He is saying that there doesn't seem to be a switcher because switchers have a constant power input characteristic within their valid input operating range and this device does not display this behavior. Say, for example, that I have a boost switcher which takes a voltage between 2V and 8V input and delivers a regulated 12V output. I then connect a load to the 12V which draws 1 Amp. Let's also say that the efficiency of the switcher is a constant 85% from 2V to 8V input (not really realistic but perhaps not too far off). Then, there is 12W being delivered to the load and 12/.85=3D14.1 W being drawn from the source, independent of the input voltage. So, at 2V, the power supply will draw 14.1/2=3D7 Amps. At 5V, it will draw 2.8 Amps. At 8V, it will draw 1.8 Amps. Notice that the current drain goes down linearly as input voltage rises. This "negative resistance" characteristic (not that V/I is negative but that dV/dI is negative) is a good sign that you are dealing with a switching power supply. A linear regulator tends to draw a constant current from the source regardless of the source voltage. This is what Olin saw in the graph. The values on the low end where the current falls off as the voltage drops are likely the "dropout" region of the linear regulator, where the output voltage is no longer being regulated but rather is following the input voltage minus an offset. Sean On Mon, Dec 6, 2010 at 6:13 PM, V G wrote: > On Mon, Dec 6, 2010 at 1:52 PM, Olin Lathrop w= rote: > >> >> I took Martin's data and made a plot of it, which is attached. >> >> This says the phone has just a linear regulator between the USB power an= d >> the LiIon battery. =A0If there was a switcher, the current would not hav= e >> been >> so completely flat from 4.8V on. =A0If the iPhone does this then just 4 = NiMH >> batteries driving the USB power will work, but leave a significant fract= ion >> of the NiMH battery capacity unused. >> > > Olin, > > Why would you say that? I suspect that there is some sort of a switching > power supply inside the iPhone from various random articles I read on the > Internet. Of course, they could be wrong. > > However, a threshold voltage of 1.15 volts/cell would not allow the pack = to > drop below 4.6 volts. At that voltage, the pack would be still be drained > above 90% (if not 95%?) according to the chart here: > > http://www.eneloop.info/home/performance-details/discharge-current.html > > I think that's a reasonable voltage to cut off at. > -- > http://www.piclist.com PIC/SX FAQ & list archive > View/change your membership options at > http://mailman.mit.edu/mailman/listinfo/piclist > --=20 http://www.piclist.com PIC/SX FAQ & list archive View/change your membership options at http://mailman.mit.edu/mailman/listinfo/piclist .