[Solved] Help Starter Capacitor Circuit

Hi Inq,

  I think the short answer is ...

That this is not the place for a simple capacitor ... I am guessing the capacitor related to heat pumps that you refer to applies to AC induction motors, which is a different game.

However, the principle of using a higher voltage to energise the relay, and a lower one to keep it latched is sound.

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The longer answer starts ...

I think it is helpful to 'set the scene', so that we can have a common starting point.

For the energising, I guess you have a circuit something like (click on picture to enlarge):

This circuit shows an NPN bipolar transistor ... it could be substituted with an n-channel MOSFET.

The resistor values for R1 and R2 would be quite different for MOSFET and bipolar transistors. For MOSFET, R1 might be replaced with a wire, and R2 might be omitted, although a high value (e.g. 100kOhms) could be beneficial.

Also, please remember that the transistor is responsible for passing current through the relay coil to energise it, whilst the diode is only active at the moments (about a second) immediately following the transistor being commanded to switch off, to de-energise the relay, when the coil will attempt to generate a high voltage and consequently risk damaging the transistor, and anything else connected to it.  This means that the diode should be physically connected, directly across the coil terminals of the relay, so that it is always connected, regardless of any other circuitry changes.

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Now, to address the original question, I suggest you need a 'circuit' which aims to apply the full battery voltage across the coil, when the relay is energised, for sufficient time for the contacts to close (say 1 second), and then reduces the voltage to a level that will keep the contacts closed. 

You have determined that the absolute minimum voltage that will keep the relay latched is 3.4V, although I would recommend adding some margin to this figure. I don't have any experience to indicate how large the margin should be ... I imagine it includes factors like environmental vibration, temperature variation, relay part variation (if you are making more than 1 of these circuits) and so on. For a start, let's imagine you pick relay coil voltage of 4.5V.

The simplest approach, would be to put a resistor in series with the relay, so that if the supply voltage is 14V, then the relay voltage is 4.5V, implying (14 - 4.5) = 9.5V is dropped across the resistor. This situation could be achieved by adding a second instance of the transistor circuit discussed above, driven by a second GPIO pin, with the additional resistor inserted into the connection between the transistor and relay coil, to drop the 9.5V. Then the microcontroller would be responsible for initially switching on both transistors, to energise the relay, and after one second (say), switching off the transistor without the dropping resistor, leaving the other transistor switched on, to maintain the relay in the energised state.

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This approach is simple, but has disadvantages, including:

• Power dropped across the resistor is wasted
• If the circuit is required to work over a substantial range of battery voltages, the dropping resistor must be chosen to maintain a high enough relay coil voltage, when the battery voltage is at it lowest, implying that the power consumption when the battery voltage is high will be considerably higher than necessary.

To overcome these disadvantages, the 'obvious' solution is to use a voltage regulator to convert the battery voltage down to the chosen relay coil voltage. Note that although a linear voltage regulator would help with the second disadvantage, it would still waste power in dropping the voltage. Thus, a more efficient approach is to employ a buck regulator to provide a source of the voltage required to maintain the relay in the energised state.

Luckily, small buck regulators are now readily available at very low cost. I am assuming that the principle of using a second GPIO pin discussed above, is adopted, then a simple circuit to link them will be required.

The next part of the discussion requires a little 'market research' and thinking on my part, whilst I have already wittered on too much, so I'll leave you a pause to digest the above, and provide any appropriate feedback. I'll aim to provide a sequel, based on your feedback.

Best wishes, Dave

@davee,

Hi Dave,

I was hoping you were still an active member.  I knew you'd be thorough and gear it to my level. 

Yes, your drawing is what I have, but I'm counting on the ESP8266 built-in pull-down resistor instead of the R2.  Maybe that's not wise???  I had some pieces on-hand and slapped them together - R1=1KΩ, N4001, and S8050 NPN.  I was prepared for crickets or smoke 😉 but it worked the first time.  

I understand your design.  I even thought about it ahead of time, but side-lined it... thinking there might be a simpler design with just a capacitor.  I remember in my childhood playing with the 201 Projects Electronic Box and remember capacitors charge up and then release energy.  I also remember putting a capacitor across a servo's power leads to handle the motor surge that the tiny buck converter couldn't deliver.  That sounded good in my head, but must be overly simplistic.  I knew your broad knowledge would know other possibilities if they existed.

As you point out duplicating the same circuit I have now driven by another pin with a different buck converter set to the ~4V makes perfect sense and it dirt cheap also.  As it might be energized continuously for days - months - years, its a small price to pay up-front to save the power and heat build-up in the relay coil.

I'll work with this.

Thank you for the advice.