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What size capacitor to use

  1. Jun 3, 2017 #1
    This is a pretty simple question, one that there is likely a simple answer to. So, I need to use a capacitor as a rapid dump battery, not a filter or anything. The required current for this application is 1.7 Amps, and only 0.23 volts is necessary. I'm not too sure as to what amount of farads a capacitor should have to supply this amount of power. If I'm missing anything, or you need more information, just let me know. Thanks in advance.
     
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  3. Jun 3, 2017 #2

    phinds

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    It is also necessary to know how long the dump is supposed to sustain 1.7amps and how low the voltage is allowed to go to be acceptable. Also it would be helpful to know what the load is. 1.7 amps at .23 volts implies an Ohmic load of .135 ohms but if the load is not purely resistive (and I suspect it is not) that can matter.
     
  4. Jun 3, 2017 #3
    The load is a very small piece of nichrome wire. The power only needs to be supplied long enough to heat the wire, so it should be around 1 to 2 seconds. If you happen to know how I could find out how fast the wire would heat up that would be great. The voltage would be able to drop down to zero, as the capacitor only has to dump it's electricity once. This capacitor is practically acting like a quick battery.
     
  5. Jun 3, 2017 #4

    phinds

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  6. Jun 3, 2017 #5
    After looking at that article, the simple example circuit it shows is exactly what I'm trying to do. It's just a switch, capacitor, and the resistor (nichrome). The purpose of this circuit is to just heat the piece of nichrome up once very fast. I understand it will have to be recharged after every power drain. Looking at the equation given on the website, I realized a few things. I'm pulling my voltages and amperage from a nichrome calculator btw. I'll assume a wattage that correlates with a higher temperature as the starting voltage, and the wattage that correlates with the lower temperature I need as the post-drain voltage. So Vc=0.23 Vo=0.48 t=1 R=0.14 and I will solve for C.
    0.23=0.48e^(-1/0.14C)
    0.48=e^(-7.14C)
    ln0.48=-7.14C
    C=0.1
    Am I on the right track here, or am I still off in lala land?
     
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