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I used a 220 ohm resistor with a 9V battery and the LED works fine. However, most sources say the forward voltage in a red LED is typically only ~2 volts. Going by this logic, the LED should have burned out, as resistors only limit current not voltage. If this is a stupid question, I’m sorry. I’m relatively new at electronics.

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    \$\begingroup\$ do more research before posting your question ... for instance, measure the voltage across the resistor \$\endgroup\$
    – jsotola
    Sep 29 '19 at 22:16
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    \$\begingroup\$ The resistor drops the voltage, so the LED only "sees" ~2 volts, as you say. Read about voltage dividers, voltage drop, and Kirchhoff's voltage law. \$\endgroup\$
    – anrieff
    Sep 29 '19 at 22:17
  • \$\begingroup\$ the Vf will rise to probably 100 mV from 20 to 30 mA on Red HB. It will probably run fine at 30 but risk reduced MTBF \$\endgroup\$ Sep 29 '19 at 22:33
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    \$\begingroup\$ Resistors limit current because they drop voltage across themselves (check on Ohms Law) thus leaving less voltage to produce current in the rest of the circuit. \$\endgroup\$
    – Barry
    Sep 30 '19 at 0:01
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Using a battery that is not brand new so its loaded voltage is 8.0V and using a 3.2V blue LED, then the current is (8V - 3.2V)/220 ohms= 21.8mA. No problem when most 5mm blue LEDs have a maximum continuous current of 30mA.

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Here is how this works -----

schematic

simulate this circuit – Schematic created using CircuitLab

I picked a 180 ohm resistor, because at 20 ohms/volt we are guaranteed a 50 milliamp intersection at the left axis. Thus we scale 20 ohms/volt by 9 volts, finding a 180 ohm resistor provides that current.

Note other LEDs will have lower forward voltage, moving solution-intersection to the left and raising the solution-current closer to 50mA.

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