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The correspondence of capacity between accumulators and capacitors
Kirill V.:
Hello every one!
I have idea about power supply with accumulators battery instead capacitors after rectifier.
I'd like to know how correlate the capacity of the battery and the capacitor?
For example: battery of 1A*h replaces X Farads of capacitors
battery of 2A*h replaces XX Farads etc.
Thanks.
soldar:
They are different things, they work differently and you can't just replace one with the other. You need a better understanding of how each of them works.
xani:
Capacitors in power supply are used to smoothen/filter the voltage, not as long term storage devices. Major differences are much higher current capability, basically infinite cycles and no charging requirements aside from "dont overvoltage it", while on other side batteries need special charging circuit, straight up irreversably die if discharged to zero.
Also if you can't convert the battery capacity to capacitor capactity at given voltage you shouldn't probably be going around trying to replace one with another for no good reason.
Kirill V.:
Thank you all.
My idea: rectifier - battery - voltage stabilizer - output. This is for low noise and ripple. Rectifier can be used for standby charge of battery and in this case battery able to smooth out ripples?
For ultimate performance AC part can be turned off and in this case battery used as storage
MrAl:
Hello there,
There is a close equivalence between a battery and a capacitor but when we compare we must take into account the voltage drop. Let me explain.
We will start with one of the definitions associated with a capacitor:
dv/dt=i/C
where
dv is the change in voltage over the time dt,
dt is the time period of the discharge in seconds.
i is the current, considered constant for this simple explanation,
C is the capacitance in Farads.
Ok lets see what we can do with this.
First lets solve
dv/dt=i/C
for C because we would like to know that.
Doing that, we get:
C=i*dt/dv
Now say we have a 12v battery that reads 12.5v when charged and 11.5v when completely discharged, and say it is a 1 ampere hour battery.
We now can plug in the values.
dv=1 because 12.5-11.5 equals 1 volt.
i=1 amp for the test to make it simple.
dt=1 hour because 1 amp for one hour gives us a 1 ampere hour 'battery' but the time must be in seconds so we use 3600 for one hour.
Plugging in the values into:
C=i*dt/dv
we do that:
C=1*3600/1=3600 (note there are 3600 seconds in one hour)
and so we get finally:
C=3600 Farads.
We can do another one.
This time we have a 20 ampere hour battery that is charged to 5v when charged and when discharged is 3v.
This means that now dv=5-3 which equals 2 volts.
We keep the current 'i' at 1 amp, and if we have a 20 ampere hour battery then it must take 20 hours to discharge at one amp so dt=20*3600=72000 seconds.
Plug in the values we know now:
C=i*dt/dv
C=1*72000/2
so:
C=72000/2=36000
so
C is 36000 Farads.
Note how high C has to be to emulate a decent size battery.
One thing we leave out is the factor that relates discharge current to capacity. We assume above that is 1 although most real batteries show a decrease in capacity as the discharge current increases. We also ignore the total charge of the capacitor for simplicity.
One more point is that if you look at good spice models for batteries you will see the main component is a large capacitor although there are other components too including a 2nd capacitor.
Try one see if you can calculate the capacitance.
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