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Re: [midatlanticretro] Re: Commodore 64 SX

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  • B Degnan
    ... What do I look like, a store? just kidding I probably could be a store, sure I will sell you one, name your price (privately). Bill
    Message 1 of 26 , Apr 8, 2009
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      Brian Cirulnick wrote:
      --- In midatlanticretro@yahoogroups.com, B Degnan <billdeg@...> wrote:
        
      I'll sell  you a Plus/4 boxed and working if you want it.  Name your price.
      
      Bill
      ---------------
          
      Bill, put me down for a Plus/4... (if you've got extras)
      Bring it to the MARCH fest, and I'll pay you then...
      
      ttyl
      Brian C.
      
        
      What do I look like, a store?  just kidding I probably could be a store, sure I will sell you one, name your price (privately).
      Bill
    • saturnine11
      ... Hi Mike, I use a simpler, more effective method. Power supply negative to cap negative. Power supply positive to 30kohm resistor to cap positive.
      Message 2 of 26 , Apr 11, 2009
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        --- In midatlanticretro@yahoogroups.com, Mike Loewen <mloewen@...> wrote:
        >
        >
        > I tore apart my IMSAI in order to remove the four large electrolytic
        > capacitors for testing/reforming. There are two 9500uf-30VDC caps and two
        > 95000uf-15VDC caps. Here's the rig I put together for bringing them back
        > to life:
        >
        > http://sturgeon.css.psu.edu/~mloewen/Oldtech/IMSAI/Reforming-L.jpg
        >
        > I put this together after reading many documents about testing old
        > capacitors. Hidden behind the cap is an 8K power resistor in series with
        > the cap. The small meter is measuring the voltage output from the the
        > power supply (a HP 6443B 0-120VDC/2.5A unit), in this case 25.0 volts.
        > The large meter is measuring the current flowing through the cap, 0.11ma.
        > The voltage across the cap is 24.7V at this point.
        >
        > I started out by raising the voltage by steps, from 3-6-9-12-15-20-25,
        > watching the current and making sure it went no higher than 0.5ma. At
        > each step, I'd let the cap charge up until the current was down to about
        > .05ma then increase the voltage. Once I got to 25.0 volts, I let it sit
        > for a while until the current was down to .03ma.
        >
        > So far I've done one of the 9500uf caps. I'm hoping this procedure is
        > effective. How about it, Dan and the rest of the electronics gurus?
        >
        >
        > Mike Loewen mloewen@...
        > Old Technology http://sturgeon.css.psu.edu/~mloewen/Oldtech/

        Hi Mike,

        I use a simpler, more effective method. Power supply negative to cap negative. Power supply positive to 30kohm resistor to cap positive. Voltmeter across 30kohm resistor. Cap voltage determines the increments I raise the supply voltage in. For low voltage caps like yours, I just raise it smoothly until voltage hitting the cap is its rated voltage... but never exceeding 1ma to get there. If the cap can't be slowly raised to rated voltage without going over 1ma, the leakage current is too high and the cap is bad.

        If the cap settles down to a low leakage at rated voltage (0.5ma or preferably much less), I'll next push the cap... by raising supply V until V hitting cap is 10% OVER the rated voltage... so 33volts and 16.5 volts for your examples. All the while keeping an eye on the leakage current as before.

        Depending on how long since use, I'll let them sit at that voltage and low leakage rate for at least 30 mins... up to several hours... Then to finish, I slowly discharge the cap with a 1k resistor.

        Since electrolytics are chemical devices, they can explode from overpressure caused from overheating from too high a *leakage current* (not the same as the charge current!) and otherwise be dangerous to work with if you don't know what you're doing. Always wear goggles.

        JS
      • Bill Dromgoole
        ... From: saturnine11 To: Sent: Saturday, April 11, 2009 11:53 AM Subject: [midatlanticretro] Re:
        Message 3 of 26 , Apr 11, 2009
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          ----- Original Message -----
          From: "saturnine11" <js@...>
          To: <midatlanticretro@yahoogroups.com>
          Sent: Saturday, April 11, 2009 11:53 AM
          Subject: [midatlanticretro] Re: Capacitor testing


          --- In midatlanticretro@yahoogroups.com, Mike Loewen <mloewen@...> wrote:
          >
          >
          > I tore apart my IMSAI in order to remove the four large electrolytic
          > capacitors for testing/reforming. There are two 9500uf-30VDC caps and two
          > 95000uf-15VDC caps. Here's the rig I put together for bringing them back
          > to life:
          >
          > http://sturgeon.css.psu.edu/~mloewen/Oldtech/IMSAI/Reforming-L.jpg
          >
          > I put this together after reading many documents about testing old
          > capacitors. Hidden behind the cap is an 8K power resistor in series with
          > the cap. The small meter is measuring the voltage output from the the
          > power supply (a HP 6443B 0-120VDC/2.5A unit), in this case 25.0 volts.
          > The large meter is measuring the current flowing through the cap, 0.11ma.
          > The voltage across the cap is 24.7V at this point.
          >
          > I started out by raising the voltage by steps, from 3-6-9-12-15-20-25,
          > watching the current and making sure it went no higher than 0.5ma. At
          > each step, I'd let the cap charge up until the current was down to about
          > .05ma then increase the voltage. Once I got to 25.0 volts, I let it sit
          > for a while until the current was down to .03ma.
          >
          > So far I've done one of the 9500uf caps. I'm hoping this procedure is
          > effective. How about it, Dan and the rest of the electronics gurus?
          >
          >
          > Mike Loewen mloewen@...
          > Old Technology http://sturgeon.css.psu.edu/~mloewen/Oldtech/

          Hi Mike,

          I use a simpler, more effective method. Power supply negative to cap
          negative. Power supply positive to 30kohm resistor to cap positive. Voltmeter
          across 30kohm resistor. Cap voltage determines the increments I raise the
          supply voltage in. For low voltage caps like yours, I just raise it smoothly
          until voltage hitting the cap is its rated voltage... but never exceeding 1ma to
          get there. If the cap can't be slowly raised to rated voltage without going
          over 1ma, the leakage current is too high and the cap is bad.

          If the cap settles down to a low leakage at rated voltage (0.5ma or
          preferably much less), I'll next push the cap... by raising supply V until V
          hitting cap is 10% OVER the rated voltage... so 33volts and 16.5 volts for your
          examples. All the while keeping an eye on the leakage current as before.

          Depending on how long since use, I'll let them sit at that voltage and low
          leakage rate for at least 30 mins... up to several hours... Then to finish, I
          slowly discharge the cap with a 1k resistor.

          Since electrolytics are chemical devices, they can explode from overpressure
          caused from overheating from too high a *leakage current* (not the same as the
          charge current!) and otherwise be dangerous to work with if you don't know what
          you're doing. Always wear goggles.

          JS



          ------------------------------------

          I don't see how your method is simpler or more effective than Mike's method.
          I probably just don't understand your method.

          If I understand you correctly you are putting a 30 Kohm resistor in series with
          the Capacitor being rejuvenated and measureing the voltage across the resistor.
          When the leakage current is equal to one milliampere the meter would read 30
          volts and at 0.5 ma it would read 15 volts, etc.
          The voltage accross the capacitor is unknown unless you have a seperate meter to
          see the output voltage from the power supply.
          You would then need to subtract the two voltages to get the capacitor voltage
          unless you have a third meter to measure capacitor voltage.
          The purpose of this exercise to to make a bad cap good again, if possible.

          As you can see, I don't fully understand your method.
        • saturnine11
          ... the Capacitor being rejuvenated and measureing the voltage across the
          Message 4 of 26 , Apr 12, 2009
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            --- In midatlanticretro@yahoogroups.com, "Bill Dromgoole" <drummy@...> wrote:
            > I don't see how your method is simpler or more effective than Mike's method.
            > I probably just don't understand your method.
            >
            > If I understand you correctly you are putting a 30 Kohm resistor in series with
            > the Capacitor being rejuvenated and measureing the voltage across the resistor.
            > When the leakage current is equal to one milliampere the meter would read 30
            > volts and at 0.5 ma it would read 15 volts, etc.
            > The voltage accross the capacitor is unknown unless you have a seperate meter to
            > see the output voltage from the power supply.
            > You would then need to subtract the two voltages to get the capacitor voltage
            > unless you have a third meter to measure capacitor voltage.
            > The purpose of this exercise to to make a bad cap good again, if possible.
            >
            > As you can see, I don't fully understand your method.
            >

            << If I understand you correctly you are putting a 30 Kohm resistor in series with > the Capacitor being rejuvenated and measureing the voltage across the resistor.>>
            ** That is correct. That configuration makes the approach simpler because of less equipment needed. You only need a variable PSU, voltmeter, and 3-5watt 30k resistor.

            << When the leakage current is equal to one milliampere the meter would read 30 volts and at 0.5 ma it would read 15 volts, etc.>>
            ** Correct.

            << The voltage accross the capacitor is unknown unless you have a seperate meter to see the output voltage from the power supply.>>
            ** Most variable PSU's have an output Volt/Ammeter built in. So you can either infer V across cap by subtracting the voltage drop across resistor from your PSU's voltage readout... OR you can just move one lead of your VDrop meter over to the cap's opposite terminal.. since you've already got one lead on it where the resistor is.


            << unless you have a third meter to measure capacitor voltage. The purpose of this exercise to to make a bad cap good again, if possible.>>
            ** Yeah, the purpose is to reform the oxide layer on the foil roll within the cap... assuming there's enough electolyte left in there to where it's not dried out, and assuming there's not shorts or other problems.

            To recap, my method is simpler just in equipment setup, and also potentially gives you a longer cap life by the 10% push (the point of which is to give you more of an oxide layer.. well within the design limits of the caps).

            jS
          • Evan Koblentz
            Hello new guy. Please introduce yourself, how you found us, etc.
            Message 5 of 26 , Apr 12, 2009
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              Hello new guy. Please introduce yourself, how you found us, etc.
              > I use a simpler, more effective method .... JS
              >
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