This is a bit of a loaded question. I say this as there are three main factors the determine the amount of HP an alternator requires. I will cover all three and try to give real world examples of each.

System Draw:

   This is probably the single biggest determining factor. An alternator changes mechanical energy, i.e. horse power and torque, into electrical energy, watts (which is amps x volts). Your alternator only produces the amount of energy required by your vehicles electrical system.
   So as you increase the load of your (turn on more electrical devices), you also increase the amount of HP to create the electrical energy to handle the increased load.
   An example of this is: You have an alternator that is capable of 200amps of output at 14.7 volts, and a pulley ratio of 2 to 1. At 25 amps of out put this alternator will require 1.97hp to turn it. Using the same alternator with a 100amp load and it will require 7.89hp to turn it. Double the load to 200amps and you have to double the HP requirements as well, making it 15.77hp.

Pulley Ratio:

   This the next biggest factor and the only variable that can be easily modified. Your alternator usually has a smaller pulley than your crank pulley. This allows the alternator to spin faster than the actual engine rpm. This is necessary because alternators have charge curves that usually start somewhere between 1200 and 1800 alternator rotor rpm. Then the faster you spin them the more amperage they produce.
   Since the alternators pulley is smaller, the crank pulley has to work harder to turn it. This causes the hp requirement to turn the alternator to increase in direct proportion to the difference in the size of the pulleys.
   To determine your pulley ratio simply divide your crank pulleys diameter by your alternators pulley diameter. Typically this will be around 3 to 1 in a stock application.
So if you have an alternator that requires 3hp to turn it at a 1 to 1 ratio, it still only requires 3hp. But if you change the pulley ratio to 2 to 1, it would then be using 6hp. Increase it again to 3 to 1, and the requirement becomes 9hp.
   It is easy to see that keeping the pulley ratio as small as possible is very desirable from a hp savings point of view. But, and that is a big but, it is more important to make sure that you do not put to small of a ratio into your system. For if the alternator is not spinning fast enough at idle, it will not have enough watts (amps x volts) to carry your system load. When this happens your voltage drops and then your system does not work efficiently.
   The bottom line of this happening is ignition spark starts to drop off in power, which reduces your motors HP. So the 2-3 HP you say by going too small with a pulley ratio is thrown way out the window when you cause your motor to lose 10-50+ HP from lack of spark. Also when you run everything at a lower than optimum voltage it cause extra heat in the device, extra heat causes damage and premature failure.To read more on pulley ratio’s please read the article “Pulley ratio” You can also read more on the subject of system voltage in the articl titled “System Voltage”

Alternator Efficiency:

   As we mention earlier, alternators change mechanical energy into electrical energy. Surprisingly they are not very efficient at doing this. Each style of alternator is slightly different in its efficiency rating, typically varying from 45-55% efficiency. This means that for every watt of mechanical power you put into the alternator, you will only get between .45 and .55 watts of DC electricity out of it. The only exception to this rule is with the new Ecoair units, which are 73% efficient. We suggest reading the article “What Does 5% mean”.What Does 5% Mean
   Where does the other 45 to 55% of the input power go? Glad you asked, it becomes watts of heat. That could actually be written as lots of heat. A typical alternator charging at 100amps output, is also producing close to 1500watts of heat. That is enough heat to heat a 10 x 12 room in the middle of winter.
     The older style of alternators, i.e. 70’s thru the mid 80’s, are typically around 50% efficient. While the newer units with avalanche diodes and such are slightly better at 55%.
   What does this really mean? An example would be comparing a 50% efficient alternator to a 55% efficient alternator. Both producing 100amps with a 2 to 1 pulley ratio. The 50% efficient alternator would require 7.89HP, while the 55% unit would require 7.17HP. Not really enough to justify the change in units on its own, but the other added advantages of the higher efficiency units do add up to make for a wise upgrade. The Ecoair unit in the same application would only require 5.33HP

Here is the actual formula:

1.) Amperage output x voltage output = watts output
     i.e. 100amps output at 14.7 volts = 1470watts

2.) Watt output x efficiency = watts input required
     for 50% divide by .5, for 55% divide by .55, for 73% divide by .73

3.) Watts input divided by 745.7 = HP requirement

4.) HP requirement x Pulley Ratio = Actual HP used in vehicle