anonymous
  • anonymous
Hi there... i`m up to make a circuit that it`s voltage booster.. i need to store 250-300V in 2 capacitor... from 14 volts. is there any one to help me??? i made some circuits.. but they are not efficient...
Engineering
  • Stacey Warren - Expert brainly.com
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schrodinger
  • schrodinger
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anonymous
  • anonymous
I'm not sure what your input and output voltages are. Is the input 14 volts? And your output between 250-300 volts? Important questions: Do you have a schematic of the circuit you want to build? It would be helpful to have one.
anonymous
  • anonymous
yes input is 14 and out put is 220-300v here i dont have my schematics .. :D i attach them later.. do U have any idea how to make this circuit??
anonymous
  • anonymous
You can't store a potential difference (voltage). What you store in a capacitor is an electric charge.

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anonymous
  • anonymous
You will have to use an op-amp configuration to amplify your low input voltage. I'm not sure if you're familiar with non-inverting op-amps...but if you aren't, here are links that will help you: http://www.wisc-online.com/objects/ViewObject.aspx?ID=SSE7906 http://www.engga.uwo.ca/people/asamani/courses/ece238a/Labs/Lab6.pdf http://www.play-hookey.com/analog/non-inverting_amplifier.html http://www.wisc-online.com/objects/ViewObject.aspx?ID=SSE3003 I did some calculations but they're based on knowledge of op-amps. I'll include them anyway. Using knowledge of a non-inverting op-amp (please reference the image I've attached), we have: Eqn 1.\[V _{out} = V_{in}[(R _{1} +R _{2})/R _{1}] = V_{in}[1 + (R _{2}/R _{1})]\]Input voltage, Vin = 14V & output voltage, Vout ~ 220 - 300V. I'll pick 266 as Vout because it's a multiple of 14 and will give us a rational number as gain (which is simpler to work with). R1 & R2 are resistors in the op-amp configuration. Using Eqn 1, here is the math!\[266 = 14(1 + (R _{2}/R _{1}))\]\[266/14 = 1 + (R _{2}/R _{1})\]\[19 = 1 +(R _{2}/R _{1})\]\[19 = (R _{1}+R _{2})/R _{1}\]\[19R _{1} = R _{1}+R _{2}\]\[19R _{1} - R _{1} = R _{2}\]\[18R _{1} = R _{2}\] Now, you have a ratio that tells you \[R _{2}\] has to be 18 times greater than\[R _{1}\]for you to have a gain of 19. Recall that gain is just Vout/Vin. I suggest you pick resistor values ranging from \[1k \Omega -100K \Omega\]Another thing, your op-amp will need a power supply (of \[\pm -- V\]For example, an LM741 op-amp would need a power supply of \[\pm15V\]

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