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!
Kirchhoff's Voltage Law! All voltage drops in a series circuit must equal the source voltage. There is no "extra" voltage.
<br>
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 1 amp value is the maximum the ps "can supply" under nominal operating conditions. If you have nothing attached to the output, i.e. infinite resistance, no current will flow. if you short circuit the output, i.e. zero ohms, it might supply more than 1 amp... till something burns out! :-)
In a series circuit, the current is the same everywhere in that circuit, that is, the current that all your combined components will draw.
Setting up your example... with a 5 volt PS rated at 3 amps, an LED, and a 470 ohm resistor, the circuit drew 6.32 milliamps. The LED dropped 1.97 volts and the resistor dropped 3.03 volts.
Changing the resistor to 220 ohms, the circuit now drew 12.96 milliamps. The LED dropped 2.02 volts and the resistor dropped 2.98 volts.
Lastly with a 100 ohm resistor the circuit drew 28.9 milliamps. The LED dropped 2.11 volts and the resistor dropped 2.88 volts.
Going to a 50 ohm resistor would probably drive the current draw to over 50 milliamps... above the typical LED current max and cause the LED to go into permanent dark mode. :-|
Take a look at this slide. If it does not make sense to you, then we will have to step you backwards a little.
After you study this, replace the "air" with a 1K Ohm resistor and measure the voltage across R1. Can you predict what you will read for voltage and current before you do this?
O.k. o.k. ..... I'm going to jump in here. I feel your frustration Yosh. I noticed your question about where did the rest of the voltage go.... so here's my attempt to help.
Your supply puts out 10V. You attach a 1K ohm resistor across it. You measure 10V across the resistor.
10V
1K
GND
Now you remove the 1K and replace it with two resistors of 500 ohms each in series (end to end.) You measure at the same place you measured the 1K resistor. You still have 10V.
10V
500 ohm
x
500 ohm GND
Now you measure from ground to the connection between the two resistors (x). You measure 5V.
Now if you measure from between the two (x) and the positive terminal of the battery (where the 10V is...) you measure 5V.
The extra voltage has not gone anywhere. There is 5V across each of the 500 ohm resistors and still total up to the 10V the supply is putting out. Both 500 ohm resistors act as (and add up to) 1K ohms. The current flow is the same in both circuits because the total resistance in each circuit is 1K ohms.
Resistors literally resist the flow of electricity. It's like they cause the electrons to back-up a bit. This back-up of electrons causes the voltage drop across the resistor.
If you only have one resistor as in the first diagram, you'll always read what the supply is putting out. Once you start dividing it up, you get different measurements according to the amount of resistance there is between the two points you're measuring from. It could be across one resistor that's in the middle of a whole circuit or from GND to some point in the middle of a circuit or whatever.
As far as current is concerned. In both examples, you'll have 10 mA flowing. If you increase the voltage to 20V. you'll have 20 mA flowing. The ability of the resistor to "resist" the flow of current hasn't changed. But, you're forcing more current through them because you've increased the voltage potential....(you're going to hit the ground harder if you jump from 20 feet up rather than 10 feet up.) Of course, now that you're applying 20V to the circuit, you'll have 10V across each of the 500 ohm resistors.
LEDs are different as they are silicone based. They basically operate on a principle that current flows once the voltage reaches a certain potential...1.2V...1.8V..whatever...depends on the LED. It's a gate that opens up under the right amount of pressure. That voltage tends to remain the same because the more you shove into it the more current the LED lets through. But, put too much through it and it burns up.
A dam will stop water. But as it rains and the amount of water behind the dam increases....eventually the water spills over the dam. the height of the dam doesn't change (like an LED will drop 1.2V all the time.), but, the amount of water flowing over the dam does. put too much water behind the dam and the dam will burst.
So you 'limit' the amount of current that can flow into an LED with a resistor.
Take a look at this slide. If it does not make sense to you, then we will have to step you backwards a little.
After you study this, replace the "air" with a 1K Ohm resistor and measure the voltage across R1. Can you predict what you will read for voltage and current before you do this?
Paul
Hi Paul,
First thing first! missing how much current this circuit draws ! So I = E/R x= 9 / (1000+1000) = .0045 (4.5 MA)
Now to get the voltage at R1, then E=IR x= .0045 X 1000 = 4.5 Volts.
Which was predictable, since we have 2 resistor of the same value!
So far so good I hope.
To: ajward
Love the statement "Permanent dark mode" I have a collection of them. Might make a lot of money on Ebay!
But a bit confused with the LED dropping more voltage as the resistor changes! Is this a normal thing?
To: Catspaw
I'm like a the speed of sound, just about to through..but missing that little extra power to go through!
Ok, i grasp the Voltage drop for the LED etc.
"LEDs are different as they are silicone based. They basically operate on a principle that current flows once the voltage reaches a certain potential...1.2V...1.8V..whatever...depends on the LED. It's a gate that opens up under the right amount of pressure. That voltage tends to remain the same because the more you shove into it the more current the LED lets through. But, put too much through it and it burns up"
That is where I stumble! Drop 1.8 volts and there is 3.2 Volts left to get rid of ! (Out of 5 Volts)
Ok So If I don't put a Resistor, what's my AMP? Since a 0 is not possible for the formula:
I'll say resistance is 1 ohm so I=E/R where 3.2/1 3.2 amps?
That is telling me that AMP is equal to Voltage !! That can't be..is it?
Since I have the ultimate PSupply, it can push 5Volts and ulimited amps (just for the fun of it) still my circuit will draw only 3.2 amps!?
Arrrrgh, that cell went pouffff, I got one left ...!
When assuming no resistance you can't use zero (there is no such thing a pure 0.000... ohms in nature). And don't use a unity value (1).
Try using 0.002 ohms (2 milli Ohm) to represent 0 ohms. Notice in my slides, there is no such thing as a perfect 0 or perfect infinity - use very low and very high values to substitue). Now try again. Also, try it with a 1K and a 2K (3K total) in series instead and see what happens; the current will drop and voltage across each resistor will be different.
So you say you don't understand how much current the circuit draws? Your math is correct!
A circuit with two 1K ohm resistors connected in _series_ results in 2K ohms total. Apply 9V of electrical pressure through that restriction and only 45 mA of current will flow. That is right, you get it from what I see. Good.
I = E / R
= 9 / 2000
= 4.5 mA
Ok, so in this circuit 4.5 mA worth of current (electrons) is flowing. No matter where you hook up your ammeter (between the battery and a resistoor, or between the two resistors for example) 4.5 mA is flowing - current at every point in the circuit it is the same. Rremember, your ammeter looks like a short so it must be connected in series as well.
Use the "Water Anology" to visualize the Current aspect:
You have a hose filled with water. At one end is the valve (your battery or power supply if you will).
At the other end is your thumb over the hole.
You turn on the (valve) water pressure (Voltage).
Your thumb is totally blocking the end of the hose (this is very high Resistance)
You feel the pressure on your thumb (the Voltage)
But no water is flowing (no Current)
Now take your thumb slightly off the end of the hose (you just lowered the Resistance)
You feel less pressure on your thumb (lower Voltage)
But water is now comming out of the hose (Current is flowing)
No matter you look in the hose, the same amount of water is passing by any point as in any other point (Current is the same everyehere in the circuit).
Now take your thumb off the end of the hose completely (Resistance is now very low)
You feel no pressure on your thumb, but a lot of water is flowing (a lot of Current).
Keeping your thumb off the hose, you close the valve a little (lower the Voltage)
Less water is now flowing out of the hose (less Current because less Voltage)
Quantity is the amount of Pressure and Flow to fill a bucket. Power = Voltage x Current.
I does not matter if your are counting electrons, water molecues, pennies, beans, whatever - the math and the concepts are essentially the same.
This is how Mr. Ohm (and others) realted to this attribute of nature.
Are you pickin up what i am puttin down (do you undrstand)?
I think you understand this more than you think.
What time zone are you in? It is hard to teach this concept using e-mail (I can't see your expressions in real time). We may have to Skipe to get the ideas to you.
Paul,
The point I'm trying hard to understand... is "Keeping your thumb off the hose, you close the valve a little (lower the Voltage)
Less water is now flowing out of the hose (less Current because less Voltage)"
The relation of Voltage and Current. Is 1 volt is equal to 1 amp ?
Like the water hose, to me it's the voltage (main valve) that pushes the water. Less water less voltage. AT THE SAME TIME, less current too!
WHich in any examples or exercises, is not give me the relation. Like the more SUN, the brighter...
I don't get the current! Especially -> Resistor resist the current! Yet I put it in series with a LED, because that 3.2 volts extra is too much!
I reduced the current to reduce the Voltage? So it's a 1 to 1 relation ?
Cheers
Yoshi
Time Zone is -5 (New York/Montreal time) Right now , at the office..hum hum...! ;-)
To: ajward
Love the statement "Permanent dark mode" I have a collection of them. Might make a lot of money on Ebay!
But a bit confused with the LED dropping more voltage as the resistor changes! Is this a normal thing?
It is... well, more to the point, as the current changes. The resistor value affects the circuit current and that's what the LED sees.
The "2 volt drop" is an average typical of LEDs.
The forward voltage (the drop) varies with the current flowing through the LED. That value can also vary based on the color of the the device.
This is a forward current vs. forward voltage chart for three Fairchild LEDs (Red. Yellow and Green).
You can see that my test values kind of follow the curve in the chart above, although at 28.9 mA the drop is less than expected, but I hope you get the idea. I don't know who made my yellow LED, perhaps my test is correct!?
That is right! The current is less because the voltage is less! I think you understand. Good!
If you have a given voltage and a given current is flowing, and then you lower the voltage, then less current will flow.
Voltage and Current are PROPORTIONAL (that is they are related and if one increases the other increases) but they are NOT THE SAME.
But Current and Resistance are INVERSERLY PROPORTIONAL (again, related but as one increases and the other decreases) but NOT THE SAME.
Voltage is "like" pressure (think of it a potential energy if that helps)
Current is "like" flow (think of it as kinetic energy it that helps)
Resistance is "like" a partial blockage (think of it as matter if that helps)
Ok, say you have a voltage and some current in a circuit (it does not matter how much of each in this example). You leave the voltage the same and then you INCREASE the RESISTANCE. What does the current do?
A) Goes up Goes down
C) Stays the same
D) Can not determine
This is what you are saying:
1 Volt = 1 Amp
E = I (WRONG - "E" does not equal "I") This is not true!
This is what it is true...
1 Volt = 1 Amp through 1 Ohm
E = I x R
You awlays need to consider three factors: Voltage (Volts), Amps (Current), and Resistance (Load)
E does not Equal I
E equals I times R
You are not ready to move on to the LED and resistor yet.
Let's stay at this level until you totally understand what is going on. You are trying to remember something you learned many years ago.
It is the pieces that you remember and the pieces you forgot that is causing your confusion.
That is where I stumble! Drop 1.8 volts and there is 3.2 Volts left to get rid of ! (Out of 5 Volts)
Ok So If I don't put a Resistor, what's my AMP? Since a 0 is not possible for the formula:
If you put no resistor, just a piece of wire, then R is very small. For a piece of copper wire, it might be 0.01. It is never zero (unless you have a superconductor!). So plug 0.01 into the formula E = I x R and you get hundreds of amps. Pfft, there goes the led!
If you leave the resistor out, you might think that R is infinite, but it isn't really because objects, things, your bench, even air does have a real resistance value, even if it is in the hundreds of megohms. So a tiny bit of current will flow.
I'll say resistance is 1 ohm so I=E/R where 3.2/1 3.2 amps?
Yes, circuits using 1 ohm resistors are rather interesting because as you say, volts = amps. Next time you are in an electronics store, see if you can buy a 5 watt 1 ohm resistor. I use mine all the time as it is a quick way to measure currents. If you have 3 amps flowing, there will be 3V across that resistor.
That is telling me that AMP is equal to Voltage !! That can't be..is it?
Yes it is but only because your resistor happened to be 1 ohm. It wouldn't be true if your resistor was 10 ohms. Then the current would be one tenth, or 0.32 amps (320mA).
And of course, your led would have gone pfft long ago with 3.2 amps. But this would apply fine to other components that draw more current, eg solenoids, motors, lightbulbs.
Since I have the ultimate PSupply, it can push 5Volts and ulimited amps (just for the fun of it) still my circuit will draw only 3.2 amps!?
By jove, you have got it! That is correct - your 5V supply, unlimited amps, and still it only draws 3.2 amps.
The only caveat is that the LED will have died, and when leds die they go open circuit.
Maybe do all your recalculations with a resistor 1000 times more. So instead of 1 ohm, use 1000 ohms (1k). And So now the current will be 3.2ma (0.0032A). And the led will be happy.
Hi All,
Wow so much to read. Just amazing the kind of response ! Just fantastic.
"
Ok, say you have a voltage and some current in a circuit (it does not matter how much of each in this example). You leave the voltage the same and then you INCREASE the RESISTANCE. What does the current do?
A) Goes up Goes down
C) Stays the same
D) Can not determine
Paul
"
B- goes down!
Pressure : Voltage
Flow : Current
Blockage : Resistance
Water : ?
A) Voltage Wire
C) Electrons
D) None of the Above
C - Electron
**"Current is "like" flow (think of it as kinetic energy it that helps)"**
Ok that I got it! That seems to trigger my last cell to calm down.
Here is my way of understanding it!
Like a SHIP on a river.
Voltage (Propeller)
Resistance (River - which water moves(flow) downstream)
Current (Actual ground speed)
So the more power to the prop, with the same state of the river, the more ground(Real) speed (amp) !
And if the river has more speed for the same power to the prop, the less ground speed (Ship fighting the faster movement of the water)
For me ,that analogy has more "tangible" elements...
So really , the AMP is the actual RESULT of the relation from "Prop and the river"
Please say yes to the above statement...! !!!!!
I know LED is maybe not the best first teaching of the Volt/Ohm/amp arena, but for the price and physically see a LED burn out, helps a lot to have a first hand LAB testing and experimentation!
Don't tell me that after 30 ++ years , that I can actually learn something?
If so, then I can go back to square one, and do stuff properly and understand the why and how.. and surely many PFFFT and ooops!...and maybe one "%$#@*&"
Ok, say you have a voltage and some current in a circuit (it does not matter how much of each in this example). You leave the voltage the same and then you INCREASE the RESISTANCE. What does the current do?
A) Goes up Goes down
C) Stays the same
D) Can not determine
B- goes down! CORRECT!
Pressure : Voltage
Flow : Current
Blockage : Resistance
Water : ?
A) Voltage Wire
C) Electrons
D) None of the Above
C - Electron CORRECT!
Ok you like the ship on the river....
Voltage (Propeller)
Resistance (River - which water moves(flow) downstream)
Current (Actual ground speed)
So the more power to the prop, with the same state of the river, the more ground(Real) speed (amp) !
Good, You are getting there!
And if the river has more speed for the same power to the prop, the less ground speed (Ship fighting the faster movement of the water)
For me ,that analogy has more "tangible" elements...
So really , the AMP is the actual RESULT of the relation from "Prop and the river"
Please say yes to the above statement...! !!!!!
YES!!!!!!! The AMPs ARE the RESULT from Pressure and Resistance..... Great Job!
Now consider this.....
So, an LED by itself has little resistance (in the foward direction - don't worry about what forward direction means right now) so if you hook up a battery (don't care about the voltge just yet - it is some value above 0V), will the LED draw a tiny bit of current or a lot of current?
Think about it... it has LOW RESISTANCE so it will draw _______ CURRENT.
Now lets say you have 5V connected to the LED. And I am telling you that the LED has a resistance of 2 ohms in this condition (You would normally find this out from the LED Data Sheet but in this case I AM YOUR DATA SHEET). Don't worry about the voltage across the Diode just yet - let's get to this point first.
Ok, 5V across 2 Ohms.... How much current is flowing in the circuit. You know the formula and what it means now.
Now I am telling you (AS YOUR DATA SHEET) that the LED can handle only 20 mA of current before it will "burn out" (go permenantly dark as you like).
So, is it OK to connect 5V directly to your LED?
Since the LED is acting like 2 ohms, is that more or less than 20 mA?
Well, I will tell you the answer because I think you now know......
Current = 5V across 2 ohms
I = E/R
2.5 Amps of current will flow in this circuit (2500 mA)
Your LED is rated to handle only 20 mA of current.
What will happen to your LED in this circuit?
Hint (it may happen so fast that you may never even see the LED light up)
If you totally understand this, then we will move on.
Getting back to and old comment you made.
You said that everyone in your class failed, that is rare, what was the common denomiator? The teacher! So from now on I do not want to hear about you losing brain cells [smile], just forget about what you were taught about electronics in the past. You have a lot of great teachers here right now. We will get you on track. You are already starting to make progress. We see it. Keep going.
Holy bukaroo!
Now I can go back to the What's a Microcontroler, and redo (for the 3rd time) and understand What I'm doing.
It is easy to follow formula and rules, but without understanding all elements involved, it does not make the process helping you to grow!
One thing I remember, in the class (which we cover very quickly- electricity) when the teacher ask: "Is it ok to have extra AMP", we all said, well more like yell "YES".... his face and body language really showed BIG Disapointment! Well then that was the end of that course!...
I can easily say that now I do grasp it all, but I may and will make mistakes, but I'll be able to go back and go --- Smile! my fault! I know why!
I'll be good with my last cell! ;-)
Many thanks to you Paul, Dr_Acula and others... Especially Jim that took the "bull by the horn" or the "noob by the hand" (lol) and guided me back to the beginning and press that REBOOT button .. took a risk there ;-)
I sincerely apologize for some snappy comments like "RTFM" , no excuse for that, but frustration took over and did not help to keep a conversation opened!
For the next beginner, I hope that I will have the same patience as many of you had towards me, and that I can somehow give help the same way I just received!
First thing I did coming back from work was to get to "The Source" store, and try to find some parts. They don't have much anymore. They replaced Radio Shack, and they are now more like Profile electronic, TV phone etc...
We do have some stores (sparse and few in between) that may supply PARTS!
Hope I can build on my NEW ACQUIRED current (Pund intented) knowledge !!!
it is easy to use, parts ship in a day and you get them in about 3 or four days (unless you want to waste money and get them overnight). You can find almost anything electronic: from resistors to test equipment. Do you have a Digital Multi Meter (DMM)? You will need one for simple troubleshooting. A decient one should cost about $40 - Radio Shack has some inexpensive ones. A good DMM (like a Fluke) will cost you 100's of dollars (you don't need that precision yet).
Once you get into electronics, you could buy an Oscillocsope (you can get used ones on E-bay), they are expensive but you can see what the signals are doing graphically. The ones I use at work cost about $185,000, the one I have at home is cheap, old and clunky but it works.
Parallax has some nifty parts that Parallax knows will work well with the Stamps. So check out the Store!!!!
Don't RESIST, order so some parts today! Its "Pun Central" around here.
guys i am learning more here thant on the books thanks.... my question: base on the idea thatthe energy its tranformed. the 3v or5v that the resistor "take" its tranformed to heat? thankssss
Yes jvrproductions, any voltage drop across a resistor is converted to heat. There are some useful formulas:
1) V = I x R. Ohms law.
2) W = I x V Watt's law ie Watts = Current times Volts
Watts measures heat, and any time you use a resistor you need to also calculate how much heat it might produce. If you use a resistor that is too small, it will start smoking and get hot and eventually burn out.
Ballpark figures - 1 watt in a little resistor will be getting pretty warm to touch. Mostly you don't want this sort of heat being wasted anyway, especially if the circuit is being run on batteries, and usually heat produced by resistors is measured in milliwatts or lower. Nothing beats grabbing a few resistors and seeing which ones get hot!
So let's say you have a 1V supply and you put a 1 ohm resistor across that supply. V = I x R, so rearranging that is I = V/R. So current is 1 amp.
Now Watts = I x V, and we know current is 1A, and volts is 1V, so that is 1 watt. A 1 watt resistor is a bit larger than the standard size 1/4 watt resistor, so this says you would need to use a larger resistor. In fact, I'd probably put in a 5 watt resistor to make sure.
In general terms with the sorts of voltages available on the typical lab bench, you don't have to worry much about the wattage of a resistor unless the value is under about 1k.
Those two formulas above are the only two you need to calculate watts, but they can be combined to one formula if you like (though then you have to remember three formulas!)
If V=IxR and W=IxV then you can substitute the first equation into the second and get W = I x I x R, or W = I^2R
Yes, all of our (the world's) IQ points have gone up quite a few points since the proliferation of the internet and the world wide web. I don't think it is cheating to use a web formula to figure something out - but doing it on a test is is another thing. But you are doing the right thing by coming here for advice.
Check out some of my slides posted earlier in this thread. They may help as well.
Comments
The 1 amp value is the maximum the ps "can supply" under nominal operating conditions. If you have nothing attached to the output, i.e. infinite resistance, no current will flow. if you short circuit the output, i.e. zero ohms, it might supply more than 1 amp... till something burns out! :-)
In a series circuit, the current is the same everywhere in that circuit, that is, the current that all your combined components will draw.
Setting up your example... with a 5 volt PS rated at 3 amps, an LED, and a 470 ohm resistor, the circuit drew 6.32 milliamps. The LED dropped 1.97 volts and the resistor dropped 3.03 volts.
Changing the resistor to 220 ohms, the circuit now drew 12.96 milliamps. The LED dropped 2.02 volts and the resistor dropped 2.98 volts.
Lastly with a 100 ohm resistor the circuit drew 28.9 milliamps. The LED dropped 2.11 volts and the resistor dropped 2.88 volts.
Going to a 50 ohm resistor would probably drive the current draw to over 50 milliamps... above the typical LED current max and cause the LED to go into permanent dark mode. :-|
HTH,
Amanda
Take a look at this slide. If it does not make sense to you, then we will have to step you backwards a little.
After you study this, replace the "air" with a 1K Ohm resistor and measure the voltage across R1. Can you predict what you will read for voltage and current before you do this?
Paul
Your supply puts out 10V. You attach a 1K ohm resistor across it. You measure 10V across the resistor.
10V
1K
GND
Now you remove the 1K and replace it with two resistors of 500 ohms each in series (end to end.) You measure at the same place you measured the 1K resistor. You still have 10V.
10V
500 ohm
x
500 ohm GND
Now you measure from ground to the connection between the two resistors (x). You measure 5V.
Now if you measure from between the two (x) and the positive terminal of the battery (where the 10V is...) you measure 5V.
The extra voltage has not gone anywhere. There is 5V across each of the 500 ohm resistors and still total up to the 10V the supply is putting out. Both 500 ohm resistors act as (and add up to) 1K ohms. The current flow is the same in both circuits because the total resistance in each circuit is 1K ohms.
Resistors literally resist the flow of electricity. It's like they cause the electrons to back-up a bit. This back-up of electrons causes the voltage drop across the resistor.
If you only have one resistor as in the first diagram, you'll always read what the supply is putting out. Once you start dividing it up, you get different measurements according to the amount of resistance there is between the two points you're measuring from. It could be across one resistor that's in the middle of a whole circuit or from GND to some point in the middle of a circuit or whatever.
As far as current is concerned. In both examples, you'll have 10 mA flowing. If you increase the voltage to 20V. you'll have 20 mA flowing. The ability of the resistor to "resist" the flow of current hasn't changed. But, you're forcing more current through them because you've increased the voltage potential....(you're going to hit the ground harder if you jump from 20 feet up rather than 10 feet up.) Of course, now that you're applying 20V to the circuit, you'll have 10V across each of the 500 ohm resistors.
LEDs are different as they are silicone based. They basically operate on a principle that current flows once the voltage reaches a certain potential...1.2V...1.8V..whatever...depends on the LED. It's a gate that opens up under the right amount of pressure. That voltage tends to remain the same because the more you shove into it the more current the LED lets through. But, put too much through it and it burns up.
A dam will stop water. But as it rains and the amount of water behind the dam increases....eventually the water spills over the dam. the height of the dam doesn't change (like an LED will drop 1.2V all the time.), but, the amount of water flowing over the dam does. put too much water behind the dam and the dam will burst.
So you 'limit' the amount of current that can flow into an LED with a resistor.
Is that helping at all?
Hi Paul,
First thing first! missing how much current this circuit draws ! So I = E/R x= 9 / (1000+1000) = .0045 (4.5 MA)
Now to get the voltage at R1, then E=IR x= .0045 X 1000 = 4.5 Volts.
Which was predictable, since we have 2 resistor of the same value!
So far so good I hope.
To: ajward
Love the statement "Permanent dark mode" I have a collection of them. Might make a lot of money on Ebay!
But a bit confused with the LED dropping more voltage as the resistor changes! Is this a normal thing?
To: Catspaw
I'm like a the speed of sound, just about to through..but missing that little extra power to go through!
Ok, i grasp the Voltage drop for the LED etc.
"LEDs are different as they are silicone based. They basically operate on a principle that current flows once the voltage reaches a certain potential...1.2V...1.8V..whatever...depends on the LED. It's a gate that opens up under the right amount of pressure. That voltage tends to remain the same because the more you shove into it the more current the LED lets through. But, put too much through it and it burns up"
That is where I stumble! Drop 1.8 volts and there is 3.2 Volts left to get rid of ! (Out of 5 Volts)
Ok So If I don't put a Resistor, what's my AMP? Since a 0 is not possible for the formula:
I'll say resistance is 1 ohm so I=E/R where 3.2/1 3.2 amps?
That is telling me that AMP is equal to Voltage !! That can't be..is it?
Since I have the ultimate PSupply, it can push 5Volts and ulimited amps (just for the fun of it) still my circuit will draw only 3.2 amps!?
Arrrrgh, that cell went pouffff, I got one left ...!
Cheers
Yosh
When assuming no resistance you can't use zero (there is no such thing a pure 0.000... ohms in nature). And don't use a unity value (1).
Try using 0.002 ohms (2 milli Ohm) to represent 0 ohms. Notice in my slides, there is no such thing as a perfect 0 or perfect infinity - use very low and very high values to substitue). Now try again. Also, try it with a 1K and a 2K (3K total) in series instead and see what happens; the current will drop and voltage across each resistor will be different.
So you say you don't understand how much current the circuit draws? Your math is correct!
A circuit with two 1K ohm resistors connected in _series_ results in 2K ohms total. Apply 9V of electrical pressure through that restriction and only 45 mA of current will flow. That is right, you get it from what I see. Good.
I = E / R
= 9 / 2000
= 4.5 mA
Ok, so in this circuit 4.5 mA worth of current (electrons) is flowing. No matter where you hook up your ammeter (between the battery and a resistoor, or between the two resistors for example) 4.5 mA is flowing - current at every point in the circuit it is the same. Rremember, your ammeter looks like a short so it must be connected in series as well.
Use the "Water Anology" to visualize the Current aspect:
You have a hose filled with water. At one end is the valve (your battery or power supply if you will).
At the other end is your thumb over the hole.
You turn on the (valve) water pressure (Voltage).
Your thumb is totally blocking the end of the hose (this is very high Resistance)
You feel the pressure on your thumb (the Voltage)
But no water is flowing (no Current)
Now take your thumb slightly off the end of the hose (you just lowered the Resistance)
You feel less pressure on your thumb (lower Voltage)
But water is now comming out of the hose (Current is flowing)
No matter you look in the hose, the same amount of water is passing by any point as in any other point (Current is the same everyehere in the circuit).
Now take your thumb off the end of the hose completely (Resistance is now very low)
You feel no pressure on your thumb, but a lot of water is flowing (a lot of Current).
Keeping your thumb off the hose, you close the valve a little (lower the Voltage)
Less water is now flowing out of the hose (less Current because less Voltage)
Quantity is the amount of Pressure and Flow to fill a bucket. Power = Voltage x Current.
I does not matter if your are counting electrons, water molecues, pennies, beans, whatever - the math and the concepts are essentially the same.
This is how Mr. Ohm (and others) realted to this attribute of nature.
Are you pickin up what i am puttin down (do you undrstand)?
I think you understand this more than you think.
What time zone are you in? It is hard to teach this concept using e-mail (I can't see your expressions in real time). We may have to Skipe to get the ideas to you.
Paul
Here is another slide that may help.
Paul
The point I'm trying hard to understand... is "Keeping your thumb off the hose, you close the valve a little (lower the Voltage)
Less water is now flowing out of the hose (less Current because less Voltage)"
The relation of Voltage and Current. Is 1 volt is equal to 1 amp ?
Like the water hose, to me it's the voltage (main valve) that pushes the water. Less water less voltage. AT THE SAME TIME, less current too!
WHich in any examples or exercises, is not give me the relation. Like the more SUN, the brighter...
I don't get the current! Especially -> Resistor resist the current! Yet I put it in series with a LED, because that 3.2 volts extra is too much!
I reduced the current to reduce the Voltage? So it's a 1 to 1 relation ?
Cheers
Yoshi
Time Zone is -5 (New York/Montreal time) Right now , at the office..hum hum...! ;-)
It is... well, more to the point, as the current changes. The resistor value affects the circuit current and that's what the LED sees.
The "2 volt drop" is an average typical of LEDs.
The forward voltage (the drop) varies with the current flowing through the LED. That value can also vary based on the color of the the device.
This is a forward current vs. forward voltage chart for three Fairchild LEDs (Red. Yellow and Green).
You can see that my test values kind of follow the curve in the chart above, although at 28.9 mA the drop is less than expected, but I hope you get the idea. I don't know who made my yellow LED, perhaps my test is correct!?
Anyhow, HTH!
Amanda
That is right! The current is less because the voltage is less! I think you understand. Good!
If you have a given voltage and a given current is flowing, and then you lower the voltage, then less current will flow.
Voltage and Current are PROPORTIONAL (that is they are related and if one increases the other increases) but they are NOT THE SAME.
But Current and Resistance are INVERSERLY PROPORTIONAL (again, related but as one increases and the other decreases) but NOT THE SAME.
Voltage is "like" pressure (think of it a potential energy if that helps)
Current is "like" flow (think of it as kinetic energy it that helps)
Resistance is "like" a partial blockage (think of it as matter if that helps)
Ok, say you have a voltage and some current in a circuit (it does not matter how much of each in this example). You leave the voltage the same and then you INCREASE the RESISTANCE. What does the current do?
A) Goes up
C) Stays the same
D) Can not determine
Paul
Right! Voltage is the pressure, but the WATER is NOT THE VOLTAGE.... WATER IS THE ELECTRONS...
Pressure is a FORCE acting on the WATER.
You are getting there!
Once you master this concept we will move on.
Water:
The PRESSURE pushes the WATERmeolecules through the hose. The water molecules flowing is the current.
Electrons:
The VOLTAGE pushes the ELECTRONS through the wire. The electrons flowing is the current.
Question:
Pressure : Voltage
Flow : Current
Blockage : Resistance
Water : ?
A) Voltage
C) Electrons
D) None of the Above
Paul
One more point. You said that 1 Volt = 1 Amp.
That is not totally complete!
This is what you are saying:
1 Volt = 1 Amp
E = I (WRONG - "E" does not equal "I") This is not true!
This is what it is true...
1 Volt = 1 Amp through 1 Ohm
E = I x R
You awlays need to consider three factors: Voltage (Volts), Amps (Current), and Resistance (Load)
E does not Equal I
E equals I times R
You are not ready to move on to the LED and resistor yet.
Let's stay at this level until you totally understand what is going on. You are trying to remember something you learned many years ago.
It is the pieces that you remember and the pieces you forgot that is causing your confusion.
Take it slow, you will get there. Ok?
Paul
Paul
If you put no resistor, just a piece of wire, then R is very small. For a piece of copper wire, it might be 0.01. It is never zero (unless you have a superconductor!). So plug 0.01 into the formula E = I x R and you get hundreds of amps. Pfft, there goes the led!
If you leave the resistor out, you might think that R is infinite, but it isn't really because objects, things, your bench, even air does have a real resistance value, even if it is in the hundreds of megohms. So a tiny bit of current will flow.
Yes, circuits using 1 ohm resistors are rather interesting because as you say, volts = amps. Next time you are in an electronics store, see if you can buy a 5 watt 1 ohm resistor. I use mine all the time as it is a quick way to measure currents. If you have 3 amps flowing, there will be 3V across that resistor.
Yes it is but only because your resistor happened to be 1 ohm. It wouldn't be true if your resistor was 10 ohms. Then the current would be one tenth, or 0.32 amps (320mA).
And of course, your led would have gone pfft long ago with 3.2 amps. But this would apply fine to other components that draw more current, eg solenoids, motors, lightbulbs.
By jove, you have got it! That is correct - your 5V supply, unlimited amps, and still it only draws 3.2 amps.
The only caveat is that the LED will have died, and when leds die they go open circuit.
Maybe do all your recalculations with a resistor 1000 times more. So instead of 1 ohm, use 1000 ohms (1k). And So now the current will be 3.2ma (0.0032A). And the led will be happy.
Wow so much to read. Just amazing the kind of response ! Just fantastic.
"
Ok, say you have a voltage and some current in a circuit (it does not matter how much of each in this example). You leave the voltage the same and then you INCREASE the RESISTANCE. What does the current do?
A) Goes up
C) Stays the same
D) Can not determine
Paul
"
B- goes down!
Pressure : Voltage
Flow : Current
Blockage : Resistance
Water : ?
A) Voltage
C) Electrons
D) None of the Above
C - Electron
**"Current is "like" flow (think of it as kinetic energy it that helps)"**
Ok that I got it! That seems to trigger my last cell to calm down.
Here is my way of understanding it!
Like a SHIP on a river.
Voltage (Propeller)
Resistance (River - which water moves(flow) downstream)
Current (Actual ground speed)
So the more power to the prop, with the same state of the river, the more ground(Real) speed (amp) !
And if the river has more speed for the same power to the prop, the less ground speed (Ship fighting the faster movement of the water)
For me ,that analogy has more "tangible" elements...
So really , the AMP is the actual RESULT of the relation from "Prop and the river"
Please say yes to the above statement...! !!!!!
I know LED is maybe not the best first teaching of the Volt/Ohm/amp arena, but for the price and physically see a LED burn out, helps a lot to have a first hand LAB testing and experimentation!
Don't tell me that after 30 ++ years , that I can actually learn something?
If so, then I can go back to square one, and do stuff properly and understand the why and how.. and surely many PFFFT and ooops!...and maybe one "%$#@*&"
Cheers
Yosh
V = IR
I = V/R
If the resistance increases the current decreases.
A) Goes up
C) Stays the same
D) Can not determine
B- goes down! CORRECT!
Pressure : Voltage
Flow : Current
Blockage : Resistance
Water : ?
A) Voltage
C) Electrons
D) None of the Above
C - Electron CORRECT!
Ok you like the ship on the river....
Voltage (Propeller)
Resistance (River - which water moves(flow) downstream)
Current (Actual ground speed)
So the more power to the prop, with the same state of the river, the more ground(Real) speed (amp) !
Good, You are getting there!
And if the river has more speed for the same power to the prop, the less ground speed (Ship fighting the faster movement of the water)
For me ,that analogy has more "tangible" elements...
So really , the AMP is the actual RESULT of the relation from "Prop and the river"
Please say yes to the above statement...! !!!!!
YES!!!!!!! The AMPs ARE the RESULT from Pressure and Resistance..... Great Job!
Now consider this.....
So, an LED by itself has little resistance (in the foward direction - don't worry about what forward direction means right now) so if you hook up a battery (don't care about the voltge just yet - it is some value above 0V), will the LED draw a tiny bit of current or a lot of current?
Think about it... it has LOW RESISTANCE so it will draw _______ CURRENT.
Now lets say you have 5V connected to the LED. And I am telling you that the LED has a resistance of 2 ohms in this condition (You would normally find this out from the LED Data Sheet but in this case I AM YOUR DATA SHEET). Don't worry about the voltage across the Diode just yet - let's get to this point first.
Ok, 5V across 2 Ohms.... How much current is flowing in the circuit. You know the formula and what it means now.
Now I am telling you (AS YOUR DATA SHEET) that the LED can handle only 20 mA of current before it will "burn out" (go permenantly dark as you like).
So, is it OK to connect 5V directly to your LED?
Since the LED is acting like 2 ohms, is that more or less than 20 mA?
Well, I will tell you the answer because I think you now know......
Current = 5V across 2 ohms
I = E/R
2.5 Amps of current will flow in this circuit (2500 mA)
Your LED is rated to handle only 20 mA of current.
What will happen to your LED in this circuit?
Hint (it may happen so fast that you may never even see the LED light up)
If you totally understand this, then we will move on.
Getting back to and old comment you made.
You said that everyone in your class failed, that is rare, what was the common denomiator? The teacher! So from now on I do not want to hear about you losing brain cells [smile], just forget about what you were taught about electronics in the past. You have a lot of great teachers here right now. We will get you on track. You are already starting to make progress. We see it. Keep going.
Paul
Holy bukaroo!
Now I can go back to the What's a Microcontroler, and redo (for the 3rd time) and understand What I'm doing.
It is easy to follow formula and rules, but without understanding all elements involved, it does not make the process helping you to grow!
One thing I remember, in the class (which we cover very quickly- electricity) when the teacher ask: "Is it ok to have extra AMP", we all said, well more like yell "YES".... his face and body language really showed BIG Disapointment! Well then that was the end of that course!...
I can easily say that now I do grasp it all, but I may and will make mistakes, but I'll be able to go back and go --- Smile! my fault! I know why!
I'll be good with my last cell! ;-)
Many thanks to you Paul, Dr_Acula and others... Especially Jim that took the "bull by the horn" or the "noob by the hand" (lol) and guided me back to the beginning and press that REBOOT button .. took a risk there ;-)
I sincerely apologize for some snappy comments like "RTFM" , no excuse for that, but frustration took over and did not help to keep a conversation opened!
For the next beginner, I hope that I will have the same patience as many of you had towards me, and that I can somehow give help the same way I just received!
Simply AWESOME !
THANKS !
Cheers
Yosh
You made my day! It is always refreshing to see someone get the "Ah Ha!" moment.
Yes, "Build on the work of others... go forth and share your knowlege"
Go explore....
Paul
First thing I did coming back from work was to get to "The Source" store, and try to find some parts. They don't have much anymore. They replaced Radio Shack, and they are now more like Profile electronic, TV phone etc...
We do have some stores (sparse and few in between) that may supply PARTS!
Hope I can build on my NEW ACQUIRED current (Pund intented) knowledge !!!
Cheers
Yosh
Try Digi-Key.
www.digikey.com
it is easy to use, parts ship in a day and you get them in about 3 or four days (unless you want to waste money and get them overnight). You can find almost anything electronic: from resistors to test equipment. Do you have a Digital Multi Meter (DMM)? You will need one for simple troubleshooting. A decient one should cost about $40 - Radio Shack has some inexpensive ones. A good DMM (like a Fluke) will cost you 100's of dollars (you don't need that precision yet).
Once you get into electronics, you could buy an Oscillocsope (you can get used ones on E-bay), they are expensive but you can see what the signals are doing graphically. The ones I use at work cost about $185,000, the one I have at home is cheap, old and clunky but it works.
Parallax has some nifty parts that Parallax knows will work well with the Stamps. So check out the Store!!!!
Don't RESIST, order so some parts today! Its "Pun Central" around here.
Paul
1) V = I x R. Ohms law.
2) W = I x V Watt's law ie Watts = Current times Volts
Watts measures heat, and any time you use a resistor you need to also calculate how much heat it might produce. If you use a resistor that is too small, it will start smoking and get hot and eventually burn out.
Ballpark figures - 1 watt in a little resistor will be getting pretty warm to touch. Mostly you don't want this sort of heat being wasted anyway, especially if the circuit is being run on batteries, and usually heat produced by resistors is measured in milliwatts or lower. Nothing beats grabbing a few resistors and seeing which ones get hot!
So let's say you have a 1V supply and you put a 1 ohm resistor across that supply. V = I x R, so rearranging that is I = V/R. So current is 1 amp.
Now Watts = I x V, and we know current is 1A, and volts is 1V, so that is 1 watt. A 1 watt resistor is a bit larger than the standard size 1/4 watt resistor, so this says you would need to use a larger resistor. In fact, I'd probably put in a 5 watt resistor to make sure.
In general terms with the sorts of voltages available on the typical lab bench, you don't have to worry much about the wattage of a resistor unless the value is under about 1k.
Those two formulas above are the only two you need to calculate watts, but they can be combined to one formula if you like (though then you have to remember three formulas!)
If V=IxR and W=IxV then you can substitute the first equation into the second and get W = I x I x R, or W = I^2R
And of course, these days you can cheat and not know any formulas and just go to the an online calculator on the internet and plug in the values! http://www.crownaudio.com/apps_htm/designtools/ohms-law.htm
Yes, all of our (the world's) IQ points have gone up quite a few points since the proliferation of the internet and the world wide web. I don't think it is cheating to use a web formula to figure something out - but doing it on a test is is another thing. But you are doing the right thing by coming here for advice.
Check out some of my slides posted earlier in this thread. They may help as well.
Paul