From: Jason White <whitewaterssoftwareinfo@gmail.com>
To: Microcontroller discussion list - Public. <piclist@mit.edu>
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Date: Thu, 6 Feb 2020 03:32:32 -0800
Subject: Re: [EE] 78L05 (SO8) overshoot on removal of over-current event?
Thread-Topic: [EE] 78L05 (SO8) overshoot on removal of over-current event?
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Hi Spehro,

Thank you for doing measurements on an actual unit. While they don't
totally confirm my observation they definitely suggest that the
overshoot is possible.

I recall the the 78L05 had only a 100nF ceramic output capacitor
(unknown voltage and dielectric) situated directly next to it. It had
perhaps 110uf of input capacitance mixed between electrolytic and
ceramic. The >100mA load was applied for 1 second and removed for 1
second with a fast rising and falling edges - on removal of the load,
micro-controller failure (smoke) would occur.

The PCB was 2 layer and measured roughly 4"x6". The regulator was all
the way to one side (bottom right) while the 5V load
(micro-controller, optocouplers, transistors, LEDs, etc.) was
distributed all over the PCB, mostly toward the other side (top left).
Being a 2 layer PCB, there was probably very little continuous
power/ground plane. Perhaps trace inductance did play a role in the
instability and overshoot.

-Jason White

On Thu, Feb 6, 2020 at 4:26 AM <speff@interlog.com> wrote:
> Hi, Jason:
>
> Just out of curiosity I tested a 78L06 with a AO3400  MOSFET applying
> and removing a 20 ohm load (causing the output voltage to drop to about 4=
..5V)
> Input capacitor 1uF 1206 ceramic X7R
>
> With an output capacitor of 100nF the overshoot was about half a volt.
> With 1uf it was less- about 300mV with a duration of about 25us.
>
> If I remove the capacitor entirely I can see a huge overshoot.
>
> Rise & fall times of the order of 75ns.
>
> Of course a fast fall-time load with layout inductance can lead to
> a voltage spike on the board if there are not bypass capacitors
> properly situated to absorb the energy stored in the inductance.
> The fault current can be in the hundreds of mA, so a relatively
> large capacitance (hundreds of nF or 1uF) would be good.
>
> Best regards,
> Spehro Pefhany
--=20
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