Calculating the Ah rating for the Defender 5101 battery backed CMOS RAM Device
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1. 5101 CMOS RAM Specs:
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Power Supply Range | 4V to 6.5V |
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Operating current | 8mA |
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Min voltage for battery | 2V |
for memory retention | |
on stand by | |
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Quiscient current | 25-200uA |
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Quiscient mode 0.95 of Vdd, presuming
this means 95% of 4v,
then 3.8v is probably
the voltage at which
quiscient mode begins
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when the MPU is running normally, normal operating current is 8mA. also, when the MPU is being shut down, either normally
or under a power failure scenario, there will be a very small period of time where data may be being written. normal
operating current during that period is 8mA
when the 5V supply rail drops to =<4.75v, the reset circuit inhibits further writes.
with a traditional 3xAA battery pack the terminal voltage is 4.5v with fresh batteries. there should be no current draw on
the Battery Pack until the main psu voltage drops below 3.8v (ie the terminal voltage of 4.5 minus the in4148 diode forward
voltage of 0.7v). As that is about the point at which the 5101 goes into quiscient mode, then the current drawn from the
battery pack is in the 25-200uA range.
with a fresh 3.6v lithium cell, there should be no current draw on the Battery Pack until the main psu voltage drops below
2.9v (ie the terminal voltage of 3.6v minus the in4148 diode forward voltage of 0.7v) As that is about the point at which
the 5101 goes into quiscient mode, then the current drawn from the battery pack is in the 25-200uA range.
so, if a 3.6v button cell is used as the CMOS RAM battery backup, less the 0.7v forward voltage of the in4148 diode, then
the chosen cell should retain its terminal voltage above (2v+0.7v) or >2.7v to ensure the supply voltage to the 5101 is
above the minimum voltage of 1.5-2v for memory retention. therefore, the cell chosen must maintain the cell terminal
voltage above 2.7v for the desired period (say a min of 12 months) at the max est current draw of 200uA. If you choose a 3v
cell, then the voltage to the 5101 is 3.0-0.7=2.3v when the battery is fresh.
if you choose a typical lower AH rating cell, say 90mAh, there is approx 1/20th the capacity and therefore may only last 3-4
weeks!
As a back of the envelope calculation, If the current drawn is x amps, the time is T hours then the capacity C in amp-hours
is C = xT. In our example, the circuit draws max of 200uA in quiscient mode, for 10,000 hours the Amp Hours (C) = 0.0002
Amps * 10,000 hours (approx 1.2 years) = 2Ah depending on the cell type, the terminal voltage will drop rapidly as it nears
the end of its capacity, so a 2Ah battery may be only good for 8000-9000 hours at 200-300uA
2. Now we need to add the extra current drawn by IC 6H and the associated 330k resistor.
4071 CMOS OR gate 14 pin package. Typical operating specifications for this:
(TBA) - 4584 uses between 30 and 120uA
and the associated 330k resistor:
i=3.8v/330,000 ohm=1.15*10exp-5=0.0000115 or 11.5uA
therefore:
if we use max values, Amp Hours (C) = 0.0003315 Amps * 10,000 hours (approx 1.2 years) = 3.3315Ah
if we use typical values, Amp Hours (C) = (0.000050+0.0000115+0.000030 Amps * 10,000 hours (approx 1.2 years) = 0.915Ah
so i estimate a 4.5v or 3.6v 2Ah battery would do the trick then.
now the ds1225 is a little different with a quoted standby current of 5-10mA (when chip not selected but presumeably in an
active circuit. therefore if we use max values as quoted,
Amp Hours (C) = 0.01 Amps * 10,000 hours (approx 1.2 years) = 100Ah. now that just doesnt look right. given the size of
the packaging, lets say it is a 2Ah battery, divided by 10000x10 (10 yrs) would yield a mem retention mode current of 20 uA.
so lets now say 20uA * 10000 hrs = 0.2Ah. now thats more like it! this est is borne out by the ds1210 spec sheet which
quotes 0.1uA for the ds1210 and 10uA for the ram, which would mean a 1Ah battery would suffice for 10 years! that is without
the 4071. maybe 40ua *10000hrs=0.4Ah. still a tiny cell, and 10yrs for 4Ah.
any flaws in the logic or methods of approximation ?