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I have designed a board and it can be supplied by USB and/or D-Sub connector. The board gets the 5V from USB or another board via D-Sub connector.

I have placed 2 Schottky diodes to prevent reverse voltage between these 2 connectors in case of they are both connected.

This is the circuit:

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USB 5 V and connector 5 V both enter diodes and they get connected as 5 V on the cathode side of diodes. My first question is, does this circuit make sense?

I have supplied the board via USB and checked the D-Sub diodes anode voltage. It was same as cathode. Don't the diodes block the voltage? I placed them for this purpose.

I assumed there is a mistake on my PCB and tried the circuit on a breadboard.

This is the first circuit I have tried:

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I have read 4.9 V on the cathode of diode.

This is the second circuit I have tried.

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I have supplied 5 V from the cathode of diode to check does it allow reverse voltage. I have read 5 V on the probe.

After that I have reversed the diode to see what happens.

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I do not understand why it behaves this way. This is the datasheet of the diode.

What I was expecting was when I applied the 5 V to the anode, I would see 4.9-4.8 V on the cathode. This part is ok. If I apply 5 V to the cathode, I should read 0 V since the diode won't allow any voltage/current to that way. Is there something wrong with my circuit or did I misunderstand the behaviour of the diode?

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  • \$\begingroup\$ I believe (and somebody will correct me) that this is correct, but the diode blocks current (in the BAT60 case, more than about 1mA at 5V). Have you tried measuring on the other side of the resistor? \$\endgroup\$
    – Ron Beyer
    Commented Jan 17, 2020 at 12:02

3 Answers 3

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Starting at the bottom with the 1.5 volts being developed across the 3k3 resistor when fed 5 volts via a reverse connected BAT60A. Look at the data sheet: -

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The reverse current might be (say) 0.5 mA and, that current forced into a 3k3 resistor produces a volt drop of 1.65 volts. Just throwing numbers together.

Prior to that you did an open circuit test of a reverse biased BAT60A and you saw the same voltage at the output and input. Should this surprise you given that schottky diodes have high leakage current and your meter might have an input impedance of 10 Mohm?

Doesn't the diodes block the voltage?

They don't block reverse current or voltage anywhere near as well as regular silicon diodes but, regular silicon diodes have a larger forward volt drop.

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  • \$\begingroup\$ Thank you so much! I could see the reverse current on the voltage source. I did not expect to see the 5V on the anode of the diode. I've assumed it doesn't block anything though it certainly did. One more tiny question. Is this circuit a good application for my purpose? \$\endgroup\$
    – Doğuhan PALA
    Commented Jan 17, 2020 at 12:36
  • \$\begingroup\$ @DoğuhanPALA the only thing to worry about is the current that passes in reverse to the "other" supply line and the small forward volt drop degrading the 5 volt rail voltage to maybe 4.7 volts (under load). \$\endgroup\$
    – Andy aka
    Commented Jan 17, 2020 at 12:56
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In you last circuit with the diode reverse biased the current is about
I = V/R = 1.5/3.3k
~= 0.45 mA.

The BAT60A datasheet on page 2 shows that
at 25 c and Vr = 5V Ireverse is

  • 0.3 mA typical and

  • 1 mA max [!!!]

While it is not what you expected, the diode is acting entirely as the spec sheet says it should!

Worse - at 80 degrees C the reverse current is typically 18 mA !
Can this be true ? !!!!!!!!!!!!!!
Sadly, yes.
Schottky diodes have significantly higher reverse leakage currents than typical silicon diodes with the same voltage and current specifications, and this is a particularly poor example.

Larger current diodes typically have higher reverse currents.
A 200 mA BAT85 has about 2uA Ir at 25C and 10 uA Ir at 50C (fig 2).

The better than most SB1H 1A 100 V Schottky diode has Irmax of 1.0 uA at 25 c and 0.5 mA at 125 c.

Some Digikey 1A Schottky diodes

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  • \$\begingroup\$ thank you for your answer. I've seen the reverse current on the voltage source. I've assumed there shouldnt be any voltage on the anode side for some reason. Does adding schottky diode serve my purpose on protection? Is this a good application? \$\endgroup\$
    – Doğuhan PALA
    Commented Jan 17, 2020 at 12:42
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    \$\begingroup\$ @DoğuhanPALA Apply ohm's law and other principles in the usual way. The data sheet shows the BAT60 is far from ideal. | It passes about 1 mA - is that what you want? [No!]. . I told you it is a bad diode to use. I gave examples of much better diodes for your purpose. I gave you a web page list of diodes to look at. | The 1N5817 8 9 Schottky diodes are low cost, avaialable 1A rated and have MUCH MUCH lower Ir. | The BAT60 is a terrible choice for your application. Using any of the many diodes with much much lower Ir will do what you want. \$\endgroup\$
    – Russell McMahon ♦
    Commented Jan 17, 2020 at 21:10
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The BAT60a is a Schottky diode, and Schottky diodes "leak" a little when they "should" be blocking.

Doing the math, it is leaking about 0.5 mA in you last picture, which is well within its specifications at 5 V, see page 2 of the data sheet under "reverse current".

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