Can I use a single resistor for multiple LED with different +ve sources?
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In the photo below I would like to use a single resistor after the LEDs, instead of having individual resistors before the LEDs. Is this problematic? Will it work?

arduino microcontroller led
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add a comment |
$begingroup$
In the photo below I would like to use a single resistor after the LEDs, instead of having individual resistors before the LEDs. Is this problematic? Will it work?

arduino microcontroller led
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No, there are tolerances inbetween one LED and another. The LED with the lowest Vf will carry a lot more current then the others.
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– Unimportant
40 mins ago
1
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Looks like Diode OR logic ;) but problematic due to sharing current and voltage drop.
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– Sunnyskyguy EE75
32 mins ago
add a comment |
$begingroup$
In the photo below I would like to use a single resistor after the LEDs, instead of having individual resistors before the LEDs. Is this problematic? Will it work?

arduino microcontroller led
$endgroup$
In the photo below I would like to use a single resistor after the LEDs, instead of having individual resistors before the LEDs. Is this problematic? Will it work?

arduino microcontroller led
arduino microcontroller led
asked 43 mins ago
DonPDonP
257
257
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No, there are tolerances inbetween one LED and another. The LED with the lowest Vf will carry a lot more current then the others.
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– Unimportant
40 mins ago
1
$begingroup$
Looks like Diode OR logic ;) but problematic due to sharing current and voltage drop.
$endgroup$
– Sunnyskyguy EE75
32 mins ago
add a comment |
$begingroup$
No, there are tolerances inbetween one LED and another. The LED with the lowest Vf will carry a lot more current then the others.
$endgroup$
– Unimportant
40 mins ago
1
$begingroup$
Looks like Diode OR logic ;) but problematic due to sharing current and voltage drop.
$endgroup$
– Sunnyskyguy EE75
32 mins ago
$begingroup$
No, there are tolerances inbetween one LED and another. The LED with the lowest Vf will carry a lot more current then the others.
$endgroup$
– Unimportant
40 mins ago
$begingroup$
No, there are tolerances inbetween one LED and another. The LED with the lowest Vf will carry a lot more current then the others.
$endgroup$
– Unimportant
40 mins ago
1
1
$begingroup$
Looks like Diode OR logic ;) but problematic due to sharing current and voltage drop.
$endgroup$
– Sunnyskyguy EE75
32 mins ago
$begingroup$
Looks like Diode OR logic ;) but problematic due to sharing current and voltage drop.
$endgroup$
– Sunnyskyguy EE75
32 mins ago
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
You can do that safely if:
- Only one LED is on at a time.
- The voltage of the circuit is less than the maximum reverse voltage of the LEDs. This often isn't specified by 5 V is generally quite safe.
Otherwise switching two or more on at a time connects them in parallel. In this situation the one with the lowest forward voltage will pass the most current. If they are different colour LEDs then the one with the lowest forward voltage, Vf, will be quite bright and the others relatively dim.

Figure 1. Variations in Vf with different colours of LEDs. Source: LED I-V curves.

Figure 2. Variations in Vf for the same type of LED due to variations in manufacture. Source: Variations in Vf and binning.
As you reminded me in the comments, I forgot to cover the current sharing aspect. The voltage at the top of R1 will be reasonably constant (because the voltage drop across the LEDs would be reasonably similar) so the current through R1 will be constant if one or more LEDs is on. That means that the current will be shared between the LEDs - fairly evenly if they're all the same colour but not so evenly if not.
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So if switching 2 on at once is connecting them in parallel, this would mean the current drop would roughly double? And the resistor if limiting to 15ma would be shared between the 2 LED and would only see 7.5ma (roughly)?
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– DonP
19 mins ago
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Correct, and I completely forgot to cover that in the answer. I'll add it in.
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– Transistor
17 mins ago
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OK. Will mark as answer. THanks for the detailed response.
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– DonP
16 mins ago
add a comment |
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It's problematic because nothing is ever identical so the LED with the lowest voltage drop will cause all the current to flow through it while the other two LEDs never turn on because the voltage drop across them isn't high enough.
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$begingroup$
even though they have different +ve signal source?
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– DonP
24 mins ago
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As the others have pointed out above, it's okay if you only ever turn on one at a time. You'll run into trouble if you try to turn on more than one at a time.
$endgroup$
– Toor
17 mins ago
add a comment |
Your Answer
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
You can do that safely if:
- Only one LED is on at a time.
- The voltage of the circuit is less than the maximum reverse voltage of the LEDs. This often isn't specified by 5 V is generally quite safe.
Otherwise switching two or more on at a time connects them in parallel. In this situation the one with the lowest forward voltage will pass the most current. If they are different colour LEDs then the one with the lowest forward voltage, Vf, will be quite bright and the others relatively dim.

Figure 1. Variations in Vf with different colours of LEDs. Source: LED I-V curves.

Figure 2. Variations in Vf for the same type of LED due to variations in manufacture. Source: Variations in Vf and binning.
As you reminded me in the comments, I forgot to cover the current sharing aspect. The voltage at the top of R1 will be reasonably constant (because the voltage drop across the LEDs would be reasonably similar) so the current through R1 will be constant if one or more LEDs is on. That means that the current will be shared between the LEDs - fairly evenly if they're all the same colour but not so evenly if not.
$endgroup$
$begingroup$
So if switching 2 on at once is connecting them in parallel, this would mean the current drop would roughly double? And the resistor if limiting to 15ma would be shared between the 2 LED and would only see 7.5ma (roughly)?
$endgroup$
– DonP
19 mins ago
$begingroup$
Correct, and I completely forgot to cover that in the answer. I'll add it in.
$endgroup$
– Transistor
17 mins ago
$begingroup$
OK. Will mark as answer. THanks for the detailed response.
$endgroup$
– DonP
16 mins ago
add a comment |
$begingroup$
You can do that safely if:
- Only one LED is on at a time.
- The voltage of the circuit is less than the maximum reverse voltage of the LEDs. This often isn't specified by 5 V is generally quite safe.
Otherwise switching two or more on at a time connects them in parallel. In this situation the one with the lowest forward voltage will pass the most current. If they are different colour LEDs then the one with the lowest forward voltage, Vf, will be quite bright and the others relatively dim.

Figure 1. Variations in Vf with different colours of LEDs. Source: LED I-V curves.

Figure 2. Variations in Vf for the same type of LED due to variations in manufacture. Source: Variations in Vf and binning.
As you reminded me in the comments, I forgot to cover the current sharing aspect. The voltage at the top of R1 will be reasonably constant (because the voltage drop across the LEDs would be reasonably similar) so the current through R1 will be constant if one or more LEDs is on. That means that the current will be shared between the LEDs - fairly evenly if they're all the same colour but not so evenly if not.
$endgroup$
$begingroup$
So if switching 2 on at once is connecting them in parallel, this would mean the current drop would roughly double? And the resistor if limiting to 15ma would be shared between the 2 LED and would only see 7.5ma (roughly)?
$endgroup$
– DonP
19 mins ago
$begingroup$
Correct, and I completely forgot to cover that in the answer. I'll add it in.
$endgroup$
– Transistor
17 mins ago
$begingroup$
OK. Will mark as answer. THanks for the detailed response.
$endgroup$
– DonP
16 mins ago
add a comment |
$begingroup$
You can do that safely if:
- Only one LED is on at a time.
- The voltage of the circuit is less than the maximum reverse voltage of the LEDs. This often isn't specified by 5 V is generally quite safe.
Otherwise switching two or more on at a time connects them in parallel. In this situation the one with the lowest forward voltage will pass the most current. If they are different colour LEDs then the one with the lowest forward voltage, Vf, will be quite bright and the others relatively dim.

Figure 1. Variations in Vf with different colours of LEDs. Source: LED I-V curves.

Figure 2. Variations in Vf for the same type of LED due to variations in manufacture. Source: Variations in Vf and binning.
As you reminded me in the comments, I forgot to cover the current sharing aspect. The voltage at the top of R1 will be reasonably constant (because the voltage drop across the LEDs would be reasonably similar) so the current through R1 will be constant if one or more LEDs is on. That means that the current will be shared between the LEDs - fairly evenly if they're all the same colour but not so evenly if not.
$endgroup$
You can do that safely if:
- Only one LED is on at a time.
- The voltage of the circuit is less than the maximum reverse voltage of the LEDs. This often isn't specified by 5 V is generally quite safe.
Otherwise switching two or more on at a time connects them in parallel. In this situation the one with the lowest forward voltage will pass the most current. If they are different colour LEDs then the one with the lowest forward voltage, Vf, will be quite bright and the others relatively dim.

Figure 1. Variations in Vf with different colours of LEDs. Source: LED I-V curves.

Figure 2. Variations in Vf for the same type of LED due to variations in manufacture. Source: Variations in Vf and binning.
As you reminded me in the comments, I forgot to cover the current sharing aspect. The voltage at the top of R1 will be reasonably constant (because the voltage drop across the LEDs would be reasonably similar) so the current through R1 will be constant if one or more LEDs is on. That means that the current will be shared between the LEDs - fairly evenly if they're all the same colour but not so evenly if not.
edited 11 mins ago
answered 39 mins ago
TransistorTransistor
85.1k784181
85.1k784181
$begingroup$
So if switching 2 on at once is connecting them in parallel, this would mean the current drop would roughly double? And the resistor if limiting to 15ma would be shared between the 2 LED and would only see 7.5ma (roughly)?
$endgroup$
– DonP
19 mins ago
$begingroup$
Correct, and I completely forgot to cover that in the answer. I'll add it in.
$endgroup$
– Transistor
17 mins ago
$begingroup$
OK. Will mark as answer. THanks for the detailed response.
$endgroup$
– DonP
16 mins ago
add a comment |
$begingroup$
So if switching 2 on at once is connecting them in parallel, this would mean the current drop would roughly double? And the resistor if limiting to 15ma would be shared between the 2 LED and would only see 7.5ma (roughly)?
$endgroup$
– DonP
19 mins ago
$begingroup$
Correct, and I completely forgot to cover that in the answer. I'll add it in.
$endgroup$
– Transistor
17 mins ago
$begingroup$
OK. Will mark as answer. THanks for the detailed response.
$endgroup$
– DonP
16 mins ago
$begingroup$
So if switching 2 on at once is connecting them in parallel, this would mean the current drop would roughly double? And the resistor if limiting to 15ma would be shared between the 2 LED and would only see 7.5ma (roughly)?
$endgroup$
– DonP
19 mins ago
$begingroup$
So if switching 2 on at once is connecting them in parallel, this would mean the current drop would roughly double? And the resistor if limiting to 15ma would be shared between the 2 LED and would only see 7.5ma (roughly)?
$endgroup$
– DonP
19 mins ago
$begingroup$
Correct, and I completely forgot to cover that in the answer. I'll add it in.
$endgroup$
– Transistor
17 mins ago
$begingroup$
Correct, and I completely forgot to cover that in the answer. I'll add it in.
$endgroup$
– Transistor
17 mins ago
$begingroup$
OK. Will mark as answer. THanks for the detailed response.
$endgroup$
– DonP
16 mins ago
$begingroup$
OK. Will mark as answer. THanks for the detailed response.
$endgroup$
– DonP
16 mins ago
add a comment |
$begingroup$
It's problematic because nothing is ever identical so the LED with the lowest voltage drop will cause all the current to flow through it while the other two LEDs never turn on because the voltage drop across them isn't high enough.
$endgroup$
$begingroup$
even though they have different +ve signal source?
$endgroup$
– DonP
24 mins ago
$begingroup$
As the others have pointed out above, it's okay if you only ever turn on one at a time. You'll run into trouble if you try to turn on more than one at a time.
$endgroup$
– Toor
17 mins ago
add a comment |
$begingroup$
It's problematic because nothing is ever identical so the LED with the lowest voltage drop will cause all the current to flow through it while the other two LEDs never turn on because the voltage drop across them isn't high enough.
$endgroup$
$begingroup$
even though they have different +ve signal source?
$endgroup$
– DonP
24 mins ago
$begingroup$
As the others have pointed out above, it's okay if you only ever turn on one at a time. You'll run into trouble if you try to turn on more than one at a time.
$endgroup$
– Toor
17 mins ago
add a comment |
$begingroup$
It's problematic because nothing is ever identical so the LED with the lowest voltage drop will cause all the current to flow through it while the other two LEDs never turn on because the voltage drop across them isn't high enough.
$endgroup$
It's problematic because nothing is ever identical so the LED with the lowest voltage drop will cause all the current to flow through it while the other two LEDs never turn on because the voltage drop across them isn't high enough.
answered 40 mins ago
ToorToor
72219
72219
$begingroup$
even though they have different +ve signal source?
$endgroup$
– DonP
24 mins ago
$begingroup$
As the others have pointed out above, it's okay if you only ever turn on one at a time. You'll run into trouble if you try to turn on more than one at a time.
$endgroup$
– Toor
17 mins ago
add a comment |
$begingroup$
even though they have different +ve signal source?
$endgroup$
– DonP
24 mins ago
$begingroup$
As the others have pointed out above, it's okay if you only ever turn on one at a time. You'll run into trouble if you try to turn on more than one at a time.
$endgroup$
– Toor
17 mins ago
$begingroup$
even though they have different +ve signal source?
$endgroup$
– DonP
24 mins ago
$begingroup$
even though they have different +ve signal source?
$endgroup$
– DonP
24 mins ago
$begingroup$
As the others have pointed out above, it's okay if you only ever turn on one at a time. You'll run into trouble if you try to turn on more than one at a time.
$endgroup$
– Toor
17 mins ago
$begingroup$
As the others have pointed out above, it's okay if you only ever turn on one at a time. You'll run into trouble if you try to turn on more than one at a time.
$endgroup$
– Toor
17 mins ago
add a comment |
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$begingroup$
No, there are tolerances inbetween one LED and another. The LED with the lowest Vf will carry a lot more current then the others.
$endgroup$
– Unimportant
40 mins ago
1
$begingroup$
Looks like Diode OR logic ;) but problematic due to sharing current and voltage drop.
$endgroup$
– Sunnyskyguy EE75
32 mins ago