Consider, the attached schematic. The supply voltage is 6 volts and the Vf is 2.4 volts. That means the LED is dropping 2.4 volts. Now most common LEDs like to run at 20 mA or less. Lets say we want this one to run at 15 mA. This means that we will have to drop (6 - 2.4)volts across the current limiting resistor.

Can you elaborate a little on the voltage drop across an LED? According to what you explained it's value is constant as long as the voltage supply is greater than Vf. So whether 15mA is flowing in the circuit (using a 240 Ohm resistor) or 20mA flowing in the circuit (using a 180 Ohm resistor) in the circuit, the voltage drop across the LED (and consequently the resistor) will be the same in both cases. If in place of the LED you connected an identical (240 or 180 Ohm resistor, as the case may be) the voltage drop across the 1st resistor would be halved from 6V to 3V regardless of the amount of current. How do you explain that to a 5th grader? If you used a non-identical resistor, the voltage drop across the 1st resistor would be relative to its size to the 2nd resistor. This is not the case for LEDs.

I guess what it comes down to is how do I explain the effect an LED has on a simple circuit with just a power supply and a resistor RATHER THAN the effect a resistor has on simple circuit with just a power supply and an LED. The latter I can explain, the former is a bit trickier.

As always, thanks for your ongoing help.

 

Can you elaborate a little on the voltage drop across an LED? According to what you explained it's value is constant as long as the voltage supply is greater than Vf. So whether 15mA is flowing in the circuit (using a 240 Ohm resistor) or 20mA flowing in the circuit (using a 180 Ohm resistor) in the circuit, the voltage drop across the LED (and consequently the resistor) will be the same in both cases.

Yes, that's exactly right! It's the nature of the beast. All diodes have a forward voltage drop. LED's are unique due to the fact that they dissipate energy in the form of light and thus have a typically higher forward voltage drop. This property is dependent on the color of the LED and to a lesser degree to the current flowing through the LED. In circuit design we typically use the average Vf published in the datasheets. See the attached chart.

If in place of the LED you connected an identical (240 or 180 Ohm resistor, as the case may be) the voltage drop across the 1st resistor would be halved from 6V to 3V regardless of the amount of current. How do you explain that to a 5th grader? If you used a non-identical resistor, the voltage drop across the 1st resistor would be relative to its size to the 2nd resistor. This is not the case for LEDs.

That's right again. The LED is a unique individual and it resembles a resistor electrically only in the fact that it has a voltage drop across it. The best way to explain this to a 5th grader is to build a breadboard and let him see for himself how the voltage drops change with resistance and what happens when you replace the LED with a resistor. And for goodness sake, teach him to manipulate Ohm's law mathematically. It's not rocket science. These little fellows are like sponges - show, tell and demonstrate. Trust me they will pick it up. Good luck!

 

Yes, that's exactly right! It's the nature of the beast. All diodes have a forward voltage drop.

What does the "forward" in "forward voltage drop imply" and what about an LED causes it to have this property?

LED's are unique due to the fact that they dissipate energy in the form of light and thus have a typically higher forward voltage drop.

Higher forward voltage drop with respect to what? How is is that power dissipation in the form of light contributes to that?

The LED is a unique individual and it resembles a resistor electrically only in the fact that it has a voltage drop across it. The best way to explain this to a 5th grader is to build a breadboard and let him see for himself how the voltage drops change with resistance and what happens when you replace the LED with a resistor.

Agree that it can be demonstrated by experimenting however I would like to be able to explain why an LED is different than a resistor if both take on a voltage drop (the former being a constant value).

And for goodness sake, teach him to manipulate Ohm's law mathematically. It's not rocket science. These little fellows are like sponges - show, tell and demonstrate. Trust me they will pick it up. Good luck!

We started that last night. Solving various examples for V, R, and I. However my guy is like the new sponge you just took out of the wrapper and doesn't want to absorb too much water yet. His younger brother is like a roll of Bounty, I can't get him to stop poking his nose around.

 

What does the "forward" in "forward voltage drop imply" and what about an LED causes it to have this property?


Higher forward voltage drop with respect to what? How is is that power dissipation in the form of light contributes to that?


Agree that it can be demonstrated by experimenting however I would like to be able to explain why an LED is different than a resistor if both take on a voltage drop (the former being a constant value).

We started that last night. Solving various examples for V, R, and I. However my guy is like the new sponge you just took out of the wrapper and doesn't want to absorb too much water yet. His younger brother is like a roll of Bounty, I can't get him to stop poking his nose around.

Bart: If you must know.........
Jim: I must, I must!
from Blazing Saddles

The basic difference between an LED (or any diode for that matter) and a resistor is that resistors are conductors and diodes are semiconductors. The nature of how semiconductors work lies in atomic physics and frankly is above your son's pay grade. However, if you must know, here are a couple of sites that do a pretty good job of explaining it in layman's terms.
http://www.explainthatstuff.com/diodes.html
https://www.youtube.com/watch?v=JBtEckh3L9Q

When a diode (any diode) is forward biased, current flows, causing a junction voltage drop. This voltage drop times the magnitude of current flow results in the amount of power dissipated in the junction. In a normal diode, this power is dissipated in the form of heat energy. LEDs dissipate this energy in the form of light. Instead of silicon, LEDs are made from elements such as phosphorus, arsenic, or gallium. When the LED is forward biased, electrons combine with holes and energy is given off in the form of light.

Remember, any time you have current flowing and a voltage drop, power will be dissipated either in the form of heat or light. The relationship is:
P = I x V

My suggestion is don't get too involved in the why and focus on what happens. I can see you son's eyes glazing over now.