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I was thinking about selecting R & C values for an RC filter, so that it consumes less power, in the overall circuit. Thinking about it, a larger value for R means a larger \$I^2R\$ loss. But a Larger value for C means requires more current to charge. Can anyone suggest on how should I go about selecting these?

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    \$\begingroup\$ Thinking about it, a larger value for R means a larger I2R loss. No! Let me correct your thinking: If we have constant voltage, then \$I=\frac{V}{R}\$. Using that, we can rewrite \$I^2R\$ as \$\frac{V^2}{R^2}R\$ or \$\frac{V^2}{R}\$. From this, we cans see that by increasing resistance, losses will decrease. If we look at the \$I^2R\$ equation, we can see that the losses are directly proportional to resistance but have square proportion to current. On the other hand, current is inversely proportional to resistance. \$\endgroup\$ – AndrejaKo Dec 28 '13 at 1:14
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    \$\begingroup\$ What's your application.? \$\endgroup\$ – Andy aka Dec 28 '13 at 1:17
  • \$\begingroup\$ Andy, its a PLL circuit \$\endgroup\$ – Sherby Dec 28 '13 at 1:44
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For a given RC time constant, you want to increase the value of R and decrease the value of C so that the product of these values is unchanged. The disadvantage is that whatever you connect to the output side of the filter must have a much higher input impedance than R or it will affect the time constant.

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  • \$\begingroup\$ Thanks! ,What is the reason for increasing R? \$\endgroup\$ – Sherby Dec 28 '13 at 1:46
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    \$\begingroup\$ You want to increase R so you can decrease C and keep the RC time constant the same. It's charging and discharging the capacitor that takes energy, so reducing C will reduce the power consumption of the filter. \$\endgroup\$ – Joe Hass Dec 28 '13 at 1:50
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For an RC low-pass filter with time constant \$\tau = RC\$, the power dissipated (lost as heat in the resistor) by the filter at some frequency \$\omega\$ is proportional to

$$P(\omega) \propto \dfrac{\omega \tau}{1 + (\omega \tau)^2}\omega C $$

Thus, decreasing \$C\$ while increasing \$R\$ in order to keep \$\tau\$ constant will result in less dissipation for a given frequency \$\omega\$.

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