Use op amps to make automatic-ORing power selector
Replace diodes with FETs to reduce power loss.
Bob Zwicker, Analog Devices, Fort Collins, CO; Edited by Martin Rowe and Fran Granville -- EDN, August 11, 2011
Many systems must select among two or more low-voltage dc-input sources, such as an ac adapter, a USB (Universal Serial Bus) port, or an onboard battery, for example. You can implement this selection using manual switches, but automatic switching is preferable. You usually want to use the highest-available input voltage to power your system. You can accomplish this task using a Schottky-diode ORing scheme (Figure 1). Unfortunately, the forward-voltage drop of a Schottky diode ranges from 300 to 600 mV. This voltage wastes power, creates heat, and decreases the voltage available to your system.
Efficient voltage ORing requires
only a P- or an N-channel MOSFET, a
suitable op amp, and a few passives. This
Design Idea describes the application
of voltage ORing to positive dc-power
rails. The P-channel-MOSFET design
is suitable for low-power, single-supply systems operating at 3.3V or higher, and
the N-channel MOSFET fits situations
in which the bus voltage is lower or the
current is higher and a suitable op-amp
bias voltage is available.Positive current flows from the MOSFET drain in an N-channel-FET design. In a P-channel design, the current flows from the MOSFET source. The MOSFET’s drain-body diode would defeat rectifier operation if the usual current flow (for switching or amplification) were used.
Your first design
task is to choose a suitable
MOSFET. The
MOSFET’s worst-case
on-resistance must be
low enough so that the
I×R (current-times-resistance)
drop with full-load
current is low enough to accomplish the design objectives. A
0.01Ω MOSFET has a 50-mV forward-voltage
drop when 5A flows through it.
Be sure to consider power dissipation
due to R×I2 and the resulting temperature
rise.Your second design task is to choose an op amp. The op amp must be able to operate with the voltages involved and to adequately drive the MOSFET’s gate voltage. The P-channel design requires a rail-to-rail I/O type. A single-supply op amp is adequate for the N-channel design. Another important consideration is the op amp’s input offset voltage, VOS. The total ±VOS window must be less than the maximum desired voltage drop across the MOSFET. For example, if you permit a 10-mV forward-voltage drop at full load, then the op amp should specify an offset voltage of ±5 mV or better.
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You can consider the MOSFET’s variable on-resistance as the element with which the op amp senses current. When you apply reverse voltage, the MOSFET de-enhances, the I×R voltage drop increases, and the op amp’s output ends up at the appropriate supply rail, driving off the MOSFET as hard as it can.
With light-load conditions and a given offset voltage, the op amp tries to servo the voltage on its power-output-sensing input to the voltage on its power-input-sensing input plus the offset voltage. With R2 open-circuited, the op amp has no intentional external offset. If the op amp’s offset voltage were in the unfavorable direction, a sizable reverse-cutoff current would occur if the input-power bus were to fall to a lower potential than the output-voltage bus.
Figure 4 shows current-voltage test data for the operating region. The complete design, including intentional offset, produces the green curve. The equivalent of an unfavorable internal offset and no intentional external offset produces the blue curve. Although the green curve sacrifices some forward-voltage drop at light-load conditions, its forward voltage is always less than the full-load maximum. The intentional offset avoids any significant reverse current in the MOSFET. This design can switch at the 0A current transition, at which the leakage-current MOSFET’s drain-body diode is likely to dominate.

Talkback
-
Bob,
First of all, you need to understand that any Op Amp could be used as a comparator. In your case, each Op Amp compares voltage at the channel input with acting output voltage.
By definition the buffer amplifier has high input and low output impedance and in case of connecting their outputs together they will load each other. You need comparators to select right channels and shot of other.
So for me really surprise that you do not know that and I am not sure that you understand how your circuitry works.
Vladimir Doubovis - 2011-7-9 16:13:55 PDT -
Vladamir:
Of course, this is a design idea which does not claim to include a thorough exploration of everything that can go wrong. However:
1) Your use of the word "comparator" strongly suggests that you do not understand the circuit. The op amp is operating as an op amp follower with the MOSFET inside the FB loop. The op amp + MOSFET tries to make the MOSFET output follow (the input - offset). The circuit will not work with a comparator.
2) Given that the MOSFET channel is resistive, you cannot have reverse current flow without reverse voltage across it. If the circuit is set up correctly, anything approaching reverse voltage across the MOSFET will cause the MOSFET to be turned off by the op amp. I hesitate to say that anything is ever impossible, but reverse current flow with this circuit is just about impossible unless it is implemented incorrectly or something is broken.
3) It is reasonable to expect the overall power system to employ adequate bypassing which should go a long way to preventing oscillation.
Bob Zwicker - 2011-3-9 18:10:18 PDT -
Bob,
I do agree that my first phrase missed word not and should be read:
The voltage drop could be reduced by paralleling a number of Schottky diodes, but of course it could not be reduced to a few millivolts.
Actually, heating up of Schottky diodes could help to reduce voltage drop across significantly and bring it close to 140mV. I did participate in such project long time ago.
The schematic employs Op Amps as comparators that means they should have some hysteresis, so applying very close values voltage sources could create situation when a couple of MOSFETs will be ON.
In case of using BJT instead of MOSFET it is clear we can get short between two channels. Maybe it will be also case there because gate voltages are present for 2 channels.
The other issue in section 4 was regarding output impedance of different voltage sources. If you need specify them and attach load to output of this circuitry. So as a result turning ON one of MOSFET, the input voltage in this channel will be less for I_load * Ro_source1.
We can have a situation that actual input voltage for this channel could be equal or less that unloaded voltage value at the second channel and the second channel will be on. So some voltage ringing seems to me should be in place.
Vladimir Doubovis - 2011-31-8 17:45:09 PDT -
Vladimir:
First I would say that using more or larger schottky diodes will only achieve a small reduction in diode Vf. Getting below 200-300 mV drop with significant current would be very difficult even with leaky low voltage schottky diodes.
If two input voltages are identical the circuit will tend to cause the inputs to share the load current(although that is not a goal per se.) The important point is that each MOSFET will pass current only in the desired direction so that you will not get reverse current. If you reduce a source voltage slightly to cause that the MOSFET will be shut off.
The other point is that the amplifiers are individually compensated. So long as you have reasonable bypassing of inputs and outputs, the control loop behavior is smooth and continous without oscillation; dare I use the word "graceful." Calculate it correctly, try it, and you will believe............
Bob Zwicker - 2011-26-8 22:02:04 PDT -
Thanks, Patrick
If I understand you correctly, I think you need current limiting. If you are saying that the intrinsic diode prevents such, you are correct but adding only back-back FETs still leaves out any control loop for regulating current. The best thing I can think of is a seperate MOSFET and associated control loop for limiting current. Of course this added FET would use the FET conventionally so that the intrinsic diode would not be able to conduct.
I do not know what your voltages are but do not be afraid to use a small real inductor-based boost (such as ADP1612/3) instead of a pump. I actually did that with my Nch demo board but I decided to exclude it from the article.
Bob Zwicker - 2011-26-8 21:52:15 PDT






















