Can I use a single resistor for multiple LED with different +ve sources?
$begingroup$
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
asked 4 hours ago
DonPDonP
307
$endgroup$
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
asked 4 hours ago
DonPDonP
307
$endgroup$
2
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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.
$endgroup$
– Unimportant
4 hours ago
1
$begingroup$
Looks like Diode OR logic ;) but problematic due to sharing current and voltage drop.
$endgroup$
– Sunnyskyguy EE75
4 hours 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
asked 4 hours ago
DonPDonP
307
$endgroup$
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 4 hours ago
DonPDonP
307
asked 4 hours ago
DonPDonP
307
asked 4 hours ago
DonPDonP
307
asked 4 hours ago
DonPDonP
307
asked 4 hours ago
DonPDonP
307
307
2
$begingroup$
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
4 hours ago
1
$begingroup$
Looks like Diode OR logic ;) but problematic due to sharing current and voltage drop.
$endgroup$
– Sunnyskyguy EE75
4 hours ago
add a comment |
2
$begingroup$
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
4 hours ago
1
$begingroup$
Looks like Diode OR logic ;) but problematic due to sharing current and voltage drop.
$endgroup$
– Sunnyskyguy EE75
4 hours ago
2
2
$begingroup$
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
4 hours 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
4 hours ago
1
1
$begingroup$
Looks like Diode OR logic ;) but problematic due to sharing current and voltage drop.
$endgroup$
– Sunnyskyguy EE75
4 hours ago
$begingroup$
Looks like Diode OR logic ;) but problematic due to sharing current and voltage drop.
$endgroup$
– Sunnyskyguy EE75
4 hours 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.
answered 4 hours ago
TransistorTransistor
85.2k784181
$endgroup$
$begingroup$
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
3 hours ago
$begingroup$
Correct, and I completely forgot to cover that in the answer. I'll add it in.
$endgroup$
– Transistor
3 hours ago
$begingroup$
OK. Will mark as answer. THanks for the detailed response.
$endgroup$
– DonP
3 hours 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.
answered 4 hours ago
ToorToor
72219
$endgroup$
1
$begingroup$
even though they have different +ve signal source?
$endgroup$
– DonP
3 hours 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
3 hours ago
add a comment |
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2 Answers
2
active
oldest
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2 Answers
2
active
oldest
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active
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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.
answered 4 hours ago
TransistorTransistor
85.2k784181
$endgroup$
$begingroup$
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
3 hours ago
$begingroup$
Correct, and I completely forgot to cover that in the answer. I'll add it in.
$endgroup$
– Transistor
3 hours ago
$begingroup$
OK. Will mark as answer. THanks for the detailed response.
$endgroup$
– DonP
3 hours 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.
answered 4 hours ago
TransistorTransistor
85.2k784181
$endgroup$
$begingroup$
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
3 hours ago
$begingroup$
Correct, and I completely forgot to cover that in the answer. I'll add it in.
$endgroup$
– Transistor
3 hours ago
$begingroup$
OK. Will mark as answer. THanks for the detailed response.
$endgroup$
– DonP
3 hours 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.
answered 4 hours ago
TransistorTransistor
85.2k784181
$endgroup$
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.
answered 4 hours ago
TransistorTransistor
85.2k784181
edited 3 hours ago
answered 4 hours ago
TransistorTransistor
85.2k784181
answered 4 hours ago
TransistorTransistor
85.2k784181
answered 4 hours ago
TransistorTransistor
85.2k784181
85.2k784181
$begingroup$
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
3 hours ago
$begingroup$
Correct, and I completely forgot to cover that in the answer. I'll add it in.
$endgroup$
– Transistor
3 hours ago
$begingroup$
OK. Will mark as answer. THanks for the detailed response.
$endgroup$
– DonP
3 hours 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
3 hours ago
$begingroup$
Correct, and I completely forgot to cover that in the answer. I'll add it in.
$endgroup$
– Transistor
3 hours ago
$begingroup$
OK. Will mark as answer. THanks for the detailed response.
$endgroup$
– DonP
3 hours 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
3 hours 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
3 hours ago
$begingroup$
Correct, and I completely forgot to cover that in the answer. I'll add it in.
$endgroup$
– Transistor
3 hours ago
$begingroup$
Correct, and I completely forgot to cover that in the answer. I'll add it in.
$endgroup$
– Transistor
3 hours ago
$begingroup$
OK. Will mark as answer. THanks for the detailed response.
$endgroup$
– DonP
3 hours ago
$begingroup$
OK. Will mark as answer. THanks for the detailed response.
$endgroup$
– DonP
3 hours 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.
answered 4 hours ago
ToorToor
72219
$endgroup$
1
$begingroup$
even though they have different +ve signal source?
$endgroup$
– DonP
3 hours 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
3 hours 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.
answered 4 hours ago
ToorToor
72219
$endgroup$
1
$begingroup$
even though they have different +ve signal source?
$endgroup$
– DonP
3 hours 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
3 hours 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.
answered 4 hours ago
ToorToor
72219
$endgroup$
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 4 hours ago
ToorToor
72219
answered 4 hours ago
ToorToor
72219
answered 4 hours ago
ToorToor
72219
answered 4 hours ago
ToorToor
72219
72219
1
$begingroup$
even though they have different +ve signal source?
$endgroup$
– DonP
3 hours 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
3 hours ago
add a comment |
1
$begingroup$
even though they have different +ve signal source?
$endgroup$
– DonP
3 hours 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
3 hours ago
1
1
$begingroup$
even though they have different +ve signal source?
$endgroup$
– DonP
3 hours ago
$begingroup$
even though they have different +ve signal source?
$endgroup$
– DonP
3 hours 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
3 hours 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
3 hours ago
add a comment |
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2
$begingroup$
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
4 hours ago
1
$begingroup$
Looks like Diode OR logic ;) but problematic due to sharing current and voltage drop.
$endgroup$
– Sunnyskyguy EE75
4 hours ago