correct voltage used in V of I x R in a simple 2 LED circuit

Thread Starter

clangray

Joined Nov 4, 2018
261
My question is whether in this circuit and when using V = IR, is Vtotal = Vsupply - (LED1fv + LED2fv) or is V just the sum of LED1fv + LED2fv? I'm pretty sure its Vsupply - (LED1fv + LED2fv) but I am getting hung up on the difference between the two. Also is Power calculation using the voltage source (Vsupply - Vforward voltages = LED1fv + LED2fv).

Screenshot Power Question.png
 

LowQCab

Joined Nov 6, 2012
3,939
Add-up all the Forward-Voltages of the LEDs,
based on the Current-level that You would like to use.

Subtract the combined Forward-Voltages from the Power-Supply-Voltage.

5-Volts minus 3.15-Volts = 1.85-Volts.

Calculate the Resistor-value for the desired Current.

1.85 / 0.020 = 92.5-Ohms

Since the Current is the same everywhere in the Circuit,
and there is only one Power-Supply,
then the Total-Wattage comes to,
5-Volts X 0.020-Amps = 0.1 Watts. ( 100-mW )

This is assuming that You actually found the Forward-Voltages by
consulting the Voltage/Current-Graphs in the Spec-Sheets for the LEDs.

The Wattage-Dissipation for each individual Component can then be calculated by
measuring the Voltage across each component, and multiplying by 0.020 Amps.

The Voltages across the 2 LEDs are obvious, and likely to be very close to the Graphs.
LED-1 = 1.25-Volts X 0.020-Amps = 0.025-Watts. ( 25-mW )
LED-2 = 1.90-Volts X 0.020-Amps = 0.038-Watts. ( 38-mW )
The Voltage across the Resistor is what is left-over, and is generally the only variable, which is 1.85-Volts.
1.85-Volts X 0.020-Amps = 0.037-Watts. ( 37-mW )

When the Temperatures of the LEDs increases,
their Forward-Voltages will tend to decrease to a small degree.
.
.
.
 
Last edited:

MrChips

Joined Oct 2, 2009
30,494
The mistake you are making is voltage is not a value at a single point.
Voltage is the potential difference between two points, A and B.
Hence when referring to voltage at any point there must be reference point.

You can talk about the voltage across a component, such as the voltage across a resistor, an LED, or a battery.

In your example, the voltage across LED1 is 1.25V and across LED2 is 1.9V.

VR1 is the voltage across resistor R1.

Thus, using Kirchhoff's Voltage Law, we can write

V1 = VR1 + VLED1 + VLED2
 

WBahn

Joined Mar 31, 2012
29,867
My question is whether in this circuit and when using V = IR, is Vtotal = Vsupply - (LED1fv + LED2fv) or is V just the sum of LED1fv + LED2fv? I'm pretty sure its Vsupply - (LED1fv + LED2fv) but I am getting hung up on the difference between the two. Also is Power calculation using the voltage source (Vsupply - Vforward voltages = LED1fv + LED2fv).
When in doubt, rely on the fundamentals.

For Ohm's Law, you need the voltage across THAT resistor, the current through THAT resistor, and the resistance of THAT resistor.

Kirchhoff's Voltage Law says that the some of the voltage drops around the circuit is zero, so start there:

Vs - Vr - Vf1 - Vf2 = 0

So the voltage across the resistor is

Vr = Vs - Vf1 - Vf2 = Vs - (Vf1 + Vf2)

Then plug this into Ohm's Law:

Vr = Ir · R

Vs - (Vf1 + Vf2) = Ir · R

If you want to know what the current is for a given resistance, solve for Ir:

Ir = [Vs - (Vf1 + Vf2)] / R

If you want to know the resistance needed for a desired current, solve for R:

R = [Vs - (Vf1 + Vf2)] / Ir

Similarly, your equation for power is referring to the power dissipated by a resistor and it needs the voltage across THAT resistor.

This will ONLY give you the power dissipated by the resistor. It will not give you the power being delivered by the source as it neglects the power being consumed by the LEDs.
 

dl324

Joined Mar 30, 2015
16,691
My question is whether in this circuit and when using V = IR, is Vtotal = Vsupply - (LED1fv + LED2fv) or is V just the sum of LED1fv + LED2fv?
\( \large I=\frac{V_{R1}}{R1} =\frac{V_{sup}-V_{LED1}-V_{LED2}}{R1}=\frac{5V-1.25V-1.9V}{90.9 \Omega} = 20mA\)
\( \large P_{R1}=I^2R1=20mA^2*90.9 \Omega=36mW\)
\( \large P_{LED1}=I*V_{LED1}=20mA*1.25V=25mW\)
\( \large P_{LED2}=I*V_{LED2}=20mA*1.9V=38mW\)
 

crutschow

Joined Mar 14, 2008
34,048
Below is an LTs simulation of the circuit showing the voltages at each node and the currents through the components for the specific LED SPICE models I have, which are reasonably close to the values you show:.
All voltages are referenced to the ground (triangle) symbol, which is by definition 0V.

The power dissipation is shown on the graph:
R1-Green trace, D1-Yellow trace, D2-Red trace.

1675566430852.png1675566465888.png
1675567752968.png
 
Last edited:

MisterBill2

Joined Jan 23, 2018
17,827
WOW!!! Certainly a lot of theory lecture for a simple question. I understand that the TS is seeking to know which voltage to use to calculate the value of the series resistor used to set the current in the LED circuit. So in this case, given that the current is known because it has already been selected as 20 mA, and given that the voltage drop required is the difference between the source voltage (5v) and the sum of the two LED voltage drops, (1.25V + 1.9V ), so the V=IR across the resistor will be 5V-(1.25V + 1.9V )/0.02A, and the power dissipated in the LEDs will be Vf x I for each, or 1.25V x 0.02A, and 1.9V x 0.02A.
 

Tonyr1084

Joined Sep 24, 2015
7,782
Add the two Vf's together.
Subtract the Vf total from the supply voltage.
That gives you a voltage drop across the resistor.
Using that voltage drop number, divide it by the amount of current you want to pass through the LED's.
That gives you the proper resistance.

Your example:
(5V - 1.25Vf - 1.9Vf) ÷ 20mA (0.02A) = 92.5Ω
This calculation assumes you want to operate your LED's at a current of 20mA. If you wanted to operate at 10mA the equation would look exactly the same but substitute 0.01A (10mA)
(5V - 1.25Vf - 1.9Vf) ÷ 0.01A = 185Ω

You can substitute any amperage the LED is either rated for or less. If an LED is rated for 300mA - obviously you substitute the numbers and come up with proper resistance. Assuming you're using an adequate voltage.

Assume you have a 12VDC supply and a 300mA LED. Off hand I don't know what the Vf might look like so I'll make something up:
(12V - 5.6Vf) ÷ 0.3A = 21.3Ω

It's all in the numbers.

[edit] Notice the supply is higher than the Vf. Using a 5V supply on a 5.6Vf LED likely will not illuminate. [end edit]

[edit #2] Don't forget to calculate the wattage for the resistor. On a 300mA at 12VDC circuit using a resistor rated for 250mW (1/4 Watt) would simply burn out. Wattage is calculated by multiplying the total voltage times the current. So 12V x 0.3A = 3.6W. In such a case you would need a 5W resistor. Or you could get away with four 1 watt resistors at a modified resistance. This example you'd need four parallel resistors with a resistance each of 85.2Ω.

Calculations for this scenario are not difficult. The formula is 1 ÷ (1/R1 + 1/R2 + 1/R3 + 1/R4)
1 ÷ (1/85.2 + 1/85.2 + 1/85.2 + 1/85.2) = 21.3Ω @ 4W
1 ÷ (0.01173 + 0.01173 + 0.01173 + 0.01173) = 21.3Ω @ 4W
[end edit #2]
{whew!}
 
Last edited:

Thread Starter

clangray

Joined Nov 4, 2018
261
the voltage drop required is the difference between the source voltage (5v) and the sum of the two LED voltage drops, (1.25V + 1.9V ), so the V=IR across the resistor will be 5V-(1.25V + 1.9V )/0.02A, and the power dissipated in the LEDs will be Vf x I for each, or 1.25V x 0.02A, and 1.9V x 0.02A.
So I thought dissipation from the LEDs was negligible (for whatever reason) but its not. Must add up the wattage of all three: resistor, LED1 and LED2. Now Power = V^2/R. Is V in this power equation V =(Vsupply) - (LED1Vf+ LED2Vf) or is (V = Vsupply)?
 
Last edited:

BobTPH

Joined Jun 5, 2013
8,665
The total power used in any circuit powered by a single battery is simply the battery voltage times the battery current.

The formula with the resistor is the power dissipated in that resistor only.
 

WBahn

Joined Mar 31, 2012
29,867
So I thought dissipation from the LEDs was negligible (for whatever reason) but its not. Must add up the wattage of all three: resistor, LED1 and LED2. Now Power = V^2/R. Is V in this power equation V =(Vsupply) - (LED1Vf+ LED2Vf) or is (V = Vsupply)?
As with Ohm's Law, power for a given component is the product of the voltage across THAT device and the current through THAT device.

If you want the power dissipated by the resistor, the voltage across the resistor is (Vsupply) - (LED1Vf+ LED2Vf).

If you want the power provided by the source, the voltage across the source is Vsupply.

If you want the power dissipated by LED1, the voltage is VfLED1.

There is also the question of whether the device is providing power (converting energy from some other form to electrical energy, such as chemical), or absorbing power (converting energy from electrical energy to some other form, such as heat). There are a couple of ways to let the math track this for you. These are known as the passive sign conventions.

If you define the voltage across a device so that the current enters whichever terminal your voltage definition declares is the positive terminal, then if the product of the two is positive, the device is absorbing electrical power, while if it is negative, the device is providing electrical power. However, this convention results in sources, which normally produce power, always having negative power. There's nothing wrong with this, but humans tend to make more mistakes if there's a lot of negative quantities running around. So a common variant on the passive sign convention is to decide up front which devices you expect to be loads and which devices you expect to be sources and define the current for sources such that it is coming out of the positive terminal of that device. Now, if you guessed correctly, all of your powers will be positive. If one of them turns out to be negative, you just know that you guessed wrong. This might be the case when a battery is acting like a load (i.e., it being charged) or a capacitor is acting like a source (i.e., is discharging into the circuit).
 

Thread Starter

clangray

Joined Nov 4, 2018
261
Add the two Vf's together.
Subtract the Vf total from the supply voltage.
That gives you a voltage drop across the resistor.
Using that voltage drop number, divide it by the amount of current you want to pass through the LED's.
That gives you the proper resistance.

Your example:
(5V - 1.25Vf - 1.9Vf) ÷ 20mA (0.02A) = 92.5Ω
This calculation assumes you want to operate your LED's at a current of 20mA. If you wanted to operate at 10mA the equation would look exactly the same but substitute 0.01A (10mA)
(5V - 1.25Vf - 1.9Vf) ÷ 0.01A = 185Ω

You can substitute any amperage the LED is either rated for or less. If an LED is rated for 300mA - obviously you substitute the numbers and come up with proper resistance. Assuming you're using an adequate voltage.

Assume you have a 12VDC supply and a 300mA LED. Off hand I don't know what the Vf might look like so I'll make something up:
(12V - 5.6Vf) ÷ 0.3A = 21.3Ω

It's all in the numbers.

[edit] Notice the supply is higher than the Vf. Using a 5V supply on a 5.6Vf LED likely will not illuminate. [end edit]

[edit #2] Don't forget to calculate the wattage for the resistor. On a 300mA at 12VDC circuit using a resistor rated for 250mW (1/4 Watt) would simply burn out. Wattage is calculated by multiplying the total voltage times the current. So 12V x 0.3A = 3.6W. In such a case you would need a 5W resistor. Or you could get away with four 1 watt resistors at a modified resistance. This example you'd need four parallel resistors with a resistance each of 85.2Ω.

Calculations for this scenario are not difficult. The formula is 1 ÷ (1/R1 + 1/R2 + 1/R3 + 1/R4)
1 ÷ (1/85.2 + 1/85.2 + 1/85.2 + 1/85.2) = 21.3Ω @ 4W
1 ÷ (0.01173 + 0.01173 + 0.01173 + 0.01173) = 21.3Ω @ 4W
[end edit #2]
{whew!}
If I have a circuit whose components have a total wattage of 0.1W or 100mw what would be more appropriate: 1/8W, 1/4W, 1/2W resistor?

What is resistor wattage level that requires a bump to the next threshold level? For example of .1W: .125 seems a rational choice. But if we steadily increase the wattage at what wattage value would we need .250W resistor?
 
Last edited:

WBahn

Joined Mar 31, 2012
29,867
If I have a circuit whose components have a total wattage of 0.1W or 100mw what would be more appropriate: 1/8w, 1/4w, 1/2w resistor?

What is resistor wattage level that requires a bump to the next threshold level? For example of .1W: .125 seems a rational choice. But if we steadily increase the wattage at what wattage value would we need .250W resistor?
A common rule of thumb is to size components so that they can handle at least twice the amount of power that is anticipated, so if you expect 100 mW (and note that watts is abbreviated with a capital W), you want to be able to dissipate at least 200 mW, so you would opt for a 1/4 W resistor. If there are reasons why you don't want to do this, such as price or availability, you might consider backing that off to as low as a 50% margin (so 150 mW in this case, which would still leave you choosing a 1/4 W resistor), but should really try to avoid that. To go any lower you want to really give it some thought. The power rating of a component is often (not always) based on the maximum temperature they can be allowed to operate at before damage starts to become likely. A common threshold is 125 °C, which is more than 250 °F. That is extremely hot, meaning that it can cause severe burns with just a moment's touch. No reason to have those kinds of temperatures in circuits that you are playing around with.
 

Thread Starter

clangray

Joined Nov 4, 2018
261
As with Ohm's Law, power for a given component is the product of the voltage across THAT device and the current through THAT device.

If you want the power dissipated by the resistor, the voltage across the resistor is (Vsupply) - (LED1Vf+ LED2Vf).

If you want the power provided by the source, the voltage across the source is Vsupply.

If you want the power dissipated by LED1, the voltage is VfLED1.

There is also the question of whether the device is providing power (converting energy from some other form to electrical energy, such as chemical), or absorbing power (converting energy from electrical energy to some other form, such as heat). There are a couple of ways to let the math track this for you. These are known as the passive sign conventions.

If you define the voltage across a device so that the current enters whichever terminal your voltage definition declares is the positive terminal, then if the product of the two is positive, the device is absorbing electrical power, while if it is negative, the device is providing electrical power. However, this convention results in sources, which normally produce power, always having negative power. There's nothing wrong with this, but humans tend to make more mistakes if there's a lot of negative quantities running around. So a common variant on the passive sign convention is to decide up front which devices you expect to be loads and which devices you expect to be sources and define the current for sources such that it is coming out of the positive terminal of that device. Now, if you guessed correctly, all of your powers will be positive. If one of them turns out to be negative, you just know that you guessed wrong. This might be the case when a battery is acting like a load (i.e., it being charged) or a capacitor is acting like a source (i.e., is discharging into the circuit).
If you want the power provided by the source, the voltage across the source is Vsupply.

When do I want power provided by the source as opposed to finding the power consumption for each component and adding them all up?
 

WBahn

Joined Mar 31, 2012
29,867
If you want the power provided by the source, the voltage across the source is Vsupply.

When do I want power provided by the source as opposed to finding the power consumption for each component and adding them all up?
If you want to know the total power being delivered by the source.

Let's say that you have a circuit with a couple dozen components in it and you have a supply rated at 5 W.

Would it not be much easier to find the total current delivered by the source and just multiply that current by the supply voltage to see if you are close to (or exceeding) the supply's ratings? Or would you really rather have to calculate the individual power for each component separately and add them all up?
 

Thread Starter

clangray

Joined Nov 4, 2018
261
Vt x I = W
So in the 5V example in the schematic at the beginning of the post, I find Vt = 5V - (1.9V+1.25V) = 1.8V. How does the resistor fit into Vt? I am wondering why the 2 forward voltages are being drawn accross the resistor? The voltage of the resistor is (LED1Vforward + LED2Vforward). But thats the same as the sum of the 2 LED forward voltages. That's what confuses me.
 
Last edited:

Thread Starter

clangray

Joined Nov 4, 2018
261
If you want to know the total power being delivered by the source.

Let's say that you have a circuit with a couple dozen components in it and you have a supply rated at 5 W.

Would it not be much easier to find the total current delivered by the source and just multiply that current by the supply voltage to see if you are close to (or exceeding) the supply's ratings? Or would you really rather have to calculate the individual power for each component separately and add them all up?
OK, this is the "art".

I capitalized w.

Thank you.
 
Top