Actually smaller circuitry has more resistance because there is less area for the charge to move, therefore increasing current density, which determines how much heat is generated. According to Ohm's Law, I = V / R, where I is current, V is voltage, and R is resistance, in a circuit where resistance doesn't change (ideal fully loaded circuit,) R becomes a constant (which isn't true, because hot metals conduct electricity less efficiently than cold metals) so any increases to voltage will increase amperage.

So lets say you have a CPU that runs stock at 1.30v (for an easy number,) with some static resistance R, that has a TDP of 90 watts. The calculated current for the CPU would be P = I * V, since we're solving for current, we get P / V = I, 90-watts / 1.3 volts = 69.2 Amps. Using the amperage we can calculate the theoretical loaded resistance of the CPU (this is all very theoretical, a lot doesn't work this way, but it will give an idea for numbers.)

Since I = V / R, we can reformulate that to be R = V / I which would be R = 1.30v / 69.2 Amps = 1.88x10^-2 Ohms. Using this resistance we can calculate the current of an over-clock at 1.42 volts.

I = 1.4v / 1.88x10^-2 = 75.5 Amps which is 6.3 amps more, a 10% increase in current.

According to Joule's first law, Q is directly proportional to (I^2)*R. So heat increases exponentially as current increases. So if you double the current, heat quadruples. You triple the current, heat increases by a factor of 9.

So increasing the voltage on our pretend CPU by 9% just increased your heat generated by 19%. Smaller circuitry has the benefit of using less power, but has the adding issue of generating more heat because of the smaller circuitry. The exponential heat problem starts earlier for IVB because it is already running hot because of the smaller circuitry.

IVB's lower stock voltage (and in turn lower current,) is what makes IVB use less electricity, not the smaller circuitry (which actually diminished some of that benefit.)