Backlight Power Measurement
Joerg Reisenweber
joerg at openmoko.org
Tue Oct 13 14:09:49 CEST 2009
[Aaron Carroll Di 13. Oktober 2009]:
> 2009/10/10 Paul Fertser <fercerpav at gmail.com>:
> > Aaron Carroll <xaaronc at gmail.com> writes:
> >> To determine backlight power, I've got a DAQ sampling the
> >> voltage drop across R1763 (Rsense for the PCF50663 LED
> >> supply) which gives me the current, and I'm sampling LED+
> >> at C1764 for the input voltage. I was hoping the product
> >> of these two would yield the power consumed by the
> >> backlight and Rsense.
> > ...
> >
> > You've got quite puzzling and unexpected results indeed. Could you
> > please provide us with the actual measurements data?
>
> Here's some data. Each point is the average of 1000 samples @ 100 kHz,
> alternating between voltage and current samples.
>
> The percentages are backlight brightness according to the position of
> the brightness slider (the phone runs Android).
>
> BL is the backlight, with I measured across R1763, and V at C1764.
> IN is measured at the battery with no other external connections,
> with the current measured via a 100mohm resistor in series with (+).
> The power is just the product of the quoted V/I.
>
> To be completely accurate, the backlight figures are not raw measurements,
> because they are scaled via an opamp-based attentuator before feeding
> into the DAQ. The values are scaled back up in software, and I have
verified
> this all happens correctly with an accurate meter.
>
> Let me know if I can provide more details or measurements.
>
>
> Thanks,
> -- Aaron
>
>
>
>
> ===========x8============
>
>
> 100%:
> BL = 14.38 V, 28.53 mA, 410.26 mW
> IN = 4.01 V, 252.89 mA, 1014.09 mW
>
> 75%:
> BL = 13.10 V, 15.91 mA, 208.42 mW
> IN = 4.04 V, 178.86 mA, 722.60 mW
>
> 50%:
> BL = 11.97 V, 6.45 mA, 77.21 mW
> IN = 4.05 V, 129.57 mA, 520.71 mW
>
> 25%:
> BL = 11.13 V, 1.82 mA, 20.26 mW
> IN = 4.06 V, 110.96 mA, 450.50 mW
>
> 0%:
> BL = 10.70 V, 0.49 mA, 5.24 mW
> IN = 4.06 V, 104.06 mA, 422.49 mW
Seems our converter is really operating with 66% efficiency :-/
I suspect L1704 and C1764 see some loss due to ESR.
You might want to test what happens if you connect a couple of low-ESR
capacitors (e.g. 5 pcs 4u7) parallel to C1764.
If bat current drops significantly on doing this, then this component really
wasn't a good choice for the purpose.
OTOH 200mW loss should be detectable anyway, just check where the temperature
rises ;-) Then you know which component is the hog
cheers
jOERG
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