Series parallel resistors LEDs
The study of series parallel resistor and LED circuits is useful in understanding many of the electronic principles that come up in many circuits.
Basic series parallel resistor and LED properties
- Parallel resistors – Increased circuit current and divides the power (heat generated) between resistors.
- Series resistors – Decreased circuit current and divides the power (heat generated).
- Parallel LEDs – Splits the current and drops a single forward voltage.
- Series LEDs – Each series LED (or parallel LEDs) adds to the total forward voltage drop.
Connecting resistive and semiconductive components in both parallel and series, gives you many ways take advantage of the current handling, power dissipation and voltage drops of multiple components, as needed to help avoid exceeding the ratings of available individual components.
I started this page after making these supercapacitor videos that use a somewhat large number of LEDs in parallel to quickly discharge a supercapacitor and to limit the discharge to more than 3 volts.
- 32 parallel LEDs allowed a maximum continuous current of about 640mA. 0.02A typical maximum indicator LED current times 32 parallel LEDs = .64A (640mA) maximum total current.
- 2 of those 32 parallel LEDs were connected in series. Each providing a minimum of about 1.5V drop, for a total of about 3V minimum voltage drop so that series supercaps don’t discharge below 3V
- 2 of my 10W10Ω resistors (videos below) are connected in series (they aren’t directly connected but still connected in series as far as the whole circuit is concerned). Up to 16V power source minus 3V min LED voltage drop = about 13V across resistor(s), which divided by 10Ω would be 1.3A. That is higher current than 32 parallel indicator LEDs are rated to handle. 2 10Ω resistors in series = 20Ω equivalent resistance. Power in watts (P=V*I) is also a major factor. A single 10Ω resistor with 13V across it is 13V/10Ω = 1.3A. When that current is multiplies by voltage for power, we get 1.3A times 13V for 16.9W. 2 equal value resistors in series split the voltage equally, so that the 2 ten ohm ten watt resistors now have about 6.5V/10Ω = 0.65A. And 0.65A times 6.5V = 4.225W of power that needs to be dissipated by each resistor.
Video below: Discharging my new 6 series supercapacitors through a bunch of LEDs and 10W resistors Click this link to watch the video directly on YouTube in a new tab!
Below: Supercapacitor powered LEDs circuit shows how RC time constant takes forever to fully discharge Click to watch directly on YouTube in a new tab!
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- This website is only intended to provide supplemental information to those already studying electronics. It takes no responsibility for how that information is used and the possible injuries or other damage that may arise.