Hello
I have a device which needs to stay on, when the mains power fails, it takes about a second before the battery kicks in, this makes the device reboot and it takes a while to get everything back together. So i wanna add a capacitor in parallel to the device, so that it could power the device for just 1 second until the battery takes on. I just want to know how to calculate its capacitance.
My device is rated at 1 amps, takes 9 volts input.
Regards!
A couple of thoughts. A capacitor is not going to supply alternating voltage like the mains do. What causes the battery circuit to kick in? Is it the drop of constant voltage, or the loss of alternating voltage. If it is the former, then anything you do to prop up the constant voltage will just delay the time the batteries kick in. You probably already have a capacitor at the output of the power supply to smooth out the ripple. Do you want to increase that capacitance more? If you do, it will increase the startup surge current. Hope the rectifier can handle the increase. You can easily figure out the amount of capacitance needed if you know the constant voltage when everything is running on the mains, the load the mains supply, the time you want to keep the constant voltage above the minimum voltage level, and the minimum voltage level.
Ratch
My device is running from 9v ac adapter, battery circuit kicks in using a set of relays, it connects/disconnects battery between the device and battery charger. Relays takes a second to switch battery from charger to device, this makes the device reboot. I wanna add capacitor parallel to the dc (9v) connection of the device. So when the main fails, while the relays connect battery to the device, during that 1 sec time the capacitor takes over. This should stop the device from reboot.
OK, let's take a typical example. Suppose you don't want to go below 7 volts. If you still want to supply 1 amp at 7 volts, then your load is around 7 ohms. Use the equation below to determine the time constant, and then the capacitance needed.
View attachment 93469
So you need over a half farad to sustain one amp for 1 sec assuming the voltage drops from 9 to 7 volts. You can see from the plot that after 1 sec at 1 amp, the voltage is down to 7 volts.
Ratch
What are the coil specs of your relays? How do you have the two relays wired?
And what is the actual measured output of your 9V adapter when loaded?
Thanks guys for all the suggestions... its much appreciated. Well after thinking about the using the 9v rechargable battery instead of capacitor. I used implemented it in the circuit today. It took a while before i stopped all the return current with diodes but the final voltage outputs were fine and the circuit was able to do what i was making it for..
@MrAl
Thanks man, your suggestion is also good, but unfortunately, i may not be able to find a 25000uf cap anywhere near. They do have 10000uf, so i can just connect three of them parallel together, and then use a voltage tripler to charge it up. Well, i already finished building the battery method you suggested and its working fine for now after a lot of tickling. Lets see how it works.
@ChrisP58
yeah, my 9v adapter powers my device as well as the two relays, which npn transistor you think i should use?
Hi again,
Well, whatever is easiest for you and you like best, that's the way to go
Also, it may be a good idea to later look into why the relay system doesnt work faster like it is, as Chris seems to be suggesting. It may be hard to fix though because they might not sense the voltage directly but by using the coil of one of the relays. That's how it is done sometimes. The problem with that is that the line voltage has to decrease low enough to let the relay open which may be too low for your device to run at. They also use a capacitor sometimes to keep the relay on a little longer so it doesnt switch by any short term sag (and a rectifier to power the DC coil relay). You could look into this though if you feel like it.
Sometimes i mod stuff rather than try to troubleshoot, because once it is modded you know exactly how it is working and dont have to figure out some hair brained scheme that somebody else thought up.
Well hello, my battery method failed way too well that i can't even explain and i don't care about it anymore. I now know why the relays takes 1 second to switch power. The issue is the power supply. Well the adapter has filtering capacitor inside it that's quite normal, it is storing enough charge to cause the relays to stay on for 1 more second when mains fail. Now i need to know how to deal with it? How to make the relays switch faster?
Hello again,
It would be good to see a schematic.
Think about this...if you connect a new 120vac relay to the line itself that will switch off within about 100ms when the mains goes out, assuming it goes down all the way or pretty much all the way. That new relay can be used to break or make any connection you need, such as what the original relay was supposed to do.
Keep in mind that it will be much faster than the old method, so it might turn on and off if there is a short term dip in the line. That shouldnt hurt anything, but you should be aware that can happen whereas with the 1 second delay that would not happen. If the battery is just taking over the load though then that shouldnt matter.
Actually any method that senses the line can be used for this, as long as it is electrically isolated from the line (small AC output transformer or wall wart with AC output for example).
If you use say a 12vdc relay and a 12vac output wall wart, you could use a rectifier and small capacitor rather than a large capacitor. That would provide faster switching too and be isolated from the line voltage.
Using an AC output wall wart, rectifier, small capacitor, small transistor, you can sense within about a quarter cycle to half cycle time which is about 4.2 to 8.3 milliseconds. The transistor can then be used to switch something in or out, or connected to a relay to switch something in or out.
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