How do resistors and forward voltage work?
izbin88
Posts: 4
If I have an adapter putting out 5v dc and 1 amp and I attach an led to it how many volts and amps will go to the led?
The led has a forward voltage of 2.0 v.
Now lets say I attach a 470 ohm resistor to the led in series. They say that the resistor will eat up the remaining 3.0 v left by the led.
And what would happen if the led had a forward voltage of 3.0v, then the resistor would absorb only 2.0 v? How does the resistor know how many volts to absorb?
And lets say I had a power supply supplying 20 v dc, the resistor would then absorb 17 v (if the led had a forward voltage of 3.0 volts)?
And back to the example of a 5v, 1 amp power supply attached to an led with a forward voltage of 2.0 v with a 470 ohm resistor.
Is the resistor absorbing the full 1 amp and only letting a couple of milliamps go to the led or is it absorbing less than that?
The led has a forward voltage of 2.0 v.
Now lets say I attach a 470 ohm resistor to the led in series. They say that the resistor will eat up the remaining 3.0 v left by the led.
And what would happen if the led had a forward voltage of 3.0v, then the resistor would absorb only 2.0 v? How does the resistor know how many volts to absorb?
And lets say I had a power supply supplying 20 v dc, the resistor would then absorb 17 v (if the led had a forward voltage of 3.0 volts)?
And back to the example of a 5v, 1 amp power supply attached to an led with a forward voltage of 2.0 v with a 470 ohm resistor.
Is the resistor absorbing the full 1 amp and only letting a couple of milliamps go to the led or is it absorbing less than that?
Comments
With a 5V power source and a 2V LED forward voltage, you have 5 - 2 = 3V to drop across the 470 Ohm resistor. By Ohm's Law, the current would have to be 3/470 which comes out to a little over 6mA. You can figure out the other cases you mentioned.
The 1A rating of the power supply is the maximum that can be supplied at the 5V specification. If you attempt to draw more, the supply voltage will probably drop low than 5V depending on the specifics of the power supply design and construction. Sometimes there's a minimum current limit as well where the output voltage will rise. It all depends on the design.
I am still trying to understand this whole thing.
Let's say I connect a led with a forward voltage of 2.0v to a 5v dc 10 mA power supply. I'm assuming it will light up. Is the led using only 2 volts? If so where are the other 3v going? Or is the led going to use all 5v and burn out?
Leds drop a fixed voltage, and they also need the current limited to under 20mA. So if you put your led directly across the 5V supply, 1A would flow and zap the led.
The resistor limits the current.
A '10mA" supply doesn't really exist here. (it does exist in the form of a current limited bench supply but these are expensive and most of us start off with much cheaper supplies, like a 5V 1A supply). When we say "1A", this is the maximum that supply will provide. In practice, the amount of current that flows is determined by the resistance of your circuit that you build.
In this case you start with the led specifications and work backwards. The led works happily at, say, 10mA, which is 0.01A. It drops 2V, so that leaves 3V left over. V=IR (as per Mike Green), and V is 3 and I is 0.01, so R is 300 ohms. (in practice 220 to 470 are often used, depending on how bright you want the led).
So the "other 3V" is across your resistor.
Keep the questions coming.
The LED is like a sprinkler hooked to the end of the nozzle. The sprinkler requires X amount of water pressure to pop up and sprinkle. Too much water pressure will cause the sprinkler head to pop off. Too little and your plants die.
This is where I'm still confused: (Yeah I know, thick head , stuck brain...name it ! ) lol
Anyway, Dr_Acula talks about resitor limits current, and M.Green talks about "The resistor has to handle the remaining 3V".
I've spent hours on Electronic 101 for really dummy idiots like me, and cannot grasp that concept!
Somewhere a patch of CELs has died and for the life of me, cannot go over that hump.
This is where some of my projects died!
All the water tank and garden hose etc... read and re-read... is not computing! Especially with a LED when voltage and current is involved!
Is there someone here with lots and lots of patience to go through it with a different approach, for teaching???
Cheers
Yoshti!
I will try, but first you have to ask some specific questions so I can understand better where you are having trouble.
Oh boy...you're so brave ! ;-)
Let me give a background : Some 35 years ago (Ouch!) we had a quick course on electronic, and the whole class failed! So maybe since then, I have a misconception! But I know if I put a 12 volts to a 5 volts device, it will not like it at all ! And Current, in french "Courant" was used , and today I know we meant Voltage, but it kind of mix the whole thing up!
So will it better to go by, some one principle at a the time?
Ex: If I put a voltmeter to a source (12 volts let say) I read 12 volts, if I add a resistor between (lets say 500 ohms) I still get 12 volts! WHy?
(By Mr. Green the resistor has to handle 3 volts) But other statement talks about current(Amps) ...
Yet I've read somewhere that the resistor, will absorbe the extra voltage...(Not according to my voltmeter!) Only if the circuit is closed ...hummmm ?
Again, thank you kindly, even to try is already a big plus for me and maybe for some noob like me browsing only the forum!
Cheers
Yoshti
Known:
Voltage source = 5V
LED voltage drop = 2V
Unknown
Resistor voltage drop?
Solve for the resistor voltage drop
5V = 2V + X
X = 3V
So the resistor drops 3V
Ohm's Law
R = V/I
If you look at the LED spec and it states the ideal LED current is 6mA then solve for resistance.
R=3V/6mA = 500 ohms
These are basic electronic 101 laws. Accept and apply the rules.
Because the 12V (battery) = 12V (resistor) [Kirchoff's Voltage Law]
Current can not flow unless the circuit is closed. Work is done when current flows.
http://en.wikipedia.org/wiki/Kirchhoff%27s_circuit_laws
Oops ... That link is just a table of contents. If you know any amateur radio operators, they may have a copy or you may find one in a library.
You will read 12v because ALL of the supply voltage is being applied across the resistor. And that is PART of the picture. Another part is how much current is flowing at the same time?
You could set your meter for current, insert it into the circuit, and read the current.
You could use Ohm's Law and solve for the current. E = I x R, or I = E / R so I = 12 v / 500 ohms = .024 Amps
So, lets say your 12v power supply is a 2 Amp supply.
Your circuit, consisting of your current meter in series with the 500 ohm resistor, when you attach your 12v supply the circuit will consume .024 Amps. And if you had a seperate voltmeter, connected to the power supply it would read 12v.
Lets not worry about the voltage the resistor is using right now, but can we agree that there would be .024 Amps flowing through the circuit?
Does this make sense?
Do you have some parts and a meter play with?
sorry for not being clear. I wanted to know what would happen if i didnt use a resistor in the circuit. Would all the 5v flow through the led? And would the led get damaged?
Yes make sense for the results. That I grasp..fairly well. I'll get my stored away box, and get parts out, to go through ...
I'll try not to jump ahead ... like the LED stuff... I got weird questions about that in regards of the formula used to calculate.
Mike G : The basic is ok, I get the same results etc. Yup I'm there!
As for wikipedia etc... Yes I did the RTM ! many times. That does not help at all. Need to grasp the inherent concept of Voltage and Current and how they work hand in hand. Too bad I've deleted my Favorites about Electronic, but I had about over 30 sites for sure, about the Electronic for dummies to Kirchhoff's law and everything in between! Name it ! I got frustrated by not grasping the basics between Volt and Amps (especially around a LED) using the formula. I got the right answer to calculate, but did not understand what I was doing! Not good...
Like Izbin88 , where that 3 volts go? Why talk about amps when I got volts too much?
And the basic formula : Indian sees Eagle over the Rabbit I=E/R is great BUT ! the formula tells me only that:
I have 12 Gallons of water to separate between 500 people... 12/500 , so each will receive .024 of the 12 gallons!
Right , so the formula tells me the result, but not what happened ! (Mostly when a LED is involved..that does not tell me nothing... The resistor is calculated only on how much each person would get parts! not How much the resistor is choking the circuit)
I guess that is where my misconception is all about.
I 'll follow exactly where JIM N8RHQ is bringing me, He may unlock my past! ;-)
No disrespect to no one, I'm just frustrated with myself, for not being able to grasp the basic! Imagine, that If I get it, the whole world will LMAO!!!
Cheers
Yoshti
Good. That I understand. off the 12, 2 is gone directly for the LED. 10 unit is what I dunno what to do with it! lol Why seperate it into 500 pieces?
But for now I need to stick with jim N8RHQ and bring the issue of the LED later !
Cheers
Great! Lets build on that. Lets replace the single 500 ohm resistor with two smaller resistors. We will call them R1 and R2. R1 will be 200 ohms and R2 will be 300 ohms, wired in series they total 500 ohms, same as before.
As in our previous circuit, the current will be the same. I = E/R, I = 12/500 = .024 A.
.................................................__|Volt.......|__............__|Volt......|____
Positive......................................|...|Meter 1 |....|...........|...|Meter 2 |......|
Supply.......................................|.....................|...........|.....................|...........Ground
|Amp |
|
R1
|
|
R2
|
................................|Meter |
Lets try to figure out what Volt Meter 1 is going to read. We know the current through R1 is .024 A, and we know the value of R1 is 200 ohms. E = I x R = .024 x 200 = 4.8v. So Volt Meter 1 should read 4.8v
Lets try Volt Meter 2. Again the current is .024 A, and R2 has a value of 300 ohms. E = I x R = .024 x 300 = 7.2v. So Volt Meter 2 should read 7.2v
Now if we add the readings from each voltmeter, 4.8 + 7.2, you will see the total 12 volts.
Sorry for the horrible ascii art, the editor makes it a little difficult.
Does this still make sense to you?
Does it make sense that we could swap the values for R1 and R2, ie R1=300 and R2=200, and if we did our voltmeter readings would change as well, ie V1=7.2v and V2=4.8v ?
Mike G, or any other moderator, please let us know if we should take this discussion off the forum and go direct to each other via email.
What do you do with the left over 10V?
Well, it depends on what you are trying to do. Let's say you want to light a blue LED at max luminosity. Reading the LED datasheet you find the blue LED shines brightest at 20mA. Now you know two things; you have 10V left over after dropping 2V on the LED and the LED is brightest at 20mA. Ohm's law states R = V / A. Therefore, 10V/20mA = 500ohms. That means, if you put a 500 ohm resistor inline with the blue LED, it will shine at its brightest.
Reading the blue LED datasheet you find that at 10mA the luminosity is half. Therefore, 10V/10mA = 1000 or 1K. If you replace the 500 ohm resistor with a 1K, the LED will be half as bright.
A resistor "resists" current flow.
As in our previous circuit, the current will be the same. I = E/R, I = 12/500 = .024 A.
.................................................__|Volt.......|__............__|Volt......|____
Positive......................................|...|Meter 1 |....|...........|...|Meter 2 |......|
Supply.......................................|.....................|...........|.....................|...........Ground
|Amp |
|
R1
|
|
R2
|
................................|Meter |
*** Problem here. My box is stuck in the shed, with ice and snow that jam the door. Put Salt to get it to move. This weekend , temp is going up, so that should free the doors.
In the mean time.... ****
Lets try to figure out what Volt Meter 1 is going to read. We know the current through R1 is .024 A, and we know the value of R1 is 200 ohms. E = I x R = .024 x 200 = 4.8v. So Volt Meter 1 should read 4.8v
**** Voltmeter 1 (VM 1) is essentially bypassing R1, shouldn't we calculate the value from R2? Because, VM 1 is taking Positive supply + , and the end of R2 to take the reading. That part does not jive with me. Although The .024 is the total with 200+300, and really by multiplying with 200, you essentially get the result of end of R2?
(*** For R2, is opposite of above)
The thinking here is that Less resistance, should be higher voltage ! R1 having 200 parts of 500 or 2 fifth (2/5) of the total voltage BECAUSE R2 is stealing 300 parts? I hope I can convey properly what I'm trying to mean here... confusing I know...
VM1 is being affected by R2 and VM2 by R1
Lets try Volt Meter 2. Again the current is .024 A, and R2 has a value of 300 ohms. E = I x R = .024 x 300 = 7.2v. So Volt Meter 2 should read 7.2v
Now if we add the readings from each voltmeter, 4.8 + 7.2, you will see the total 12 volts.
Sorry for the horrible ascii art, the editor makes it a little difficult.
*** You should see my drawing....blurk!
Does this still make sense to you?
**** By formula , its working.
Does it make sense that we could swap the values for R1 and R2, ie R1=300 and R2=200, and if we did our voltmeter readings would change as well, ie V1=7.2v and V2=4.8v ?
**** Yes it does !
Mike G, or any other moderator, please let us know if we should take this discussion off the forum and go direct to each other via email.
VM2 is 12 V - 4.8 V = 7.2 **** Source - Reading VM1=VM2
VM1 is the result of R2
Ho Man I have a hard time to explain! 8-(
But I hope I'm right !
http://www.facstaff.bucknell.edu/mastascu/elessonshtml/Basic/Basic5Kv.html
The voltage drop across resistors in a serial circuit is proportional to the resistance.
12V(200/500) = 4.8V
12V(300/500) = 7.2V
Therefore,
12V = 12V(200/500) + 12V(300/500)
You should consider picking up an Electronics 101 textbook and study.
BTW, Mike Green is a moderator not I.
I got the CD Electronic for dummies, I'm steadily on ALLABOUTCIRCUITS forum etc....
Even the PDF file: http://www.allaboutcircuits.com/pdf/DC.pdf is openned every day and compare it to others at the same time just in case it will eventually connect the dots !
It's all great. But They simply put out the formula, and tell you how to calculate etc.
But they do not actually break it down to a logic where it shows you , just like you did! Which is much more appealing to me than straight I=E/R.
jim N8RHQ is actually ramping me up to where I will break down and from there will try get over the issu where I can't grasp.
No disrespect to you or anyone, but I want to let JIM finish his crash course, and slap me silly if he has to! but since he volunteers to help me out... he has the first shot at it! ;-)
Cheers !
Edit: This a kind of analogy that I can relate to: http://www.newton.dep.anl.gov/askasci/phy99/phy99297.htm
Kind of 10 000 cars getting there. If the traffic lights is choking a lot, less cars can go through, (less amps)
But if everything is in sync, lots of cars go through traffic lights (more amps). Hummm... that make sense!
Paul, great slides, thank you.
Yoshti, if I understand your question, I think you have got it.
First, about the volt and amp meters in our circuit, lets pretend they have no affect on the circuit.
The meters have no influence on what's going on, they just tell us what is happening.
"VM2 is 12 V - 4.8 V = 7.2 **** Source - Reading VM1=VM2"
"VM1 is the result of R2"
Yes! Except VM1 is the result of R1 and R2 working together with the source.
They are "dividing the voltage between them"
You've got it to the point where we can change the total R, R1+R2+R3+Rn,
and I think it will be useful to have three this time.
Plus 12v
|Amp|
A
R1
B
R2
C
R3
D
Ground
R1 = 200 ohm
R2 = 300 ohm
R3 = 500 ohm
R=R1+R2+R3=200+300+500=1000 ohms
I=E/R=12/1000=.012 Amps
VM1 = Voltmeter connected to A and B
Often, we might call this "the voltage across R1"
E=IxR1=.012x200=2.4v
VM2 = voltmeter connect to B and C
the voltage across R2
E=IxR2=.012x300=3.6v
VM3 = voltmeter connected to C and D
the voltage across R3
E=IxR3=.012x500=6v
VM4 = voltmeter connected to B and D
the voltage across R2 AND R3
E=IxR3+R2=.012X800=9.6v
VM5 = voltmeter connected to A an C
the voltage across R1 AND R2
E=IxR1+R2=.012x500=6v
Now all THREE resistors are working together to divide the voltage between them.
Notice that because R1+R2=R3, that the total 12v is divided in half at the point between R2 and R3
as shown by VM5 and VM3.
If in our previous circuit we had used two 250 ohm resitiors instead of 200 and 300, the voltage across
each resistor would have been 6v.
A---R1---B---R2---C
VM6= voltmeter to A and B
VM7= voltmeter to B and C
VM6 will equal VM7 when R1=R2
How we doing Yoshti?
So far so good. I'm following. (Can wait to get that box out... and apply the fundementals... again)
But You know I'm dying to ask: Voltage and Current. The resistor is for current, how can I read a different voltage?
When (at the beginning..) I put my voltmeter at the end of a resistor, from a 9 volts Bat, I read 9 v ! I digged it because it affects the current , not the voltage.
But When the same resitance is in a closed circuit, it no longer showed me the same voltage. What!?!
I understand, current is proportional to voltage. The opposite is not necessarly true?
I have the same issue with Izbin88. Where that extra voltage go?
The resistor is limiting the Current. From a statement "The other 3V is across the resistor." I don't dig that part!
I know it's kind of odd that not understanding that part is weird. But that is where my show stopper is !
Cheers
Yosh