a photovoltaic panel and a 12V DC voltage source. The schematic: enter image description here

The step source represents the photovoltaic panel and the two resistors represent two bilge pumps of 12 VDC 30 W.

It's done in a way that the normal closed contact is fed by the net. But when there's sun, there is voltage and then the relay (12 V rated) changes its contacts. Pumps are then fed with power coming from the panel.

The problem that I got, was that the release/decoupling voltage (voltage to change the the contacts back to normal) isn't the same as the operating voltage of the relay. The relay comes back to normal when the voltage is approximately 3 V, so when it's going dark and the panel is generating between 3 V and 12 V, my pumps aren't going to work properly or at all. They have to be ON 24hrs/day.

Only when it reaches so dark that the voltage is lower than 3 V do the pumps get fed with power from the net again.

What can I do to solve this problem? Is it possible to keep the relay strategy or do I have to change it to something else?

Thanks in advance.

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Another option is using ORing diodes instead of the relay.

schematic

simulate this circuit – Schematic created using CircuitLab

In the circuit above the transient between the solar panel and the 12V DC power source will be smooth and at about 11.0 V (assuming a 0.5 V voltage drop). You could play with the transition point by adding more Schottky diodes in series with D2 and D3.

A drawback is that each Schottky will dissipate power. If possible, you could use an ORing configuration for each pump. This will reduce the power dissipation a little.

Illustration the right circuit dissipates less:
The current per pump is 30 W / 12 V = 2.5 A.
Using for example the VS-12TQ035-M3 for every Schottky diode, the voltage drop per diode in the left circuit will be about 0.45 V @ 5 A, so a 2.25 W dissipation. In case only D1 is conduction, the wasted power is thus 2.25 W.
Using the right circuit, the voltage drop 0.35 V @ 2.5 A, so a 0.875 W dissipation per diode. In case only D4 and D5 are conduction, the wasted power is a little less: 2 * 0.875 W = 1.75 W. enter image description here

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  • \$\begingroup\$ thank you very much!! \$\endgroup\$ – João Victor Aug 23 '19 at 18:23
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Indeed, relays have a bit if hysteresis. They will require a certain voltage to turn on, but will not turn off again until the voltage is significantly lower.

Operating a relay in this "mid-region" of coil voltage is not recommended, as the contact force may not be enough to provide a good connection. This may significantly affect the relays current carrying capability negatively.

Also, the "release" voltage may not be stable as, among other things, microscopic fusing may occur on the contacts requiring a higher force to pull them apart (lower voltage). You can not rely on this in a well designed circuit.

For a setup like this you should construct a voltage comparator circuit that drives a transistor to provide either the full 12V when on, or 0V when off, to the relay depending on the measured voltage. This circuit should also have some hysteresis, or the circuit may oscillate near the transition voltage causing premature relay failure. There are also solid state solutions that will require no mechanical relay.

To find example circuits and even pre-fabricated modules, do a google search for voltage comparator relay. Add in terms like module, hysteresis and schematic to find more specific results.

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  • \$\begingroup\$ thank you very much!! \$\endgroup\$ – João Victor Aug 23 '19 at 18:23
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I also faced the same problem while designing one of my circuits. Ronald has pointed out some good ways to resolve this. But if you are not much concerned about the life of the relay here is a what you can try. Instead of powering the relay directly from the panel, use a series arrangement of Light Dependent Resistor (LDR) and a 1k ohm potentiometer.

enter image description here

A typical 12V relay has a resistance of around 400ohms. Now adding more resistance to the branch may or may not provide enough current from the panel to pass through relay in order to turn it ON. Firstly you can check out this point with your circuit by setting the potentiometer at 0 ohm. If there's no problem in the switching just adjust the potentiometer and use the circuit. But if it's not switching even in daytime you can replace the 12V relay with a 5V relay. Or you can increase the supply voltage to the relay by using some other voltage source in series with the relay. However the latter approach is not recommended because of the extra hardware and power requirement.

A typical LDR may have resistance range from less than 100ohm in the sunlight to 1Mohm in the dark. So as the sun goes down the resistance of the LDR increases exponentially. Even if it's inside a well lightened room in the daytime it's resistance can be more than 1000ohm. This circuit, by adjusting the potentiometer, should cutoff the supply from the 12V DC source at an acceptable light intensity range.

The method is not full proof and I have not tested it but you can play with the values and try something different. However if you can use a microcontroller instead, it would be easier to resolve the issue.

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  • \$\begingroup\$ thank you very much!! \$\endgroup\$ – João Victor Aug 23 '19 at 18:23

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