Hello,
I am in the first steps of learning electronics as a hobby, but I always encounter a lot of confusing things that need to be explained. For example, in a simple electronic circuit that contains a resistance (300 ohms), a LED, and a 9-volt battery. To calculate the current in this circuit according to Ohm's law. I =V/R, I = 9/300 = 0,03 mA. But when making measurements on the circuit, the value of the current is slightly different from the result of the calculation. What I want to know is that the reason for the difference is because we did't take in account the resistance value of the LED? In real circuits analysis we have to take into account the resistance value of electronic components (other than resistors) when calculating the value of the current,? and how is this done in practice?
Thanks in advance for your appreciated HELP.
Walid

 

Hi there and welcome to AAC and the wonderful world of electronics.

You are correct in using Ohm's Law, but you have assumed there is 9v across the 300ohm resistor. But there is a voltage drop across the diode too - its 'forward voltage' or Vf in the datasheet - typically 1.5 - 2v for a red LED at normal brightness (10 - 30mA typically). So the actual voltage across the resistor is:

9 - Vf = 9 - 1.8 = 7.2v​

​

and therefore the current is 7.2/300 = 24mA as shown by this simulation....

As a beginner, some advice - always read and understand the datasheet of any device you use or plan to use.
Spend time learning and understanding how generic components - both passive (R,L,C) and active (diode, transistor, MOSFET) behave. You'll quickly be able to grasp basic circuit functionality.

 

Yes, you have to take into account all the components. Different components behave differently, you cannot assign a resistance to the LED because it does nit act like a resistor. So you cannot simply use Ohms law to analyze a circuit. Explaining how you analyzing circuits is way beyond what can be done in a forum post. You need to get an elementary text in basic electronics and study it from the beginning if you way to make progress toward your goal.

 

Explaining how you analyzing circuits is way beyond what can be done in a forum post.

While @BobTPH is correct in a generic "how to analyze circuits" sense - its too broad a subject. But if you have a specific circuit you've tried to analyse and need further clarification or help, then by all means post it here with your efforts so far and we'll be happy to assist.

 

You have selected a rather complex example for learning how to analyze simple circuits using Ohm's Law.
There is no simple solution and one usually has to make assumptions in order to arrive at an estimated answer.

A diode or LED is a non-linear device and does not follow Ohm's Law. You cannot use the resistance of the diode because it changes with current.
You need to know both the current through the diode and the voltage across the diode. For this you need to know the I-V characteristics of the diode.

1) You can make an assumption that the voltage across the diode is about 1.8V as @Irving did in post #2. This is the quick solution.

2) You can do the numerical analysis if you knew the I-V characteristics of the diode.


Again, this requires knowledge of particular parameters of the diode equation, for example, the reverse saturation current, Is.
Furthermore, in order to be more precise, you need to know the ambient temperature of the diode since the forward current varies with temperature. You can quickly see how this becomes much more complex than originally thought.


3) You can find a graphical solution knowing the I-V curve of the diode and applying "load-line" analysis.
From this, you find the operating point (or "Q-point") which is the intersection of the diode I-V curve and the resistor load line.
This gives the same solution as for solving with the diode equation in (2) above.

 

Hi there and welcome to AAC and the wonderful world of electronics.

You are correct in using Ohm's Law, but you have assumed there is 9v across the 300ohm resistor. But there is a voltage drop across the diode too - its 'forward voltage' or Vf in the datasheet - typically 1.5 - 2v for a red LED at normal brightness (10 - 30mA typically). So the actual voltage across the resistor is:

9 - Vf = 9 - 1.8 = 7.2v​

​

and therefore the current is 7.2/300 = 24mA as shown by this simulation....

As a beginner, some advice - always read and understand the datasheet of any device you use or plan to use.
Spend time learning and understanding how generic components - both passive (R,L,C) and active (diode, transistor, MOSFET) behave. You'll quickly be able to grasp basic circuit functionality.

View attachment 278214

Thank you very much, Irving, you have given me exactly the answer I wanted. I studied the resistors in series and parallel well, but the confusion that occurs when starting to learn electronics seems to prevent me from knowing the correct answer. Thanks

 

Yes, you have to take into account all the components. Different components behave differently, you cannot assign a resistance to the LED because it does nit act like a resistor. So you cannot simply use Ohms law to analyze a circuit. Explaining how you analyzing circuits is way beyond what can be done in a forum post. You need to get an elementary text in basic electronics and study it from the beginning if you way to make progress toward your goal.

Thanks,
Can you Pls. suggest me a good text to start?

 

You have selected a rather complex example for learning how to analyze simple circuits using Ohm's Law.
There is no simple solution and one usually has to make assumptions in order to arrive at an estimated answer.

A diode or LED is a non-linear device and does not follow Ohm's Law. You cannot use the resistance of the diode because it changes with current.
You need to know both the current through the diode and the voltage across the diode. For this you need to know the I-V characteristics of the diode.

1) You can make an assumption that the voltage across the diode is about 1.8V as @Irving did in post #2. This is the quick solution.

2) You can do the numerical analysis if you knew the I-V characteristics of the diode.
View attachment 278218

Again, this requires knowledge of particular parameters of the diode equation, for example, the reverse saturation current, Is.
Furthermore, in order to be more precise, you need to know the ambient temperature of the diode since the forward current varies with temperature. You can quickly see how this becomes much more complex than originally thought.


3) You can find a graphical solution knowing the I-V curve of the diode and applying "load-line" analysis.
From this, you find the operating point (or "Q-point") which is the intersection of the diode I-V curve and the resistor load line.
This gives the same solution as for solving with the diode equation in (2) above.


View attachment 278219

Many thanks for your answer. Very helpful.

 

Hello,
I am in the first steps of learning electronics as a hobby, but I always encounter a lot of confusing things that need to be explained. For example, in a simple electronic circuit that contains a resistance (300 ohms), a LED, and a 9-volt battery. To calculate the current in this circuit according to Ohm's law. I =V/R, I = 9/300 = 0,03 mA. But when making measurements on the circuit, the value of the current is slightly different from the result of the calculation. What I want to know is that the reason for the difference is because we did't take in account the resistance value of the LED? In real circuits analysis we have to take into account the resistance value of electronic components (other than resistors) when calculating the value of the current,? and how is this done in practice?
Thanks in advance for your appreciated HELP.
Walid

The underlying mistake you are making is that you are abusing Ohm's Law. It's critically important to remember two things about Ohm's Law:

1) Ohm's Law relates the current through a resistor to the voltage across THAT resistor and the resistance of THAT resistor.
2) Ohm's Law only applies to components for which Ohm's Law is valid (referred to as "ohmic devices").

Your 9 V is NOT the voltage appearing across your resistor, but rather the voltage appearing across the combination of the resistor and the LED.

The LED is not an ohmic device. For it, the voltage across it is largely constant for typical current levels that you would want it to have. The actual voltage depends on color, but values range, roughly, from about 1.8 V to 3 V. So this needs to be subtracted from your 9 V source to get the voltage across just the resistor.