Assuming that a single green LED with 10mA forward current should have a constant operating voltage of 5V, the series resistor R V equals (5V -V F,10mA )/10mA = 300Ω. The forward voltage is 2V, as indicated by a graph of typical operating conditions found in the data sheet (Figure 2). Figure 1. Standard red, green, and yellow LEDs have forward The soldering is more reliable. SMT has proven to be more reliable when performing in conditions of vibration and shake. Disadvantages. For components that will be under mechanical stress (e.g., connectors) SMT can be unreliable when used as the only method to attach components on a PCB.

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Assuming you do the right thing and use current limiting resistors for them all then the resistor has to drop 8.6 volts at 16mA. This is a resistor value of about 537 ohms. However, you could wire 3 in series to produce a combined LED voltage of 10.2 volts then use a resistor of 112 ohms. It will be a more efficient way of driving the LEDs .
100pcs 0603 (1616) Green And Red Surface Mount Chip SMT Bicolor SMD LED Bead Light Emitting Diode LED Lamp, $4.86/100. their forward voltage drops are 2.0–2.2 V for red, and 3.0–3.2 V for

If you pick a resistor form the SMD power rating chart that can dissipate 0.025W then dissipating 0.0025W will obviously not be a problem. Regarding your question about your circuit with R2,R3 and high current you calculated: R2 and R3 are current sense resistors (hence their low value). And yes, the power dissipated will be large.

Two smaller resistors are cheaper to mass-produce than one larger, higher-power resistor. 200k (1.2mA) is enough load to significantly "clamp" a lot of the noise and cross-talk coupled between AC conductors. Of course 5mA would work better, but then you'd need more resistors and then more power is wasted as heat.
As per the datasheet of the 5mm White LED, the Forward Voltage of the LED is 3.6V and the Forward Current of the LED is 30mA. Therefore, VS = 12V, VLED = 3.6V and ILED = 30mA. Substituting these values in the above equation, we can calculate the value of Series Resistance as. RSERIES = (12 – 3.6) / 0.03 = 280Ω.
D1 is a visible LED that indicates when the input is pressed. R1 and R2 are 1/4 of a 8-resistor SOIC-16 SMD package (Bourns 4816P-1). Since there are three of these circuits, two of the resistors in the package are left unconnected. Question 1: Is there any reason why this circuit would need two 2 kΩ resistors, instead of one 4 kΩ resistor? NXtlhDr.
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    • do smd leds need resistors