# LTC4353 secondary input voltage drop

I'm experiencing a strange behaviour on a board where I'm using LTC4353 as a "12V Main and Auxiliary Supply Diode-OR" . The circuit I'm working on is depicted below (it's pretty the same circuit found on page 11 of the original LTC4353 datasheet)

When VIN exceed 9V OUT value it correctly feeds OUT (with a value around VIN minus the voltage drop on Q2 RDSOn). When VIN falls below 9V U5 correctly switch the power source (shutting off Q2 and enabling Q1) and +BATT feeds OUT. BUT In this condition (Q1 enabled) I can measure an unexpected voltage drop around 0.6V on OUT (I was assuming being near +BATT minus voltage drop on Q1 RDSon ... far less than 0.6V). I cannot understand why this's not working and if this's a standard behaviour why it's not symmetrical.. BTW the circuit have been tested with a light load conneted between OUT and GND (1kOhm).

On page 8 of the datasheet I found "The disadvantage of these approaches is the diode’s significant forward-voltage drop and the resulting power loss. The LTC4353 solves these problems by using an external N-channel MOSFET as the pass element. The MOSFET is turned on when power is being passed, allowing for a low voltage drop from the supply to the load. When the input source voltage drops below the output common supply voltage it turns off the MOSFET, thereby matching the function and performance of an ideal diode".

In my opinion this device should work as an Ideal Diode power supply OR circuit but it seems not to work properly ... someone suggested me that the voltage drop should be caused by the body diode effect (which one?) but I cannot understand how it can lead to such a voltage drop on OUT since both MOSFETs drains are equipotential to OUT.

• This chip uses a pump charge boost voltage converter to create a 12V gate voltage for turning on the FET high side with low resistance and thus low mV drop . Check all cap voltages are correct for gate bias voltage. – Tony Stewart Sunnyskyguy EE75 Jul 17 '18 at 12:26
• in short you suggest that Q1 it's not fully saturating? – weirdgyn Jul 17 '18 at 12:31
• Vgs must be >2.5x Vgs(th) see what you did wrong – Tony Stewart Sunnyskyguy EE75 Jul 17 '18 at 12:35
• Record all pin voltages where it fails in your test with above – Tony Stewart Sunnyskyguy EE75 Jul 17 '18 at 12:47
• Ciss of Q1/Q2 is around 3450pF charge pump capacitors are about 56nF (they should fit). I'm going to measure levels but I was wondering if this's a really weird behaviour or maybe it's just a feature! :-) – weirdgyn Jul 17 '18 at 12:52

I think the problem is with your logic. If you go to the datasheet, on page 5, this is what it says:

With the voltage divider you currently have, it's not possible to enable the MOSFET which switches on +BATT (Q1 on your schematic).

You have a divider ratio of:

$$\dfrac{165k\Omega}{165k\Omega+10k\Omega} \approx 0.94$$

(Are the resistors swapped? This could fix the issue)

So whatever your input voltage is (Vin) gets scaled by that and depending on whether that value is greater than 0.6V, you can enable or not the +BATT MOSFET. So, even if Vin drops to 3.3V, you still can't turn that on (would have about $3.3V(0.94)=3.1V$ at the enable pin—far away from turning on the gate since it needs to be $\leq 0.6V$).

The reason why you see a voltage drop of about 0.6V is because of the body diode. Since +BATT in your scenario is greater than Vin, +BATT still drives the output but instead of having a path through RDSon of the corresponding transistor, it does it through the body diode of your MOSFET.

I'll show you what is happening:

simulate this circuit – Schematic created using CircuitLab

Hopefully this helps.

• Ohhhh god YESSSS the resistors are swapped......thnx... I think I checked more than once but..... – weirdgyn Jul 18 '18 at 9:40
• I'll swap them and test again... I will probably have to recalibrate also because 165k doesn't seem to be the correct value to have 0.6V (with 9V input) – weirdgyn Jul 18 '18 at 9:41
• @weirdgyn correct. You may have to decrease to, say, 145k and keep the other one (10k). That way, the ratio would be ~0.065. So if Vin fell below 9V (e.g 8.95V), then the EN pin would see 8.95V(0.065)=0.58V—enough to turn on the aux battery. – Big6 Jul 18 '18 at 16:02
• Yep I will give a try with 150k... – weirdgyn Jul 18 '18 at 16:04