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 Author: Shemale [ Tue Feb 19, 2008 2:33 am ] Post subject: DVDRW Samsung SH-202H idle power consumption findings My Philips DVDR1660P died (problem recognizing disks), and i replaced with a Samsung SH-202H . I was curious about power consumption at idle, and using a multimeter obtained these results: Philips DVDR1660P : on 5V 0.46A ; on 12V 0.7mA Samsung SH-202H : on 5V 0.8mA ; on 12V 0.2mA So the Philips consume 2.3W and Samsung 0.0064W , it's a very big difference. Some time ago i have a Ricoh MP5240A and a Toshiba DVD (don't know the model, player only, reading at 16X), and i remember the power consumption was between 1W and 2W. Don't know about power consumption at idle of newer model from other manufacturers, but seems that Samsung is one step ahead. BTW, on the label of the drive say Toshiba-Samsung

 Author: jhhoffma [ Tue Feb 19, 2008 8:23 am ] Post subject: Uhh, you may want to check your math. I see where you get 2.3W (thought you should check your units, is it A or mA) on the Philips, but the .0064W is a little out of whack.

 Author: drees [ Tue Feb 19, 2008 5:11 pm ] Post subject: 5V * 0.46A = 2.3w 12V * 0.007A = 0.08w Total = ~2.4w 5V * 0.008A = 0.04w 12V * 0.002A = 0.024w Total = ~0.06w He's only off by a factor of 10 with the Samsung measurement, but either way, it's significantly better than the Philips drive, power draw is just about negligible. Edit: These numbers are wrong, too!

 Author: jhhoffma [ Tue Feb 19, 2008 7:20 pm ] Post subject: Actually, he's off by a factor of 100 for three of those measurements from what I can see. The .46A is the only one that makes sense to me.

 Author: drees [ Tue Feb 19, 2008 10:48 pm ] Post subject: jhhoffma wrote:Actually, he's off by a factor of 100 for three of those measurements from what I can see. The .46A is the only one that makes sense to me. So you think his measurements should be this? Philips DVDR1660P : on 5V 0.46A ; on 12V 0.07A Samsung SH-202H : on 5V 0.08A ; on 12V 0.02A I actually converted his numbers incorrectly previously, I'm off by a factor of 10, hah!

 Author: Shemale [ Wed Feb 20, 2008 1:41 am ] Post subject: drees wrote:5V * 0.46A = 2.3w12V * 0.007A = 0.08wTotal = ~2.4w5V * 0.008A = 0.04w12V * 0.002A = 0.024wTotal = ~0.06wSo, here 1A=100mA ? drees wrote:Philips DVDR1660P : on 5V 0.46A ; on 12V 0.07ASamsung SH-202H : on 5V 0.08A ; on 12V 0.02ASo, here 1A=10mA ? I think 1A=1000mA.

 Author: drees [ Wed Feb 20, 2008 2:35 am ] Post subject: Shemale wrote:I think 1A=1000mA. I know. I made a mistake in my first post. My 2nd post was an attempt to guess what jhhoffma thinks the numbers should be since he's keepng them to himself.

 Author: jhhoffma [ Wed Feb 20, 2008 6:20 am ] Post subject: I was mistaken thinking that this was total power draw, not idle. This is what the results should be for a modern drive (result of a whitepaper by Intel). From this you can see that idle power is pretty much zero, and continuous read draw is around 2.5W. Regarding the .46A, what I meant was why is this one in amps and the rest are milliamps?

 Author: Shemale [ Wed Feb 20, 2008 9:20 am ] Post subject: Sooooory, i made a mistake, for 5V measurement for Samsung i forgot to change the connection for proper scale on multimeter. 200mA and 20A scale have different connection on multimeter. The value for 5V on Samsung drive is 83.8mA I remeasure the consumption and obtain approx. the same values for other measurements. Philips on 5v: Philips on 12V: Samsung on 12V: So, the difference is not so big, but it is 1,937W I think the whitepaper from Intel contains values from a laptop drive, after a few spin-up with Samsung drive i measure a maximum of 1.06A on 12V line, but i considered the measurement invalid because the multimeter is not able to measure a peak value, a few second of constant drain is required for proper measurement.

 Author: jhhoffma [ Wed Feb 20, 2008 9:56 am ] Post subject: Shemale wrote:I think the whitepaper from Intel contains values from a laptop drive, after a few spin-up with Samsung drive i measure a maximum of 1.06A on 12V line, but i considered the measurement invalid because the multimeter is not able to measure a peak value, a few second of constant drain is required for proper measurement. You would be correct, but the values would still be in the same ballpark, but maybe a bit higher.

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