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I bought a 10000mAh (actual 4000mAh) solar power bank from Victsing few days ago. After having solved a short-circuit problem that affected the solar panel, I noticed that it took an extremely long time to charge from direct sunlight: in six very sunny days just about one quarter. The device presents four blue indicator LEDs that sequentially turn on when the power bank is charging (both from sun or microUSB). The problem is that the LEDs are very brights and seems to consume a lot of energy. From some calculations and measurements, it seems that they consume about 10-30% of the energy harvested by the solar panel, a lot in my opinion. The four LEDs are controlled by an unidentified microcontroller. The question is: is it possible to reduce in someway the consumption of the LEDs, by modifying the circuit, the microcontroller or the LEDs?

Thanks a lot.

Alessio

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    \$\begingroup\$ What calculations have you done which have led you to these conclusions? Have you tried either desoldering the LEDs or putting a resistor in series with them? \$\endgroup\$ – stefandz Sep 2 '15 at 13:57
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Most LEDs have a nominal current of 20mA. By choosing a larger resistor, the current can be reduced, but their light becomes dimmer. There are LEDs for just 2mA, saving 90% of current / energy without being too dim. So, changing the LEDs may also be an option, if the device doesn't already use low-currrent LEDs.

However, what's the size of your solar panel? If it's 10cm x 10cm = 0.1m x 0.1m = 0.01 square meters, it catches only 10W of sunlight. (The sun gives us a maximum of 1kW per square meter) The efficiency is as low as 20%, resulting in just 2W electrical power gained. At 5V, that's just 400mA.

So, one to three 20mA LEDs are noticeable consumers, but 2mA LEDs are not.

Now you say your powerbank holds a charge of 4000-10000mAh. It takes 10-25 hours of bright sunlight to collect this charge.

In reality, there are power conversion losses and sunlight isn't always as intense as in the Sahara at 12 o'clock. You can for sure add 20% to the charging time, maybe more.

So, don't expect too much from the solar module.

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  • \$\begingroup\$ Agreed, in fact, some of these powerbanks etc,, use the less efficient amorphous panels giving more like 8% resulting in only 0.8 watts. At 5 volts that's only 160mA. Add in approximately 15% loss for the conversion to and from the panel to 5v and then to the 3-4.2 volts of the battery and you get 136mA! It will take a long time to charge! The LED may not be the problem! \$\endgroup\$ – Filek Sep 3 '15 at 6:33
  • \$\begingroup\$ The solar panel is 60cm^2, so considering a realistic scenario of 800W per square meter provioded by sun, the solar panel capture about 5W. It's efficiency seems to be very low, it's a very bad quality thin film solar panel, so I guess an efficiency of 10% at best, providing 500mW as output. The battery is 3.7V so the output max current is about 135mA. Considering 15% of internal loss, I get 115mA. This means that a 4000mAh battery will be fully charged in about 34h, or about 4 sunny days. In 6 sunny days I just obtained a 25% charged battery, so which is the problem? I guess the LEDs. \$\endgroup\$ – Alessio Sep 4 '15 at 12:19
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There should be some resistors placed between the microcontroller and the LEDs or the LEDs and ground to limit the current through the LEDs.

If you increase the resistance of those (that is: replace them with higher value ones), the current through and brightness of the LEDs will decrease.

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Are you sure about that 10-30% figure? A typical blue indicator LED would require around 20mA at 3V - around 0.06W.

As Arsenal says, if you want them dimmer, just stick a resistor in series with each LED. LEDs will glow visibly with only very small currents through them.

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  • \$\begingroup\$ Adding a resistor, I will reduce the current through the LEDs and then their brightness, but I won't reduce the power consumption because the energy consumed by the LEDs will be simply converted in heat. Am I wrong? \$\endgroup\$ – Alessio Sep 4 '15 at 12:21
  • \$\begingroup\$ No, because as the current falls, so will the heat dissipated by the resistor. Power = V²/R. If the supply voltage minus the LED voltage drop is roughly constant, then as you increase the resistance, the power drops. \$\endgroup\$ – Simon B Sep 4 '15 at 12:38

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