>How the mobile phones retain Correct Date and Time while they may be switched off for days? >they still remains all data if we removed the battery?
There is in nearly every phone an additional battery or a goldcap(a kind of big capacitor)
>is mobile has CMOS Battery like our desktop had?
That's the point
But not every phone has it (i know about a few cheap Nokias (no Symbian devices)) that will lose their settings, if you remove the batterie for a couple of minutes.
>goldcap(a kind of big capacitor) for how long it provides backup for mobile? I think they use very small goldcaps (small for goldcaps not for real capacitors). I think max 1/2 Farad
Q = I * t = C * U (Q = charge, I = current, t = time, C = capacity, U = voltage)
so you get t = (C * U) / I
You must know the voltage inside the goldcap and need the current that the 'rest'-electronic needs. I think for an oszillator at low frequency (usually 32.768 kHz) which will drive only a few CMOS-arrays you need only a few uA (I mean micro). I think half a year is no problem (only my estimation)...
>coz capacitors (as i know ) leak the power after very short time. goldcaps are no real capacitors. They are something between capacitors and accumulators/batteries. They have very small leak-current.
Forum posts: 176
>they still remains all data if we removed the battery?
There is in nearly every phone an additional battery or a goldcap(a kind of big capacitor)
>is mobile has CMOS Battery like our desktop had?
That's the point
But not every phone has it (i know about a few cheap Nokias (no Symbian devices)) that will lose their settings, if you
remove the batterie for a couple of minutes.
regards;
CG
Forum posts: 118
goldcap(a kind of big capacitor) for how long it provides backup for mobile?
coz capacitors (as i know ) leak the power after very short time.
Prashant {prashu}
Forum posts: 176
I think they use very small goldcaps (small for goldcaps not for real capacitors). I think max 1/2 Farad
Q = I * t = C * U (Q = charge, I = current, t = time, C = capacity, U = voltage)
so you get t = (C * U) / I
You must know the voltage inside the goldcap and need the current that the 'rest'-electronic needs. I think for an oszillator at low frequency (usually 32.768 kHz) which
will drive only a few CMOS-arrays you need only a few uA (I mean micro).
I think half a year is no problem (only my estimation)...
>coz capacitors (as i know ) leak the power after very short time.
goldcaps are no real capacitors. They are something between capacitors and accumulators/batteries. They have very small leak-current.
regards;
CG
Forum posts: 118
really i got more then i expected.
Thanx.
Prashant {prashu}