Ohm's Law

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Ohm's Law

Ohm's Law is a formula that relates the basic electrical properties of Voltage, Current and Resistance. These properties can be measured with a multimeter.
How to use a multimeter
Multimeter.jpg
The common formula for Ohm's Law is:

V=IR
Voltage = Current x Resistance


Ohm's Law Diagram

The following diagram shows the relationship between the various electrical properties.
Ohms-law.jpg

Voltage (V)

Voltage is measured in volts (V). Voltage is a measure of the electric potential difference between two points.
There are two basic type of voltage:

AC Voltage

  • AC Alternating Current
    • Common values:
      • The voltage of a typical wall power outlet in the United States is 120 volts AC (VAC).

DC Voltage

  • DC Direct Current
    • Common values:
      • The voltage of a typical NiMh AA battery is 1.25 volts DC (VDC).
      • Many integrated circuits require 5 volts DC (VDC) to power them.

Voltage Formulas

  • V=IR
  • V=√PR
  • V=P/I

Voltage Examples

  1. What is the voltage drop across a 4 Ω resistor that has a current of 0.3A?
    V=IR, V=(0.3)(4)=1.2V
  2. What is the voltage across a 5W lamp with a resistance of 2 Ω?
    V=√PR=√(5)(2)=√10=3.16V
  3. What is the voltage across a 1 W LED that has a current of 0.35A running thru it?
    V=P/I, V=1/0.35= 2.86V


Current (I)

Current is measured in amps (A). Current is a measure of the flow of electric charge. Current is often measured in milliamps (ma) in many electronic circuits. A milliamp is one thousandth of an amp (0.001A).

  • Common values:
    • A typical LED draws 20 milliamps (0.02A or 20mA).
    • A typical string of 100 count incandescent lights draws 0.33A.

Current Formulas

  • I=V/R
  • I=P/V
  • I=√P/R
  • IParallel= I1+I2+I3+...In

Current Examples

  1. How much current flows thru a 680 Ω resistor that has 5V applied to it?
    I=V/R, I=5/680=0.007A (7 ma)
  2. How much current is used by a 75W floodlight using 120V?
    I=P/V, I=75/120=0.625A
  3. What is the current thru a 5W lamp with a resistance of 2 Ω?
    V=√P/R=√(5)/2=2.24/2=1.12A
  4. How much current is required by 1 RGB pixel that draws 20ma (0.02A) each per color?
    IParallel= I1+I2+I3=0.02+0.02+0.02=0.06A, or 60mma
  5. How much current is required by 4 RGB pixels that draw 60ma (0.06A) each when lit white?
    IParallel= I1+I2+I3+I4=0.06+0.06+0.06+0.06=0.24A, or 240 ma

Resistance (R)

Resistance is measured in ohms (Ω). Resistance is a measure of the opposition to the passage of an electric current through a circuit component. Resistors are the most common form of resistance in circuits.

  • Common values:
    • A common current limiting resistor for driving optoisolators is 680 ohms (Ω).

Resistance Formulas

  • R=V/I
  • R=V²/P
  • R=P/I²

Resistance Examples

  1. What value resistor do you need to drop 3V and allow 20ma thru the circuit?
    R=V/I=3/0.020=150Ω
  2. What is the resistance of a 120V light string that draws 40W?
    R=V²/P=(120)²/40=14400/40=360Ω
  3. What is the resistance of a 1W LED that draws 350ma?
    R=P/I²=1/(0.35)²=1/0.123)=8.16Ω

Power (P)

Power is measured in watts (W). Power is a measure of the rate of doing work.
The common formula for Power is:

P=VI
Power=Voltage x Current
  • Common values:
    • A common table lamp is 100 watts (100w).

Power Formulas

  • P=VI
  • P=I²R
  • P=V²/R

Power Examples

  1. How much power does a 100 count string of 120V incandescent lights that use 0.33A use?
    P=VI=(120)(0.33)=39.6 W
  2. How much power does a 4 Ω resistor dissipate as heat that has 0.3A running thru it?
    P=I²R=(0.35)²(4)= (0.123)(4)=0.49 W
  3. How much power does a 10 Ω resistor dissapate as heat that has 12V across it?
    P=V²/R=(12)²/10=144/10=14.4W


LED Resistor Calculator

By using Ohm's Law you can calculate the current limiting resistor needed to include in a circuit with a LED. LEDs need a resistor in a circuit to limit the amount of current flowing though it to prevent it from being destroyed. The current used by LEDs depends on the type of LED being used. Common values are:

LED TypeTypical Current I
Optoisolator LED6ma
Typical 5mm LED20ma
Piranha Superflux LED20ma
"1/3 W" LED100ma
1 W LED350ma

All LEDs have a characteristic voltage drop across them called the VF that depends on the color of the LED. Common values are:

LED Color Typical VF
Optoisolator1.2-1.6 V
Red1.9-2.3 V
Green2.9-3.5 V
Blue2.9-3.5 V
White3.2-3.4 V

LED Resistor Example

  1. What resistor would you use with a 5VDC power supply and a Blue 5mm LED with a VF of 3.0V?
    First you need to calculate the voltage that the resistor needs to drop. That is equal to
    VDrop=VPowerSupply - VF=5-3=2V
    The table above recommends a current I of 20 ma for a 5mm LED. The formula for resistance gives us:
    R=V/I=2/0.02=100Ω Generally you choose a resistor for that value or the next higher standard resistor value.
    To calculate the wattage resistor needed you use the formula for Power:
    P=VI=(2)(0.02)= 0.04W. You would use the next standard size up resistor, 1/8 W.
  2. What resistor would you use with a 5VDC power supply and an optoisolator with a VF of 1.2V?
    First you need to calculate the voltage that the resistor needs to drop. That is equal to
    VDrop=VPowerSupply - VF=5-1.2=3.8V
    The table above recommends a current I of 6 ma for an optoisolator. The formula for resistance gives us:
    R=V/I=3.8/0.006=633Ω Generally you choose a resistor for that value or the next higher standard resistor value, 680Ω.
    To calculate the wattage resistor needed you use the formula for Power:
    P=VI=(3.8)(0.006)= 0.02W. You would use the next standard size up resistor, 1/8 W.

Single LED Calculator

A simple calculator for calculating resistors can be found at: LED Calculator for 1 LED.

Array of LEDs Calculator

A simple calculator for calculating resistors for an array of LEDs can be found at:LED Calculator for an array of LEDs.

Pixel Voltage Drop Calculator

Due to the high current used by pixels and the small guage wires used by many people, there can be issues due to voltage drops caused by the wire resistance.

Voltage Drop Example

  1. What is the voltage drop when using 5V and drawing 500ma over a 10M pair of Cat5 wire?
    Since Cat5 wire has a loop resistance of about 0.19 Ω/M, the 10 M length will have a resistance of (10)(0.19)=1.9 Ω
    By Ohm's Law we know that V=IR, then
    V=IR=(0.5)(1.9)=0.95 V drop, so only 4.05 V will be at the far end of the wire.
  2. What is the voltage drop when using 5V and drawing 30A over a 2ft length of 18ga wire?
    Since 18 ga wire has a loop resistance of about 0.013 Ω/ft, the 2ft wire will have a resistance of (2)(0.012) = 0.024 Ω
    By Ohm's Law we know that V=IR, then
    V=IR=(30)(0.024)=0.72 V drop, so only 4.28 V will be at the far end of the wire. The wire will get hot because it is dissipating P=VI=(0.72)(30)= 21W! You must use larger wire for high currents!

This is a thread that discusses voltage drop.

This is a Voltage Drop Calculator usefull for calculating the effects of different wire types on pixel strings.