Do output LEDs here need resistors?
up vote
1
down vote
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Consider:
Here is the schematic I'm trying to work around. My major question is about the LED bar graph I want to use, on the right side; does it need resistors?
I saw three different versions of this microcontroller use ... sometimes with, sometimes without, and the datasheet did not really help. On this point, and it also displays an example of use without a resistor.
Bonus question: Is there something wrong in this?
led-driver lm3915
edited Nov 25 at 11:51
Peter Mortensen
1,58031422
asked Nov 24 at 15:25
francois P
13817
add a comment |
up vote
1
down vote
favorite
Consider:
Here is the schematic I'm trying to work around. My major question is about the LED bar graph I want to use, on the right side; does it need resistors?
I saw three different versions of this microcontroller use ... sometimes with, sometimes without, and the datasheet did not really help. On this point, and it also displays an example of use without a resistor.
Bonus question: Is there something wrong in this?
led-driver lm3915
edited Nov 25 at 11:51
Peter Mortensen
1,58031422
asked Nov 24 at 15:25
francois P
13817
3
The 3915 is not a microprcessor (UC.)
– JRE
Nov 24 at 15:56
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Consider:
Here is the schematic I'm trying to work around. My major question is about the LED bar graph I want to use, on the right side; does it need resistors?
I saw three different versions of this microcontroller use ... sometimes with, sometimes without, and the datasheet did not really help. On this point, and it also displays an example of use without a resistor.
Bonus question: Is there something wrong in this?
led-driver lm3915
edited Nov 25 at 11:51
Peter Mortensen
1,58031422
asked Nov 24 at 15:25
francois P
13817
Consider:
Here is the schematic I'm trying to work around. My major question is about the LED bar graph I want to use, on the right side; does it need resistors?
I saw three different versions of this microcontroller use ... sometimes with, sometimes without, and the datasheet did not really help. On this point, and it also displays an example of use without a resistor.
Bonus question: Is there something wrong in this?
led-driver lm3915
led-driver lm3915
edited Nov 25 at 11:51
Peter Mortensen
1,58031422
asked Nov 24 at 15:25
francois P
13817
edited Nov 25 at 11:51
Peter Mortensen
1,58031422
asked Nov 24 at 15:25
francois P
13817
edited Nov 25 at 11:51
Peter Mortensen
1,58031422
edited Nov 25 at 11:51
Peter Mortensen
1,58031422
edited Nov 25 at 11:51
Peter Mortensen
1,58031422
1,58031422
asked Nov 24 at 15:25
francois P
13817
asked Nov 24 at 15:25
francois P
13817
asked Nov 24 at 15:25
francois P
13817
13817
3
The 3915 is not a microprcessor (UC.)
– JRE
Nov 24 at 15:56
add a comment |
3
The 3915 is not a microprcessor (UC.)
– JRE
Nov 24 at 15:56
3
3
The 3915 is not a microprcessor (UC.)
– JRE
Nov 24 at 15:56
The 3915 is not a microprcessor (UC.)
– JRE
Nov 24 at 15:56
add a comment |
3 Answers
3
active
oldest
votes
up vote
12
down vote
accepted
As always, READ THE DATASHEET!
The chip you use has programmable constant current outputs, so you don't need external resistors to determine the current, the chip does this for you.
The current sunk by each output is ~ 10 x the current out of the Vref.
In your case 10 * ( 1.2 V / (1k2 + 680 ) Ohm ) ~= 6 mA, which is OK for a normal (20 mA max) LED.
answered Nov 24 at 15:51
Wouter van Ooijen
44k150117
add a comment |
up vote
5
down vote
The LM3915 regulates the current to the LEDs. You do not need a current limiting resistor for each LED.
In your circuit, R9 and R10 set the LED current.
Vref is 1.25V. The current drawn through that pin controls the LED current.
According to the datasheet, the LED current is 10* the current drawn through Vref.
In your circuit, that'd be 10*(1.25V/(680 ohms + 1200 ohms))
So, about 6.5 mA.
To make the LEDs brighter, you reduce R9 and R10 (but keep the ratio the same.) So, if you changed to R9=340 ohms and R10= 600 ohms, you'd get about 13 mA for the LEDs.
The given current is per LED.
I've simplified the calculation. The datasheet says you are supposed to take into account the current through the internal divider chain. That amounts to about 0.5 mA, so I just ignored it for simplicity's sake.
answered Nov 24 at 15:53
JRE
20.2k43767
add a comment |
up vote
2
down vote
No, you don't need resistors with LM3915.
The LEDs need to be driven with the correct current to achieve the desired brightness. The resistors are used with LEDs to adjust the current when they are driven with constant voltage. LM3915 drives the LEDs with a constant current, which you can program, so you don't need the resistors.
edited Nov 25 at 11:50
Peter Mortensen
1,58031422
answered Nov 24 at 15:45
TemeV
665
New contributor
TemeV is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
12
down vote
accepted
As always, READ THE DATASHEET!
The chip you use has programmable constant current outputs, so you don't need external resistors to determine the current, the chip does this for you.
The current sunk by each output is ~ 10 x the current out of the Vref.
In your case 10 * ( 1.2 V / (1k2 + 680 ) Ohm ) ~= 6 mA, which is OK for a normal (20 mA max) LED.
answered Nov 24 at 15:51
Wouter van Ooijen
44k150117
add a comment |
up vote
12
down vote
accepted
As always, READ THE DATASHEET!
The chip you use has programmable constant current outputs, so you don't need external resistors to determine the current, the chip does this for you.
The current sunk by each output is ~ 10 x the current out of the Vref.
In your case 10 * ( 1.2 V / (1k2 + 680 ) Ohm ) ~= 6 mA, which is OK for a normal (20 mA max) LED.
answered Nov 24 at 15:51
Wouter van Ooijen
44k150117
add a comment |
up vote
12
down vote
accepted
up vote
12
down vote
accepted
As always, READ THE DATASHEET!
The chip you use has programmable constant current outputs, so you don't need external resistors to determine the current, the chip does this for you.
The current sunk by each output is ~ 10 x the current out of the Vref.
In your case 10 * ( 1.2 V / (1k2 + 680 ) Ohm ) ~= 6 mA, which is OK for a normal (20 mA max) LED.
answered Nov 24 at 15:51
Wouter van Ooijen
44k150117
As always, READ THE DATASHEET!
The chip you use has programmable constant current outputs, so you don't need external resistors to determine the current, the chip does this for you.
The current sunk by each output is ~ 10 x the current out of the Vref.
In your case 10 * ( 1.2 V / (1k2 + 680 ) Ohm ) ~= 6 mA, which is OK for a normal (20 mA max) LED.
answered Nov 24 at 15:51
Wouter van Ooijen
44k150117
edited Nov 24 at 21:32
answered Nov 24 at 15:51
Wouter van Ooijen
44k150117
answered Nov 24 at 15:51
Wouter van Ooijen
44k150117
answered Nov 24 at 15:51
Wouter van Ooijen
44k150117
44k150117
add a comment |
add a comment |
up vote
5
down vote
The LM3915 regulates the current to the LEDs. You do not need a current limiting resistor for each LED.
In your circuit, R9 and R10 set the LED current.
Vref is 1.25V. The current drawn through that pin controls the LED current.
According to the datasheet, the LED current is 10* the current drawn through Vref.
In your circuit, that'd be 10*(1.25V/(680 ohms + 1200 ohms))
So, about 6.5 mA.
To make the LEDs brighter, you reduce R9 and R10 (but keep the ratio the same.) So, if you changed to R9=340 ohms and R10= 600 ohms, you'd get about 13 mA for the LEDs.
The given current is per LED.
I've simplified the calculation. The datasheet says you are supposed to take into account the current through the internal divider chain. That amounts to about 0.5 mA, so I just ignored it for simplicity's sake.
answered Nov 24 at 15:53
JRE
20.2k43767
add a comment |
up vote
5
down vote
The LM3915 regulates the current to the LEDs. You do not need a current limiting resistor for each LED.
In your circuit, R9 and R10 set the LED current.
Vref is 1.25V. The current drawn through that pin controls the LED current.
According to the datasheet, the LED current is 10* the current drawn through Vref.
In your circuit, that'd be 10*(1.25V/(680 ohms + 1200 ohms))
So, about 6.5 mA.
To make the LEDs brighter, you reduce R9 and R10 (but keep the ratio the same.) So, if you changed to R9=340 ohms and R10= 600 ohms, you'd get about 13 mA for the LEDs.
The given current is per LED.
I've simplified the calculation. The datasheet says you are supposed to take into account the current through the internal divider chain. That amounts to about 0.5 mA, so I just ignored it for simplicity's sake.
answered Nov 24 at 15:53
JRE
20.2k43767
add a comment |
up vote
5
down vote
up vote
5
down vote
The LM3915 regulates the current to the LEDs. You do not need a current limiting resistor for each LED.
In your circuit, R9 and R10 set the LED current.
Vref is 1.25V. The current drawn through that pin controls the LED current.
According to the datasheet, the LED current is 10* the current drawn through Vref.
In your circuit, that'd be 10*(1.25V/(680 ohms + 1200 ohms))
So, about 6.5 mA.
To make the LEDs brighter, you reduce R9 and R10 (but keep the ratio the same.) So, if you changed to R9=340 ohms and R10= 600 ohms, you'd get about 13 mA for the LEDs.
The given current is per LED.
I've simplified the calculation. The datasheet says you are supposed to take into account the current through the internal divider chain. That amounts to about 0.5 mA, so I just ignored it for simplicity's sake.
answered Nov 24 at 15:53
JRE
20.2k43767
The LM3915 regulates the current to the LEDs. You do not need a current limiting resistor for each LED.
In your circuit, R9 and R10 set the LED current.
Vref is 1.25V. The current drawn through that pin controls the LED current.
According to the datasheet, the LED current is 10* the current drawn through Vref.
In your circuit, that'd be 10*(1.25V/(680 ohms + 1200 ohms))
So, about 6.5 mA.
To make the LEDs brighter, you reduce R9 and R10 (but keep the ratio the same.) So, if you changed to R9=340 ohms and R10= 600 ohms, you'd get about 13 mA for the LEDs.
The given current is per LED.
I've simplified the calculation. The datasheet says you are supposed to take into account the current through the internal divider chain. That amounts to about 0.5 mA, so I just ignored it for simplicity's sake.
answered Nov 24 at 15:53
JRE
20.2k43767
answered Nov 24 at 15:53
JRE
20.2k43767
answered Nov 24 at 15:53
JRE
20.2k43767
answered Nov 24 at 15:53
JRE
20.2k43767
20.2k43767
add a comment |
add a comment |
up vote
2
down vote
No, you don't need resistors with LM3915.
The LEDs need to be driven with the correct current to achieve the desired brightness. The resistors are used with LEDs to adjust the current when they are driven with constant voltage. LM3915 drives the LEDs with a constant current, which you can program, so you don't need the resistors.
edited Nov 25 at 11:50
Peter Mortensen
1,58031422
answered Nov 24 at 15:45
TemeV
665
New contributor
TemeV is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
up vote
2
down vote
No, you don't need resistors with LM3915.
The LEDs need to be driven with the correct current to achieve the desired brightness. The resistors are used with LEDs to adjust the current when they are driven with constant voltage. LM3915 drives the LEDs with a constant current, which you can program, so you don't need the resistors.
edited Nov 25 at 11:50
Peter Mortensen
1,58031422
answered Nov 24 at 15:45
TemeV
665
New contributor
TemeV is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
up vote
2
down vote
up vote
2
down vote
No, you don't need resistors with LM3915.
The LEDs need to be driven with the correct current to achieve the desired brightness. The resistors are used with LEDs to adjust the current when they are driven with constant voltage. LM3915 drives the LEDs with a constant current, which you can program, so you don't need the resistors.
edited Nov 25 at 11:50
Peter Mortensen
1,58031422
answered Nov 24 at 15:45
TemeV
665
New contributor
TemeV is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
No, you don't need resistors with LM3915.
The LEDs need to be driven with the correct current to achieve the desired brightness. The resistors are used with LEDs to adjust the current when they are driven with constant voltage. LM3915 drives the LEDs with a constant current, which you can program, so you don't need the resistors.
edited Nov 25 at 11:50
Peter Mortensen
1,58031422
answered Nov 24 at 15:45
TemeV
665
New contributor
TemeV is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited Nov 25 at 11:50
Peter Mortensen
1,58031422
edited Nov 25 at 11:50
Peter Mortensen
1,58031422
edited Nov 25 at 11:50
Peter Mortensen
1,58031422
1,58031422
answered Nov 24 at 15:45
TemeV
665
New contributor
TemeV is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered Nov 24 at 15:45
TemeV
665
answered Nov 24 at 15:45
TemeV
665
665
New contributor
TemeV is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
TemeV is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
TemeV is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
add a comment |
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3
The 3915 is not a microprcessor (UC.)
– JRE
Nov 24 at 15:56