1
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Currently, I have a circuit with two motors (200 W, 24V, 8A max), motor drivers, DC DC converters, controllers and other components powered by a 24V Li-ion battery. The problem with current setup is whenever the battery drains, we need to replace it with the fully charged battery which forces us to shut down the whole system while swapping batteries.

The proposed idea for this issue is to use two batteries and design a power management system such that if one battery drains below the threshold voltage, the circuit automatically switches power supply to the other battery and vise versa. Before the second battery depletes, the user will be prompted to replace the first battery and vice versa. The battery management circuit should also measure the voltage level via Arduino Analog pins so that user can be prompted.

Though, i found many useful information from similar posts, i am still feeling little bit lost.

How to use diode oring to switch between two power sources

From the above link, i found 2 simple circuits. enter image description here

The main problem with the circuit above is that if we use batteries for both sources (V1 and V2) , they deplete concurrently which is not preferred in my case.


enter image description here

In the circuit above, When V1 is cut off or drops below threshold voltage, the relay switches two V2. But, the problem is once V1 is connected back, the circuit switches back from V2 to V1 which again is not desired.


What I am looking for is a solution where the circuit runs on first battery and when the first battery drops below the threshold value, the circuit should switch to second battery and the user will later replace the first battery. Again, when the second battery voltage drops below the threshold value, the circuit should switch back to the first battery. This is to ensure that the battery is fully drained before charging.

I am open to use diodes, relays, transistors and arduino (it's already in the circuit). I am afraid there might by huge power loss if we use diodes.

Please, feel free to advise and comment :)

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  • 1
    \$\begingroup\$ 1) How much current would have to be switched? 2) Must switchover occur with no dropout? \$\endgroup\$ – EM Fields Aug 8 '16 at 7:55
  • \$\begingroup\$ @EMFields 1) 5A to 10A may have to be switched. 2) Yes, the switchover must occur without a drop out (as the CPU turns off immediately if there's a dropout) \$\endgroup\$ – prasanth_ntu Aug 8 '16 at 8:31
  • 1
    \$\begingroup\$ What happens when the 2nd battery becomes depleted and the first battery is not in position or recharged. You need to logically cover the what-ifs. \$\endgroup\$ – Andy aka Aug 8 '16 at 8:55
  • \$\begingroup\$ @Andyaka Thanks for the prompt. Before the 2nd battery depletes & when the first battery is not recharged, user will be notified (via buzzers or screen) to manually replace the first battery. The same applies for the other battery. \$\endgroup\$ – prasanth_ntu Aug 8 '16 at 9:15
  • 1
    \$\begingroup\$ No, you are missing my point. What if the battery is not replaced - you want a solution that has to cope with the what ifs. It's no good saying it will be replaced - what if it isn't - how does the system cope, what should the system do. Think about it more. \$\endgroup\$ – Andy aka Aug 8 '16 at 9:17
2
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The battery that is currently switched in must keep the other battery switched out until the first battery is drained, and once switched over must stay that way until the other battery drains (even after the first battery is replaced). This implies a symmetrical bistable circuit.

Here is a possible solution using relays:-

schematic

simulate this circuit – Schematic created using CircuitLab

Each battery has a relay with a normally closed contact that keeps the other battery (and its relay) switched off. Whichever battery is connected first will have priority - until it is drained, then its relay will turn off and give control to the other battery.

This circuit has a few potential problems.

  1. When the drained battery's relay releases and connects the other battery the voltage will rise, which may cause the relay to pick up again before the other relay has time to operate. If this happens the relays will chatter instead of switching over cleanly.

  2. The relays may become sensitive to mechanical shock or vibration when close to the release point, and the cutoff voltages may not be well defined.

  3. Relay contacts increase resistance and get noisier with use. Eventually the circuit may become unreliable.

For these reasons I would consider using a solid state circuit (comparators and logic gates or an MCU) to monitor the battery voltages and control the switchover, and MOSFETs to switch between batteries.

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  • 1
    \$\begingroup\$ Thanks for the detailed explanation including narrating the potential problems. Could you explain more on using the solid state circuit (MCU, i already have an arduino) to monitor the battery voltages and control the switchover, and MOSFETs to switch between batteries. I am more interested to explore this option. Note: I am using a 24V battery. For monitoring the battery voltage using an Arduino, would it be advisable to use resistors to do a voltage divider (from 24V to 5v) or is there any better way to do it? \$\endgroup\$ – prasanth_ntu Aug 10 '16 at 4:59
  • \$\begingroup\$ You would use P channel MOSFETs (rated for at least 30V and 30A) with level shifting gate drivers, or DC solid state relays. The Arduino continuously measures the voltage on each battery, and operates the FETs/SSRs to switch each battery in and out as required. Using a voltage divider is fine. Only requirement is that the resistor going to ground should be 10k or less. \$\endgroup\$ – Bruce Abbott Aug 10 '16 at 6:56
  • \$\begingroup\$ @BruceAbott: I added the circuit that i have came up with in the link shared below. I am quite new to electrical circuits and I am pretty sure that there are numerous mistakes in it. Any help/ suggestion/ advise would be highly appreciated. The intended functionality of the circuit is that if battery1 voltage is greater than threshold value, it will turn on NMOS1 using Arduino Digital Pin 1 and vise versa. **[Click here to access the Link for circuit diagram]**(drive.google.com/file/d/0B0F55gMRFtkWU1kxN1UzRmZtdm8/…). \$\endgroup\$ – prasanth_ntu Aug 11 '16 at 3:55
  • \$\begingroup\$ Your circuit makes no sense. The load has two battery inputs and two switches going to a common ground, but how does it distinguish between them? The Arduino measures positive voltages only so you should common the battery negatives (= Arduino ground) and switch the positives using P channel MOSFETs. Level translators will be required to change the Arduino's 0~5V output to 24V~12V. These can just be a transistor and two resistors, like this:- electronics.stackexchange.com/questions/70729/… \$\endgroup\$ – Bruce Abbott Aug 11 '16 at 4:40
  • \$\begingroup\$ Thanks again. Any reason for recommending to use P channel MOSFETs and not N Channel? I assume, for N channel, we don't need a level translators \$\endgroup\$ – prasanth_ntu Aug 11 '16 at 4:56
1
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From your comment question, here are some high side and some low side switched circuits:

enter image description here

From your subject question, and as has has already been suggested, at first glance it might seem attractive to use a couple of P-channel MOSFETs, as shown below, to switch between the batteries.

Unfortunately, the MOSFET substrate diodes will create a problem when either battery voltage drops about 0.7 volt lower than the other one.

When that happens, the lower voltage diode will no longer be reverse biased and the higher voltage battery will begin pumping current into the lower voltage one when the higher voltage one is turned on.

In the schematic included below I've shown external diodes, for clarity, and have indicated the path the conventonal current will follow, with R1 representing the 10 ampere load you mentioned.

Note that, in this instance, Q1 is completely turned off and Q2 is in saturation.

I've included the LTspice circuit list so you can play with the circuit if you want to.

enter image description here

The same thing happens if you do low side switching using NMOS.

R2 is the 10 ampere load, and R1 represents control stuff that needs to be ON all the time.

enter image description here

Version 4
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TEXT -3392 -32 Left 2 !.tran 2

UPDATE:

While it appears that MOSFETS won't work because of interactions between their substrates which allow reverse currents to flow when the MOSFET is in forward cutoff, it seems like a couple of relays, with thousands of megohms between their contacts when they're open, might.

To that end, here's a circuit which seems to work pretty well in a simulator, with one small caveat: during switchover there's about a 3 microsecond load dropout which doesn't affect the circuit's control circuitry, but only the heavy switched current to the load.

With proper "tuning" that dropout can be brought under control and a seamless switchover achieved, but why should I have all the fun? ;)

Here's the schematic, the plot, and the LTspice circuit list so you can play with/optimize the circuit if you want to.

enter image description here

Version 4
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\$\endgroup\$
  • \$\begingroup\$ Again, thanks for the insight. I have modified a circuit that was shared by another user. Things looked fine until i realised that when one battery is cut off or voltage drops than other, the reverse current from higher voltage battery flows to the lower voltage battery which is kind of troublesome (As if what you have mentioned in the answer above). The update you have provided is beyond my knowledge & needs lot of components. Is there any fix or alternative solution for this issue within existing circuit that i have shared below: Link \$\endgroup\$ – prasanth_ntu Aug 16 '16 at 8:19
  • \$\begingroup\$ @prasanth_ntu: Not that I know of. \$\endgroup\$ – EM Fields Aug 16 '16 at 9:47

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