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Understanding “Current Draw” in terms of “Ohm's Law”
What determines a load's current requirements?Constant current power supply and 2 LED - series or parellel?Why are these readings violating ohm's law? (Are they?)adapting 60 LED battery operated string-light to work with wall wartCalculating LED current using ohms lawOhms law and voltageResistance in Ohm's lawWhy are voltage and current directly proportional?Connecting a single LED to Motherboard CHA Fan HeaderIf amps and volts can vary independently, how does Ohm's law actually work?
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$begingroup$
I'm having trouble understanding Current Draw as it relates to Ohm's Law (V = IR).
I have a power supply rated at 5V and 500mA. I'm trying to power a string of LED lights. According to my datasheets, each LED draws ~50mA of current. If I have a string of 5 LEDs, that would draw 250mA. If I have a string of 10 LEDs, that will draw 500mA of current.
Here's my problem:
According to my understanding of Ohm's law, Current = Voltage / Resistance. As I understand it, when I add additional LEDs to my circuit, each LED should contribute resistance to the circuit. If this is the case, according to Ohm's Law, I would expect that current would DECREASE with each additional LED. In other words, as Resistance increases and as Voltage stays the same... when I divide V by an increasing R, this equation results in a lower I.
Instead, it is INCREASING!
Am I misunderstanding something here?
EDIT:
Yes, I was misunderstanding something there. The LEDs on my LED strip are wired in parallel, not in series. Resistances wired in parallel lower the total resistance through the circuit, so current would INCREASE coming from my power supply.
Additionally, LEDs don't function exactly like a regular resistors. As Voltage increases, current increases non-linearly, so Ohm's Law cannot apply perfectly.
led current current-measurement led-strip ohms-law
asked Jun 7 at 14:33
Jcb RbJcb Rb
233
New contributor
Jcb Rb is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
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add a comment |
$begingroup$
I'm having trouble understanding Current Draw as it relates to Ohm's Law (V = IR).
I have a power supply rated at 5V and 500mA. I'm trying to power a string of LED lights. According to my datasheets, each LED draws ~50mA of current. If I have a string of 5 LEDs, that would draw 250mA. If I have a string of 10 LEDs, that will draw 500mA of current.
Here's my problem:
According to my understanding of Ohm's law, Current = Voltage / Resistance. As I understand it, when I add additional LEDs to my circuit, each LED should contribute resistance to the circuit. If this is the case, according to Ohm's Law, I would expect that current would DECREASE with each additional LED. In other words, as Resistance increases and as Voltage stays the same... when I divide V by an increasing R, this equation results in a lower I.
Instead, it is INCREASING!
Am I misunderstanding something here?
EDIT:
Yes, I was misunderstanding something there. The LEDs on my LED strip are wired in parallel, not in series. Resistances wired in parallel lower the total resistance through the circuit, so current would INCREASE coming from my power supply.
Additionally, LEDs don't function exactly like a regular resistors. As Voltage increases, current increases non-linearly, so Ohm's Law cannot apply perfectly.
led current current-measurement led-strip ohms-law
asked Jun 7 at 14:33
Jcb RbJcb Rb
233
New contributor
Jcb Rb is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
1
$begingroup$
Are you adding LEDs in series or parallel? The current through a LED is non-linear and ohms law doesn't necessarily apply.
$endgroup$
– R.Joshi
Jun 7 at 14:43
$begingroup$
Tip: LED and other semiconductor devices are called "non-linear devices". Consider the term "linear", what does it mean? It means the current is proportional to the voltage, in other words, it obeys Ohm's Law. Hence, "non-linear electronics devices" basically means, "electronic devices that do not obey Ohm's Law" (although in many cases there are techniques allowing you to perform a linear approximation within a small operating range and pretend Ohm's Law works).
$endgroup$
– 比尔盖子
Jun 7 at 17:08
add a comment |
$begingroup$
I'm having trouble understanding Current Draw as it relates to Ohm's Law (V = IR).
I have a power supply rated at 5V and 500mA. I'm trying to power a string of LED lights. According to my datasheets, each LED draws ~50mA of current. If I have a string of 5 LEDs, that would draw 250mA. If I have a string of 10 LEDs, that will draw 500mA of current.
Here's my problem:
According to my understanding of Ohm's law, Current = Voltage / Resistance. As I understand it, when I add additional LEDs to my circuit, each LED should contribute resistance to the circuit. If this is the case, according to Ohm's Law, I would expect that current would DECREASE with each additional LED. In other words, as Resistance increases and as Voltage stays the same... when I divide V by an increasing R, this equation results in a lower I.
Instead, it is INCREASING!
Am I misunderstanding something here?
EDIT:
Yes, I was misunderstanding something there. The LEDs on my LED strip are wired in parallel, not in series. Resistances wired in parallel lower the total resistance through the circuit, so current would INCREASE coming from my power supply.
Additionally, LEDs don't function exactly like a regular resistors. As Voltage increases, current increases non-linearly, so Ohm's Law cannot apply perfectly.
led current current-measurement led-strip ohms-law
asked Jun 7 at 14:33
Jcb RbJcb Rb
233
New contributor
Jcb Rb is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
I'm having trouble understanding Current Draw as it relates to Ohm's Law (V = IR).
I have a power supply rated at 5V and 500mA. I'm trying to power a string of LED lights. According to my datasheets, each LED draws ~50mA of current. If I have a string of 5 LEDs, that would draw 250mA. If I have a string of 10 LEDs, that will draw 500mA of current.
Here's my problem:
According to my understanding of Ohm's law, Current = Voltage / Resistance. As I understand it, when I add additional LEDs to my circuit, each LED should contribute resistance to the circuit. If this is the case, according to Ohm's Law, I would expect that current would DECREASE with each additional LED. In other words, as Resistance increases and as Voltage stays the same... when I divide V by an increasing R, this equation results in a lower I.
Instead, it is INCREASING!
Am I misunderstanding something here?
EDIT:
Yes, I was misunderstanding something there. The LEDs on my LED strip are wired in parallel, not in series. Resistances wired in parallel lower the total resistance through the circuit, so current would INCREASE coming from my power supply.
Additionally, LEDs don't function exactly like a regular resistors. As Voltage increases, current increases non-linearly, so Ohm's Law cannot apply perfectly.
led current current-measurement led-strip ohms-law
led current current-measurement led-strip ohms-law
asked Jun 7 at 14:33
Jcb RbJcb Rb
233
New contributor
Jcb Rb is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Jun 7 at 14:33
Jcb RbJcb Rb
233
New contributor
Jcb Rb is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited Jun 7 at 15:24
Jcb Rb
asked Jun 7 at 14:33
Jcb RbJcb Rb
233
New contributor
Jcb Rb is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Jun 7 at 14:33
Jcb RbJcb Rb
233
asked Jun 7 at 14:33
Jcb RbJcb Rb
233
233
New contributor
Jcb Rb is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Jcb Rb is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
1
$begingroup$
Are you adding LEDs in series or parallel? The current through a LED is non-linear and ohms law doesn't necessarily apply.
$endgroup$
– R.Joshi
Jun 7 at 14:43
$begingroup$
Tip: LED and other semiconductor devices are called "non-linear devices". Consider the term "linear", what does it mean? It means the current is proportional to the voltage, in other words, it obeys Ohm's Law. Hence, "non-linear electronics devices" basically means, "electronic devices that do not obey Ohm's Law" (although in many cases there are techniques allowing you to perform a linear approximation within a small operating range and pretend Ohm's Law works).
$endgroup$
– 比尔盖子
Jun 7 at 17:08
add a comment |
1
$begingroup$
Are you adding LEDs in series or parallel? The current through a LED is non-linear and ohms law doesn't necessarily apply.
$endgroup$
– R.Joshi
Jun 7 at 14:43
$begingroup$
Tip: LED and other semiconductor devices are called "non-linear devices". Consider the term "linear", what does it mean? It means the current is proportional to the voltage, in other words, it obeys Ohm's Law. Hence, "non-linear electronics devices" basically means, "electronic devices that do not obey Ohm's Law" (although in many cases there are techniques allowing you to perform a linear approximation within a small operating range and pretend Ohm's Law works).
$endgroup$
– 比尔盖子
Jun 7 at 17:08
1
1
$begingroup$
Are you adding LEDs in series or parallel? The current through a LED is non-linear and ohms law doesn't necessarily apply.
$endgroup$
– R.Joshi
Jun 7 at 14:43
$begingroup$
Are you adding LEDs in series or parallel? The current through a LED is non-linear and ohms law doesn't necessarily apply.
$endgroup$
– R.Joshi
Jun 7 at 14:43
$begingroup$
Tip: LED and other semiconductor devices are called "non-linear devices". Consider the term "linear", what does it mean? It means the current is proportional to the voltage, in other words, it obeys Ohm's Law. Hence, "non-linear electronics devices" basically means, "electronic devices that do not obey Ohm's Law" (although in many cases there are techniques allowing you to perform a linear approximation within a small operating range and pretend Ohm's Law works).
$endgroup$
– 比尔盖子
Jun 7 at 17:08
$begingroup$
Tip: LED and other semiconductor devices are called "non-linear devices". Consider the term "linear", what does it mean? It means the current is proportional to the voltage, in other words, it obeys Ohm's Law. Hence, "non-linear electronics devices" basically means, "electronic devices that do not obey Ohm's Law" (although in many cases there are techniques allowing you to perform a linear approximation within a small operating range and pretend Ohm's Law works).
$endgroup$
– 比尔盖子
Jun 7 at 17:08
add a comment |
4 Answers
4
active
oldest
votes
$begingroup$
Well first off I would say that an LED is an active device, which means voltage and current are not linearly related like a passive device such as a resistor. The resistance of an LED is not static, it changes according to the device's IV (current-voltage) relationship. That is more of an aside, since Ohm's Law is not as straight forward as it may appear when you are dealing with active devices.
However the explanation to your problem is fairly straight forward. You are adding LEDs in parallel I assume, each LED has a fairly static resistance since current and voltage are stable. I am assuming you are stringing the LED's in parallel. When you add resistance in parallel it decreases the total network's resistance. Look up effective resistance of parallel resistors and you will see.
answered Jun 7 at 14:53
Shamus MacDonaldShamus MacDonald
334
$endgroup$
$begingroup$
Thanks. This was the clearest and most concise answer to my problem.
$endgroup$
– Jcb Rb
Jun 7 at 15:21
add a comment |
$begingroup$
You have a number of misunderstandings.
Ohm's law applies to resistors. LEDs are non-linear elements and therefore Ohm's law doesn't apply.
Elements in series have the same current flowing through them. If you have 3 LEDs in series and one has 50mA flowing in it, so do the other two. Total current?
50mA. Your source has to have enough voltage to "overcome" the sum of the forward voltages of the LEDs plus some headroom for whatever is regulating or limiting the current.
Elements in parallel have the same voltage across them, and may have different currents flowing through them. The sum of the currents through each element is the total current drawn by the elements from the source. Three parallel LEDs with 50mA flowing in each would draw 150mA from the supply.
You have to control the current through LEDs by using a constant current drive or a resistor in series to set the appropriate operating point.
So the draw will depend on how you design your lighting system. Series, parallel, drive scheme and target current.
answered Jun 7 at 14:45
John DJohn D
12.9k12132
$endgroup$
$begingroup$
Thank you for this answer. Turns out, my problem was just a lack of basic knowledge about circuits wired in parallel. The LED strip I was using had the LEDs wired in parallel on the strip. Resistors in parallel lower total resistance of the circuit, therefore as I add LEDs to the circuit, the resistance lowers, the voltage stays the same, and the current increases from the power supply.
$endgroup$
– Jcb Rb
Jun 7 at 15:20
add a comment |
$begingroup$
A string of LEDs is a bunch of LEDs in series and this means that they share the same current. If the forward volt drop across each LED is (say) 2 volts, then 5 LEDs requires a supply of at least 10 volts by the way.
answered Jun 7 at 14:46
Andy akaAndy aka
248k11190437
$endgroup$
add a comment |
$begingroup$
It can often help to flip it upside down and think of conductance instead of resistance. Conductance has a unit of Siemens or S.
Conductance in Siemens is 1 / resistance in ohms.
C E = I Conductance * voltage = current
E = I / C
So for instance a 10 ohm resistor has a conductance of 0.1 Siemens. A 5 ohm resistor has a conductance of 0.2 Siemens.
What happens if you parallel them? Their conductance is - hold onto your hat - 0.3 Siemens! In parallel, conductances just add.
Boy, that's easier to work with.
answered Jun 7 at 23:06
HarperHarper
7,0821027
$endgroup$
add a comment |
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Well first off I would say that an LED is an active device, which means voltage and current are not linearly related like a passive device such as a resistor. The resistance of an LED is not static, it changes according to the device's IV (current-voltage) relationship. That is more of an aside, since Ohm's Law is not as straight forward as it may appear when you are dealing with active devices.
However the explanation to your problem is fairly straight forward. You are adding LEDs in parallel I assume, each LED has a fairly static resistance since current and voltage are stable. I am assuming you are stringing the LED's in parallel. When you add resistance in parallel it decreases the total network's resistance. Look up effective resistance of parallel resistors and you will see.
answered Jun 7 at 14:53
Shamus MacDonaldShamus MacDonald
334
$endgroup$
$begingroup$
Thanks. This was the clearest and most concise answer to my problem.
$endgroup$
– Jcb Rb
Jun 7 at 15:21
add a comment |
$begingroup$
Well first off I would say that an LED is an active device, which means voltage and current are not linearly related like a passive device such as a resistor. The resistance of an LED is not static, it changes according to the device's IV (current-voltage) relationship. That is more of an aside, since Ohm's Law is not as straight forward as it may appear when you are dealing with active devices.
However the explanation to your problem is fairly straight forward. You are adding LEDs in parallel I assume, each LED has a fairly static resistance since current and voltage are stable. I am assuming you are stringing the LED's in parallel. When you add resistance in parallel it decreases the total network's resistance. Look up effective resistance of parallel resistors and you will see.
answered Jun 7 at 14:53
Shamus MacDonaldShamus MacDonald
334
$endgroup$
$begingroup$
Thanks. This was the clearest and most concise answer to my problem.
$endgroup$
– Jcb Rb
Jun 7 at 15:21
add a comment |
$begingroup$
Well first off I would say that an LED is an active device, which means voltage and current are not linearly related like a passive device such as a resistor. The resistance of an LED is not static, it changes according to the device's IV (current-voltage) relationship. That is more of an aside, since Ohm's Law is not as straight forward as it may appear when you are dealing with active devices.
However the explanation to your problem is fairly straight forward. You are adding LEDs in parallel I assume, each LED has a fairly static resistance since current and voltage are stable. I am assuming you are stringing the LED's in parallel. When you add resistance in parallel it decreases the total network's resistance. Look up effective resistance of parallel resistors and you will see.
answered Jun 7 at 14:53
Shamus MacDonaldShamus MacDonald
334
$endgroup$
Well first off I would say that an LED is an active device, which means voltage and current are not linearly related like a passive device such as a resistor. The resistance of an LED is not static, it changes according to the device's IV (current-voltage) relationship. That is more of an aside, since Ohm's Law is not as straight forward as it may appear when you are dealing with active devices.
However the explanation to your problem is fairly straight forward. You are adding LEDs in parallel I assume, each LED has a fairly static resistance since current and voltage are stable. I am assuming you are stringing the LED's in parallel. When you add resistance in parallel it decreases the total network's resistance. Look up effective resistance of parallel resistors and you will see.
answered Jun 7 at 14:53
Shamus MacDonaldShamus MacDonald
334
answered Jun 7 at 14:53
Shamus MacDonaldShamus MacDonald
334
answered Jun 7 at 14:53
Shamus MacDonaldShamus MacDonald
334
answered Jun 7 at 14:53
Shamus MacDonaldShamus MacDonald
334
334
$begingroup$
Thanks. This was the clearest and most concise answer to my problem.
$endgroup$
– Jcb Rb
Jun 7 at 15:21
add a comment |
$begingroup$
Thanks. This was the clearest and most concise answer to my problem.
$endgroup$
– Jcb Rb
Jun 7 at 15:21
$begingroup$
Thanks. This was the clearest and most concise answer to my problem.
$endgroup$
– Jcb Rb
Jun 7 at 15:21
$begingroup$
Thanks. This was the clearest and most concise answer to my problem.
$endgroup$
– Jcb Rb
Jun 7 at 15:21
add a comment |
$begingroup$
You have a number of misunderstandings.
Ohm's law applies to resistors. LEDs are non-linear elements and therefore Ohm's law doesn't apply.
Elements in series have the same current flowing through them. If you have 3 LEDs in series and one has 50mA flowing in it, so do the other two. Total current?
50mA. Your source has to have enough voltage to "overcome" the sum of the forward voltages of the LEDs plus some headroom for whatever is regulating or limiting the current.
Elements in parallel have the same voltage across them, and may have different currents flowing through them. The sum of the currents through each element is the total current drawn by the elements from the source. Three parallel LEDs with 50mA flowing in each would draw 150mA from the supply.
You have to control the current through LEDs by using a constant current drive or a resistor in series to set the appropriate operating point.
So the draw will depend on how you design your lighting system. Series, parallel, drive scheme and target current.
answered Jun 7 at 14:45
John DJohn D
12.9k12132
$endgroup$
$begingroup$
Thank you for this answer. Turns out, my problem was just a lack of basic knowledge about circuits wired in parallel. The LED strip I was using had the LEDs wired in parallel on the strip. Resistors in parallel lower total resistance of the circuit, therefore as I add LEDs to the circuit, the resistance lowers, the voltage stays the same, and the current increases from the power supply.
$endgroup$
– Jcb Rb
Jun 7 at 15:20
add a comment |
$begingroup$
You have a number of misunderstandings.
Ohm's law applies to resistors. LEDs are non-linear elements and therefore Ohm's law doesn't apply.
Elements in series have the same current flowing through them. If you have 3 LEDs in series and one has 50mA flowing in it, so do the other two. Total current?
50mA. Your source has to have enough voltage to "overcome" the sum of the forward voltages of the LEDs plus some headroom for whatever is regulating or limiting the current.
Elements in parallel have the same voltage across them, and may have different currents flowing through them. The sum of the currents through each element is the total current drawn by the elements from the source. Three parallel LEDs with 50mA flowing in each would draw 150mA from the supply.
You have to control the current through LEDs by using a constant current drive or a resistor in series to set the appropriate operating point.
So the draw will depend on how you design your lighting system. Series, parallel, drive scheme and target current.
answered Jun 7 at 14:45
John DJohn D
12.9k12132
$endgroup$
$begingroup$
Thank you for this answer. Turns out, my problem was just a lack of basic knowledge about circuits wired in parallel. The LED strip I was using had the LEDs wired in parallel on the strip. Resistors in parallel lower total resistance of the circuit, therefore as I add LEDs to the circuit, the resistance lowers, the voltage stays the same, and the current increases from the power supply.
$endgroup$
– Jcb Rb
Jun 7 at 15:20
add a comment |
$begingroup$
You have a number of misunderstandings.
Ohm's law applies to resistors. LEDs are non-linear elements and therefore Ohm's law doesn't apply.
Elements in series have the same current flowing through them. If you have 3 LEDs in series and one has 50mA flowing in it, so do the other two. Total current?
50mA. Your source has to have enough voltage to "overcome" the sum of the forward voltages of the LEDs plus some headroom for whatever is regulating or limiting the current.
Elements in parallel have the same voltage across them, and may have different currents flowing through them. The sum of the currents through each element is the total current drawn by the elements from the source. Three parallel LEDs with 50mA flowing in each would draw 150mA from the supply.
You have to control the current through LEDs by using a constant current drive or a resistor in series to set the appropriate operating point.
So the draw will depend on how you design your lighting system. Series, parallel, drive scheme and target current.
answered Jun 7 at 14:45
John DJohn D
12.9k12132
$endgroup$
You have a number of misunderstandings.
Ohm's law applies to resistors. LEDs are non-linear elements and therefore Ohm's law doesn't apply.
Elements in series have the same current flowing through them. If you have 3 LEDs in series and one has 50mA flowing in it, so do the other two. Total current?
50mA. Your source has to have enough voltage to "overcome" the sum of the forward voltages of the LEDs plus some headroom for whatever is regulating or limiting the current.
Elements in parallel have the same voltage across them, and may have different currents flowing through them. The sum of the currents through each element is the total current drawn by the elements from the source. Three parallel LEDs with 50mA flowing in each would draw 150mA from the supply.
You have to control the current through LEDs by using a constant current drive or a resistor in series to set the appropriate operating point.
So the draw will depend on how you design your lighting system. Series, parallel, drive scheme and target current.
answered Jun 7 at 14:45
John DJohn D
12.9k12132
answered Jun 7 at 14:45
John DJohn D
12.9k12132
answered Jun 7 at 14:45
John DJohn D
12.9k12132
answered Jun 7 at 14:45
John DJohn D
12.9k12132
12.9k12132
$begingroup$
Thank you for this answer. Turns out, my problem was just a lack of basic knowledge about circuits wired in parallel. The LED strip I was using had the LEDs wired in parallel on the strip. Resistors in parallel lower total resistance of the circuit, therefore as I add LEDs to the circuit, the resistance lowers, the voltage stays the same, and the current increases from the power supply.
$endgroup$
– Jcb Rb
Jun 7 at 15:20
add a comment |
$begingroup$
Thank you for this answer. Turns out, my problem was just a lack of basic knowledge about circuits wired in parallel. The LED strip I was using had the LEDs wired in parallel on the strip. Resistors in parallel lower total resistance of the circuit, therefore as I add LEDs to the circuit, the resistance lowers, the voltage stays the same, and the current increases from the power supply.
$endgroup$
– Jcb Rb
Jun 7 at 15:20
$begingroup$
Thank you for this answer. Turns out, my problem was just a lack of basic knowledge about circuits wired in parallel. The LED strip I was using had the LEDs wired in parallel on the strip. Resistors in parallel lower total resistance of the circuit, therefore as I add LEDs to the circuit, the resistance lowers, the voltage stays the same, and the current increases from the power supply.
$endgroup$
– Jcb Rb
Jun 7 at 15:20
$begingroup$
Thank you for this answer. Turns out, my problem was just a lack of basic knowledge about circuits wired in parallel. The LED strip I was using had the LEDs wired in parallel on the strip. Resistors in parallel lower total resistance of the circuit, therefore as I add LEDs to the circuit, the resistance lowers, the voltage stays the same, and the current increases from the power supply.
$endgroup$
– Jcb Rb
Jun 7 at 15:20
add a comment |
$begingroup$
A string of LEDs is a bunch of LEDs in series and this means that they share the same current. If the forward volt drop across each LED is (say) 2 volts, then 5 LEDs requires a supply of at least 10 volts by the way.
answered Jun 7 at 14:46
Andy akaAndy aka
248k11190437
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add a comment |
$begingroup$
A string of LEDs is a bunch of LEDs in series and this means that they share the same current. If the forward volt drop across each LED is (say) 2 volts, then 5 LEDs requires a supply of at least 10 volts by the way.
answered Jun 7 at 14:46
Andy akaAndy aka
248k11190437
$endgroup$
add a comment |
$begingroup$
A string of LEDs is a bunch of LEDs in series and this means that they share the same current. If the forward volt drop across each LED is (say) 2 volts, then 5 LEDs requires a supply of at least 10 volts by the way.
answered Jun 7 at 14:46
Andy akaAndy aka
248k11190437
$endgroup$
A string of LEDs is a bunch of LEDs in series and this means that they share the same current. If the forward volt drop across each LED is (say) 2 volts, then 5 LEDs requires a supply of at least 10 volts by the way.
answered Jun 7 at 14:46
Andy akaAndy aka
248k11190437
answered Jun 7 at 14:46
Andy akaAndy aka
248k11190437
answered Jun 7 at 14:46
Andy akaAndy aka
248k11190437
answered Jun 7 at 14:46
Andy akaAndy aka
248k11190437
248k11190437
add a comment |
add a comment |
$begingroup$
It can often help to flip it upside down and think of conductance instead of resistance. Conductance has a unit of Siemens or S.
Conductance in Siemens is 1 / resistance in ohms.
C E = I Conductance * voltage = current
E = I / C
So for instance a 10 ohm resistor has a conductance of 0.1 Siemens. A 5 ohm resistor has a conductance of 0.2 Siemens.
What happens if you parallel them? Their conductance is - hold onto your hat - 0.3 Siemens! In parallel, conductances just add.
Boy, that's easier to work with.
answered Jun 7 at 23:06
HarperHarper
7,0821027
$endgroup$
add a comment |
$begingroup$
It can often help to flip it upside down and think of conductance instead of resistance. Conductance has a unit of Siemens or S.
Conductance in Siemens is 1 / resistance in ohms.
C E = I Conductance * voltage = current
E = I / C
So for instance a 10 ohm resistor has a conductance of 0.1 Siemens. A 5 ohm resistor has a conductance of 0.2 Siemens.
What happens if you parallel them? Their conductance is - hold onto your hat - 0.3 Siemens! In parallel, conductances just add.
Boy, that's easier to work with.
answered Jun 7 at 23:06
HarperHarper
7,0821027
$endgroup$
add a comment |
$begingroup$
It can often help to flip it upside down and think of conductance instead of resistance. Conductance has a unit of Siemens or S.
Conductance in Siemens is 1 / resistance in ohms.
C E = I Conductance * voltage = current
E = I / C
So for instance a 10 ohm resistor has a conductance of 0.1 Siemens. A 5 ohm resistor has a conductance of 0.2 Siemens.
What happens if you parallel them? Their conductance is - hold onto your hat - 0.3 Siemens! In parallel, conductances just add.
Boy, that's easier to work with.
answered Jun 7 at 23:06
HarperHarper
7,0821027
$endgroup$
It can often help to flip it upside down and think of conductance instead of resistance. Conductance has a unit of Siemens or S.
Conductance in Siemens is 1 / resistance in ohms.
C E = I Conductance * voltage = current
E = I / C
So for instance a 10 ohm resistor has a conductance of 0.1 Siemens. A 5 ohm resistor has a conductance of 0.2 Siemens.
What happens if you parallel them? Their conductance is - hold onto your hat - 0.3 Siemens! In parallel, conductances just add.
Boy, that's easier to work with.
answered Jun 7 at 23:06
HarperHarper
7,0821027
answered Jun 7 at 23:06
HarperHarper
7,0821027
answered Jun 7 at 23:06
HarperHarper
7,0821027
answered Jun 7 at 23:06
HarperHarper
7,0821027
7,0821027
add a comment |
add a comment |
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Are you adding LEDs in series or parallel? The current through a LED is non-linear and ohms law doesn't necessarily apply.
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– R.Joshi
Jun 7 at 14:43
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Tip: LED and other semiconductor devices are called "non-linear devices". Consider the term "linear", what does it mean? It means the current is proportional to the voltage, in other words, it obeys Ohm's Law. Hence, "non-linear electronics devices" basically means, "electronic devices that do not obey Ohm's Law" (although in many cases there are techniques allowing you to perform a linear approximation within a small operating range and pretend Ohm's Law works).
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– 比尔盖子
Jun 7 at 17:08