Calculate LED Current Draw w/ Resistor & 5V Supply

  • Thread starter Thread starter tangodirt
  • Start date Start date
  • Tags Tags
    Current Led
AI Thread Summary
To calculate the current through an LED with a 5V power source and a resistor, apply Kirchhoff's Voltage Law to find the voltage across the resistor, which is the supply voltage minus the LED's forward voltage. Using Ohm's Law, the current can be determined by the equation I = (Vin - Vf) / R, where Vin is the supply voltage, Vf is the forward voltage, and R is the resistance. For example, with a 330-ohm resistor and a forward voltage of 3.2V, the current through a standard LED is calculated to be 20mA. Additionally, it's important to ensure the resistor can handle the power dissipation, calculated using P = I^2 * R. Always refer to the LED datasheet for accurate forward voltage specifications.
tangodirt
Messages
51
Reaction score
1
Simple question for you guys. I am trying to drive an LED with a 5V power source and a resistor in series.

I understand how to calculate the minimum resistance needed to drive the LED with maximum current ((voltage supply - forward voltage) / maximum current) but I can't for the life of me figure out how to calculate the current through an LED with a given resistor.

Say, we have our 5V source --> 330 ohm resistor --> LED with forward voltage of 3.15V --> ground. What's the current flowing through the circuit?
 
Engineering news on Phys.org
By applying Kirchoff's Voltage Law, you know the voltage across the resistance (supply voltage - forward voltage).

With this, you can apply Ohm's Law to calculate the current through the resistor (you know the voltage across it and the resistance). Since it is a series circuit, the current through the resistor = the current through the LED.
 
mdjensen22 said:
By applying Kirchoff's Voltage Law, you know the voltage across the resistance (supply voltage - forward voltage).

With this, you can apply Ohm's Law to calculate the current through the resistor (you know the voltage across it and the resistance). Since it is a series circuit, the current through the resistor = the current through the LED.

Ugh, of course. Too simple. I knew I had enough information to back that number out, I suppose it has just been a long day and my brain already went into weekend mode. Thanks!
 
You'll want to check the LED datasheet for typical operating voltage @ specified current.

Assuming a standard 5mm LED: White @ 20mA

(Vin - Vf) / R = I

Vin = 5 V
Vf = 3.2V (found from LED datasheet @ 20mA)
I = 0.02A

5V - 3.2 / R = 0.02A

R = 90 Ohm

Dont burn up your resistor! Calculate power in R:
P = I^2 * R
P = (0.02 * .02) * 90
P = .036W
 
Very basic question. Consider a 3-terminal device with terminals say A,B,C. Kirchhoff Current Law (KCL) and Kirchhoff Voltage Law (KVL) establish two relationships between the 3 currents entering the terminals and the 3 terminal's voltage pairs respectively. So we have 2 equations in 6 unknowns. To proceed further we need two more (independent) equations in order to solve the circuit the 3-terminal device is connected to (basically one treats such a device as an unbalanced two-port...
suppose you have two capacitors with a 0.1 Farad value and 12 VDC rating. label these as A and B. label the terminals of each as 1 and 2. you also have a voltmeter with a 40 volt linear range for DC. you also have a 9 volt DC power supply fed by mains. you charge each capacitor to 9 volts with terminal 1 being - (negative) and terminal 2 being + (positive). you connect the voltmeter to terminal A2 and to terminal B1. does it read any voltage? can - of one capacitor discharge + of the...
Thread 'Weird near-field phenomenon I get in my EM simulation'
I recently made a basic simulation of wire antennas and I am not sure if the near field in my simulation is modeled correctly. One of the things that worry me is the fact that sometimes I see in my simulation "movements" in the near field that seems to be faster than the speed of wave propagation I defined (the speed of light in the simulation). Specifically I see "nodes" of low amplitude in the E field that are quickly "emitted" from the antenna and then slow down as they approach the far...

Similar threads

Back
Top