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My circuit is very simple: it is a 12 VDC battery, a 14.4 VDC charger, 12 VDC power supply and finally devices on load.

What I intended to do is to make a UPS system here on DC between the battery and it's charger from one side and the 12 V adapter from the other side so I bought a 12 VDC relay (10 ms 90A) that has 5 terminals and I connected it this way: 2 for coils (from +/-12 V power supply,) 2 for positive inputs (battery/charger +12 V power supply) and 1 for positive output that goes to the load.

Because I wanted the positive to be switched from the 12 V power supply to the battery once the 12 V adapter power lost then I had to connect relay's coil to the 12 V power supply.

Of course I connected all the grounds together.

What should be happening is that once the 12 V adapter is turned on, the relay should be activated and power then should flow from the adapter and once disconnected the relay should be deactivated and power goes passively through the battery positive.

What is happening is that once the 12 V adapter is disconnected the relay waits for about 1 to 3 seconds to deactivate shutting off the load. It is kind of like the power is still flowing from the 12V power supply for a brief of time.

Once the power supply is connected again, the relay gets activated immediately and everything works just fine.

I want to know why the relay is getting delayed once the power supply power gets lost and suggestions to fix my circuit issue without paying much of the power loss on the circuit.

I didn't mention how much power have to go through the load (it is 1..15 Ah on the 12 V) so that is not a little relatively speaking of course.

enter image description here

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    \$\begingroup\$ Relays, typically have a much lower hold voltage than pull-in voltage. Thus, a 12 volt relay may pull-in (activate) at 10 volts but may not drop out (deactivate) until the voltage drops to 5-6 volts. I suspect that your load stops pulling current as the voltage drops. Then with only the relay as load, the filter caps in the power adapter keep the voltage up for a long time. You may need to drive the relay from a voltage comparator. \$\endgroup\$
    – DoxyLover
    Commented Oct 21, 2021 at 7:11
  • \$\begingroup\$ @DoxyLover Well, understood, but what kind of a voltage comparator I should be using?. I mean the name of electronic parts or something that can be usefull I can look for and learn about. \$\endgroup\$ Commented Oct 21, 2021 at 7:23
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    \$\begingroup\$ please provide a simple schematic of your circuit, rather than expecting others to parse your verbal description. The process of drawing it might even lead you to ideas about the isse. \$\endgroup\$
    – danmcb
    Commented Oct 21, 2021 at 7:57
  • \$\begingroup\$ There are probably big capacitors in the circuit holding the relay. Also a relay is slow anyway, you need to decide if about 10ms of switch time are sufficient for your purpose \$\endgroup\$ Commented Oct 21, 2021 at 12:41
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    \$\begingroup\$ Please show a schematic. \$\endgroup\$ Commented Oct 21, 2021 at 23:29

3 Answers 3

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The solution would be to use a 12V ~ DPDT relay.

enter image description here

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  • \$\begingroup\$ The difference between using 12V~ DPDT relay and my existing 12 V DC relay is: 1- charger is disconnected from the battery if there was no power from the wall outlet (this does not change anything) 2- the relay would then be energised from another 12 V ~ power supply separating it (this sounds like a possible solution). I will have to try this \$\endgroup\$ Commented Oct 22, 2021 at 19:46
  • \$\begingroup\$ This solution fixed the issue partially. The switching time has been significantly reduced, but there is still some delay enough to break my work on over 50% load (7A) and restart my electronic device so I will consider adding capacitors on the load input which may hopefully be enough not to drop the voltage on my device input \$\endgroup\$ Commented Oct 24, 2021 at 15:32
  • \$\begingroup\$ Why did you choose an alternating current relay type? \$\endgroup\$ Commented Oct 25, 2021 at 9:36
  • \$\begingroup\$ Islam Ghunym, The AC relay should then be of the 'make-before-break' (MBB) type. The AC relay was chosen to avoid the slow turn-off of a DC relay, caused by capacitors in the DC power supply. \$\endgroup\$
    – vu2nan
    Commented Oct 25, 2021 at 9:43
  • \$\begingroup\$ Understood, but why not adding a 220V AC relay instead of adding a 12V AC power supply along with the 12V ~ DPDT relay. Or just adding a separate 12V DC power supply for the existing DC relay coil. It would be more cost effective any of these, isn't it? \$\endgroup\$ Commented Oct 25, 2021 at 14:58
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Do you need to use a solution with slow relays?

You should consider using a LTC4418, which has the ability to select 2 different power inputs via P channel mosfets

For this, you need to connect your primary power supply to the first channel, and configure the under voltage (UV) threshold to trip if the voltage is lower than 11.5V.

Once the under voltage trips, it tries to switch to the next power input, the one where you have connected your battery. It does this quickly , so your load doesn't see the voltage dropping very far.

Once power comes back on the primary feed, it will automatically switch over. This chip also monitors voltage levels, and will only switch on a lower input voltage once the output voltage has dropped to that level, so you don't have to worry about capacitors in your load charged to 14V suddenly connecting to your just started 12V supply.

enter image description here

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  • \$\begingroup\$ This chip does not meet my power needs. It can't drive more than some uA while I mentioned that my load can go up to 15 A \$\endgroup\$ Commented Oct 23, 2021 at 3:49
  • \$\begingroup\$ The chip drives p channel mosfets which then handle your main power. You have to select a proper mosfet that can handle the 15A at the around 6V mark \$\endgroup\$
    – Ferrybig
    Commented Oct 23, 2021 at 11:12
  • \$\begingroup\$ How much the expected power loss by adding a circuit like this on 15A load? (only battery power efficiency matters). Currently my simple circuit efficiency is 98.7% at 15A 12V DC load (only very little lost in wires, switches and relays) \$\endgroup\$ Commented Oct 24, 2021 at 15:25
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    \$\begingroup\$ +1 for the unusual amount of work in that schematic. The circuit editor is painful for schematics that large. \$\endgroup\$
    – DKNguyen
    Commented Nov 23, 2021 at 2:02
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    \$\begingroup\$ @DKNguyen The worst part was the editor bugging out, where the save button stopped working. (according to the browser console I saved to many circuits recently) I had no choose to make a screenshot and then press discard in order to add it to the answer \$\endgroup\$
    – Ferrybig
    Commented Nov 23, 2021 at 10:58
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I see some issues that were not addressed and have a circuit suggestion for you to work on it.

What I intended to do is to make a UPS system here on DC between the battery and it's charger from one side and the 12 V adapter from the other side so I bought a 12 VDC relay (10 ms 90A) ... connected it this way: 2 for coils (from +/-12 V power supply) ...

Here I see you are using the Relay to be always energized by the PSU, where the NO contact is connected to the PSU and the NC to the battery.

Issue #1 - Increased commutation time using the Release configuration: Most electromechanical relays have two states: Operate and Release.

  • Operate time is usually shorter than release. The Operate Voltage is about 70% of nominal coil voltage.
  • Release time is longer, being reflected also on a lower Release Voltage, of about 30% to return the relay to a de-energized state.

Using your original configuration, the PSU needs to be as low as 3V ~ 4V for the relay then start the contacts' changeover due to coil de-energization. So, you should prioritize the Energize Operation to do quickly what you need to be done: Engage the battery as a standby energy source.

Issue #2 - Aggravation of Release time: Someone could decide to maintain the relay already energized and use a voltage comparator to de-energize quickly the relay as soon as the PSU is lower than say, 11V, for instance. They would use a relay for this fast relay switching. Then, 99% of chance they would use a diode (as 1N4004) in anti-parallel to the coil. This causes another problem: It increases the solenoid's current decay time, as summarized in Zettler's AN and highlighted the mentioned case:
enter image description here

schematic

simulate this circuit – Schematic created using CircuitLab

Features and Comments:

  1. Voltages in PSU and SLA Battery are represented by Nodes "PSU" and "Batt".
  2. Commutation voltage is measured at "LOAD" and current is measured at "R5", where commutation spike is smoothed by C2 = 20K uF.
  3. Power failure is simulated by Switches SW1 and SW2, turned-off @ 1s.
  4. LOAD is sized to consume about 10A.
  5. SLA Battery discharging behavior is simulated as C1 + R_Disch depleting here in about 5s.
  6. Relay commutation time in simulation seems to be "1x simulation step time". Then, step size shall be 0.01s for the purchased relay (this behavior is better observed in the 2 final simulations, at the end).
  7. When PSU voltage is above 10.6V (ZD2=10V + VbeQ1=0.6V), Q1 is turned-on and inhibits any driving of transistor Q2.
  8. When PSU voltage lowers, Q1 is turned-off, and Q2 bias is able to work.
  9. When Batt voltage is above 10V (ZD1=9V + VbeQ1=0.6V), Q1 is energized and Operate relay RLY1.
  10. When Batt is discharged lower than 10V, Q1 is de-energized, the load is disengaged and the circuit avoids the SLA battery from being over-discharged.

The following graphs show this behavior in the first 6 seconds:

enter image description here enter image description here

Commutation of relay and transients of voltage and current are zoom-in around 1s mark, where time-step in simulation is 0.01s: enter image description here enter image description here

Observe that for another simulation at +/- 10A of LOAD (R=1.2ohm), in 140ms the voltage dropped from 12.0V to 10.6V with C2 = 1F. The following picture is the configuration simulated: enter image description here enter image description here What would be the sizing guideline we could guess?

  • For each Ampere of loading at 12V and every 10 thousand uF in C2,
  • With Relay "Operate time" = 10ms,
  • LOAD has a voltage drop ~ 1Volt/A/10ms.

It was a nice opportunity to interact with the CircuitLab integrated here, and fun to observe the simulation results. If anyone wishes you can "play" with the simulation too!

I hope it helps and good luck!

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