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I’m building a simple DC-UPS using two power sources and Schottky diodes for OR-ing.

The diode model I’m using is MBR2545CT (dual Schottky, common-cathode, TO-220).

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The problem • Source 1: 20 V • Source 2 (disconnected): I still see ~17 V on the second input • Load and capacitors are disconnected • Both diodes behave the same

This means reverse isolation is failing — the second input receives almost the full voltage from the first source.

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Measurements

Using the multimeter diode-test mode: • Cathode ↔ Anode drop = 0.165–0.170 V on both diodes inside the package • Reverse direction shows no open circuit • Two separate MBR2545CT modules show the exact same behavior

Does this indicate defective or counterfeit MBR2545CT parts? Has anyone seen Schottky modules with such low diode-test Vf (0.16 V) and reverse leakage strong enough to pass 17 V backwards?

Is there anything I might be missing, or should I just replace these diodes?

scheme design

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    \$\begingroup\$ In general, for diodes there's an inverse relationship between forward voltage and reverse leakage. The lower the forward voltage, the greater the reverse leakage and vice-versa. That is the tradeoff. This is a fundamental concept to know when designing with diodes. A silicon PN diode will generally have lower leakage than a Schottky diode, but you can get Schottky diodes that have lower leakage than other Schottky diodes. Again, the tradeoff will be forward voltage. If you want the best of both worlds, you'll need a MOSFET ideal diode controller or power multiplexer. \$\endgroup\$ Commented 2 days ago

2 Answers 2

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Schottky diodes leak current in reverse much more than standard diodes.

So what you measure with a high impedance multimeter is normal.

The diodes are not faulty.

The data sheet says it can leak 0.2mA at room temperature when there is the maximum 45V voltage in reverse over it. So it can be approximated as 225 kohm resistor. The multimeter impedance is unknown but likely 1 Mohm or above, so it will explain also why 20V results into leakage that measures about 17V.

Now, another thing is, since they leak current in reverse, they might not be the correct solution for your circuit if the current leaks into a lithium battery, possibly overcharging it, which is an abnormal condition that a battery protection circuit needs to handle constantly, instead of only when something has failed. If your battery pack even implements protection of any kind.

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Totally fine.

Multimeter in diode mode measures with very small current where the schottky has very low drop.

Regarding voltage you measure, the power-schottky especially low-voltage diodes have huge leakage (reverse) current, sometimes even 20mA. With 100k input impedance of multimeter it can develop a huge voltage drop across multimeter input you measured.

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    \$\begingroup\$ so i will always see voltage on second anode right? like very small current only- i just scared since second anode is lithium bat and its can fire my home) \$\endgroup\$ Commented 2 days ago
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    \$\begingroup\$ Yes, the D2 passes the current from cathode to anode to Li-ion. In MBR2545CT this current is about 0.2mA (at 25degC) and 40mA (at 125 degC). If the Li-ion has a balancer with overcharge protection it shouldn’t cause a problem. \$\endgroup\$ Commented 2 days ago
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    \$\begingroup\$ @MichalPodmanický It shouldn't be a problem but it will push charge into the battery while it shouldn't in the first place, and then the overcharge protection circuit intended to be the last line of protection from abnormal operation has to deal with it as if abnormal conditions are now the normal conditions where battery will always overcharge. Simply relying on it is not a very good idea. \$\endgroup\$ Commented 2 days ago
  • \$\begingroup\$ @Justme if the diode in reverse looks like ~200kΩ (as in your answer), wouldn't something like 400kΩ to ground divide this down to just less than the full charge voltage of the battery without wasting noticeable power? That would then mean the overcharge protection circuit wouldn't have to do anything \$\endgroup\$ Commented 2 days ago
  • \$\begingroup\$ @ChrisH It would then always discharge the battery, albeit slowly, and the overdischarge protection is necessary to cut the battery off so it does not damage it with undervoltage. The 0.2mA leakage is a typical value at 25°C. It may leak much more at higher temperatures. Perhaps something else than a Schottky diode is just more suitable here. \$\endgroup\$ Commented 2 days ago

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