rises by more than 100mV above the other, the MOSFET
across that battery is switched on.
If the battery is being charged, this has the effect of shunt-
ing some of the charge current around that battery so that it
receives a lower charging current than the other, decreas-
ing the voltage differential over time, as the battery with
the lower voltage is then receiving more charging current.
If the unit is operating while the battery is not being
charged, the effect is to slightly discharge the battery with
the higher voltage until they are closer in voltage.
It’s a linear circuit, so the shunt current is proportional
to the difference in voltage. As the imbalance rises, so
does the shunt current until the limit of around 300mA
is reached. This is to prevent the MOSFET and resistor
from overheating.
Detecting a voltage difference
A resistive divider comprising two 10MΩ resistors and a
200kΩ trimpot VR1 is connected across the battery, before
diodes D1 and D2 so that their forward voltage does not
affect the calculation of the difference in voltages.
VR1 is adjusted so that the voltage at its wiper is exactly
half that of the total battery. This half-battery voltage is
buffered by voltage-follower op amp IC1a. This op amp has
a very high input resistance of around 40GΩ, resulting in a
low input bias current of approximately 250pA, so the high
values of these resistors (chosen to minimise the quiescent
current) will not result in a large error voltage. The other
half of the dual op amp (IC1b) compares the voltage at the
junction of the two batteries (from pin 2 of CON1) to the
output voltage from IC1a. If the upper battery has a higher
voltage than the lower battery then the half-battery voltage
will be higher than the voltage at pin 2 of CON1. That means
that the voltage at non-inverting input pin 5 will be higher
than at the inverting input, pin 6.
As a result, IC1b’s output will swing positive. The ratio
of the 390kΩ feedback resistor to the 10kΩ resistor that goes
to the battery junction (ie, 39:1) means that the output will
increase by 40mV for each 1mV difference in battery voltages.
Once the voltage at output pin 7 has risen by a couple
of volts, N-channel MOSFET Q1a will switch on because
its gate is being driven above its source, which connects
to pin 2 of CON1 via a low-value shunt resistor (47mΩ).
So current will flow from the positive terminal of the
upper battery, through diode D1, the 27Ω 3W resistor,
MOSFET Q1a and then the 47mΩ resistor to the negative
terminal of the upper battery.
Once this current starts to flow, it will also develop a
voltage across the 47mΩ resistor, which will increase the
voltage at pin 6 of IC1b, providing negative feedback. This
feedback is around 1mV/20mA, due to the shunt value.
This prevents Q1a from switching fully on. Rather, its gate
voltage will increase until the current through the 47mΩ
resistor cancels out the difference in the two voltages.Fig.1: the circuit for the Battery Balancer, shows the balancing section at top and low-voltage cut-out at bottom, based
around dual micropower op amps IC1 and IC2 respectively. IC1 drives dual MOSFETs Q1 and Q2 to perform balancing
when necessary; IC2 drives the indicator LEDs and disables IC1 using MOSFET Q3 when the battery voltage is low.2x 12V Battery Balancer
Reproduced by arrangement with
SILICON CHIP magazine 2019.
http://www.siliconchip.com.au