# Do resistor-based fan regulators save no power at all?

I have heard that the traditional resistor-based fan speed regulators are inefficient. In fact, I have noticed that such regulators tend to get hot when the fan is set to low speed. However, does that mean that they save no power at all at low speeds, or is it just that power is saved but the amount of power saved is far less than ideal? I was expecting that since power P = V / R^2, the overall power consumption of the fan plus regulator system would be comparatively lower at lower speeds.

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With no resistance, the full voltage is applied to the fan, and you get mechanical work done, at whatever efficiency the fan itself is capable of. Never minding the fan itself, so far as the electrical aspect goes, you could say it's 100% efficient.

With resistance in the system, for example about equal to the resistance of the fan, you have less current flowing through the system. Half as much. Same voltage. So the system is using half the power. The fan, now one part of a voltage divider circuit, is getting half the voltage, and getting half the current. It runs slower, doing less, doing about 1/4 the work. 1/4 of the fan action divided by 1/2 power going into the system means that the system is 50% efficient.

Imagine a lot of resistance in the system, megohms. You'll have only a trickle of current, say a microamp. The fan barely moves. The system will be using very little power, but most of that little power is just making the resistor hot. Or rather, a small fraction of a degree warmer than ambient temperature. The system efficiency is close to zero.

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There are two types of energy source. 1. Voltage source 2. Current source

In voltage source, source voltage remains constant in spite of varying load resistance connected to it. In current source, current remains constant through the load in spite of its resistance.

The main line is a voltage source. so the power consumption (electric energy) used bye the system (regulator + fan) is inversely proportional to resistance of the system (because $P= V^2/R$ and $V$ is constant)

NOTE: Here we if we think the formula $P= I^2 R$, it also gives the power consumption. But here $P$ is NOT proportional to $R$ , because we are using voltage source and $I$ is not constant (instead $I$ depends on $R$)

So, if fan/motor speed is slow, total power consumption (electric energy) will be slower than its full speed, but efficiency of the system will be less because we do not want heat from regulator.

Thus, the smaller the final fan speed, the smaller the electric bill.

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You can use normal LaTeX notation inside single and double dollar signs. –  Emilio Pisanty Jan 14 '13 at 19:26

You answer your own question! P = V/r^2. Add r, you reduce P (total power consumed). There is no addition of r that could -increase- P. So, you save -a little- power at low speed, even though the resistor gets hot.

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Looking at this analytically, let's model the fan, for simplicity's sake, as a resistance, $R_F$. We'll model the source as an ideal source and place a series resistance, $R_S$ between the source and fan.

The power delivered by the source is:

$p_S = \dfrac{V^2}{R_S + R_F}$

So, yes, as the series resistance is increased the total power delivered by the source decreases, i.e., power is "saved".

But, the power delivered to the fan is:

$p_F = \dfrac{V^2}{(R_S + R_F)^2}R_F$

Taking the ratio of fan power to source power gives an efficiency measure:

$\eta = \dfrac{R_F}{R_S + R_F}$

Clearly, for $R_F < R_S$, more power is delivered to the series resistance than to the fan.

So, while you're reducing total power delivered by the source with a resistor divider, you're also wasting more power (unless you actually use the considerable heat from the series resistance in some way...).

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