I don't understand how I am "calculating for the voltage drop"

  • #1
Assume I have connected a single light-emitting diode (2V, 30mA) and resistor to a standard 9V battery. Ohm's Law is used to calculate the value for the resistor. The observed relationship between voltage, current, and resistance is as follows, V=IR.

I intuitively understand there will be a voltage drop across the diode because it takes a certain amount of energy (in joules) for electrons to "jump" from cathode to anode, and voltage is defined as joules/coulomb. Therefore, the potential difference will now be 9-2=7V.

So, 7=.03R, R=233Ω


What's the problem? I understand everything I just did, and yet I am oblivious to the fact that when I solved for value R, I was also solving for a 7V drop across the resistor.

Would it be better to disregard the diode and just use a single resistor to help me understand? So, 9=I(233). -I randomly chose that value for R. I'm not sure how you figure voltage drop using only 9=I(233). From that, I can only derive current, nothing else...
 
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  • #2
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If you put your 233 ohm resistor across a 9 volt battery, it drops 9 volts. Voltage drop = voltage across resistor.
Resistor "see" 9V across his terminals because you connected a 9V battery directly across a single resistor.
So in this case resistor will only change the current in the circuit. Because how can a resistor change the battery voltage?
attachment.php?attachmentid=72628&stc=1&d=1409587758.png


We have a different situation when we add another resistor in series
attachment.php?attachmentid=72629&stc=1&d=1409588458.png
 

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  • #3
berkeman
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Assume I have connected a single light-emitting diode (2V, 30mA) and resistor to a standard 9V battery. Ohm's Law is used to calculate the value for the resistor. The observed relationship between voltage, current, and resistance is as follows, V=IR.

I intuitively understand there will be a voltage drop across the diode because it takes a certain amount of energy (in joules) for electrons to "jump" from cathode to anode, and voltage is defined as joules/coulomb. Therefore, the potential difference will now be 9-2=7V.

So, 7=.03R, R=233Ω


What's the problem? I understand everything I just did, and yet I am oblivious to the fact that when I solved for value R, I was also solving for a 7V drop across the resistor.

Would it be better to disregard the diode and just use a single resistor to help me understand? So, 9=I(233). -I randomly chose that value for R. I'm not sure how you figure voltage drop using only 9=I(233). From that, I can only derive current, nothing else...
Your first calculation is correct. Start with the datasheet for the LED to see what Vf is at your desired current (brightness). Then the rest of the battery voltage needs to drop across the current-limiting resistor. Different color LEDs and different efficiency LEDs have different Vf values, so you pretty much need to always check the datasheet to find the right Vf to use for the calculation.


EDIT -- Nice post by Jony! :smile:
 
  • #4
jim hardy
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I intuitively understand there will be a voltage drop across the diode because it takes a certain amount of energy (in joules) for electrons to "jump" from cathode to anode, and voltage is defined as joules/coulomb. Therefore, the potential difference will now be 9-2=7V.

So, 7=.03R, R=233Ω


What's the problem? I understand everything I just did, and yet I am oblivious to the fact that when I solved for value R, I was also solving for a 7V drop across the resistor.

Would it be better to disregard the diode and just use a single resistor to help me understand? So, 9=I(233). -I randomly chose that value for R. I'm not sure how you figure voltage drop using only 9=I(233). From that, I can only derive current, nothing else...
Sounds to me like an algebra problem not an electronic one.
Remember beginning algebra - one equation per unknown?

If A = B X C, you must know two things to figure out the third one.
Ditto for V = I X R .
I am oblivious to the fact that when I solved for value R, I was also solving for a 7V drop across the resistor.
No, you weren't solving for 7 volts - you figured out earlier that 7 was the voltage across resistor. And you correctly wrote 7 in for V. You got 7 by subtracting 2V diode drop from 9V battery supply per Kirchoff.

Then you wrote in .03 for I. You got .03 from LED datasheet.

So you were left with one equation and one unknown, which is solvable. And you solved it.

Keep it simple. Talk your way through these things - talking stimulates different parts of the brain than does just that "inner voice"..
 
  • #5
I thought voltage drop was strictly the result of a component actively using voltage. Remember this?

"I intuitively understand there will be a voltage drop across the diode because it takes a certain amount of energy (in joules) for electrons to "jump" from cathode to anode, and voltage is defined as joules/coulomb. Therefore, the potential difference will now be 9-2=7V."


So when you told me there was a 7V drop across the resistor, I immediately asked myself how in the world a resistor "uses" 7V and hence the last part of my initial post,

"Would it be better to disregard the diode and just use a single resistor to help me understand? So, 9=I(233). -I randomly chose that value for R. I'm not sure how you figure voltage drop using only 9=I(233). From that, I can only derive current, nothing else..."


***There is a difference between voltage drop as a result of a component actively using voltage VS voltage drop by simply shorting out the leads of a 9V -of course the drop is going to be 9V, that's the EMF of the battery. In my case, it was easiest for me to think of the 7V as "left-over" voltage ...having absolutely nothing to do with the value of the resistor.


I have one additional relevant question,

Why should a light-emitting diode have a minimum current rating such as 30mA? As long as there is enough voltage supplied -so electrons can jump from cathode to anode (or vise versa), then there should be no reason why electrons shouldn't flow even if there's only 0.00000000000001 amps supplied.

Is the 30mA some kind of threshold for the naked eye?
 
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  • #6
berkeman
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I thought voltage drop was strictly the result of a component actively using voltage. Remember this?

"I intuitively understand there will be a voltage drop across the diode because it takes a certain amount of energy (in joules) for electrons to "jump" from cathode to anode, and voltage is defined as joules/coulomb. Therefore, the potential difference will now be 9-2=7V."


So when you told me there was a 7V drop across the resistor, I immediately asked myself how in the world a resistor "uses" 7V and hence the last part of my initial post,

"Would it be better to disregard the diode and just use a single resistor to help me understand? So, 9=I(233). -I randomly chose that value for R. I'm not sure how you figure voltage drop using only 9=I(233). From that, I can only derive current, nothing else..."


***There is a difference between voltage drop as a result of a component actively using voltage VS voltage drop by simply shorting out the leads of a 9V -of course the drop is going to be 9V, that's the EMF of the battery. In my case, it was easiest for me to think of the 7V as "left-over" voltage ...having absolutely nothing to do with the value of the resistor.


I have one additional relevant question,

Why should a light-emitting diode have a minimum current rating such as 30mA? As long as there is enough voltage supplied -so electrons can jump from cathode to anode (or vise versa), then there should be no reason why electrons shouldn't flow even if there's only 0.00000000000001 amps supplied.

Is the 30mA some kind of threshold for the naked eye?
The 30mA is the value for that LED to give good brightness, apparently. That's actually pretty high for modern LEDs, BTW. It's more common to see 5-10mA on datasheets currently.

As you learn more about basic electronics, you will stop using terms like "voltage used up" and so on. Just thing about voltage drops due to PN junctions, resistors, and so on. Later you will learn about reactive elements like inductors and capacitors, which can have AC voltage drops across them that vary with frequency...
 
  • #7
phinds
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I thought voltage drop was strictly the result of a component actively using voltage.
But the resistor IS actively using the voltage. It is a little bitty heater. How do you think BIG electric heaters work? Do you think that they are not "actively" using the voltage that provides the current that creates the heat that they radiate?
 
  • #8
But the resistor IS actively using the voltage. It is a little bitty heater. How do you think BIG electric heaters work? Do you think that they are not "actively" using the voltage that provides the current that creates the heat that they radiate?
My chemistry teacher told us there was such things as insulators and conductors. I think those terms are for people who don't understand that you can pass electric current across any substance granted you supply enough voltage. I believe whatever the resistor is composed of has a threshold voltage; that is the required voltage of a substance for current to pass through it (assuming normal conditions like temp. etc). So yes, you are correct in saying that the resistor IS actively using voltage. I do, however, think that it is irrelevant here because it's so insignificant and it is better to think of that 7V as "left-over" voltage.


I may be very wrong, but I'm letting you know how I see it.



Edit: For instance, maybe the resistor does require a tenth of a volt for current to pass through it, but that figure is so insignificant compared to the 7V that has been left-over. If that figure would be more significant, then we would have to start dropping voltages on every resistor as we did with the 2V drop on the light-emitting diode.



I thought about this^ over dinner and it's (partially) incorrect, but mostly! That 233Ω resistor IS using all 7V granted there's .03 amperes of current. I forgot about Ohm's Law...

I say partially because when I read your post, it made it seem like or at least to me, that all 233Ω resistors require 7V. That isn't true. You can double your power (14V, .06 amperes) and it still works out, hence 14=(.06)(233) or 14=14.
 
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  • #9
analogdesign
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My chemistry teacher told us there was such things as insulators and conductors. I think those terms are for people who don't understand that you can pass electric current across any substance granted you supply enough voltage. I believe whatever the resistor is composed of has a threshold voltage; that is the required voltage of a substance for current to pass through it (assuming normal conditions like temp. etc). So yes, you are correct in saying that the resistor IS actively using voltage. I do, however, think that it is irrelevant here because it's so insignificant and it is better to think of that 7V as "left-over" voltage.
The terms insulator and conductor are very useful for people that understand electronics and electricity because they classify solids and how and when you can/should use them. A true resistor is a linear device, and has no such thing as a threshold voltage. It will obey Ohm's law down to the point where you can no longer reliably measure current. The only threshold effect you could see would be due to the interface between the resistor material and its contacts to whatever metal is used in its terminal, but that would be indicative of a poor design. Indeed, resistor bulk is coupled to metal using what is called an "ohmic" contact just to avoid any such thresholding effect.


I may be very wrong, but I'm letting you know how I see it.

Edit: For instance, maybe the resistor does require a tenth of a volt for current to pass through it, but that figure is so insignificant compared to the 7V that has been left-over. If that figure would be more significant, then we would have to start dropping voltages on every resistor as we did with the 2V drop on the light-emitting diode.
Yes, you're very wrong so you've come to the right place! :) The idea that a resistor would require a 10th of a volt for current to pass through it but 7V would be "left over" makes no sense and indicates you don't understand how a resistor works. It's rough, but people have found the water analogy useful to get started in electronics. Imagine a garden hose connect to a faucet on the side of your house. Basically, the water pressure is the voltage, the flow of water through the hose is the current and the resistance analogous to the inverse of the diameter of the hose.

V = IR, or water pressure = flow*(1/diameter).

Think about it, if you constrict the hose with constant pressure the flow increases (you use this to blast suds off your car, for instance). If you imagined a fat hose with ten times the diameter the water would just trickle out, since the volume of water (or the "power") would be constant.

If you don't like the water analogy, ignore it (plenty of people hate it because it can be misleading at times).... in that case remember that for resistor circuits V = IR. Manipulate this equation to see how volts, amps, and ohms are related when it comes to flow and power. As you can see in the equation, the idea of "left over" volts makes no sense. All the voltage is dropped across the total resistance.
 
  • #10
Thank you analogdesign. You caught me while I was editing my post (#8). Here

I thought about this over dinner and it's (partially) incorrect, but mostly! That 233Ω resistor IS using all 7V granted there's .03 amperes of current. I forgot about Ohm's Law...

I say partially because when I read your post, it made it seem like or at least to me, that all 233Ω resistors require 7V. That isn't true. You can double your power (14V, .06 amperes) and it still works out, hence 14=(.06)(233) or 14=14.


The only time I like to see water analogies is with "valves" (diodes/triodes) because you can take it literally:smile: Oh, and I don't mean to knock the whole insulator/conductor thing; I'm just one of those 0.0000000001 isn't 0 kinna guys.
 
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  • #11
davenn
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Yes

I was in the middle of drawing a pic when you replied ....

attachment.php?attachmentid=72636&stc=1&d=1409618668.gif



seeing a circuit often helps visualising the problem :smile:

Dave
 

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  • #12
How did you make that schematic? I don't see anything underneath "reply." Is that another program you're using and you're just hitting the paperclip icon to copy it?
 
  • #13
davenn
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so we know there is 9V across the resistor and LED ... ie. the voltage measured between A and C = 9V

the datasheet tells us that there will be a 2V drop across the LED ... between B and C
that means there must be a 7V drop across the resistor ... between A and B
BECAUSE all the voltage drops must add up to the total voltage applied to the circuit :smile:

Dave
 
  • #14
analogdesign
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The only time I like to see water analogies is with "valves" (diodes/triodes) because you can take it literally:smile: Oh, and I don't mean to knock the whole insulator/conductor thing; I'm just one of those 0.0000000001 isn't 0 kinna guys.
Ha! I'm an engineer so I'm a "0.0000000001 = 0" kind of guy when we're talking about amps, but I'm a "0.0000000001 isn't 0" kind of guy when we're talking about nanoamps. It's all about context.
 
  • #15
davenn
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How did you make that schematic? I don't see anything underneath "reply." Is that another program you're using and you're just hitting the paperclip icon to copy it?
I just drew it quickly in MS paint

If you click on the Go Advanced button below the text box, it gives you the option to manage attachments
you can use that to upload files from your puter
 
  • #16
davenn
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so some things you can now work out

here's a problem for you to solve ...

in a series DC circuit, the current is measured the same everywhere, therefore there is 30mA through the LED, through the resistor and through the wires coming from and to the battery

Using Ohms Law --- R = V/I

knowing that the voltage drop across the LED is 2V and the current is 30mA ---
What is the resistance of the LED ?

knowing that the voltage drop across the resistor is 7V and the current is 30mA ---
What is the resistance of the resistor ?

Dave
 
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  • #17
phinds
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I do, however, think that it is irrelevant here because it's so insignificant and it is better to think of that 7V as "left-over" voltage.
I'd suggest that you not try to become an Electrical Engineer.
 
  • #18
davenn
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I'd suggest that you not try to become an Electrical Engineer.
awww don't be too harsh :wink:

At least he came to the right place to learn the error of his ways and is starting to get the hang of it :smile:

Dave
 
  • #19
I'd suggest that you not try to become an Electrical Engineer.
Am I supposed to be embarrassed for attempting to come up with a logical solution rather than have something be spoon fed to me & then pretend I'm smart? Right, because that takes a lot of thought... You must be one of those guys who got their BA in Chemistry w/out ever thinking about the "everyday" stuff around them. Rofl, yeah what an achievement.

Resistance of R1 was already calculated (233Ω). Total resistance of system is 300Ω, hence 9=.03(300), so if you subtract the two, you get the resistance of the L.E.D. which is around 67Ω.

Internal resistance was ignored, but most 9V put out 750mA, so that gives a value of 12Ω. Can be ignored here.

"Shut up & calculate." -Feynman. Always wondered who/what he was talking about. I guess I know now. Please refrain from posting in my threads, especially when you're going to be that discouraging. And by the way, George Ohm is dead so I don't know who I'm talking to, but it sure isn't the guy who came up with Ohm's Law, so you're in the same boat as I am. Now am I supposed to tell you to find another field of study?
 
  • #20
davenn
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don't get too upset mate

read the PM I sent you

is there anything else concerning this that you are unsure of ?
if so, ask :smile:

cheers
Dave
 
  • #21
davenn
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Internal resistance was ignored, but most 9V put out 750mA, so that gives a value of 12Ω. Can be ignored here.
Yes we usually ignore the internal resistance of the battery/power supply

You do understand that it is the battery's internal resistance that limits the current from it ?
which is what stops an infinite current flowing under short circuit conditions

Dave
 
  • #22
jim hardy
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Mr Light - these concepts become second nature after you work a few dozen exercises solving circuits by laws of Kirchoff and Ohm.
I learn in the sequence "What" then "Why".

I suggest you practice until the concepts of loop current, node voltage, mesh currents seem as natural as riding a bicycle.

We all started out awkward.

old jim
 
  • #23
sophiecentaur
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Resistance of R1 was already calculated (233Ω). Total resistance of system is 300Ω, hence 9=.03(300), so if you subtract the two, you get the resistance of the L.E.D. which is around 67Ω.
It is true that the ratio of volts to current is resistance but an LED does not exhibit the same resistance as the current varies - it's non-Ohmic because it's not a piece of metal. If you try to use that value of resistance you (correctly) calculated anywhere else, you will fall over. The voltage drop across an LED is more or less the same over a range of currents - so that Voltage is what you have to use in calculations. In this case, Resistance of a non-Ohmic component is like a wet bar of soap in the bath - you can't latch onto it.
 
  • #24
It is true that the ratio of volts to current is resistance but an LED does not exhibit the same resistance as the current varies - it's non-Ohmic because it's not a piece of metal. If you try to use that value of resistance you (correctly) calculated anywhere else, you will fall over. The voltage drop across an LED is more or less the same over a range of currents - so that Voltage is what you have to use in calculations. In this case, Resistance of a non-Ohmic component is like a wet bar of soap in the bath - you can't latch onto it.
What happens when I pull the resistor leaving only the L.E.D. across the 9V battery? I want to calculate both amperage & resistance, but it's a little confusing. I know there's a 2V drop across the L.E.D., but I also know at those same points there'll be a 9V drop across the terminals of the battery.
 
  • #25
analogdesign
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What happens when I pull the resistor leaving only the L.E.D. across the 9V battery? I want to calculate both amperage & resistance, but it's a little confusing. I know there's a 2V drop across the L.E.D., but I also know at those same points there'll be a 9V drop across the terminals of the battery.
It's confusing because an LED is an active device and doesn't follow Ohm's Law.

If you pull the resistor leaving only the LED you will burn out the LED because the source resistance of the battery is probably small. The 2V LED will have 9 V across it, so the battery current will get really high and kill the LED.

The LED will either fail closed in which case the battery will be drained by sourcing current to ground, or it will fail open, in which case the current will drop to zero because there won't be a closed circuit. Whether it fails open or closed depends on what fails first, the semiconductor junction or the bondwires/case connectin.
 

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