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pumping current

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  • Jim Mullins
    OK, I ve been doing some investigating on the net and am still hitting a wall here. I m trying to find a circuit to increase the current output of one of the
    Message 1 of 7 , Jun 29, 2003
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      OK, I've been doing some investigating on the net and am still hitting a wall here. I'm trying to find a circuit to increase the current output of one of the thin film solar cells. I've found voltage pumps, DC-DC converters, and capacitive multipliers. Most all increase the incoming voltage at the outputs. I've been told that the cell which produces 3-4VDC @ 50-60mA is netting me 150mW and I'm trying to get around 300mW out of it. Impossible I'm told. But, I know we have some circuits that amplify current when turning controller pulses into motor movement. Or, am I wrong and we are increasing voltage to the motors? Is there no way to turn about 2VDC into 100 or more mA of current? It seems to me that using the same formula would provide the same mW rating with lower voltage and higher current. I've still got questions floating around the net, but thought I'd see if there are any ideas here.
    • Martin Jay McKee
      As long as the product of voltage and current ( V x I ) are the same you re perfectly withing the relm of the possible. I don t have a circuit at the moment
      Message 2 of 7 , Jun 30, 2003
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        As long as the product of voltage and current ( V x I ) are the same you're perfectly withing the relm of the possible.  I don't have a circuit at the moment but it should be possible to turn a voltage multiplier into a voltage divider pretty eaisly.  A ladder that charges two capacitors in series that then are made parallel, and charge a third capacitor in parallel the the other two.  This would give you half the current, but the same amount of energy transfered (minus a small amount for the switching circuitry) in the same amount of time and thus give the same power.  As a result you'd get double the current, and half the voltage.  It wouldn't be very easy to adapt this to other ratios but for your 4v to 2v example it should work fairly well.
         
        Martin Jay McKee
      • Jim Mullins
        Thanks for the reply Martin. What you ve outlined sounds like a capacitive charge pump of sorts. I ve been discussing these on a couple of EE sites, but
        Message 3 of 7 , Jun 30, 2003
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          Thanks for the reply Martin. What you've outlined sounds like a capacitive charge pump of sorts. I've been discussing these on a couple of EE sites, but haven't yet come up with a circuit. I'm gonna fiddle around a bit on my breadboard with the idea you present here and see if anything comes of it. There is a virtual plethera of voltage pumps out there, but pumping current seems to be harder to nail down.
          ----- Original Message -----
          Sent: Monday, June 30, 2003 4:35 AM
          Subject: RE: [beam] pumping current

          As long as the product of voltage and current ( V x I ) are the same you're perfectly withing the relm of the possible.  I don't have a circuit at the moment but it should be possible to turn a voltage multiplier into a voltage divider pretty eaisly.  A ladder that charges two capacitors in series that then are made parallel, and charge a third capacitor in parallel the the other two.  This would give you half the current, but the same amount of energy transfered (minus a small amount for the switching circuitry) in the same amount of time and thus give the same power.  As a result you'd get double the current, and half the voltage.  It wouldn't be very easy to adapt this to other ratios but for your 4v to 2v example it should work fairly well.
           
          Martin Jay McKee


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        • Wilf Rigter
          Hey Jim, The circuit you want is called a buck converter. It delivers output higher current at a lower output voltage. For a current pump example see :
          Message 4 of 7 , Jun 30, 2003
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            Hey Jim,

            The circuit you want is called a buck converter. It delivers output higher
            current at a lower output voltage. For a "current pump" example see :

            http://www.national.com/pf/LM/LM3352.html

            The 2.5V version is a bucking type only but the higher output voltage
            versions go one step further, maintaining a constant output voltage while
            automatically switching to buck or boost mode depending on whether the input
            voltage is higher or lower than the output voltage.

            With the 2.5V part you can use two NiCds or a super cap at the output for
            energy storage. Max efficiency is over 80%, so with 5V in and 2.5V output
            it can deliver a maximum output current of 20mA with an input current of
            about 12mA . At higher current the best input voltage range is about 4.5V
            where 100mA out requires 60mA input current. This device delivers up to
            200mA output current.

            Since there is no free lunch input power is always greater than output
            power. But sometimes this type of conversion allows other components to run
            near their maximum efficiency and so the total efficiency may be increased
            despite the conversion losses.

            Magnetic type buck converters (using inductors) are generally more efficient
            than charge pumps.

            wilf


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          • Jim Mullins
            Thanks Wilf! A buck converter aye? No one has mentioned one of these in all of my forum searches. Leave it to the master to pull the rabbit out of the hat.
            Message 5 of 7 , Jun 30, 2003
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              Thanks Wilf! A buck converter aye? No one has mentioned one of these in all of my forum searches. Leave it to the master to pull the rabbit out of the hat. I'll check this out tonight! Thanks again.
              ----- Original Message -----
              Sent: Monday, June 30, 2003 10:40 PM
              Subject: Re: [beam] pumping current

              Hey Jim,

              The circuit you want is called a buck converter. It delivers output higher
              current at a lower  output voltage. For a  "current pump" example see :

              http://www.national.com/pf/LM/LM3352.html

              The 2.5V version is a bucking type only but the higher output voltage
              versions go one step further, maintaining a constant output voltage while
              automatically switching to buck or boost mode depending on whether the input
              voltage is higher or lower than the output voltage.

              With the 2.5V part you can use two NiCds or a super cap at the output for
              energy storage.  Max efficiency is over 80%, so with 5V in and 2.5V output
              it can deliver a maximum output current of 20mA with an input current of
              about 12mA . At higher current the best input voltage range is about 4.5V
              where 100mA out requires 60mA input current. This device delivers up to
              200mA output current.

              Since there is no free lunch input power is always greater than output
              power. But sometimes this type of conversion allows other components to run
              near their maximum efficiency and so the total efficiency may be increased
              despite the conversion losses.

              Magnetic type buck converters (using inductors) are generally more efficient
              than charge pumps.

              wilf


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            • Jesse Jimenez
              aye? you must be a Canadian.....heh heh heh...I guess thats comparable to the far superior Texan y all ..... heh heh heh Jesse Jim Mullins
              Message 6 of 7 , Jun 30, 2003
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                aye? you must be a Canadian.....heh heh heh...I guess thats comparable to the far superior Texan "y'all".....
                 
                heh heh heh
                 
                Jesse

                Jim Mullins <botart@...> wrote:
                Thanks Wilf! A buck converter aye? No one has mentioned one of these in all of my forum searches. Leave it to the master to pull the rabbit out of the hat. I'll check this out tonight! Thanks again.
                ----- Original Message -----
                Sent: Monday, June 30, 2003 10:40 PM
                Subject: Re: [beam] pumping current

                Hey Jim,

                The circuit you want is called a buck converter. It delivers output higher
                current at a lower  output voltage. For a  "current pump" example see :

                http://www.national.com/pf/LM/LM3352.html

                The 2.5V version is a bucking type only but the higher output voltage
                versions go one step further, maintaining a constant output voltage while
                automatically switching to buck or boost mode depending on whether the input
                voltage is higher or lower than the output voltage.

                With the 2.5V part you can use two NiCds or a super cap at the output for
                energy storage.  Max efficiency is over 80%, so with 5V in and 2.5V output
                it can deliver a maximum output current of 20mA with an input current of
                about 12mA . At higher current the best input voltage range is about 4.5V
                where 100mA out requires 60mA input current. This device delivers up to
                200mA output current.

                Since there is no free lunch input power is always greater than output
                power. But sometimes this type of conversion allows other components to run
                near their maximum efficiency and so the total efficiency may be increased
                despite the conversion losses.

                Magnetic type buck converters (using inductors) are generally more efficient
                than charge pumps.

                wilf


                ---
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                Checked by AVG anti-virus system (http://www.grisoft.com).
                Version: 6.0.492 / Virus Database: 291 - Release Date: 6/24/2003



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                beam-unsubscribe@egroups.com



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              • Jim Mullins
                ... From: Jesse Jimenez To: beam@yahoogroups.com Sent: Tuesday, July 01, 2003 12:38 AM Subject: Re: OT.[beam] pumping current aye? you must be a
                Message 7 of 7 , Jun 30, 2003
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                  ----- Original Message -----
                  Sent: Tuesday, July 01, 2003 12:38 AM
                  Subject: Re: OT.[beam] pumping current

                  aye? you must be a Canadian.....heh heh heh...I guess thats comparable to the far superior Texan "y'all".....
                   
                  heh heh heh
                   
                  Jesse
                   
                  Nah, American through and through. Though I once worked in a place right off I-95. Lots of Canadians come down that road on their way to Florida to vacation. I loved talking to them as they filled their RVs with water and fuel. Sweetest people I ever met.
                  And, they say y'all a lot in Georgia too. In Kentucky, it's you'ins, But, I think the Canadian "aye" is more akin to the American "ain't it". Like "It's a shame, ain't it." And I've met some Americans that would say, "It's a shame, huh." Amazing the different ways to express the same conversational connections between two folks.
                  And, BTW Wilf, I"m getting free samples of the 2.5 version of the LM3352 as it seems the best choice for this application. There's even a schematic on the first page of the data sheets. I know my application isn't exactly BEAM. But, I see some possibilities here to power up BEAM circuits while using the smallest possible solar cell area. Just everyone think about what a solar roller or photovore could do in the BEAM competitions with one of these little buck boys in their guts.
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