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I want to use 100 to 150 SMD LEDs of 1w 3.3v in a project and light them, how can I convert 3.7v to 3.3v to use them from a single lipo battery keeping it as cheap as possible |
| Answer» <html><body><p><strong>Answer:</strong></p><p></p><p><strong>Explanation:I keep receiving requests from the readers who seem confused with the connection details of 5mm LEds with a 3.7V Li-ion cell. The requests inspired me to write this post, hopefully it <a href="https://interviewquestions.tuteehub.com/tag/would-3285927" style="font-weight:bold;" target="_blank" title="Click to know more about WOULD">WOULD</a> answer the many related queries.</strong></p><p><strong></strong></p><p><strong>Using a Cellphone Li-ion Cell</strong></p><p><strong>Since standard 3.7V Li-Ion cells which are normally used in cell phones are rated at around 800 to 1100mAh, are quite capable of supporting a few 5mm LEDs, and would be able to keep them illuminated for quite sometime.</strong></p><p><strong></strong></p><p><strong>A normal 5mm white LED requires about 20mA current at 3.3V for getting illuminated optimally.</strong></p><p><strong></strong></p><p><strong>The circuit involved for illuminating 5mm LEds through a 3.7V Li-Ion cell is actually too simple, primarily because the parameters are closely matched with each other.</strong></p><p><strong></strong></p><p><strong>Here, connecting the 5mm LEDs in series wouldn't be feasible because the maximum volts from the cell is just 3.7V while even two LEDS in series would call for above 6V.</strong></p><p><strong></strong></p><p><strong>Therefore the only option left is putting them in parallel.</strong></p><p><strong></strong></p><p><strong>Ideally when parallel connections are involved, a series limiting <a href="https://interviewquestions.tuteehub.com/tag/resistor-622654" style="font-weight:bold;" target="_blank" title="Click to know more about RESISTOR">RESISTOR</a> becomes imperative with each LED in the array. This helps ensure uniform light distribution or emission from the LEDs.</strong></p><p><strong></strong></p><p><strong>However it's not an absolute requirement, <a href="https://interviewquestions.tuteehub.com/tag/especially-975202" style="font-weight:bold;" target="_blank" title="Click to know more about ESPECIALLY">ESPECIALLY</a> when the driving voltage is close to the forward voltage of the LEDs.</strong></p><p><strong></strong></p><p><strong>Also taking the simplicity factor into account, a single limiting resistor may be used in such cases and therefore here too we have eliminated individual resistors.</strong></p><p><strong></strong></p><p><strong></strong></p><p><strong> </strong></p><p><strong>How to Connect the LEDs</strong></p><p><strong>The circuit diagram below shows a simple configuration comprising of a 3.7V Li-ion cell, 5nos 5mm LEDs and a limiting resistor R1. The procedure shows how simply a Li-ion cell may be used for illuminating 5mm LEDs for a reasonably long period of time.</strong></p><p><strong></strong></p><p><strong>Each LED is supposed to consume 20mA current, therefore 5nos would together consume around 100mA, therefore R1 may be calculated as follows:</strong></p><p><strong></strong></p><p><strong>The Formula</strong></p><p><strong>R = (Supply voltage - LEd forward voltage)/LED current</strong></p><p><strong>= (3.7 - 3.3)/<a href="https://interviewquestions.tuteehub.com/tag/100-263808" style="font-weight:bold;" target="_blank" title="Click to know more about 100">100</a> = 0.4/0.1 = 4 ohms.</strong></p><p><strong>The required wattage would be 0.4 x 0.1 = 0.04W, so a 1/4 watt resistor would be more than enough.</strong></p><p><strong></strong></p><p><strong>Assuming the cell to be rated at 800mAH, with 5 LEDs, the <a href="https://interviewquestions.tuteehub.com/tag/approximate-882889" style="font-weight:bold;" target="_blank" title="Click to know more about APPROXIMATE">APPROXIMATE</a> back up time available from the cell could be calculated using the following cross-multiplication.</strong></p><p><strong></strong></p><p><strong>800/100 = x/1100x = 800x = 800/100 = 8 hours ideally.</strong></p><p><strong></strong></p><p><strong>However practically you would find the above calculated back up time to be considerably less due to many inherent inefficiencies associated with the system or the circuit.</strong></p><p></p></body></html> | |