Power-OR Circuit Help Please

I am in need of a circuit that allows me to have a 5v power supply and a 15v power supply feed the same Vin node of a SMPS.

I used a common cathode diode (Vishay SS8P4C-M3/86A) and simply tied the 5v to anode 1 and 12v to anode 2, with the cathode going to the SMPS input.

When powering from 12v (anode 2), the 5v supply pin (anode 1) goes to 12v. I realize there is back currents through the diode, but they are NOT negligible. I fried another device (supplying the 5v) when I plugged it in and had it already powered by 12v.

So I would like for this to never happen again obviously. My research shows that I need to make a "power ORing" circuit which makes use of a FET in an ideal diode type situation.

Does anyone have a basic schematic for reference?

Are there a class of chips out there that do this? Linear makes some chips but Vin minimum is 9v on the ones that can handle 15v, and the Vin maximum is less than 15v on ones that can handle 5v.

Why did this diode allow such dangerous backfeeding? Is it just a poor choice of component, but the design is ok? I see power-oring diodes all the time in schematics.

Any help would be appreciated. Thankyou!

NVergunst said:

... there is back currents through the diode, but they are NOT negligible. I fried another device ...

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Depends on what you consider as "NOT negligible". Schottky diodes of this forward current rating typically provide reverse currents of several to hundreds of µAs (@ room temperature; mAs @ high temperatures).

NVergunst said:

Why did this diode allow such dangerous backfeeding? Is it just a poor choice of component, but the design is ok?

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Yes, the design is ok. You should either care for your power supply to be able to stand this reverse current (e.g. by using an appropriate Z-diode in parallel to its output, i.e. at the anode side of your OR-ing schottkys), or you take the higher voltage loss of std junction diodes, which reward you with much less reverse current.

Power switching by FETs is preferable, if you need near to zero voltage drop. But FETs are conducting bidirectionally, so you they can't operate self-controlled.

erikl said:

Depends on what you consider as "NOT negligible". Schottky diodes of this forward current rating typically provide reverse currents of several to hundreds of µAs (@ room temperature; mAs @ high temperatures).


Yes, the design is ok. You should either care for your power supply to be able to stand this reverse current (e.g. by using an appropriate Z-diode in parallel to its output, i.e. at the anode side of your OR-ing schottkys), or you take the higher voltage loss of std junction diodes, which reward you with much less reverse current.

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I would need a zener diode with a low enough turn on current that it should "always" be on, yet high enough to withstand maximum reverse current correct?

I do not really need a low forwards voltage drop, but backfeeding is not acceptable. So I need really low reverse current. If the 5v supply is on as is the 12v supply, and I get say 10uA of 12v back through the diode how does this effect the 5v supply? What does it actually do? From my previous experiment, the 5v rail was brought up to 12v. So a 5v zener that turns on at < 10uA is required correct? And since the zener will never conduct unless it is being backfed from the power supply ORing, I do not really need a current limiting resistor on the zener right?

I do not really need a low forwards voltage drop, but backfeeding is not acceptable.

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You can use regular Si rectifier diodes instead of schottky. They have a very low reverse current.

So what actually happens when 10uA of the 12v rail passes back into the 5v rail? Does the 5v rail try and go to 12v, or what happens? How do you model this? Ideally I want it to always be 5v, just somehow throw that current to ground, which I guess is what the zener is for. But what if the zener turns on at 10uA and only 9uA comes through?

HI, i think your zener diode should solved this problem, try and connect as a voltage regulator your output voltage should'nt go beyond the zener voltage.

The possibly most simple solution is to modify the 5V supply to handle 10 or 100 uA reverse current safely. You didn't tell anything about the involved power supply units, so noone can tell, how to model the effect.

If you connect a 1K resistor across the output of the 5V supply, the 10uA will drop 10mV across this so will be ignored by virtually everything electronic.
Frank

FvM said:

The possibly most simple solution is to modify the 5V supply to handle 10 or 100 uA reverse current safely. You didn't tell anything about the involved power supply units, so noone can tell, how to model the effect.

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The 5v supply is from a USB bus. I can only modify what comes out the cable for the peripheral. The 12v supply can be anything like a wall wart or battery between 5 and 18 volts.

chuckey said:

If you connect a 1K resistor across the output of the 5V supply, the 10uA will drop 10mV across this so will be ignored by virtually everything electronic.
Frank

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And by across, do you mean in series with the 5v? 10uA * 1kohm = 10mV.

So:

5v ---> 1k ---> anode of ORing diode --> cathode of ORing diode --> internal power system --> load --> ground

NVergunst said:

And by across, do you mean in series with the 5v? 10uA * 1kohm = 10mV.

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No. In parallel to the 5V supply is meant. I'd use a 5.6V (5V6) Z-Diode instead (in parallel, too).

So coming off of the 5v anode would be a 1k to ground? I see this working if the 5v supply is floating or disconnected. Then all the backwards current will be directed to ground and everyone is happy. However, the 5v supply will almost never be inactive. So it is pushing between 100mA and 500mA @ 5v down this path even when I do not need it to. It is a very picky supply, and cannot withstand much apparently. If this is the case, then how will the resistor work?

Yesterday I modified my schematic to:

**broken link removed**

I should change the diodes to be 2 individual Si diodes. However after reviewing my internal power subcircuit, I don't think I could handle a full 0.7v drop. Perhaps a Si diode on the wall power, and a FET on the USB power?

The 1k resistor will consume 5 mA at 5V, which far above expectable diode leakage currents at enviromental and elevated temperatures.

So it is pushing between 100mA and 500mA @ 5v down this path even when I do not need it to.

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That's nonsense. The VUSB supply will only source current, if the voltage is 5V or below because it's voltage regulated.

Typically, the resistor will absorb the 100 uA leakage current and draw 4.9 mA from the USB supply.

What I meant was that if the 12v power is not connected, then the 5v USB will be sourcing 5mA just to power this resistor to ground. When the 12v is connected, how is what current goes where modeled? What is to stop the 5v supply from supplying 5mA for the resistor and that 10uA at 12v to have nowhere to go...

So just use the 5V6 Z-Diode solution (instead of the 1kΩ resistor) and all is fine!

The Z-Diode won't draw current from your USB supply, and it will easily swallow your 1 .. 10 .. 100µA reverse current via the Schottky diode and still keep the USB output at ≈5.6V .

---------- Post added at 18:23 ---------- Previous post was at 17:55 ----------

NVergunst said:

... I don't think I could handle a full 0.7v drop. Perhaps a Si diode on the wall power, and a FET on the USB power?

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A FET needs controlling, as FvM pointed out above. Why not stay with the Schottky diode? The suggested Z-Diode at the USB output will keep it at a safe level.