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I am hoping that someone can help me to calculate voltage and current to the LED socket. I see that there are 3 batteries in series, plus a resistor (?). Thank you very much.

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You will have a max open-circuit voltage of about 5V at the LED socket.

As for current, that becomes far more complex & depends as much on your LED as anything else:

The 'forward voltage' of the LED is a fairly fixed value. So, once current is flowing through a given LED, the total current flow is determined by the resistance of any series resistor and the 'overvoltage' of the supply.

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  • \$\begingroup\$ Thank you very much. The LED is a 700mA and 2.9-3.25 recommended forward Voltage. So, in this circuit is the resistor working to limit the voltage? We ultimately want to replace this with a rechargeable Li-Ion pack. Will this be possible? Thank you again for your help. \$\endgroup\$ – lagold Mar 1 '16 at 16:59
  • \$\begingroup\$ The LED technically limits its own voltage, as it presents effectively a short-circuit to any voltage above its "knee" voltage. However, using this property means allowing the LED to self-destruct by drawing a huge overcurrent. The resistor serves to limit the current, based on the overvoltage available in the circuit. For example, with an LED voltage of 2.9V & a battery pack voltage of 4.5V you'll have around 1.6V of overvoltage. Placing an 8ohm (I think that's what I see labeled on that resistor) resistor in series with the LED means that 1.6V can drive 200mA through the resistor and LED. \$\endgroup\$ – Robherc KV5ROB Mar 1 '16 at 17:42
  • \$\begingroup\$ @lagold If you exchanged the 1.5V(nominal) alkaline batteries with a 3.6V-4.2V(nominal) lithium-chemistry cell, the overvoltage at full charge would be reduced, so the current across the resistor & led would reduce along with it. This may pose no problem at all, or it may reduce the light output of the LED too much for your purposes, in which case you would want to reduce the value lf the xeries resistor to bring the current back up. \$\endgroup\$ – Robherc KV5ROB Mar 1 '16 at 17:46
  • \$\begingroup\$ Thank you for the great explanation! So the series resistor helps to stabilize the LED current. Makes sense. So we are driving this LED at roughly 200mA and its recommended rating is 700mA. So we could make it 'brighter' if we wanted. This is a spotlight for a 'telescope' that attaches to your glasses (like a dentist uses) to help with fine work .... for those of us who don't have young eyes anymore. And finally, what current would the LED see without the 8ohm resistor? \$\endgroup\$ – lagold Mar 1 '16 at 19:17
  • \$\begingroup\$ @lagold with no series resistor, the current would increase exponentially with voltage above the LED's 'knee' voltage, until you reached a point where the batteries' internal resistance cancelled the overvoltage, or the LED explodes. LEDs are diodes, not resistors; once you overcome the voltage necessary to forward-bias the diode, they don't do much anything to control their own current past that, which is why all led drive circuits need either a series resistor, or a current-controlled power source. \$\endgroup\$ – Robherc KV5ROB Mar 1 '16 at 19:59
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"Double A" batteries each have a voltage of 1.5v. There are 3, making the total voltage going through the circuit 4.5v. The current is defined by the load, or what current the LED is able to draw. There's a limit to what the batteries can supply, but any LED you'd put in that socket would be well within said limit. Good luck!

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