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Re: [Electronics_101] Re: newbie question of voltage dividers

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  • Ed Jacobson
    Yes, what is this voltage divider good for? It wouldn t be any good to power another circuit that requires current, for the circuit would load down the
    Message 1 of 13 , Sep 29, 2001
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      Yes, what is this voltage divider good for?

      It wouldn't be any good to power another circuit that requires current, for
      the circuit would load down the voltage. It would be good, however, at
      the input of a circut with high imput impedance, such as an input of an
      opamp.

      Ed

      <html></html>



      >From: mrostamy@...
      >Reply-To: Electronics_101@yahoogroups.com
      >To: Electronics_101@yahoogroups.com
      >Subject: [Electronics_101] Re: newbie question of voltage dividers
      >Date: Sat, 29 Sep 2001 16:37:20 -0000
      >
      >Hi!
      >Tako placed a very basic question, and Larry very nicely put the
      >answer...where do you get the art for designing circuits in TXT??
      >
      >However I would suggest the following question to Tako and all other
      >begginers. We have a voltage divider like the one Larry designed. So
      >the 3 different voltages (refered to GND) are present. But what if we
      >connect something to them, thus drawing current from the circuit.
      >Would we still have a voltage divider? Maybe a current divider? Surely
      >if we darw current from any of the three voltages their voltage would
      >change! So what is this Voltage divider stuff good for???
      >
      >have a nice weekend!
      >
      >
      >--- In Electronics_101@y..., Larry Hendry <hendrysr@y...> wrote:
      > > Tako, What you have read is true. What you have drawn below is not a
      > > parallel circuit. In a circuit of parallel resistors, both ends of
      > > each resistor would be tied together. voltage is the same across
      > > all, and current divides based on resistance.
      > >
      > > Like this:
      > >
      > > +9V ---+----+----+
      > > | | |
      > > 9K 9K 18K
      > > | | |
      > > Gnd ---+----+----+
      > >
      > >
      > > In this parallel circuit, each resistor has 9 volts across it. the 9
      > > K resistors have .001 amp or 1 milliamp each. The 18 K resistor has
      > > 0.5 milliams. Total current is 2.5 milliamps. 9/9,000=.001 and
      > > 9/18,000 = .0005
      > >
      > > A series circuit is used to divide voltages. The current is the same
      > > in all resistors. like this:
      > >
      > > +9V ----1k---2k---6k---ground
      > >
      > > total current flow = 9/(1000+2000+6000) = 0.001 amps
      > > Voltage across each individual resistor
      > > = 1,000 * .001 amps = 1 volt
      > > = 2,000 * .001 amps = 2 volts
      > > = 6,000 * .001 amps = 6 volts
      > >
      > > The sum of the voltage drops (1+2+6) must equal the total voltage
      > > across the resistors. The both equal 9 volts.
      > >
      > > So, if I took a voltage reading in reference to ground starting at
      > > the 9 volt point, I would read 9V, then 8V then 6V.
      > >
      > > Hope this helps.
      > > Larry H.
      > >
      >


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    • mehdi Rostami
      Hi Ed.. Come on! let them think a little!!! ehehe doesnt harm them at all! C U Ed Jacobson wrote: Yes, what is this voltage divider good
      Message 2 of 13 , Sep 29, 2001
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        Hi Ed..

        Come on! let them think a little!!! ehehe doesnt harm them at all!

         C U

         

         

          Ed Jacobson <enj99@...> wrote:

        Yes, what is this voltage divider good for?

        It wouldn't be any good to power another circuit that requires current, for
        the circuit would load down the voltage.    It would be good, however, at
        the input of a circut with high imput impedance, such as an input of an
        opamp.

        Ed

        <html></html>



        >From: mrostamy@...
        >Reply-To: Electronics_101@yahoogroups.com
        >To: Electronics_101@yahoogroups.com
        >Subject: [Electronics_101] Re: newbie question of voltage dividers
        >Date: Sat, 29 Sep 2001 16:37:20 -0000
        >
        >Hi!
        >Tako placed a very basic question, and Larry very nicely put the
        >answer...where do you get the art for designing circuits in TXT??
        >
        >However I would suggest the following question to Tako and all other
        >begginers. We have a voltage divider like the one Larry designed. So
        >the 3 different voltages (refered to GND) are present. But what if we
        >connect something to them, thus drawing current from the circuit.
        >Would we still have a voltage divider? Maybe a current divider? Surely
        >if we darw current from any of the three voltages their voltage would
        >change! So what is this Voltage divider stuff good for???
        >
        >have a nice weekend!
        >
        >
        >--- In Electronics_101@y..., Larry Hendry <hendrysr@y...> wrote:
        > > Tako, What you have read is true.  What you have drawn below is not a
        > > parallel circuit.  In a circuit of parallel resistors, both ends of
        > > each resistor would be tied together.  voltage is the same across
        > > all, and current divides based on resistance.
        > >
        > > Like this:
        > >
        > > +9V ---+----+----+
        > >        |    |    |
        > >       9K   9K   18K
        > >        |    |    |
        > > Gnd ---+----+----+
        > >
        > >
        > > In this parallel circuit, each resistor has 9 volts across it.  the 9
        > > K resistors have .001 amp or 1 milliamp each.  The 18 K resistor has
        > > 0.5 milliams.  Total current is 2.5 milliamps.  9/9,000=.001 and
        > > 9/18,000 = .0005
        > >
        > > A series circuit is used to divide voltages. The current is the same
        > > in all resistors.  like this:
        > >
        > > +9V ----1k---2k---6k---ground
        > >
        > > total current flow = 9/(1000+2000+6000) = 0.001 amps
        > > Voltage across each individual resistor
        > > = 1,000 * .001 amps = 1 volt
        > > = 2,000 * .001 amps = 2 volts
        > > = 6,000 * .001 amps = 6 volts
        > >
        > > The sum of the voltage drops (1+2+6) must equal the total voltage
        > > across the resistors.  The both equal 9 volts.
        > >
        > > So, if I took a voltage reading in reference to ground starting at
        > > the 9 volt point, I would read 9V, then 8V then 6V.
        > >
        > > Hope this helps.
        > > Larry H.
        > >
        >


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      • mehdi Rostami
        Hi Ed, Come on...you should have left them thinking for a while! Would not hurt at all I suppose!! Ed Jacobson wrote: Yes, what is this
        Message 3 of 13 , Sep 29, 2001
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           Hi Ed,

          Come on...you should have left them thinking for a while! Would not hurt at all I suppose!!

            Ed Jacobson <enj99@...> wrote:

          Yes, what is this voltage divider good for?

          It wouldn't be any good to power another circuit that requires current, for
          the circuit would load down the voltage.    It would be good, however, at
          the input of a circut with high imput impedance, such as an input of an
          opamp.

          Ed

          <html></html>



          >From: mrostamy@...
          >Reply-To: Electronics_101@yahoogroups.com
          >To: Electronics_101@yahoogroups.com
          >Subject: [Electronics_101] Re: newbie question of voltage dividers
          >Date: Sat, 29 Sep 2001 16:37:20 -0000
          >
          >Hi!
          >Tako placed a very basic question, and Larry very nicely put the
          >answer...where do you get the art for designing circuits in TXT??
          >
          >However I would suggest the following question to Tako and all other
          >begginers. We have a voltage divider like the one Larry designed. So
          >the 3 different voltages (refered to GND) are present. But what if we
          >connect something to them, thus drawing current from the circuit.
          >Would we still have a voltage divider? Maybe a current divider? Surely
          >if we darw current from any of the three voltages their voltage would
          >change! So what is this Voltage divider stuff good for???
          >
          >have a nice weekend!
          >
          >
          >--- In Electronics_101@y..., Larry Hendry <hendrysr@y...> wrote:
          > > Tako, What you have read is true.  What you have drawn below is not a
          > > parallel circuit.  In a circuit of parallel resistors, both ends of
          > > each resistor would be tied together.  voltage is the same across
          > > all, and current divides based on resistance.
          > >
          > > Like this:
          > >
          > > +9V ---+----+----+
          > >        |    |    |
          > >       9K   9K   18K
          > >        |    |    |
          > > Gnd ---+----+----+
          > >
          > >
          > > In this parallel circuit, each resistor has 9 volts across it.  the 9
          > > K resistors have .001 amp or 1 milliamp each.  The 18 K resistor has
          > > 0.5 milliams.  Total current is 2.5 milliamps.  9/9,000=.001 and
          > > 9/18,000 = .0005
          > >
          > > A series circuit is used to divide voltages. The current is the same
          > > in all resistors.  like this:
          > >
          > > +9V ----1k---2k---6k---ground
          > >
          > > total current flow = 9/(1000+2000+6000) = 0.001 amps
          > > Voltage across each individual resistor
          > > = 1,000 * .001 amps = 1 volt
          > > = 2,000 * .001 amps = 2 volts
          > > = 6,000 * .001 amps = 6 volts
          > >
          > > The sum of the voltage drops (1+2+6) must equal the total voltage
          > > across the resistors.  The both equal 9 volts.
          > >
          > > So, if I took a voltage reading in reference to ground starting at
          > > the 9 volt point, I would read 9V, then 8V then 6V.
          > >
          > > Hope this helps.
          > > Larry H.
          > >
          >


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        • Larry Hendry
          ... Larry Hendry writes: Obviously, these questions were written by a guy that already knows the answer. :) I enjoy seeing this type of discussion on the
          Message 4 of 13 , Sep 29, 2001
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            --- mrostamy@... wrote:
            > However I would suggest the following question
            > to Tako and all other begginers. We have a voltage
            > divider like the one Larry designed.
            > So the 3 different voltages (refered to GND) are
            > present. But what if we connect something to them,
            > thus drawing current from the circuit.
            > Would we still have a voltage divider?
            > Maybe a current divider? Surely if we draw
            > current from any of the three voltages their
            > voltage would change! So what is this Voltage
            > divider stuff good for???

            Larry Hendry writes:
            Obviously, these questions were written by a guy that already knows
            the answer. :) I enjoy seeing this type of discussion on the list
            as getting people "thinking" is really what E101 should be about.

            Now, It just so happens that my last DIY synthesizer project used a
            voltage divider as a precision source for octave and interval
            switching on the 1 volt per octave oscillator control standard. And,
            as others (who already knew the answers pointed out) the key is
            drawing only very small amount of currents with hi impedance inputs
            to op amps.

            Below is a link to the project I am writing about with both
            schematics and a very detailed analysis of the circuit operation.
            This might be fun reading for those interested in voltage dividers
            and op amps.

            http://www.wiseguysynth.com/larry/jlh-822/822.htm

            Now, I want to suggest another use for a voltage divider circuit
            where high currents can be drawn and the voltage divider does exactly
            what we want.

            Let's say we have a 15 volt source in our power supply, but we want
            to also power something in our ciruit with 5 volts. Certainly, we
            can use a 3-terminal 5 volt regulator like a 7805. And, if the
            current is low, that's all you need. But, what if the current is
            higher. We may have problems with the 5 volt regulator overheating
            since the heating of the regulator is (Vin-Vout)*I. What if we sized
            a power resistor in series with the input of the voltage regulator so
            that at miximum current draw, the voltage drop across the resistor in
            the divider ciruit was 7 volts. Now, at low currents regulator
            Vin-vout = nearly 10. At max current Vin-Vout= 3. So, as current
            goes up, voltage across the regulator goes down, and much of the
            power that would have added heating to the regulator is now
            harmlessly heating the resistor (if you sized it right).

            Is my thinking right here? I just thought it would be fun to discuss
            how varying outputs from voltage dividers is another aspect of their
            usefulness.

            Larry Hendry



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          • mdwilliams@chartertn.net
            Hi folks, I also am a newbie here. I know there must be a way to do what I need, but I got lost on this thread. Let me explain, perhaps clearly what my need
            Message 5 of 13 , Oct 4, 2001
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              Hi folks,

              I also am a newbie here. I know there must be a way to do what I
              need, but I got lost on this thread.

              Let me explain, perhaps clearly what my need is.

              I have a 12v 5AH sealed lead acid battery. I need to split the
              voltage output for multiple circuits. First is a need for 4.8v to run
              electronic circuitry associated with a digital camera. Secondly is a
              circuit to charge a flash unit needing 7.2v.

              As it happens, the two circuits will be drawing current at the same
              time onlt rarely (the flash of course uses a large capacitor for
              storing current before firing whereas the camera uses current while
              firing the shutter).

              OK so now the question is do I need to build a couple of step down
              transformers, or can I utilize a voltage splitting circuit. Of course
              my preference would be a splitter if possible.

              Incidentally, I haven't taken any circuit theory in 30 years so
              please forgive my assumptions if they are in error. Any help
              greatfully accepted.

              TIA
              Mark Williams
              --- In Electronics_101@y..., mehdi Rostami <mrostamy@y...> wrote:
              >
              > Hi Ed,
              > Come on...you should have left them thinking for a while! Would not
              hurt at all I suppose!!
              > Ed Jacobson <enj99@h...> wrote: Yes, what is this voltage divider
              good for?
              >
              > It wouldn't be any good to power another circuit that requires
              current, for
              > the circuit would load down the voltage. It would be good,
              however, at
              > the input of a circut with high imput impedance, such as an input
              of an
              > opamp.
              >
              > Ed
              >
              > <html></html>
              >
              >
              >
              > >From: mrostamy@y...
              > >Reply-To: Electronics_101@y...
              > >To: Electronics_101@y...
              > >Subject: [Electronics_101] Re: newbie question of voltage dividers
              > >Date: Sat, 29 Sep 2001 16:37:20 -0000
              > >
              > >Hi!
              > >Tako placed a very basic question, and Larry very nicely put the
              > >answer...where do you get the art for designing circuits in TXT??
              > >
              > >However I would suggest the following question to Tako and all
              other
              > >begginers. We have a voltage divider like the one Larry designed.
              So
              > >the 3 different voltages (refered to GND) are present. But what if
              we
              > >connect something to them, thus drawing current from the circuit.
              > >Would we still have a voltage divider? Maybe a current divider?
              Surely
              > >if we darw current from any of the three voltages their voltage
              would
              > >change! So what is this Voltage divider stuff good for???
              > >
              > >have a nice weekend!
              > >
              > >
              > >--- In Electronics_101@y..., Larry Hendry <hendrysr@y...> wrote:
              > > > Tako, What you have read is true. What you have drawn below is
              not a
              > > > parallel circuit. In a circuit of parallel resistors, both
              ends of
              > > > each resistor would be tied together. voltage is the same
              across
              > > > all, and current divides based on resistance.
              > > >
              > > > Like this:
              > > >
              > > > +9V ---+----+----+
              > > > | | |
              > > > 9K 9K 18K
              > > > | | |
              > > > Gnd ---+----+----+
              > > >
              > > >
              > > > In this parallel circuit, each resistor has 9 volts across it.
              the 9
              > > > K resistors have .001 amp or 1 milliamp each. The 18 K
              resistor has
              > > > 0.5 milliams. Total current is 2.5 milliamps. 9/9,000=.001 and
              > > > 9/18,000 = .0005
              > > >
              > > > A series circuit is used to divide voltages. The current is the
              same
              > > > in all resistors. like this:
              > > >
              > > > +9V ----1k---2k---6k---ground
              > > >
              > > > total current flow = 9/(1000+2000+6000) = 0.001 amps
              > > > Voltage across each individual resistor
              > > > = 1,000 * .001 amps = 1 volt
              > > > = 2,000 * .001 amps = 2 volts
              > > > = 6,000 * .001 amps = 6 volts
              > > >
              > > > The sum of the voltage drops (1+2+6) must equal the total
              voltage
              > > > across the resistors. The both equal 9 volts.
              > > >
              > > > So, if I took a voltage reading in reference to ground starting
              at
              > > > the 9 volt point, I would read 9V, then 8V then 6V.
              > > >
              > > > Hope this helps.
              > > > Larry H.
              > > >
              > >
              >
              >
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            • Larry Hendry
              ... Mark, voltage dividers are not what you seek as something like this would rely on a fairly constant current draw. While I doubt that the flash unit
              Message 6 of 13 , Oct 4, 2001
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                --- mdwilliams@... wrote:
                > I have a 12v 5AH sealed lead acid battery.
                > I need to split the voltage output for multiple
                > circuits. First is a need for 4.8v to run
                > electronic circuitry associated with a digital
                > camera. Secondly is a circuit to charge a flash
                > unit needing 7.2v. As it happens, the two circuits
                > will be drawing current at the same time only
                > rarely (the flash of course uses a large capacitor
                > for storing current before firing whereas the camera
                > uses current while firing the shutter).

                Mark, voltage dividers are not what you seek as something like this
                would rely on a fairly constant current draw. While I doubt that the
                flash unit voltage is that critical, I expect the other is.

                If I were trying this, believing that the current draws are not very
                much, I would just use voltage regulators. I think the LM-317
                regulator can be set to have an adjustable output in the range that
                you are seeking. Just use two -- one for your 4.8 and one for your
                7.2 volts. I don't have a schematic handy, but these are common and
                I bet the data sheet is all you will probably need. Anyone know who
                makes those or have alink to a data sheet or sample schematic? The
                parts count is quite low.

                > OK so now the question is do I need to
                > build a couple of step down transformers.

                Remember that transformers are a device for AC. They wil not be of
                any use to you for straight DC to DC appications.

                Larry Hendry (not an expert on the subject by any means)


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              • Larry Hendry
                ... Mark, voltage dividers are not what you seek as something like this would rely on a fairly constant current draw. While I doubt that the flash unit
                Message 7 of 13 , Oct 4, 2001
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                  --- mdwilliams@... wrote:
                  > I have a 12v 5AH sealed lead acid battery.
                  > I need to split the voltage output for multiple
                  > circuits. First is a need for 4.8v to run
                  > electronic circuitry associated with a digital
                  > camera. Secondly is a circuit to charge a flash
                  > unit needing 7.2v. As it happens, the two circuits
                  > will be drawing current at the same time only
                  > rarely (the flash of course uses a large capacitor
                  > for storing current before firing whereas the camera
                  > uses current while firing the shutter).

                  Mark, voltage dividers are not what you seek as something like this
                  would rely on a fairly constant current draw. While I doubt that the
                  flash unit voltage is that critical, I expect the other is.

                  If I were trying this, believing that the current draws are not very
                  much, I would just use voltage regulators. I think the LM-317
                  regulator can be set to have an adjustable output in the range that
                  you are seeking. Just use two -- one for your 4.8 and one for your
                  7.2 volts. I don't have a schematic handy, but these are common and
                  I bet the data sheet is all you will probably need. Anyone know who
                  makes those or have alink to a data sheet or sample schematic? The
                  parts count is quite low.

                  > OK so now the question is do I need to
                  > build a couple of step down transformers.

                  Remember that transformers are a device for AC. They wil not be of
                  any use to you for straight DC to DC appications.

                  Larry Hendry (not an expert on the subject by any means)


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                • mehdi Rostami
                  Hi mark, Is 4.8V+7.8 V coincidentialy equal to 12V? Well...Using a voltage divider is not adequate in this stituation when you need real power supply
                  Message 8 of 13 , Oct 5, 2001
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                    Hi mark,

                    Is 4.8V+7.8 V coincidentialy equal to 12V?

                    Well...Using a voltage divider is not adequate in this stituation when you need real power supply situation. The voltage divider is mainly used for instrumentation, sensors, and analog to digital converters where one really deals with signals and not power. The rheostat was and still is sometimes used but in very specific situations where the current is well  known or at least easily assertable. the reason for it not being a good ideia to use resistive voltage dividers is that a very large part of the power you are draining from the source is lost (dissipated) as heat...that is obviously not very nice...

                    The voltage regulators like LM7806, or LM317 although are also very atractive do have their drawbacks. Mark did not tell us how much current his application needs. Imagine that you are want to draw 200mA for the 4.8V aplication. Well that means that the LM317 would have to dissipate 200mA x (12-4.8)V ....now thats 1.44W dissipation. It may seem to be small but you are actually dissipating more power than the power your device is requiring....I dont appreciate that very much when i have battery operated situations. If you were using mains power and a transformer, well, that would be ok, but if portability is a desire i suggest more efficient ways such as "switching power supplies" ...

                    Maxim semiconductor has a lot of ready to use IC´s in this field, you only need an extra inductor, a pair of capacitors and sometimes a special diode. If you are interested in this subject, we can discuss it better! I would be glad to do so! The principle behind is that these systems are not quite DC systems...ther rather deliver power in packets of energy thus achieving efficiency rates higher than 80 percent. thats almost twice the efficiency of the LM317 we saw before!

                     



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                  • Larry Hendry
                    ... Certainly, you have hit the nail on the head. The right solution greatly depends on the anticipated current draw. Larry H
                    Message 9 of 13 , Oct 5, 2001
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                      --- mehdi Rostami <mrostamy@...> wrote:
                      > The voltage regulators like LM7806, or LM317 although are also very
                      > atractive do have their drawbacks. Mark did not tell us how much
                      > current his application needs. Imagine that you are want to draw
                      > 200mA for the 4.8V aplication. Well that means that the LM317 would
                      > have to dissipate 200mA x (12-4.8)V ....now thats 1.44W
                      > dissipation. It may seem to be small but you are actually
                      > dissipating more power than the power your device is requiring....I
                      > dont appreciate that very much when i have battery operated
                      > situations.

                      Certainly, you have hit the nail on the head. The right solution
                      greatly depends on the anticipated current draw.
                      Larry H

                      __________________________________________________
                      Do You Yahoo!?
                      NEW from Yahoo! GeoCities - quick and easy web site hosting, just $8.95/month.
                      http://geocities.yahoo.com/ps/info1
                    • darkbone@sentra.net
                      Sorry. Testing. ... $8.95/month.
                      Message 10 of 13 , Oct 12, 2001
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                        Sorry. Testing.
                        >
                        > __________________________________________________
                        > Do You Yahoo!?
                        > NEW from Yahoo! GeoCities - quick and easy web site hosting, just
                        $8.95/month.
                        > http://geocities.yahoo.com/ps/info1
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