Hello,

I am starting to get myself into electronics and build a simplest circuit including a power source (3V, drawn from a microbit powered by a 2x 1.5 V battery pack) a red led and 100 Ohm resistor connected in series.

I now wanted to measure the current in the circuit and used a multi-meter. Based on ohms law i expected a current of I = 3V / 100 Ohm = 0.0.3 Amp.

However, putting the multi-meter in series between the diode and the resistor i measured 0.01 Ampere.

Calculating backwards it seems that the circuit experiences R = 3V / 0.01A = 300 Ohm resistance.

Does this mean that the diode in fact has a resistance of 200 Ohm ...

Does this make sense? I thought that a diode in forward direction would have legible resistance

Do i miss something?

any guidance is much appreciated,

Dan

Hello,

I am starting to get myself into electronics and build a simplest circuit including a power source (3V, drawn from a microbit powered by a 2x 1.5 V battery pack) a red led and 100 Ohm resistor connected in series.

I now wanted to measure the current in the circuit and used a multi-meter. Based on ohms law i expected a current of I = 3V / 100 Ohm = 0.0.3 Amp.

However, putting the multi-meter in series between the diode and the resistor i measured 0.01 Ampere.

Calculating backwards it seems that the circuit experiences R = 3V / 0.01A = 300 Ohm resistance.

Does this mean that the diode in fact has a resistance of 200 Ohm ...

Does this make sense? I thought that a diode in forward direction would have legible resistance

Do i miss something?

I think you're looking at the LED's operation incorrectly. What you have is a 3V source which goes through the LED which requires about 2V to operate, so by the time the resistor is encountered the effective voltage across it is only 1 Volt.

LEDs are usually considered as voltage drop instead of resistance value. You can calculate the effective resistance but it won't be linear over voltage. That is, if you pass1V through the LED it probably won't light at all and if you put 100V through it, then everybody on the block will notice it 🙂

Does this explain it ?

Usually you measure voltage relative to ground at the terminals of components. Breaking the circuit to put in an amp meter isn't usually practical. It is called ground because it used to be a connection to the real ground, the earth, but it is just a common reference point and is usually the negative terminal which means all the voltage (pressure) readings will be positive.

There is not always a linear resistance for components for all voltages. In the case of the LED or any DIODE for that matter.

http://lednique.com/current-voltage-relationships/resistance-of-an-led/

@will thank you,

To try things out -- i exchanged to 100 Ohm resistor with a 1K Ohm one -- and the led still caused a 2 V drop, with the rest going to 1K Ohm Resistor.

However, when i exchanged the led with a 100 Ohm Resistor while keeping the 1K Ohm resistor as well, this lead to a 100 Ohm and 1K Ohm resistor in series powered by the 3V -- then the 100 Ohm resistor (now instead of the led) had a 0.27 V drop, with the rest going to the 1K Ohm Resistor -- and the current dropping as well to a number i can't measure anymore.

So while a led can have an effective resistance -- it does not behave like a resistor --

I guess this is what you mean with i am not looking at the operation of the led correctly ...

Confusing and interesting 🙂

Dan

@robotbuilder thank you ... its interesting in the reference that a led is modeled

*"If we look at a typical LED IV curve we can see that it is approximately linear over much of its useful range. This allows us to model the LED as a resistor and voltage source."*

So, its not only offers resistance but behaves like a voltage source as well -- hence, perhaps the reason why the led keeps to a 2V drop no matter what sized resistor i am adding in series.

As I said, the effective resistance of the LED (or any diode) is not linear with respect to voltage. With using LEDs we're interested in 2 values - the voltage drop across the LED and the current we want through the LED (which affects the brightness and longevity of the component.

E = IR for a resistor, but this relation does not hold true for all components. Calculations for LEDs and regular diodes normally use the voltage drop across the component (which is more-or-less constant) instead of their resistance (which is not constant). Normally LEDs drop 2-3V depending on their colour, silicon diodes drop about 0.6V and some other types of diodes can drop as little as.1V to .3V.

Consider the case where you want to calculate the current limiting resistor to use a blue diode (3V drop) to indicate some condition. You need a resistor to drop the current to a safe value for the blue diode. Let's suppose you're using a 5V supply and you want to run 10 mA through the LED.

Ohm's law says R = V/I. Taking the source as 5V and using the fact that the blue diode will drop 3V across it, that leaves 5-3 = 2V across the resistor. so R = (5V-3V)/(0.010Amp) = 200 ohms.So you'll need a 200 ohm resistor to have 10mA flowing through the LED.

Suppose you want to use as 12V source instead of 5V. Now the voltage across the resistor is (12-3)=9V. So using the same R=V/I as above, we have (9V)/(.010) = 900 ohms.

If we consider the case where we want to use a red LED (2V drop) instead of a blue LED using 5 Volts supply and run again at 10mA, then R=(5-2)/(0.010) = 300ohms.

Is this helping ?

Just to complicate things there are also voltage dependent resistors,

https://eepower.com/resistor-guide/resistor-types/varistor/#

My own children had ZERO interest in electronics or programming computers although they loved playing computer games.

Perhaps let them choose an electronic project/s so they own the idea.

There's a simple graphic method for selecting the series resistor for a led. Check out the attached example

@foxy Thank you -- yes, that's what i was wondering about -- how to calculate the resistor needed to adequately limit the current.

But, i was also curious to learn what all quantitative values of the circuit are -- and learn a lot from the replies here.

thank you,

Dan

Happy this was useful. Actually this approach works nicely for many problems involving non-linear devices: transistors, mosfets, etc and anything else for which you can get a linear or non-linear VI curve or family of curves

Foxy

@foxy I just read that a led is in fact a semi conductor device -- and hence whether there is a flow of electricity or not is dependent on a min voltage threshold applied to the led, but the led does not behave like a resistor -- where voltage drop increases as more voltage is applied.

Instead, once a threshold voltage level is achieved it stays the same on the led -- hence additional voltage drops occur in the connected circuit -- while the led if not connected elsewhere will eventually burn out.

A LED is, in fact a semiconductor diode; you could use it as diode as long as you stayed below the reverse breakdown voltage which is quite low so there is no reason to do this. Also an LED has a relatively high threshold voltage of approximately 2 or 3 volts while a normal diode has a threshold voltage of about .8 volt for a silicon diode and .2 volt for a germanium diode. (germaniums aren't much used any more because of high reverse leakage). Be careful in assuming that the LED forward voltage stays constant above breakdown--it simply behaves like a low, not zero resistance so if you are designing close to the LED maximum current rating it's better to get it's curve from the data sheet and use the graphical method I gave.

Foxy