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I'm a mechanical engineering student and I have a little home project I'm hoping to work on. Unfortunately the only electric circuits class I had to take taught plenty of theory but very little about practicality so I am still at a loss when it comes to designing more complex circuits than the basic ones to prove class concepts.

I'm trying to create a simple circuit that will have multiple leds connected to it (those leds will have fiber optics running off their tops). It seems like connecting them in parallel would be the way to go. If I have a prototype board,a power supply, and a method of turning on and off the power, could I simply solder those leds to wires and create a parallel circuit without any additional elements? I've heard I should include a resistor before each led, but how would I go about choosing which resistor to use, the smallest possible? Would this circuit need any diodes or bridge rectifiers?

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    \$\begingroup\$ Are all the LEDs supposed to be on/off at the same time? What power supplies are available for this? How many LEDs? How much current thru each LED? No, don't connect them in parallel, but what to do instead depends on answers to the above questions. \$\endgroup\$ – Olin Lathrop Feb 7 '17 at 19:20
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If you don't need to control the LEDs individually, you can connect the LEDs in parallel, but you need a current-limiting resistor in series with each LED, like so:

schematic

simulate this circuit – Schematic created using CircuitLab

The resistor value can be calculated using Ohm's Law. You will need to decide on a suitable current for the LED (NOT the Absolute Maximum value from the datasheet! - 10 mA is safe for common LEDs). Determine the forward voltage drop of the LED, and subtrract that from the supply voltage to get the voltage across the resistor.

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  • \$\begingroup\$ Thanks. Also, are those diodes in the picture you included? Why would I need them? \$\endgroup\$ – UALHunter Feb 7 '17 at 19:31
  • \$\begingroup\$ Those "diodes" are LED's (Light Emitting Diodes) \$\endgroup\$ – JonRB Feb 7 '17 at 19:33
  • \$\begingroup\$ Oh haha. Thanks. I swear my professor never made the connection between the two. \$\endgroup\$ – UALHunter Feb 7 '17 at 20:08
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Using the equation:

$$V_B - I_RR - V_D = 0$$ Where: $$V_B \text{ is the battery voltage}$$ $$I_R \text{ is current in each LED branch}$$ $$R \text{ is the value of the resistor connected to each LED}$$ $$V_D \text{ is the diode forward voltage}$$

Solving for the resistance R and using a typical diode forward voltage of 0.7 V and current value of 10 mA and a battery voltage of 5 V:

$$R = {V_B - V_D \over I_R} => R = {5 - 0.7 \over 10^{-3}} = 430 \text{ }\Omega$$

Components are not perfect and this equation assumes some idealities so this is an estimation that should get you in the ball park. To reinforce, this equation applies to each branch individually since they are in parallel. You can play around with resistance values to get brighter and dimmer LEDs in different branches.

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The simple way is to use a constant current power supply and drive up to 16 LEDs in serial. No resistors needed.

Additionally you can dim the LEDs with a PWM signal, a single resistor 10K to 100K = 10% to 100%, or a dc voltage (1-10Volts = 10% to 100%.

A top end power supply would be a Meanwell HLG-40H (40 Watt) sell for about $30. Great power supply, 90% efficiency, 7 year warranty, no fan (even their 400W does not need a fan).

Meanwell also makes a LDD series for about $6-7 from 300mA up to over an AMP.

Mouser sells them.

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