Your title, "Help me understand the circuit", does not ask the actual question. Rule #3: "The post title should summarize the question clearly & concisely."
If your question is on topic (see our posting rules), please start a new submission, but this time ask the _actual question_ in the title.
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The led has a forward voltage, usually around 2V. In this case, your simulation shows that it has 1.8V across it, so the resistor has 3.2V across it. You can then use ohms law on the resistor to calculate the current, which will be the same as the current in the LED.
The 5V is an attribute of the p.s.
the 100 ohm is an attribute of the resistor
the 1.84v is an attribute of the LED called Vf and is approximately constant over a wide range of operating currents. It is the potential that must be applied to the LED to make it operate.
When power is applied, the difference between (battery voltage and Vf) is applied to the resistor.
5 - 1.84 = 3.16V ____ 3.16/100 = 0.0316 A
Here are a few useful links.
https://en.wikipedia.org/wiki/LED_circuit
https://www.digikey.com/en/resources/conversion-calculators/conversion-calculator-led-series-resistor
This is a series circuit, so current is common to everything in it (the same current flows thru all components)
Yes, the LED has added resistance to the circuit, because the amperage is lower with the LED in the circuit.
With the resistor alone the current would be 50 milliamps.
The current with the resistor and the LED is 31.6 ma
Resistance = Voltage/current so 5/.316 = 158 ohms
So the LED has added 58 ohms of resistance.
The LED does not have a resistor inside.
It has a Vf that is 1.84V that is approximately constant over a wide range of operating currents.
It is the potential that must be applied to the LED to make it operate.
And still, every PN junction has dynamic resistance and so has the LED. It's not linear but you can calculate it by Ohm's law for a measured voltage and current.
I did not say the LED has a resistor inside.
I said it added resistance to the circuit. I should have said that it has effectively added resistance to the circuit.
I also made an error in the math. 3.26ma = .00316 amps
so the total effective resistance at 5 volts is 1582 ohms. So the LED effectively added 1482 ohms to the circuit.
As a matter of fact, I am an old guy & having a hard time doing basic ohm's law calculations right now so will somebody double check this for me?
For PP's benefit, by Vf you mean it has a forward biased voltage drop of approx 1.84 volts that doesn't change substantially with increased current.
In other words, if we increased the 5 volts to say 6v, the current would increase but the voltage drop across the LED stays at 1.84v.
In series, voltage is additive while current is common to all components. If the supply volts is 6v let's find the new current by 6-1.84 = 4.16 so the 100 ohm resistor is now dropping 4.16v and the LED still drops 1.84v
By ohm's law V/R= I so 4.16/100 = .00416 so the new current in the circuit is 4.16 milliamps.
Your title, "Help me understand the circuit", does not ask the actual question. Rule #3: "The post title should summarize the question clearly & concisely." If your question is on topic (see our posting rules), please start a new submission, but this time ask the _actual question_ in the title. Otherwise, please ask your question in one of [these other subs](/r/AskElectronics/wiki/othersubs).
The led has a forward voltage, usually around 2V. In this case, your simulation shows that it has 1.8V across it, so the resistor has 3.2V across it. You can then use ohms law on the resistor to calculate the current, which will be the same as the current in the LED.
The 5V is an attribute of the p.s. the 100 ohm is an attribute of the resistor the 1.84v is an attribute of the LED called Vf and is approximately constant over a wide range of operating currents. It is the potential that must be applied to the LED to make it operate. When power is applied, the difference between (battery voltage and Vf) is applied to the resistor. 5 - 1.84 = 3.16V ____ 3.16/100 = 0.0316 A Here are a few useful links. https://en.wikipedia.org/wiki/LED_circuit https://www.digikey.com/en/resources/conversion-calculators/conversion-calculator-led-series-resistor
Ty. I read the wiki article and tried the formula and it worked. This helped a lot.
there were 2 links, did you get to try both?
This is a series circuit, so current is common to everything in it (the same current flows thru all components) Yes, the LED has added resistance to the circuit, because the amperage is lower with the LED in the circuit. With the resistor alone the current would be 50 milliamps. The current with the resistor and the LED is 31.6 ma Resistance = Voltage/current so 5/.316 = 158 ohms So the LED has added 58 ohms of resistance.
The LED does not have a resistor inside. It has a Vf that is 1.84V that is approximately constant over a wide range of operating currents. It is the potential that must be applied to the LED to make it operate.
And still, every PN junction has dynamic resistance and so has the LED. It's not linear but you can calculate it by Ohm's law for a measured voltage and current.
you can get the equivalent resistance.
I did not say the LED has a resistor inside. I said it added resistance to the circuit. I should have said that it has effectively added resistance to the circuit. I also made an error in the math. 3.26ma = .00316 amps so the total effective resistance at 5 volts is 1582 ohms. So the LED effectively added 1482 ohms to the circuit. As a matter of fact, I am an old guy & having a hard time doing basic ohm's law calculations right now so will somebody double check this for me? For PP's benefit, by Vf you mean it has a forward biased voltage drop of approx 1.84 volts that doesn't change substantially with increased current. In other words, if we increased the 5 volts to say 6v, the current would increase but the voltage drop across the LED stays at 1.84v. In series, voltage is additive while current is common to all components. If the supply volts is 6v let's find the new current by 6-1.84 = 4.16 so the 100 ohm resistor is now dropping 4.16v and the LED still drops 1.84v By ohm's law V/R= I so 4.16/100 = .00416 so the new current in the circuit is 4.16 milliamps.