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This is a Sequential LED flasher circuit i found in you tube.

He uses 11 leds in 5v power supply source and he connect them in parallel with a resistor 470ohms.

I try figure out how he measure this resistor but i can not.

If we assume red leds are 2.1v and their current 20ma

11 red leds=220ma 5v-2.1=2.9v 2,9/0.220=13ohms resistor

i know i am wrong , so how he calculate the 470ohms resistor in the kind of circuits? (knight rider , led chaser etc.)

enter image description here

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  • \$\begingroup\$ you are not lighting all the LEDs at the same time \$\endgroup\$ – jsotola Feb 11 '18 at 18:35
  • \$\begingroup\$ The designer may not have designed a particular resistor in, instead she may have experimented and found the one that worked best \$\endgroup\$ – Taniwha Feb 11 '18 at 19:34
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The 4017 is a decade counter with 10 decoded outputs - only one of the ten outputs can be active at a time, so the resistor only has to handle the current of one LED at a time.

With 2 V LEDs, the 470 Ohm resistor will allow an LED current of about 6 mA.

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  • \$\begingroup\$ Actually with that circuit, two LEDs will be on at any given time. There are extra diodes. \$\endgroup\$ – Tom Carpenter Feb 11 '18 at 18:42
  • \$\begingroup\$ @TomCarpenter: yes, now that I look closely... - so the actual current in each lit LED will be about 3 mA. \$\endgroup\$ – Peter Bennett Feb 11 '18 at 18:51
  • \$\begingroup\$ The center LED is the only one not paired up, and there is no series diode on it. So that one LED will see the highest current. \$\endgroup\$ – Nedd Feb 11 '18 at 19:32
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First your assumption of 20ma per LED is not a requirement. The LEDs might be rated at 20ma (max) but that does not say they must be run with that amount of current. Using a lower current just results in a reduced brightness. Also as others have pointed out not all LEDs are on at the same time. (In this case being the knight rider arrangement the sequence starts at the center then fans out and back in pairs.)

Secondly, your calculation of 2.9V on each LED is only good for the LEDs that do not have the extra series diodes, the other LEDs are turned on in pairs through a diode so the available voltage on those LEDs is further reduced by about 0.7V.

Further, the current in each paired LED will actually be even less then the center LED as the current divides between them. Also note that running two LEDs in parallel does not guarantee they share equal currents, but to keep things simple it may not be so important.

If you're wondering about the need for the series diodes, these are needed since the 4017 outputs are not hi-z capable (they are always a high or a low), so directly shorting two outputs would cause problems. The diodes over come this problem by preventing current from the single high output from being shorted by the paired low output pin.

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  • \$\begingroup\$ thank you for your answer. how do you suggest the circuit should be after the diodes , or what resistors shall i edit to each led to be correct ? \$\endgroup\$ – panayiotis kasapis Feb 11 '18 at 19:54
  • \$\begingroup\$ It may be that the center LED was meant to be brighter. \$\endgroup\$ – Nedd Feb 17 '18 at 11:28
  • \$\begingroup\$ If you wanted to make all LEDs an equal brightness you could use a separated resistor on every LED, (though that would defeat the simplicity of the circuit). A compromise would be to use 3 separate resistors for the 3 LEDs that do not have the series diodes, then group the 4 remaining left LEDs together using one resistor, and group the 4 remaining right LEDs together with one resistor. So you could get away with using a total of 5 resistors. Just remember that when calculating the currents that the LEDs with no series diode will not have the extra 0.7V reduction. \$\endgroup\$ – Nedd Feb 17 '18 at 12:05

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