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I am currently working on a low power battery backed data logger. An Intersil ICL7673 backup switch is used to switch between the battery and external power. Because the circuit draws more than 30mA I have to use two mosfets (instead of transistors) as suggested in the datasheet:

schematic

simulate this circuit – Schematic created using CircuitLab

One of the problems I ran into is that the mosfets let current flow into both directions. This makes it difficult to check if the supply is turned on. Two addional diodes could be a solution for this, but do I even need the intersil backup switch then? Wouldn't a simple diode-or be a more efficient solution?

Because the battery input already has a diode I am thinking about just putting another diode in series with the power supply. That would make the whole circuit cheaper and consume less power.

EDIT: Just for clarification: The circuit draws a bit more than 30mA once every few minutes for a few milliseconds. The rest of the time it will be at under 100uA. I tried the setup with resistors as described in the datasheet but it seemed to draw around 500uA (not mA, sorry for confusing that in my first comment). If I am using a 10k and 1k resistor souldn't that alone use around 0.3mA?

schematic

simulate this circuit

EDIT 2: After trying the solution with two mosfets, I took out the ICL7673 and put a schottky diode between power supply and vcc. It works perfectly. As long as the supply voltage is greater than the battery voltage, the circuit will only draw current from the power supply. On the battery side I needed a diode for polarity protection anyway, so the voltage drop is unavoidable.

schematic

simulate this circuit

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1) You need a power manager IC like the ICL7673. Summing power using diodes such as a Schottky 1N5817 will cause a Vdrop of .40 to .50 volts. That is too much of a penalty when your main supply is only 3.3 volts. You must use mosfets that do NOT have a reverse protection diode built in. That is why you have current flow in the reverse direction on each mosfet.

2) If the design called for PNP transistors, what was wrong in using them? 30mA can be handled by any PNP such as a 2N3906. Bjt transistors will block reverse currents up to the voltage limit of the transistor. The current being consumed is so low that even bjt transistors cause very little power loss or voltage drop.

3) For PNP transistors you need a 1K to 3.3K resistor on the transistors base in series with the Pbar and Sbar outputs, else a high base current will flow to the IC. Vcc should not draw more current than expected regardless of the transistor type.

4) I looked at the IC datasheet and it shows what could be a 10K resistor from the base to the emitter of each PNP transistor. This is to make sure the transistors turn OFF when the drive signal (Pbar or Sbar) is OFF. The same resistors should be used if P-channel mosfets are used (with no reverse diode) , except the 10K resistor can be increased to 100K.

5) I suggest you re-think your reasons for the change you made, and which transistor type is the easiest to install with no weird behavior such as reverse current flow.

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  • \$\begingroup\$ Thanks for your quick answer! I tried the circuit with transistors but they seemed to draw a lot of current (several 100 mA). So I switched to mosfets. Do you think a PNP transistor such as the 2N3906 can be used with only a few uA continuously? And what would be the downside of just using a diode-or to switch between battery and power supply? \$\endgroup\$ – MaxMKA Jun 10 '16 at 21:52
  • \$\begingroup\$ For PNP transistors you need a 1K to 3.3K resistor on the transistors base in series with the Pbar and Sbar outputs, else a high base current will flow to the IC. Vcc should not draw more current than expected regardless of the transistor type. \$\endgroup\$ – Sparky256 Jun 10 '16 at 21:59
  • \$\begingroup\$ I actually had a 1k Resistor in series and got a high current consumption. Maybe something else was wrong with the wiring. I will try again and see if that works. Thanks again for your help! \$\endgroup\$ – MaxMKA Jun 10 '16 at 22:09
  • \$\begingroup\$ You should not have such high current consumption unless the IC is faulty or you have a IC pin shorted. Is Vcc consuming the excess current or is it the IC? \$\endgroup\$ – Sparky256 Jun 10 '16 at 22:15
  • \$\begingroup\$ I looked at the IC datasheet and it shows what could be a 10K resistor from the base to the emitter of each PNP transistor. This is to make sure the transistors turn OFF when the drive signal (Pbar or Sbar) is OFF. \$\endgroup\$ – Sparky256 Jun 10 '16 at 22:21
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If you really want to use MOSFETs, then you need to use two on each node:

schematic

simulate this circuit – Schematic created using CircuitLab

as the internal diodes are opposing each other, there will not be any diode conduction provided the drain to source voltage is smaller than \$V_{BR(DSS)}\$

Replacing M1 and M2 in your diagram with this arrangement solves the issue of reverse current flow.

There are very few MOSFETs sold today with the substrate exposed (which permits you to reverse bias the body diode to prevent conduction).

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