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schematic

  1. Initial situation:

    • Q1 is HIGH, Q3 is HIGH, Q5 is LOW
    • D1 is forward biased, D2 is forward biased, D3 is reverse biased
    • Therefore, Q3 is protected against shorting by D2

  2. Next step:

    • Q1 is changed to LOW, Q3 remains HIGH, Q5 remains LOW
    • Nothing changed with Q3 and Q5, so, of course D2 is still forward biased and D3 is still reverse biased
    • Q1 did change, however, since only a very short period of time has passed, D1 didn't switch yet and is still forward biased as well
    • Therefore, for a very short period of time, current can flow from Q3 HIGH, through forward biased D1 into Q1 LOW.

Is this true? Is this a problem? If not, why not?

I thought about setting every output pin to LOW for the necessary switching time according to the diodes' data sheet.

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    \$\begingroup\$ This doesn't make sense; these are parallel outputs. What are you ultimately trying to accomplish? \$\endgroup\$
    – Blair Fonville
    Jul 11, 2017 at 0:29
  • \$\begingroup\$ @BlairFonville Hi :-) This is only part of the full schematic. The parallel outputs are connected to a single microcontroller digital input pin. I need to read a HIGH if any of the shift register's output pins is HIGH. And a LOW only if all output pins are LOW. Should work like this, right? \$\endgroup\$
    – Liam
    Jul 11, 2017 at 0:38
  • \$\begingroup\$ @BlairFonville Like this \$\endgroup\$
    – Liam
    Jul 11, 2017 at 0:40
  • \$\begingroup\$ There's your answer then: Use an OR gate. \$\endgroup\$
    – Blair Fonville
    Jul 11, 2017 at 0:40
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    \$\begingroup\$ Diodes don't work that way. You'd have to have very high levels to reach the breakdown voltage for the diodes to conduct in reverse. \$\endgroup\$
    – Blair Fonville
    Jul 11, 2017 at 1:07

1 Answer 1

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No need to add further protection. You will be hard pressed to see any effect from the reverse recovery time of a small signal diode; the reverse recovery times are often in the single-digit nanoseconds and also somewhat proportional to the current at switching time, which will be small, and the recovery time will be much less than the switching time of the HC part.

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  • \$\begingroup\$ Thank you for your answer. What do you mean with "switching time of the HC part"? \$\endgroup\$
    – Liam
    Jul 12, 2017 at 21:26
  • \$\begingroup\$ Digital parts come in different families that have different input and output characteristics. For instance, the part you are using comes in both an HC and HCT version; the HC have input levels compatible with CMOS logic while HCT operates at %V logic and is compatible with the older TTL. HC family parts have propagation delays from input to output in the hundreds of nanoseconds, so any effect of a few nanoseconds of reverse recovery will not be noticed in your circuit. \$\endgroup\$
    – John Birckhead
    Jul 13, 2017 at 3:49
  • \$\begingroup\$ Let me try to say that in my own words: It takes a certain time (related to the propagation delay) for the output pin to change from HIGH to LOW, which is far greater than the time the diode needs to switch from forward bias to reverse bias. So the diode will already be reverse biased when possible harmful current would flow into the (now LOW) output pin. Did I understand you correctly? \$\endgroup\$
    – Liam
    Jul 13, 2017 at 22:26
  • \$\begingroup\$ I would say it like this: The amount of current you will get during reverse recovery of a small body diode is negligible - it is more akin to a very small (pf) capacitor being charged by your output than anything else. It is not damaging by any means. The reverse recovery time is so short that it is dwarfed by other parameters, notably propagation time, and so need not be accounted for in your design. In summary, it can be ignored. \$\endgroup\$
    – John Birckhead
    Jul 13, 2017 at 23:25
  • \$\begingroup\$ Well, okay. For me, the key argument would be that you're saying the amount of current during reverse recovery is negligible - and I'd be very happy with this - however, I'm only asking this question at all, because I found contrary statements on this: Here, it says "The current through the diode will be fairly large in a reverse direction during this small recovery time." \$\endgroup\$
    – Liam
    Jul 13, 2017 at 23:41

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