My laptop is running off my e-bike battery as I type

[swbluto]

I plugged my 48V 10AH ping 1.0 battery directly into the laptop's power supply and... it worked.

Oddly enough, though, it only seemed to work for one polarity. I thought if it were to convert AC into DC before the switching part, it would work on a full rectifier(Graetz junction, perhaps) and any polarity should work. But, alas, it doesn't. Maybe the power supply is more complex than I thought.

Here are some stats:

PC battery charge 77%.
48V Battery voltage: 52V
At brightest setting, its "resting" current was 1 amp. "working current" varied between 1.1-1.4(peak)(as measured from the ping battery-side), with an average of 1.2 amps. The dimmest setting on the LCD seemed to decrease current by .1 amp

I charged up the laptop's battery to 100% and redid the tests.
48V battery voltage: 52V
At brightest setting, its "resting" current was .6 amp. "working current" varied between .6-1 amp(peak)(as measured from the ping battery-side), with an average of .69 amps. The dimmest setting on the LCD seemed to decrease current by .1 amp.

I also checked standby current usage and for the 77% battery state, it was .4 amps.
The standby usage for the fully charged PC battery was .04 amps.

So, it seems like my laptop normally consumes .65 amps while plugged in(with the PC battery fully charged). At 52V DC, that's 33.8 watts, just 38.2 watts away from the power supply's top power rating of 72 watts(16V ~4.5A). With a 480wh battery, that's 14.2 hours of computing power. With its current rating of ~7.3 AH(Just a few bad cells I need to replace), I realistically have ~10.5 hours of mobile computing power. Add in my laptop's battery's duration of 30 minutes, I have about 11 hours of computing power currently available to me, or ~14.5 hours in the near future. I think that'll be enough for those long trips.

(I just found out that this periodic shrilling from the laptop that happens when the laptop is plugged in, but it thinks it's not(it's draining from the PC battery at that point, according to the battery indicator icon), actually seems to "short" the battery through the power supply as current through the Ping battery was about 1-amp. This heavily suggests that this shrill means something is being shorted out near the laptop's power supply connection. I have not a clue what, though.)

 

[Link]

Oddly enough, though, it only seemed to work for one polarity. I thought if it were to convert AC into DC before the switching part, it would work on a full rectifier(Graetz junction, perhaps) and any polarity should work. But, alas, it doesn't. Maybe the power supply is more complex than I thought.

It probably uses a half-bridge rectifier, then. That is to say, a single diode in line with the input and probably a cap to smooth out the resulting waveform.

(I just found out that this periodic shrilling from the laptop that happens when the laptop is plugged in, but it thinks it's not(it's draining from the PC battery at that point, according to the battery indicator icon), actually seems to "short" the battery through the power supply as current through the Ping battery was about 1-amp. This heavily suggests that this shrill means something is being shorted out near the laptop's power supply connection. I have not a clue what, though.)

That doesn't sound good and I have no idea what it could be. :?

 

[swbluto]

I guess it could be a half-bridge rectifier. Just thought everybody "standard" did everything "completely" or some vague notion like that.

Anyways, *gets math geeky*, the average voltage of a sine curve with a peak voltage of V is 2*V/pi. With a half-bridge rectifier, the average voltage would be V/pi. So since the lower threshold of AC voltage, according to my power supply, is 100V, then that means it should run off of down to (100/pi) or ~31.9 volts if the "average voltage", or some physical equivalent, is the threshold. But I think the average power may be the threshold, and average power for one sine crest is...(P = V^2/R. Int(P) = int(sin^2(t) dt)/R from 0 to pi is... = ??? Where's my calculator? :lol: Looking at "The intregrator's result", it seems to be)...pi/(2R*pi) = 1/(2R) or 1/2 the peak power... I think?(I need to learn not to do math when I'm sleepy) So, then, 1/2 the peak power for one sine crest extends to 1/4 the peak power for one sine crest over the whole period of a half-bridge rectifier, meaning the minimum average power is... 100^2/(4R) or 2500/R, meaning 1/4 the DC voltage is necessary or 100/4 being 25V dc. But... I have my doubts that average power or average voltage determine the DC voltage minimum.

 

[TylerDurden]

Too complicated for me... I run my laptop with my 24VDC emolis (direct to the DC input), when the grid goes down.

"Big-blue" draws 6A. (2003 17" Toshiba; P4 HT)

 

[PaulM]

So since the lower threshold of AC voltage, according to my power supply, is 100V

That 100V minimum will be an RMS voltage, the peak is actually 100 * sqrt 2, or 141.4.

http://www.practicalphysics.org/go/Guidance_107.html

 

[swbluto]

So since the lower threshold of AC voltage, according to my power supply, is 100V

That 100V minimum will be an RMS voltage, the peak is actually 100 * sqrt 2, or 141.4.

http://www.practicalphysics.org/go/Guidance_107.html

How do you know the power supply's indicator is expressed by the RMS voltage? I can't say I know, so I'm curious how you do.

Ahhhh.. I just realized that AC power is expressed by RMS voltage. I thought it was peak. :lol:

At http://en.wikipedia.org/wiki/Mains_electricity, it states "The voltages are generally in the range 100–240 V (always expressed as root-mean-square voltage)."

Since the minimum peak voltage is 141.4, it seems the minimum DC voltage is 141.4/pi or 44.9 volts. That does seem kind of close to 48V, and that flickering at the beginning of being plugged in might be explained by that closeness. But since the United States power supply is 120 volts, that should mean it's running of an equivalent 120*1.41/pi volts or 53.8 DC volts which seems "close enough" so it shouldn't have problems until possibly the last 10% of battery capacity. That's great because I then would have TWO "LVC"'s protecting the battery!