I recently saw a circuit that implemented a BAV99, which is just two diodes, between +3.3V and GND. How does this work? Based on the spec sheet, the fwd voltage varies based on the current. Am I to interpret it as the first diode connected to ground dictating that the voltage abide by the fwd voltage (that is, pin 3 shows 1V given 50mA), and that the upper diode to +3.3V doesn't come into effect until the pin 3 voltage is at (+3.3V - 1V [upper forward voltage])= +2.3V? But how is +2.3V ever reached if the bottom diode clamps it to its own forward voltage?

Any help is greatly appreciated - thanks!

  • \$\begingroup\$ Please provide the circuit. Until you do, nobody can help you, since we don't know for sure what you're talking about. \$\endgroup\$ – WhatRoughBeast Apr 23 '15 at 1:17

The BAV 99 is two diodes connected in series (cathode of one connected to anode of the other).

I think you are referring to it used like this:


simulate this circuit – Schematic created using CircuitLab

With the diodes oriented this way, they will not conduct as long as the signal remains between Vcc and Gnd. If the signal goes more positive than Vcc + 0.7 volts, then D1 will conduct, preventing the signal from rising above Vcc + 0.7 volts.

Similarly, if the signal tries to go more than 0.7 volts below ground, D2 will conduct.

This arrangement is used to ensure that inputs to an IC do not go outside the safe voltage range of the IC.

| improve this answer | |
  • \$\begingroup\$ This not only makes great sense because of your explanation, but is also a pretty cool trick to use. Thanks! \$\endgroup\$ – EE_padowan Apr 24 '15 at 2:09

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