EEVblog Electronics Community Forum
Electronics => Projects, Designs, and Technical Stuff => Topic started by: GeorgeTsak on May 11, 2017, 03:58:29 pm
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Hello everyone! During a power shortage, my modem turned off and the internet connectivity got lost. Then I connected 3 lithium batteries in series (arround 11-12V) and the modem turned on and I could browse the internet again. (I already know phone lines are not affected by electricity). So I wanted to make something simple to prevent something like that occuring in the future and I made a very simple circuit. (see image). I tried it and it worked ok, but here is my issue.
If the voltage of the normal power source < battery's voltage then the battery will "push" electricity in the normal power supply. I thought adding a second diode to the normal power supply's positive, would solve the issue but it didn't as the load would be split into battery and the normal power source. (Previously because of the voltage drop only the normal power source would provide power). Is there any way arround this? I know I could simply buy a UPS which is plug n play, but I want to try and make it by myself.
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You can solve this problem with one relay
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You can solve this problem with one relay
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Yeah, but will it be fast enough that the modem doesn't reset?
Depends how complicated the OP wants it to be, but I would run the modem from a battery pack, possibly via a step up converter, and use the mains PSU to charge the pack
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Ye its the simplest way and relay is proppably fast enough becose cappacitors in the modem will keep it on long enough....Sorry for my bad english
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Ye its the simplest way and relay is proppably fast enough becose cappacitors in the modem will keep it on long enough....Sorry for my bad english
Hmm yes I agree. Even if the modem's capacitors don't keep it turned on, I could always add my own capacitors.
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I recently did some ghetto-grade reverse-engineering of a security-oriented UPS board (12V 10A) with a similar configuration to what you were talking about, but with a 39-ohm resistor in parallel with the battery supply diode, so that the battery can charge. No relays in this board, but there was another diode in line with the output of the switching regulator.
I haven't explored this possibility myself, but this UPS board used common-cathode double-diodes, and each double-diode had the anode legs tied together. I am not sure this reduces the voltage drop as I haven't tried it out or tested it to confirm this.
If the draw of your modem is small enough, the total power loss of the added diode in series will be 200-400mW or so. Modems often have no trouble with a bit of under-voltage.
So you can add the diode in series as you described, but I would also add a 1000uF cap in parallel with the modem.
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I recently did some ghetto-grade reverse-engineering of a security-oriented UPS board (12V 10A) with a similar configuration to what you were talking about, but with a 39-ohm resistor in parallel with the battery supply diode, so that the battery can charge. No relays in this board, but there was another diode in line with the output of the switching regulator.
I haven't explored this possibility myself, but this UPS board used common-cathode double-diodes, and each double-diode had the anode legs tied together. I am not sure this reduces the voltage drop as I haven't tried it out or tested it to confirm this.
If the draw of your modem is small enough, the total power loss of the added diode in series will be 200-400mW or so. Modems often have no trouble with a bit of under-voltage.
So you can add the diode in series as you described, but I would also add a 1000uF cap in parallel with the modem.
This circuit is great for using it with a lead acid battery, as these batteries can be abused a lot, but on the other hand those lithium ion batteries are very delicate. No overcharging, no overvoltage, no undevoltage, no charging for a long time, charging balancing when using multiple cells and so on. Thank you very much for sharing this with me!
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No worries George.
I would be interested to see if this kind of simple circuit works for a Li-Ion pack with a decent battery management system built into it. The resistor will limit the max. current to 350mA (at 13.65V) but a good BMS should prevent overcharging anyway.
I am probably going to experiment with my LiFePO4 pack, which is advertised as a SLA battery replacement and has an integrated BMS. LiFePO4 are meant to have a higher resilience than lead acid batteries of the same capacity rating.
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No worries George.
I would be interested to see if this kind of simple circuit works for a Li-Ion pack with a decent battery management system built into it. The resistor will limit the max. current to 350mA (at 13.65V) but a good BMS should prevent overcharging anyway.
I am probably going to experiment with my LiFePO4 pack, which is advertised as a SLA battery replacement and has an integrated BMS. LiFePO4 are meant to have a higher resilience than lead acid batteries of the same capacity rating.
I completely forgot about cells with built in protection circuitry. Yes these should work fine, but you have to buy good quality one as cheap ultrafire crap batteries from ebay can really ruin your day!
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Careful though, many of these "protected" cells have a circuit that is just that, protection, it is intended as a last ditch effort to keep the cell from going thermal, it is not a battery management system. Relying on the protection circuit is a bit like relying on your seatbelt and not bothering to fit brakes.