# RC Filter for Low Power consumption

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?

• 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. – AndrejaKo Dec 28 '13 at 1:14
• What's your application.? – Andy aka Dec 28 '13 at 1:17
• Andy, its a PLL circuit – Sherby Dec 28 '13 at 1:44

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 $\omega$.