14 May 2016
n WITH TIM BROWN
• Low Battery Indicator/Battery Life
Low Battery Indicator/Battery Life
QThere was a Q&A reprint in a recent N&V newsletter titled Low Battery Indicator. Regarding the circuit shown in figure 1, it states “The circuit draws a mere 120 microamps,
which is just slightly more than the self-discharge current
of the same battery sitting on a shelf.” I am not sure this is
I calculate that at nine volts the threshold adjustment
divider will consume 37 microamps, the reference divider
43 microamps, and from the spec sheet, Icc typical for
the comparator is 60 microamps. This all adds up to
140 microamps, which is reasonably close to the 120
microamp figure. According to Duracell, the capacity of an
MN1604 battery at low drain rates is about 600 milliamp
hours. If you divide 600 mAh by 140 microamps, you
come out with a battery lifetime of about 4285.7 hours,
which is about a half a year.
Duracell guarantees a shelf life of five years, after
which the battery should still provide acceptable service.
So, I think the self-discharge current in the battery must be
lower than the current consumed by the circuit by at least
a factor of 10 and probably closer to a factor of 50 to 100.
Does this all make sense to you or did I screw up my math
North Saanich, BC
AThis question and circuit (from the January 16, 2016 N&V weekly newsletter) is based on a Q&A column by my predecessor, T.J. Byers. From the Texas Instruments datasheet, Iq
(current with IC not sourcing current to load and not in
a switched mode) is 25 microamps per channel (we are
only using one of the four comparator channels). For the
2N3906 transistor, the collector cutoff current, ICco, is
50 nanoamps (0.050 microamps) which I will use 0.100
microamps for the two transistors in parallel (the LED will
limit the current, but this is a rough estimate).
Adding in the 37 microamps for the threshold
adjustment divider and 43 microamps for the reference
voltage divider (I calculated the same values you did),
the total current when the LED is not illuminated is 105.1
microamps which is lower than Mr. Byers’ calculation.
However, he may have used a different datasheet or had
different operating conditions. Since the original calculation
is approximately halfway between your calculation and
mine, it is reasonable to use the 120 microamp current
draw for the circuit. (See Q&A SIDELINES for links to the
Q & A
n FIGURE 1.
n FIGURE 2.
In this column, Tim answers questions about
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