Which consumes more power: hard drive or SD card/card reader?
Wikipedia's article about Secure Digital (SD) says this :
The power consumption of microSD cards varies by manufacturer, but appears to be in the range of 66-330 mW (20-100 mA at a supply voltage of 3.3 V). Specifications from TwinMos technologies list a maximum of 149 mW (45 mA) during transfer.[21] Toshiba, on the other hand, lists 264-330 mW (80-100 mA).
In the article Hard Disk Drive Power Consumption Measurements: X-bit’s Methodology Indepth the consumption of a hard disk was very carefully measured to be :
the multimeter reported the following: 1.06A average consumption, 1.13A maximum consumption. The oscilloscope data reads: 1.04A average consumption and 2.71A maximum consumption. As you can see, the multimeter managed to get the average value pretty closely, but failed to catch any of the consumption peaks.
Any hard disk review will normally claim between 6-10 W, which is an order of magnitude more consumption than the SD card.
Regarding SSD, there is a surprise in the artice SSD hard drives tank laptop battery life which shows that :
not only are they not as energy efficient, SSDs actually use more power than conventional hard drives.
That said, although SD cards are much more energy efficient, it is not recommended to use them as hard disks. SD cards are much more prone to failure than hard disks, and should really be used with great caution. Their performance also is quite abysmal.
Short answer: Programs should go on Hard Drives not SD cards even though the SD card pulls significantly less power. A majority of power is used by an SD card reader is used to power the SD card itself.
Explanation:
Simply having a hard drive plugged in can pull anywhere from 6-10 watts. When reading data, an additional 3-5 watts are used. SD cards use about 1 watt at max load and are much slower than a normal hard drive. The performance hit you are going to take by putting your programs on the SD card is not really worth the power savings of 2-4 watts. That of course assumes that the hard drive remains idle while the program runs. If anything accesses the hard drive, the power saving is lost since the drive has to spin up to fetch the data.
The better way to lower your power consumption is to switch a spinning disk (6-10 Watts) to an SSD (2 Watts). This reduced the power load but doesn't sacrifice performance.
But you don't have to take my word for it, you can test this yourself. All you need is a power monitor like a Kill-A-Watt. Use the Kill-A-Watt like a power strip and run your laptop off of AC power. Capture the amount of power used in each scenario over the course of at least an hour. The lower number wins the test.
[EDIT]
I did some digging and found two interesting sources:
MSDN Blog entry on power consumption - This page shows the power usage breakdown by component. Written in 2009, it shows that a hard drive takes up about 5% of the total load of a system. That math puts hard drive consumption at 1.5W at 30W load and 4.5W at a 90W load. Knowing what we know about hard drive, that is a very generous number that probably takes an average of power usage over multiple idle events. The cart is still good to understand the general power consumption.
uiuc.edu research paper hosted at PSU - This PDF document has a component by component breakdown of power consumption. This is working with an old Pentium M, but it backups up the ratios of power draw. The hard drive is a minor part of total power consumption. This ranged from 3% to 15% of total power depending on what was going on. That 15% only happened when the system was doing a hard drive stress test. The total wattage at this high percentage was still only 3W out of 18W total draw.
So what does this mean? That trying to idle your system will only save on average only 5% of your total power draw. Even though you are technically right when it comes to power usage, it will slow down you machine and the power savings is minor.
[End EDIT]
Hope this helps