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Re: Restart on Power Failure?!

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  • James Stewart
    ... I agree with this. I would start with about a 100 ohm resistor. If it works, great, you can try larger values if you want. If it doesn t then try a 50
    Message 1 of 14 , Jan 3, 2006
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      --- In LinkStation_General@yahoogroups.com, Derek Taubert
      <taubert@g...> wrote:
      >
      > On Sun, Jan 01, 2006 at 04:24:16PM +0100, Schelte Bron wrote:
      > > On Wednesday 24 August 2005 18:50, aaldefaria wrote:
      > > > I have a question, is there way to set the Linkstation 160 to
      > > > resume after power failure meaning if I lose power in my
      > > > house I want the unit to turn itself back on after power
      > > > failure.
      > > >
      > > I have just published an article on the wiki showing how I
      > > achieved this through a very simple hardware modification.
      > > http://linkstationwiki.org/Articles/GeneralAutostart
      >
      > Interesting. A few concerns:
      >
      > 1) When the power button is pressed, the capacitor will drain
      > directly to ground through the switch contacts. That's going to be
      > quite a zap from a 100uF cap. At minimum, I would suggest putting
      > a resistor in series with the cap (something several times less
      > resistive than whatever the weak pullup to 3.3V is) and then using
      > a smaller cap.

      I agree with this. I would start with about a 100 ohm resistor. If
      it works, great, you can try larger values if you want. If it doesn't
      then try a 50 ohm or even a 10 ohm. Adding resistence to this circuit
      would also add delay to the startup which would allow you to use a
      smaller capacitor if you wanted to get the same turn on timing.

      > 2) Applying voltage (the charged cap) to an IC input after device
      > power is removed is a great way to cause latchup and permanent
      > damage over time. Have you looked closely at this trace to
      > determine what devices are connected to it?

      Yes, it would be nice to see what this is controlling. My guess would
      be a CMOS switch, but it could be a bipolar transistor.

      > I would advise against using this particular mod as is. A better
      > design would duplicate the switch behavior (0 ohms vs. Meg ohms)
      > rather than the resulting voltage on the pin (0V vs. 3.3V).

      Okay, you could but the charging capacitor in front of a transistor
      that is bridged across the switch. A little more complicated, but
      safer. I could draw up a simple schematic if there is interest.

      Of couse if no one has blown up their LS with the circuit as-is, then
      that might speak for itself.
    • Schelte Bron
      ... I don t have the impression you can create sparks with a 100uF capacitor charged to 3.3V, so I don t think this is a big problem. On the other hand I
      Message 2 of 14 , Jan 3, 2006
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        On Tuesday 03 January 2006 01:25, Derek Taubert wrote:
        > Interesting. A few concerns:
        >
        > 1) When the power button is pressed, the capacitor will drain
        > directly to ground through the switch contacts. That's going
        > to be quite a zap from a 100uF cap. At minimum, I would
        > suggest putting a resistor in series with the cap (something
        > several times less resistive than whatever the weak pullup to
        > 3.3V is) and then using a smaller cap.
        >
        I don't have the impression you can create sparks with a 100uF
        capacitor charged to 3.3V, so I don't think this is a big
        problem. On the other hand I wonder if it will still work with
        a smaller cap with a resistor in series. Anyway, the switch is
        not going to be used much after this modification. The LS will
        boot when the power is applied and will be shutdown with the
        shutdown command.

        > 2) Applying voltage (the charged cap) to an IC input after
        > device power is removed is a great way to cause latchup and
        > permanent damage over time. Have you looked closely at this
        > trace to determine what devices are connected to it?
        >
        Good point. I haven't checked what kind of circuitry the switch
        is connected to. I assumed it would be some part of the power
        supply. That should be able to handle a 3.3V reverse voltage,
        but I'm not sure. Should a diode in series with the cap not be
        able to prevent this reverse voltage? The disadvantage would be
        that the cap may take a longer time to discharge.

        > I would advise against using this particular mod as is. A
        > better design would duplicate the switch behavior (0 ohms vs.
        > Meg ohms) rather than the resulting voltage on the pin (0V
        > vs. 3.3V).
        >
        Thanks for sharing your concerns.


        Schelte.
      • Schelte Bron
        ... I am interested. ... Mine has at least survived a couple of cycles :-) Schelte.
        Message 3 of 14 , Jan 3, 2006
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          On Tuesday 03 January 2006 19:41, James Stewart wrote:
          > > 2) Applying voltage (the charged cap) to an IC input after
          > > device power is removed is a great way to cause latchup and
          > > permanent damage over time. Have you looked closely at
          > > this trace to determine what devices are connected to it?
          >
          > Yes, it would be nice to see what this is controlling. My
          > guess would be a CMOS switch, but it could be a bipolar
          > transistor.
          >
          > > I would advise against using this particular mod as is. A
          > > better design would duplicate the switch behavior (0 ohms
          > > vs. Meg ohms) rather than the resulting voltage on the pin
          > > (0V vs. 3.3V).
          >
          > Okay, you could but the charging capacitor in front of a
          > transistor that is bridged across the switch. A little more
          > complicated, but safer. I could draw up a simple schematic
          > if there is interest.
          >
          I am interested.

          > Of couse if no one has blown up their LS with the circuit
          > as-is, then that might speak for itself.
          >
          Mine has at least survived a couple of cycles :-)


          Schelte.
        • KeepIt SimpleStupid
          Power on resets in general are a PITA. A good control element to use for push buttons etc. is a LED/FET like the VACTEC VTL5C1. It s a nice package with
          Message 4 of 14 , Jan 3, 2006
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            Power on resets in general are a PITA. A good control
            element to use for push buttons etc. is a LED/FET like
            the VACTEC VTL5C1. It's a nice package with leads in
            and leads out.

            Other FET/LED DIP packages are available from
            www.digikey.com.

            RC circuits with a schmidt trigger are the de-facto
            standard for power-on reset, but they are not
            reliable.

            I designed a really good circuit for a car, but the
            timer part doesn't exist anymore (obsoleted by
            manufacturer). 0-10 sec, very fast reset, probably
            less than 10 mS.

            The hardest part is to get good reset once the timer
            has timed out.

            I'd try a delay on make timer from SSAC combined with
            the vactrol p/n. For giggles, I'd add an LM334 and a
            diode for polarity protection. When you do this you
            get a polarity protected 3-30V DC capable 2-wire
            isolated line that needs 10 mA that you can pretty
            much do anything with.

            What you need to do, is discharge the timing element
            when the device is off (no an easy feat) and discharge
            the timing element after it's been used for timing as
            well. Again, not necessarily an easy feat.

            That CAP thing is the primary reason many devices
            don't work after a brief power failure. It takes
            $10-$15 to do it right.
            I also learned the hard way, that it depands on how
            fast the power supply powers up.

            In a microprocessor based system I worked on, if I
            used a linear supply the standard RC and gate didn't
            work. If I used a switching supply, it worked. When
            I changed to a schmidt trigger, they both worked.

            Most timers, including commercial have very crummy
            reset times on the order of 100mS. Now you know why
            they say: unplug, wait 1 minute and plug back in.
            It's probably the reason it's not in the LS - cost!



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          • Schelte Bron
            ... I would really like to keep it simple, so how about a cap and a resistor in parallel connected to ground on one side and the other side connected via a
            Message 5 of 14 , Jan 5, 2006
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              On Tuesday 03 January 2006 01:25, Derek Taubert wrote:
              > 1) When the power button is pressed, the capacitor will drain
              > directly to ground through the switch contacts. That's going
              > to be quite a zap from a 100uF cap. At minimum, I would
              > suggest putting a resistor in series with the cap (something
              > several times less resistive than whatever the weak pullup to
              > 3.3V is) and then using a smaller cap.
              >
              > 2) Applying voltage (the charged cap) to an IC input after
              > device power is removed is a great way to cause latchup and
              > permanent damage over time. Have you looked closely at this
              > trace to determine what devices are connected to it?
              >
              I would really like to keep it simple, so how about a cap and a
              resistor in parallel connected to ground on one side and the
              other side connected via a diode to the on/off button signal?

              The diode will prevent any current flowing back into the LS
              circuitry. The cap will only be able to discharge through the
              resistor. That should take care of both concerns you raised.

              I have done some experiments and it works on my LS with a 6k8
              resistor and a 22uF cap. The LS reboots if the power has been
              disconnected for as little as 3 seconds.

              Do you see any more problems with this design?

              Actually, in my experiments I found that I could get all the
              functionality I need by just shorting the switch completely.
              The LS would boot when power is applied and I could shut it
              down with the shutdown command. It doesn't seem to shutdown or
              reboot automatically at other times. I expect the disk sleep
              function will also still work. Of course the on/off button is
              completely disabled in this situation.


              Schelte.
            • KeepIt SimpleStupid
              Not knowing exactly what the LS has you can try something like wire oring. Lets try some ascii art. ***************** Ground ! ! ! ! !
              Message 6 of 14 , Jan 5, 2006
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                Not knowing exactly what the LS has you can try
                something like wire oring.

                Lets try some ascii art.

                ***************** Ground
                ! !
                ! !
                ! !
                ****** *****
                / \ / \
                ! !
                cap switch
                ! !
                ******************** To whatever needs reseting
                by a contact closure to ground

                This way either one of the signals can be interpreted
                as a
                keypress. It's just that power-up reset is
                unreliable.
                Add your resistor if you care. I'm assuming that the
                LS uses a contact closure to ground, but it doesn't
                have to be.
                PS: It's drawn upside down.

                What chip/pin is the power switch going to. Is it
                connected to anything else like a pull up or pull down
                resistor?


                --- Schelte Bron <sbron@...> wrote:

                > On Tuesday 03 January 2006 01:25, Derek Taubert
                > wrote:
                > > 1) When the power button is pressed, the capacitor
                > will drain
                > > directly to ground through the switch contacts.
                > That's going
                > > to be quite a zap from a 100uF cap. At minimum, I
                > would
                > > suggest putting a resistor in series with the cap
                > (something
                > > several times less resistive than whatever the
                > weak pullup to
                > > 3.3V is) and then using a smaller cap.
                > >
                > > 2) Applying voltage (the charged cap) to an IC
                > input after
                > > device power is removed is a great way to cause
                > latchup and
                > > permanent damage over time. Have you looked
                > closely at this
                > > trace to determine what devices are connected to
                > it?
                > >
                > I would really like to keep it simple, so how about
                > a cap and a
                > resistor in parallel connected to ground on one side
                > and the
                > other side connected via a diode to the on/off
                > button signal?
                >
                > The diode will prevent any current flowing back into
                > the LS
                > circuitry. The cap will only be able to discharge
                > through the
                > resistor. That should take care of both concerns you
                > raised.
                >
                > I have done some experiments and it works on my LS
                > with a 6k8
                > resistor and a 22uF cap. The LS reboots if the power
                > has been
                > disconnected for as little as 3 seconds.
                >
                > Do you see any more problems with this design?
                >
                > Actually, in my experiments I found that I could get
                > all the
                > functionality I need by just shorting the switch
                > completely.
                > The LS would boot when power is applied and I could
                > shut it
                > down with the shutdown command. It doesn't seem to
                > shutdown or
                > reboot automatically at other times. I expect the
                > disk sleep
                > function will also still work. Of course the on/off
                > button is
                > completely disabled in this situation.
                >
                >
                > Schelte.
                >




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              • Schelte Bron
                ... The ascii picture is a bit hard to make out, but that looks exactly like the setup I used and described on the wiki. With my more complicated proposal I m
                Message 7 of 14 , Jan 6, 2006
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                  On Friday 06 January 2006 05:48, KeepIt SimpleStupid wrote:
                  > Not knowing exactly what the LS has you can try
                  > something like wire oring.
                  >
                  The ascii picture is a bit hard to make out, but that looks
                  exactly like the setup I used and described on the wiki. With
                  my more complicated proposal I'm trying to address the concerns
                  some people raised about that first setup.

                  > Add your resistor if you care. I'm assuming that the
                  > LS uses a contact closure to ground, but it doesn't
                  > have to be.
                  >
                  > What chip/pin is the power switch going to. Is it
                  > connected to anything else like a pull up or pull down
                  > resistor?
                  >
                  The LS indeed uses a contact closure to ground. I haven't been
                  able to determine where the signal is going to. I lost track of
                  the circuit board trace when it disappeared under the IDE
                  connector. From my measurements it appears like the signal is
                  pulled up with a resistance of around 7K.


                  Schelte.
                • KeepIt SimpleStupid
                  ... Try this on for size: Disconnect the switch from the non ground side. Label the switch point (b) and the circuit board side point (b). Connect the anodes
                  Message 8 of 14 , Jan 6, 2006
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                    --- Schelte Bron <sbron@...> wrote:

                    > On Friday 06 January 2006 05:48, KeepIt SimpleStupid
                    > wrote:
                    > > Not knowing exactly what the LS has you can try
                    > > something like wire oring.
                    > >
                    > The ascii picture is a bit hard to make out, but
                    > that looks
                    > exactly like the setup I used and described on the
                    > wiki. With
                    > my more complicated proposal I'm trying to address
                    > the concerns
                    > some people raised about that first setup.
                    >
                    > > Add your resistor if you care. I'm assuming that
                    > the
                    > > LS uses a contact closure to ground, but it
                    > doesn't
                    > > have to be.
                    > >
                    > > What chip/pin is the power switch going to. Is it
                    > > connected to anything else like a pull up or pull
                    > down
                    > > resistor?
                    > >
                    > The LS indeed uses a contact closure to ground. I
                    > haven't been
                    > able to determine where the signal is going to. I
                    > lost track of
                    > the circuit board trace when it disappeared under
                    > the IDE
                    > connector. From my measurements it appears like the
                    > signal is
                    > pulled up with a resistance of around 7K.
                    >
                    >
                    > Schelte.


                    Try this on for size:

                    Disconnect the switch from the non ground side. Label
                    the switch point (b) and the circuit board side point
                    (b).

                    Connect the anodes of two 1n4001 diodes together and
                    connect to point (b).

                    Connect the cathode of one of the diodes to point (a).

                    Connect the cathode of the other to your cap and the
                    negative of the cap to ground.

                    In this case the diodes create isolation and there is
                    no issue of a cap across the switch.

                    If you cannot unsolder both ends of the switch, then
                    go to plan (b).

                    Obtain p/n CNC1H001CT-ND from www.digi-key.com.
                    Connect pins 4 and 6 across the switch. Polarity
                    doesn't matter.

                    Make R=(Vcc-1.2)/10mA. Vcc could be 3.3, 5 or even
                    12V and pull this resistor to this voltage. Connect
                    the other end of this resistor to pin #1 and a cap
                    from pin 2 to ground observing polarity. Make 5 * R *
                    C about equal to (0.1 to 0.5 where R is in ohms, and C
                    is in Farads.

                    KISS








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                  • KeepIt SimpleStupid
                    I was a little hasty before I replied. There is a problem with both designs. The reset time would be really bad because the charge on the capacitor would not
                    Message 9 of 14 , Jan 6, 2006
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                      I was a little hasty before I replied. There is a
                      problem with both designs. The reset time would be
                      really bad because the charge on the capacitor would
                      not drain. I'd suggest a starting value of a parallel
                      resistor of ten times the value of R used.

                      KISS

                      --- KeepIt SimpleStupid <keepitsimplestupid@...>
                      wrote:

                      >
                      >
                      > --- Schelte Bron <sbron@...> wrote:
                      >
                      > > On Friday 06 January 2006 05:48, KeepIt
                      > SimpleStupid
                      > > wrote:
                      > > > Not knowing exactly what the LS has you can try
                      > > > something like wire oring.
                      > > >
                      > > The ascii picture is a bit hard to make out, but
                      > > that looks
                      > > exactly like the setup I used and described on the
                      > > wiki. With
                      > > my more complicated proposal I'm trying to address
                      > > the concerns
                      > > some people raised about that first setup.
                      > >
                      > > > Add your resistor if you care. I'm assuming
                      > that
                      > > the
                      > > > LS uses a contact closure to ground, but it
                      > > doesn't
                      > > > have to be.
                      > > >
                      > > > What chip/pin is the power switch going to. Is
                      > it
                      > > > connected to anything else like a pull up or
                      > pull
                      > > down
                      > > > resistor?
                      > > >
                      > > The LS indeed uses a contact closure to ground. I
                      > > haven't been
                      > > able to determine where the signal is going to. I
                      > > lost track of
                      > > the circuit board trace when it disappeared under
                      > > the IDE
                      > > connector. From my measurements it appears like
                      > the
                      > > signal is
                      > > pulled up with a resistance of around 7K.
                      > >
                      > >
                      > > Schelte.
                      >
                      >
                      > Try this on for size:
                      >
                      > Disconnect the switch from the non ground side.
                      > Label
                      > the switch point (b) and the circuit board side
                      > point
                      > (b).
                      >
                      > Connect the anodes of two 1n4001 diodes together and
                      > connect to point (b).
                      >
                      > Connect the cathode of one of the diodes to point
                      > (a).
                      >
                      > Connect the cathode of the other to your cap and the
                      > negative of the cap to ground.
                      >
                      > In this case the diodes create isolation and there
                      > is
                      > no issue of a cap across the switch.
                      >
                      > If you cannot unsolder both ends of the switch, then
                      > go to plan (b).
                      >
                      > Obtain p/n CNC1H001CT-ND from www.digi-key.com.
                      > Connect pins 4 and 6 across the switch. Polarity
                      > doesn't matter.
                      >
                      > Make R=(Vcc-1.2)/10mA. Vcc could be 3.3, 5 or even
                      > 12V and pull this resistor to this voltage. Connect
                      > the other end of this resistor to pin #1 and a cap
                      > from pin 2 to ground observing polarity. Make 5 * R
                      > *
                      > C about equal to (0.1 to 0.5 where R is in ohms, and
                      > C
                      > is in Farads.
                      >
                      > KISS
                      >
                      >
                      >
                      >
                      >
                      >
                      >
                      >
                      > __________________________________________
                      > Yahoo! DSL � Something to write home about.
                      > Just $16.99/mo. or less.
                      > dsl.yahoo.com
                      >
                      >




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                    • Schelte Bron
                      ... Thanks for thinking along with me, but I think there s a slight miscommunication. The setup I proposed works, is about as simple as it gets and has a
                      Message 10 of 14 , Jan 7, 2006
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                        On Saturday 07 January 2006 00:11, KeepIt SimpleStupid wrote:
                        > Try this on for size:
                        >
                        > Disconnect the switch from the non ground side. Label
                        > the switch point (b) and the circuit board side point
                        > (b).
                        >
                        > Connect the anodes of two 1n4001 diodes together and
                        > connect to point (b).
                        >
                        > Connect the cathode of one of the diodes to point (a).
                        >
                        > Connect the cathode of the other to your cap and the
                        > negative of the cap to ground.
                        >
                        > In this case the diodes create isolation and there is
                        > no issue of a cap across the switch.
                        >
                        > If you cannot unsolder both ends of the switch, then
                        > go to plan (b).
                        >
                        > Obtain p/n CNC1H001CT-ND from www.digi-key.com.
                        > Connect pins 4 and 6 across the switch. Polarity
                        > doesn't matter.
                        >
                        > Make R=(Vcc-1.2)/10mA. Vcc could be 3.3, 5 or even
                        > 12V and pull this resistor to this voltage. Connect
                        > the other end of this resistor to pin #1 and a cap
                        > from pin 2 to ground observing polarity. Make 5 * R *
                        > C about equal to (0.1 to 0.5 where R is in ohms, and C
                        > is in Farads.
                        >
                        Thanks for thinking along with me, but I think there's a slight
                        miscommunication. The setup I proposed works, is about as
                        simple as it gets and has a reasonable reset time (3 seconds).
                        My only question was if anyone sees any problems with that
                        design.

                        If you look at the picture I included in the wiki article you
                        will see that your plan A is not going to work because
                        unsoldering the switch cannot be done without a big risk of
                        damaging the LS.

                        Your plan B doesn't sound simpler than my proposal and I don't
                        see how it could work. The part number you mentioned is a 16
                        pin SMD 4-port opto-isolator and you seem to be using pins from
                        different ports. I assume you meant something like p/n H11F1-ND
                        or similar, but then the circuit is still more complicated than
                        mine and I don't see any advantages.

                        So unless you can point me to any dangers for the LS using my
                        setup, I'm going to go with that.


                        Thanks,
                        Schelte.
                      • KeepIt SimpleStupid
                        ... Your correct with the part number that you selected. So one cap, 1 IC and 2 resistors you should have something that works and keeps the functionality of
                        Message 11 of 14 , Jan 7, 2006
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                          --- Schelte Bron <sbron@...> wrote:

                          > On Saturday 07 January 2006 00:11, KeepIt
                          > SimpleStupid wrote:
                          > > Try this on for size:
                          > >
                          > > Disconnect the switch from the non ground side.
                          > Label
                          > > the switch point (b) and the circuit board side
                          > point
                          > > (b).
                          > >
                          > > Connect the anodes of two 1n4001 diodes together
                          > and
                          > > connect to point (b).
                          > >
                          > > Connect the cathode of one of the diodes to point
                          > (a).
                          > >
                          > > Connect the cathode of the other to your cap and
                          > the
                          > > negative of the cap to ground.
                          > >
                          > > In this case the diodes create isolation and there
                          > is
                          > > no issue of a cap across the switch.
                          > >
                          > > If you cannot unsolder both ends of the switch,
                          > then
                          > > go to plan (b).
                          > >
                          > > Obtain p/n CNC1H001CT-ND from www.digi-key.com.
                          > > Connect pins 4 and 6 across the switch. Polarity
                          > > doesn't matter.
                          > >
                          > > Make R=(Vcc-1.2)/10mA. Vcc could be 3.3, 5 or
                          > even
                          > > 12V and pull this resistor to this voltage.
                          > Connect
                          > > the other end of this resistor to pin #1 and a cap
                          > > from pin 2 to ground observing polarity. Make 5 *
                          > R *
                          > > C about equal to (0.1 to 0.5 where R is in ohms,
                          > and C
                          > > is in Farads.
                          > >
                          > Thanks for thinking along with me, but I think
                          > there's a slight
                          > miscommunication. The setup I proposed works, is
                          > about as
                          > simple as it gets and has a reasonable reset time (3
                          > seconds).
                          > My only question was if anyone sees any problems
                          > with that
                          > design.
                          >
                          > If you look at the picture I included in the wiki
                          > article you
                          > will see that your plan A is not going to work
                          > because
                          > unsoldering the switch cannot be done without a big
                          > risk of
                          > damaging the LS.
                          >
                          > Your plan B doesn't sound simpler than my proposal
                          > and I don't
                          > see how it could work. The part number you mentioned
                          > is a 16
                          > pin SMD 4-port opto-isolator and you seem to be
                          > using pins from
                          > different ports. I assume you meant something like
                          > p/n H11F1-ND
                          > or similar, but then the circuit is still more
                          > complicated than
                          > mine and I don't see any advantages.
                          >
                          > So unless you can point me to any dangers for the LS
                          > using my
                          > setup, I'm going to go with that.

                          Your correct with the part number that you selected.
                          So one cap, 1 IC and 2 resistors you should have
                          something that works and keeps the functionality of
                          the button.

                          Many years ago, like 20, I used this technique to turn
                          off an audio processor when the AMP turned off. I
                          also had the ability to turn the audio system on via
                          the processors remote, Still works.
                          Weird switches (Dedicated off)
                          ON (one push -20 db, 2nd push 0 db)
                          Since the circuitry was external, isolation was
                          important,

                          Thanks for catching the errors.

                          KISS
                          >
                          >
                          > Thanks,
                          > Schelte.
                          >




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