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Here is the atached schematic of the clamping diode condition. Isolated the connection to contoller and kept the end open. When USB input is varied above 5V the expected output is diode cathode voltage (5V) + diode drop, which I was getting when im simulating the same.

Tried to test the same condition on board level buy varying the USB input above 5V ,I'm seeing the same USB input voltage at output (diode anode). Please let me know what is the problem in real case.enter image description here

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    \$\begingroup\$ what's the problem? This sounds right. \$\endgroup\$ – Marcus Müller Jul 1 '20 at 12:08
  • \$\begingroup\$ Testing result it not same as simulation. If USB input is 10 V(continuous input), the voltage is at diode anode is 10V, but the actual to be 5v + diode drop. Let us know how to test this. \$\endgroup\$ – Rohan Kharvi Jul 1 '20 at 12:38
  • \$\begingroup\$ The diode is named after the German physicist Walter H. Schottky. It is also called Schottky (with a capital 'S'). \$\endgroup\$ – Transistor Jul 1 '20 at 13:09
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I suspect it's because your +5V supply can't sink current. In simulation, that 5V source will stay at 5V no matter what, but on the bench that 60 ohm resistor can probably provide plenty of current for your 5V circuitry and then some. Unless your power supply has the capability to sink current in order to maintain the output voltage at 5V, that 5V node will drift up to the input voltage until something dies and shorts. You can verify if this is the case by measuring your +5V supply to see if it tracks USB VCC.

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  • \$\begingroup\$ How can I test this clamping condition on bench. \$\endgroup\$ – Rohan Kharvi Jul 1 '20 at 13:16
  • \$\begingroup\$ As mentioned above, check your +5V supply line and see if it's tracking the USB VCC. \$\endgroup\$ – Cristobol Polychronopolis Jul 1 '20 at 13:19

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