There is nothing wrong with that circuit. A stepper motor is just a motor with two AC coils. When used as a generator, it is like have two AC motors being spun by one shaft. As far as efficiency compared to a DC brushed motor: I can't say, that requires measuring and testing. All I do know is that a DC brushed motor is simpler in construction, and the rectifiers will drop the input voltage by about 1.2 Volts. However, I don't think efficiency is the issue, power draw is.
The rectifiers are needed to just convert the alternating current to DC current. In the video he only gets about 6 volts when the rectifiers are wired in parallel, but he gets about 15 volts when wired in series. This is exactly the same as when you wire two batteries in parallel, versus series.
In parallel you get twice the amount of potential current you can use, but less volts. In series you can get more volts, but less current. The total amount of actual power (watts) in both circuits is the actually same.
The biggest thing to focus on is total available power. Your generator has a fixed power of some sort. Efficiencies here won't make much difference, your not trying to get more miles out of an electric car.
In this case you have a 25V capacitor. The 25V is a measure of the maximum voltage it can handle without damaging the capacitor. The reason it is called a
capacitor is that is a container that can store a certain
capacity of potential energy. Capacity is measured in
Farads or MicroFarads (uF). To use water as an analogy, imagine how a bucket can hold more water volume than a pint glass. Hence why physically larger capacitors hold more energy.
The voltage inside the capacitor will never go about the voltage of the generator. Instead the voltage in the capacitor will end up being an average of all the voltages over time that were driven by the generator. The stored energy inside the capacitor is a direct measure of that voltage, and the amount of amperage that has passed through the capacitor. However, just like a glass of water, it can only hold so much energy.
To use a water analogy again: Think of a capacitors like a
local water tower. They are placed up high on a hill, so that any water stored inside can flow down via gravity. This is equivalent to voltage. The capacity of the water tower is fixed, by the volume of water inside. The amount of usable power in the tower is a measure of how much water is inside, and how high that water is above you. In electronics capacitors are actually used almost exactly how a city uses water towers. Capacitors provide instant local demanded power, and smooth out the voltage fluctuations. In your circuit they are smoothing out the voltage that the generator is putting out, as well as storing the energy it is making.
I assume you are trying to run an LED which most likely needs 20mA, and a volt meter. Lets assume this needs 10mA. So you need 30mA of current to run this circuit. (5V * 0.03A = 0.15Watts) That is a pretty small amount of power to generate.
The volt-meter you have is what 0-24V? Lets assume the volt-meter is made from a 10mA amp-meter, with a resistor added. (24V / 0.01A = 2400Ω). So the resistor inside is probably a 2400Ω resistor. (2.4kΩ). This also means it takes 0.24Watts of power to operate that meter at 24V. (24V * 0.01 = 0.24W)
You need the meter to instead react from 0 to 5V. (5V / 0.01A = 500Ω). So replacing the 2400Ω resistor with a 500Ω would make that amp meter work from 0-5V instead of 0-24V. The current required to move the amp-meter stays the same, but since voltage dropped, so did the amount of power required. (5V * 0.01A = 0.05Watts). By replacing that resistor, the meter is now four times as efficient. It's maximum voltage level has dropped, but amps are what count in your circuit.