Efficient LED Circuit Design

This project follows the Engineering Design Process. Confirm with your teacher if this is acceptable for your project, and review the steps before you begin.

Planning Your Circuit

1. Think about how you want to use LED lighting. Do you want to have ambient LED lighting in your room? Do you want to build a portable device that you can bring to a party? A flashlight? Lighting for the inside of a car?
2. Decide how many and what color LEDs you want for your circuit. This procedure will walk you through an example circuit design. You can then apply the same process to your own circuit.
3. This is an engineering design project, so think about the design constraints and criteria for your circuit.
   1. At minimum, you need to make sure you do not burn out any of the resistors or LEDs in your circuit by passing too much current through them.
   2. Consider the total power consumption and efficiency of your circuit. How much power is drawn from the power supply? How much of that power is "wasted" in the resistors instead of being delivered from the LEDs?
   3. You may need to work within a certain budget, or you may only have certain options available for a power supply. Depending on where and how you want to use the device, will a wall outlet or car charger/adapter be available? Will it need to be battery powered?
4. Watch the following three videos to make sure you are familiar with the basic calculations needed for this project.

https://www.youtube.com/watch?v=JkFkUTxMkCk

https://www.youtube.com/watch?v=OgRzJmu03VY

https://www.youtube.com/watch?v=xx5ou6n0GEU

Parallel Circuit

1. Let us start with an example circuit that has one red, one green, and one blue LED. The parameters for each LED are shown in Table 1 (note that these are not the same values as the example shown in the video).
   
   

   LED Color	Forward Voltage	Current
   Red	1.8 V	30 mA
   Green	2.2 V	20 mA
   Blue	3.8 V	20 mA

   Table 1. LED parameters.
2. For now, assume we have a 9 V battery available and decide to use that for our power supply. One simple circuit design is just to give each LED its own current-limiting resistor, and then put the three branches of the circuit in parallel as shown in Figure 5.
   
   Image Credit: Ben Finio, Science Buddies / Science Buddies
   Figure 5. Circuit diagram for the red, green, and blue LEDs, each with its own current-limiting resistor, connected in parallel to the 9 V battery.

https://www.youtube.com/watch?v=9ZQz2IUEtO8

1. For this configuration, since the three branches of the circuit are in parallel, the voltage drop over each branch is equal to the battery voltage. We can apply Equation 4 from the introduction to calculate the current-limiting resistor value for each LED individually:

   Equation 7 (red LED):   $R_{red} = \frac{9V \;-\; 1.8V}{0.030 A} = 240\Omega$

   Equation 8 (green LED):   $R_{green} = \frac{9V \;-\; 2.2V}{0.020 A} = 340\Omega$

   Equation 9 (blue LED):   $R_{blue} = \frac{9V \;-\; 3.8V}{0.020 A} = 260\Omega$

2. Our resistor kit does not have these exact values available, so we choose the next largest resistance value for each LED. That gives 270 Ω resistors for the red and blue LEDs and a 470 Ω resistor for the green LED.
3. Changing the resistor value will change the current through the circuit. So, before we calculate the power dissipated by each resistor, we need to calculate the actual current through each branch of the circuit. We do that by rearranging Equation 4 to solve for current:

   Equation 10:   $I = \frac{V_{batt} \;-\; V_{LED}}{R}$

   Using this equation, we can find the current through each LED and its corresponding resistor:

   Equation 11 (red LED):   $I_{red} = \frac{9V \;-\; 1.8V}{270\Omega} = 0.0266 A = 26.6 mA$

   Equation 12 (green LED):   $I_{green} = \frac{9V \;-\; 2.2V}{470\Omega} = 0.0144 A = 14.4 mA$

   Equation 13 (blue LED):   $I_{blue} = \frac{9V \;-\; 3.8V}{270\Omega} = 0.0192 A = 19.2 mA$

4. Now, using Equation 6, we calculate the power dissipated by each resistor and make sure it is less than 250 mW. (Our resistors are rated for 1/4 W.)
   

   Equation 14 (red LED's resistor):   $P = (0.0266A)^2 \times 270\Omega = 0.1910W = 191.0mW \lt 250mW$

   Equation 15 (green LED's resistor):   $P = (0.0144A)^2 \times 470\Omega = 0.0974W = 97.4mW \lt 250mW$

   Equation 16 (blue LED's resistor):   $P = (0.0192A)^2 \times 270\Omega = 0.0995W = 99.5mW \lt 250mW$

5. So far, it looks like we are OK! Each LED is below its rated current and each resistor is below its rated power level. So this circuit meets the requirement for not burning out any of the LEDs or resistors. However, what about the power consumption and efficiency of the circuit?

   First, we can calculate the total power dissipated in all three resistors by adding them together:

   Equation 17:   $P_{resistors} = 191.0mW + 97.4mW + 99.5mW = 387.9mW$

   We can also calculate the total power delivered to the circuit by the battery. We do this by multiplying the battery's voltage by the total current drawn from the battery. Since the three circuit branches are in parallel, their currents add. The total current drawn from the battery is:

   Equation 18:   $I_{total} = I_{red} + I_{green} + I_{blue} = 26.6mA + 14.4mA + 19.2mA = 60.2mA$

   The total power drawn from the battery is:

   Equation 19:   $P_{batt} = I_{total} V_{batt} = 0.0602A \times 9V = 0.5148W = 514.8mW$

   The amount of power delivered to the LEDs is equal to the total power from the battery minus the power dissipated by the resistors:

   Equation 20:   $P_{LEDs} = P_{batt} - P_{resistors} = 514.8mW - 387.9mW = 126.9mW$

   However, the LEDs themselves are not 100% efficient. They do not convert 100% of the input electrical power to visible light—some of it is also lost as heat. LED datasheets will not always include an efficiency value, but you can use 50% as a rough estimate. That means that the total efficiency (η) of the circuit—the percentage of electrical power from the battery that is converted to visible light—is:

   Equation 21:   $\eta = \frac{0.5 \;\times\; P_{LEDs}}{P_{batt}} = \frac{0.5 \;\times\; 126.9mW}{514.8mW} = 0.123 = 12.3\%$

   So our circuit draws just over half a watt of power from the battery and is only 12.3% efficient. Just over three-quarters of the power drawn from the battery is wasted as heat dissipated in the resistors:

   Equation 22:   $\%P_R = \frac{P_{resistors}}{P_{batt}} = \frac{387.9mW}{514.8mW} = 0.753 = 75.3\%$

   Can we do anything to decrease overall power consumption or improve the efficiency of our circuit? Let us find out!

Series Circuit

1. You may have noticed that a lot of voltage from the 9 V battery is "dropped" over the resistors in the circuit in Figure 5. The voltage drop over the individual LEDs is only about 2–4 V, depending on the color. That means that 5–7 V is dropped over the resistors, resulting in a lot of wasted power.

   There are several ways we could improve this. One would be to choose a lower-voltage power supply (that is still high enough in voltage to power our LEDs). For example, you could use a 4xAA battery pack to put four 1.5 V AA batteries in series, giving a total voltage of 6 V. Then you could redo the calculations, starting back at step 5, to solve for new resistor values. However, let us assume for now that we only have a 9 V battery available to work with. Another option is to put the LEDs in series, as shown in Figure 6. In this configuration the circuit only requires one resistor. Since the LEDs are in series, their voltage drops add up, and less voltage is dropped over the resistor.

   Image Credit: Ben Finio, Science Buddies / Science Buddies
   Figure 6. Circuit diagram for the red, green, and blue LEDs connected in series with a single current-limiting resistor

https://www.youtube.com/watch?v=qZSP5-jYW34

1. We can still calculate a resistor value using Equation 4, but this time we need to subtract the voltage drops over all three LEDs. We also need to choose a single value for the current, since the LEDs are in series and will all have the same current. To avoid burning out the blue and green LEDs, we will choose 20 mA, since that is the lowest value. That gives a resistor value of:

   Equation 23:   $R = \frac{V_{batt} \;-\; V_{red} \;-\; V_{green} \;-\; V_{blue}}{I} = \frac{9V \;-\; 1.8V \;-\; 2.2V \;-\; 3.8V}{0.020A} = 60\Omega$

2. Again, our resistor kit does not have a resistor with a value of exactly 60 Ω, so we go up to the next biggest value, which is 68 Ω. We then calculate the actual current through the resistor:

   Equation 24:   $I = \frac{V_{batt} \;-\; V_{red} \;-\; V_{green} \;-\; V_{blue}}{R} = \frac{9V \;-\; 1.8V \;-\; 2.2V \;-\; 3.8V}{68\Omega} = 0.0176A = 17.6mA$

3. Now we can calculate the power dissipated by the resistor, the power delivered by the battery, the total power delivered to the LEDs, the efficiency of the circuit, and the percentage of power dissipated in the resistors, just like we did before:

   Equation 25:   $P_{resistor} = I^2 R = (0.0176A)^2 \times 68\Omega = 0.0210W = 21.0mW \lt 250mW$

   Equation 26:   $P_{batt} = IV = 0.0176A \times 9V = 0.1584W = 158.4mW$

   Equation 27:   $P_{LEDs} = P_{batt} - P_{resistor} = 158.4mW - 21.0mW = 137.4mW$

   Equation 28:   $\eta = \frac{0.5 \;\times\; P_{LEDs}}{P_{batt}} = \frac{0.5 \;\times\; 137.4mW}{158.4mW} = 0.433 = 43.3\%$

   Equation 29:   $\%P_R = \frac{P_{resistors}}{P_{batt}} = \frac{21.0mW}{158.4mW} = 0.132 = 13.2\%$

   You can see that we have dramatically reduced the power consumption of our circuit and improved its efficiency! Just by rearranging the parts, we have reduced the total power consumption of the circuit from 514.8 mW to 158.4 mW, a savings of nearly 70%. The overall efficiency of the circuit improved from only 12.3% to 43.3%, increasing by a factor of 3.5!

Combination Series-Parallel Circuit Design

1. You might now think, "OK, great, so to design an LED circuit, you just put all the LEDs in series with a single resistor, right?" But it is not that simple. Here are a few things to consider:
   1. What if you want a circuit with dozens or even hundreds of LEDs? Putting them all in series would require a very high-voltage supply, out of range of most common battery packs and wall adapters. So it is not practical to just put all of the LEDs in series.
   2. What about properly tuning the brightness of the LEDs? We did not talk about this when we designed our series circuit and our parallel circuit. The brightness of an LED depends on how much current flows through it. In our series circuit, the green LED only had about 14 mA through it, since we had to choose a larger resistor. In the parallel circuit, the red LED only had about 20 mA, even though it was rated for 30 mA. In either case, the LED with less current might appear dimmer than the other LEDs.
2. How, then, do you design a circuit with more than a few LEDs? You can build a combination series-parallel circuit, as shown in Figure 7. This circuit has multiple branches in parallel. Each individual branch consists of one or more LEDs and a resistor. You can analyze each branch individually and choose the resistor size based on the corresponding LED voltage drops using Equation 24.
   Image Credit: Ben Finio, Science Buddies / Science Buddies
   Figure 7. A combination series-parallel circuit with nine LEDs and three resistors.
3. Now that you have seen some examples, it is time for the open-ended engineering design part of this project. You will need to design a circuit based on the number and color of LEDs you want. For example, say that you want to build the circuit shown in Figure 7, with three each of red, green, and blue LEDs. Should each branch contain LEDs of all one color, or should each branch have one of each color LED? Should you choose a lower-voltage power supply and only put one or two LEDs in each branch instead of three? Which configuration will maximize efficiency and minimize overall power consumption from the battery? We will not do the math for you this time—it is up to you to figure out the best design for your circuit!
4. Another tip: So far, we have only chosen single resistors. This can limit your choices, since common resistors are only available with certain values. You can combine resistors in series and/or parallel to create an equivalent resistor value according to Equations 30 (for series) and 31 (for parallel). For example, you can combine two 100 Ω resistors in series for an equivalent resistance of 200 Ω, and you can combine two 100 Ω resistors in parallel for an equivalent resistance of 50 Ω. Combining resistors to create different resistance values can allow you to more precisely tune the amount of current through your LEDs. However, adding more parts also increases the cost and complexity of a circuit, so there is a design trade-off. Remember that there is no right or wrong answer, since this is an engineering design project.

   Equation 30:   $R_{series} = R_1 + R_2$

   Equation 31:   $\frac{1}{R_{parallel}} = \frac{1}{R_1} + \frac{1}{R_2}$

5. Something else we have not addressed yet: In all of our circuits so far, the resistors go before the LEDs. What happens if you put the resistor after the LED? Does it matter? Does it affect the circuit design or analysis at all? You can do the math yourself to find out, but we also cover it in this video:
   

   https://www.youtube.com/watch?v=NUKD9qESO58
6. There are many other things you can consider when designing your circuit. See the Variations section for more ideas.