Add extra output to a boost converter
The added output has its own voltage regulation.
Vladimir Oleynik, Moscow, Russia; Edited by Paul Rako and Fran Granville -- EDN, November 17, 2011
Designers use step-up-converter ICs in battery-powered portable equipment. These chips usually provide one output with a fixed or an adjustable voltage. Some chips contain an LBI/LBO (low-battery-in/low-battery-out) function. The chip manufacturer intends for these pins to be used for monitoring a low-battery condition and to warn gadget owners when a battery goes flat. You can instead use this function to provide an extra voltage output.The Maxim MAX756 boost converter
provides a fixed output of 3.3 or
5V at 300 and 200 mA, respectively
(Figure 1). The input voltage can
range from 0.7 to 5.5V. For low-battery
detection, the part has on-chip circuitry
comprising a comparator, a reference,
and an open-drain MOSFET. When
the voltage at the LBI input is lower
than its threshold level of 1.25V, the
MOSFET at the LBO output sinks current
to ground.
You can use these components to
make a second output with a regulated
voltage (Figure 2). R1 and R2 determine
the secondary output voltage according
to the following equation: Output 2
=VREF(R1+R2)/R2, where VREF is the reference
voltage, which is 1.25V for this
chip.
You can also use the LBI/LBO function
to make a second boost converter
(Figure 3). The CD4093 quad Schmitt-triggered
NAND gates, inductor L2, R2
through R4, Q1, D1, C1, and C2 compose
this boost converter. Add C1 and R2 to
IC1B to make a free-running oscillator
that IC1A gates on. For the values of
R2 and C1 in the figure, the oscillator
frequency is approximately 17 kHz. R1
pulls up the open-drain LBO output.
When the voltage at the LBI pin
is lower than 1.25V, the LBO pin is
low, thus allowing operation of the
IC1B oscillator. IC1C and IC1D drive power MOSFET transistor Q1. When Q1 is on, it pulls
current from inductor L1. When Q1 is off, this energy
charges capacitor C2 through flyback diode D1. You apply
feedback with resistor divider R3 and R4 to determine
the Output 2 voltage, according to the following equation:
Output 2=1.25V×(R3+R4)/R4. IC1 gets power from
Output 1.
The voltage at Output 2 is a function of the output current
and the input voltage (Figure 4). If you have adequate input
voltage, the output graph shows a flat section where the IC’s
regulation is effective (figures 5 and 6).
Talkback
-
Duane,
You can add-in current sensing with a series resistor between the diode and output cap, a single transistor and two resistors will allow you to limit the voltage drop to about 0.2V peak under normal operation, though the tricky part may be how to feed this signal into this simple controller - from first blush I only see a shutdown input and a reference output that might be used to disable the output in an overload situation (short the reference to ground or push the controller into shutdown). The internal switching FET may already have overcurrent detection, but this may be a way to add it to the extra output.
Regarding the first example - we are essentially building an LDO there. What is wrong with using a single LDO part instead of using a discrete FET and resistors to construct an LDO? If the input voltage is lower than the output (boost operation) and guaranteed higher than the min LDO input voltage, then it is much more efficient to run the LDO directly from the input instead of adding its current to the boosted output, doubling up the losses.
Cor van de Water - 2011-21-11 13:24:05 PST -
The regulation problem at low voltage as well as the limited current is due to the low switching frequency and the low inductance value, causing this output to run in highly dis-continuous mode.
My guess is that at 4V input and 40mA output, the current in the inductor ramps up to about 1/2 Amp until the series resistance of the inductor limits the current during most of the approx 5us ON-time, then when the FET is turned off, the current ramps down in about 2us and then stays at zero for the remaining 90% of the cycle, where the output capacitor will have to supply the current. If you increase the inductor value and/or increase switching frequency, the current peaks will reduce, the operation may even become continuous so the ripple reduces and it will regulate at lower input voltages, if necessary while also reducing the FET and inductor series resistance if that still dominates the max ON current.
Cor van de Water - 2011-21-11 12:36:27 PST -
This is a very clever use for the LBI/LBO pins, but by using it you give up any current limiting and other protection circuitry normally present in a boost converter, so it should be used with caution.
Duane Hall - 2011-21-11 12:35:59 PST
