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How does 1N4148 Diode work

Started by Sherldonnnn, July 11, 2018, 11:34:18 AM

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I have 2 PWM ports, one generates 5V PWM and another at 3V. But only 1 will be generating at a time. However, their output is going to a same wire or pin. So, to avoid the current to pass into each other, I want to put diodes on each port as protection. For this, I would need fast switching diodes as PWM frequency is KHz or few MHz. I searched for fast switching diodes and found 1N4148. It does what I need but its forward voltage is 1V. So, for 3V and 5V, it will blow up ? or am I understanding the datasheet wrong ?
Right,as we see in the page name " Ten common electronic components that electronic engineers love" ,With the development of semiconductor materials and technology, various semiconductor materials, doping distribution and geometric structure have been developed, and many kinds of crystal diodes with various kinds of structure and different functional applications have been developed.Crystal diode is a semiconductor devices at both ends of solid-state electronic devices. The main features of these devices are nonlinear current voltage characteristics.

I read in this answer that forward voltage is voltage drop ? so if its voltage drop, whats the max voltage that it can support ? also 3V will come out as 2V, so I will be required to level it up to 3V again, right ?


Some comments:

  • If you want the circuit to work, you'll need to put a pulldown resistor between your output and GND.  Else there's nothing that will pull the output signal 'low', as the diodes can only pull the signal 'high'.
  • Foward voltage is indeed voltage drop. Whether 1V drop is a problem, depends on the receiving circuit.  If 2V is enough for the receiving circuit to detect it as a 'high', your diode circuit is perfectly fine, and there is no need to level it up to 3V again.
  • If you want to avoid the 1V drop, I'd recommend to implement the below twice (and connect the outputs together). A standard transistor BC547 should do the trick. Mind however that the circuit inverts the signal.