I must be misunderstanding ohms laws

Im a first year engineering student and i think i am misunderstanding something very basic here.

I simplified the circuit to just one led and resistor to wrap my brain around it better

my issue is this

I am getting ~4.7v from 5v to the NPN collector
so in theory the rest of my circuit see a voltage source of 4.7 , you know Thevenin and all.

so with a 220 ohm resistor i should expect a current of ~20mA or so… but i am reading 9mA .
and i do not understand why… is my led like a resistor too?
Voltage drop of the led reads 2.9

that leaves the resistor with about a 2.0v drop.

so i dont understand how my led will ever “see” 5v and 20mA

and voltage drop across one single resistor and a voltage source… how do you measure the voltage drop of a resistor is in essence what your measuring is the voltage source?

I know my theory is all messed. First year engineering student i just finished my DC class going into AC in the summer. I mean i learned a lot because it all made sense in class and measuring this and that but just resistors with other resistors, the led is throwing my expected values off, rather i am not sure how to satisfy the led to get it 5v from 5v voltage source and at the same time limit the current it sees to 20mA if after my resistor the led no longer sees 5v

schematic in the attachment below

eddiea6987: I simplified the circuit to just one led and resistor

Uh, you may have to revisit kindergarten. The circuit you posted has a few more than "one LED and resistor".

ill meet you there , the second sentence clearly says

"I simplified the circuit to just one led and resistor to wrap my brain around it better "

and that post was perhaps to rack up post count? provided very little substance in reference to my questions.

eddiea6987: and that post was perhaps to rack up post count?

It was to get you to post a [u]correct[/u] schematic so those on this side of your monitor don't waste our time on something that is clearly out-of-date.

eddiea6987: ill meet you there

You won't be meeting me anywhere until you post a corrected schematic.

The LED is like a resistor in that it does have a voltage drop. Unlike a resistor, that drop is not closely related to current. You already seem to know that the LED has a fixed voltage. So subtract that voltage from 4.7 to find out what the voltage drop across the resistor must be. Now how much current does Ohm's law say flows through the resistor?

Or build the circuit on a breadboard. Measure the voltages. Play with different resistors until the LED is "bright enough."

eddiea6987: and that post was perhaps to rack up post count? provided very little substance in reference to my questions.

Sure, Coding Badly has only 16182 posts, he clearly needs one or two more :grinning:

anyway, to make things clearer, the answer is in your question, the only thing you have to do is to use your brain (the non agressive part of it ;-) ) :

Voltage drop of the led reads 2.9

then there is 5-2.9 = 2.1 V left for the resistor U = RI I= U/R 2.1/220= 0.0095

..but i am reading 9mA

which is correct ;-)

alnath:
Sure, Coding Badly has only 16182 posts, he clearly needs one or two more :grinning:

:grinning:

“misunderstanding Ohm’s Laws” plural? I must have been sick the day they did the others :slight_smile:

Anyhoo…

You may find it useful to look at the photos I attached from #7 onwards in this thread.

… ummm, make that 16183 counting the smiley.

edit: I’m attaching OP’s pic here for others’ benefit. (Adding to existing post lest I be accused of racking up my count. Damn, can’t wait to be a member.)

dbbfb492b6633d4ceb85db0216c6bd158789d346.jpg

eddiea6987: I am getting ~4.7v from 5v to the NPN collector so in theory the rest of my circuit see a voltage source of 4.7 , you know Thevenin and all.

so with a 220 ohm resistor i should expect a current of ~20mA or so.... but i am reading 9mA . and i do not understand why.. is my led like a resistor too? Voltage drop of the led reads 2.9

that leaves the resistor with about a 2.0v drop.

2V / 220ohms = 9mA, just as expected. Ohms law applies to conductors and resistors, not semiconductor devices which are highly non-linear. It doesn't matter that the supply is 5V or 4.7V, the resistor is seeing 2.0V, so the current is 9mA. In a series circuit each component takes the same current.

so i dont understand how my led will ever "see" 5v and 20mA

If your LED actually sees 5V it will vaporize as it tries to carry many amps. The point of the series resistor is to limit the current so that the LED isn't overloaded, and at its normal operating point the forward voltage happens to be 2.9V as you've measured. But current in any semiconductor diode is an exponential function of voltage, and forward voltage is sensitive to temperature, so you normally set the current, not the voltage.

Or put another way the current in a semiconductor diode is a highly sensitive function of voltage and temperature.

Just to clarify what is being said about the non-linearity of an LED (which is actually similar to other diodes), here is a cute graph I found that shows how the current increases dramatically as soon as you hit the forward voltage threshold: Notice how he correctly shows what happens if you try to drive the LED with a voltage instead of a current ...

Ohms law applies to conductors and resistors, not semiconductor devices which are highly non-linear. It doesn't matter that the supply is 5V or 4.7V, the resistor is seeing 2.0V, so the current is 9mA. In a series circuit each component takes the same current.

Wrap your brain around that.

Also, FYI, it is unwise to try to spar with a Global Moderator. They have absolute power. They can make you disappear in an instant...

PHTT!! The sound of the top of the LED when it hits your head ;)

.

I've never seen that happen with an led. I have seen a TO92 2N3906 explode with a loud bang , propelling a cone shaped piece onto the wall at high velocity leaving a crater in the flat face of tge TO92 package.

Ohms law applies to conductors and resistors, not semiconductor devices which are highly non-linear.

:astonished: Like a logarithmic pot? Ohm's Law doesn't apply? Really?

I've never seen that happen with an led.

I have, got a load of RGB LEDs from China and the red one made a noise as I applied normal power to it.

Try applying a certain variable voltage to water. What you'll find is that the resistance changes with voltage. Air also doesn't obey Ohm's Law - you've got gigantic voltages floating in the air. But there's almost no current until the voltage reaches a certain level. What you observe then is a spark in the form of a lighting. Ohm's law applies only to resistive materials - by definition. What does not obey Ohm's Law is not a resistor

Ohms's law does not apply to non-linear devices.

A logarithmic potentiometer IS a linear device - it is a simple passive resistor - the logarithmic relationship is purely the relation between resistance and mechanical movement.

Ohm's law relates to the [u]proportional and linear[/u] relationship between current and voltage with respect to resistance. For example, if you double the voltage across a fixed resistor you get double the current flow through it (a proportional linear relationship)

For non-linear devices, such as semiconductors, the relationship of proportionality does not apply. For example if you double the voltage across a conducting diode you get more than twice the current flow. Similarly if you perform the same exercise on a reverse biased diode you do not get twice the reverse current flow if the reverse voltage is doubled.

However, the equations of ohms law can be used to determine "equivalent resistance" of non-linear devices. For example if you know the voltage across a device and the current flowing through it you can determine the "equivalent resistance" of the device at those conditions and [u]only[/u] at those conditions. Change the voltage and you get a non-proportional change in current.

Using the equations determined by Ohm does not mean "Ohm's law applies"

A logarithmic potentiometer IS a linear device

OK, thanks. That's all I needed to know.

In fact Ohm's law only applies to certain kinds of conductors where charge-carrier drift dissipates into heat. Metals, some intrisic semiconductors, electrolytes with inert electrodes that aren't depolarizing. Its an effect due to the electric field accelerating charges, and the charge movement being randomized by collisions and turning into thermal energy - in other words resistance to charge carrier flow that scales with charge carrier drift velocity. A bit like viscous friction.

A vacuum allows charge carriers to travel unhindered, as do superconductors, and things like tunneling junctions and semiconductor junctions have very different physics.