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

by BeautifulLight
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BeautifulLight
#1
Sep1-14, 09:56 AM
P: 14
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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Jony130
#2
Sep1-14, 11:21 AM
P: 410
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?


We have a different situation when we add another resistor in series
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berkeman
#3
Sep1-14, 11:23 AM
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Quote Quote by BeautifulLight View Post
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!

jim hardy
#4
Sep1-14, 12:52 PM
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I don't understand how I am "calculating for the voltage drop"

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"..
BeautifulLight
#5
Sep1-14, 05:48 PM
P: 14
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?
berkeman
#6
Sep1-14, 05:59 PM
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Quote Quote by BeautifulLight View Post
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...
phinds
#7
Sep1-14, 06:28 PM
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Quote Quote by BeautifulLight View Post
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?
BeautifulLight
#8
Sep1-14, 06:51 PM
P: 14
Quote Quote by phinds View Post
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.
analogdesign
#9
Sep1-14, 07:20 PM
P: 535
Quote Quote by BeautifulLight View Post
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.


Quote Quote by BeautifulLight View Post
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.
BeautifulLight
#10
Sep1-14, 07:36 PM
P: 14
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 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.
davenn
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Sep1-14, 07:40 PM
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Yes

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




seeing a circuit often helps visualising the problem

Dave
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BeautifulLight
#12
Sep1-14, 07:47 PM
P: 14
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?
davenn
#13
Sep1-14, 07:48 PM
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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

Dave
analogdesign
#14
Sep1-14, 07:48 PM
P: 535
Quote Quote by BeautifulLight View Post
The only time I like to see water analogies is with "valves" (diodes/triodes) because you can take it literally 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.
davenn
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Sep1-14, 07:48 PM
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Quote Quote by BeautifulLight View Post
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
davenn
#16
Sep1-14, 08:02 PM
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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
phinds
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Sep1-14, 08:09 PM
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Quote Quote by BeautifulLight View Post
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.
davenn
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Sep1-14, 08:25 PM
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Quote Quote by phinds View Post
I'd suggest that you not try to become an Electrical Engineer.
awww don't be too harsh

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

Dave


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