# A noob resistance question

Maybe if running at more than half the maximum current one LED will have too much current and burn out a lot quicker.

Yeah, that's the baddie. Looking at this d/s, there is 33% tolerance on forward voltage, could be bad.

http://www.surplustronics.co.nz/library/Mega-White-LED.pdf

Grumpy_Mike, Telecommando, and oric_dan(333),

Thank you all for your help. I would like to say that I believe you have helped me to understand this subject better. I have read and reread your responses, followed your links, and bit by bit I THINK I may be getting a handle on this.

I know I can go to different websites and read what other people have done to set up, let's say an RTC, and I can mimic what they have done but I really want to understand WHY you put a capacitor here or need a resistor there or why the oscillator should be 32Mhz (forgive the capitalized M if it is incorrect please) instead of 16Mhz. Is there a good resource for electronic components that explains 1)this is what it is 2) this is what it does 3) this is how it works and 4) this is why you would want to use it? Not a cookbook as much as a compendium of ingredients and a cooks insight on how you might want to combine them.

32Mhz (forgive the capitalized M if it is incorrect please) instead of 16Mhz

Yep capitalisation is correct. :)

It is tricky because all that knowledge often comes down to simple experience, and knowing what the particular circumstances are. I usually say that given N engineers there will always be at least N+1 strongly held opinions about the correct design decisions.

That is why there are engineers who produce good designs and those who produce not so good ones. There is such a lot to know and I think the trick is knowing just how much to take account of.

I have had many engineers work under me and they would all produce different designs but some were better than others. They all knew the same stuff but how they weighed it in the balance would be different. At some levels it is as much an art as a science.

I was about to start a thread on this very topic, but apparently it has been asked before! :)

1. When performing Ohm's Law calculations, what specifies the values for V and I? V, to me, should be the supply voltage (3.3V or 5V would be what I'm using in this case). And should I be the maximum allowable current to pull from the pin? Just wanting to make sure... :)

2. What is the maximum allowable current draw for the 5V pin on an Arduino Uno? From this page, I see 50mA for the 3.3V pin, but nothing is listed for the 5V pin.

3. Can one model a switch/pushbutton similarly to an LED? I ask this because a switch also has "infinite" resistance when not activated, but near 0 resistance when pressed.

At some levels it is as much an art as a science.

Except artist aren't generally held to target costs and time to market goals. Hard to shine when the clouds of management call the shots. ;)

Lefty

anorton: I was about to start a thread on this very topic, but apparently it has been asked before! :)

1. When performing Ohm's Law calculations, what specifies the values for V and I? V, to me, should be the supply voltage (3.3V or 5V would be what I'm using in this case). And should I be the maximum allowable current to pull from the pin? Just wanting to make sure... :)

40ma is absolute maximum, but 20-30 ma is a better maximum limit standard to stick to. Keep in mind due to internal resistance of the output pins the output voltage will sag as current draw from the output pin is increased. Just shows that there is a difference from fundamental principal and electrical reality of real world components. That is why the datasheet for devices are so important to study and understand a chip limits and capabilities.

1. What is the maximum allowable current draw for the 5V pin on an Arduino Uno? From this page, I see 50mA for the 3.3V pin, but nothing is listed for the 5V pin.

Because there is no exact answer. If the board is being powered via USB then there is a 500ma thermofuse that sets a maximum limit for the board and anything you wire up to it. But if powered from an external DC voltage source then there might be somewhat more but it depends on the exact DC voltage input as there is a heat dissipation limit for the on-board 5vdc voltage regulator. So hard answers are desirable but often allusive in the real world.

1. Can one model a switch/pushbutton similarly to an LED? I ask this because a switch also has "infinite" resistance when not activated, but near 0 resistance when pressed.

No, totally different components, and a switch doesn't share all the characteristic of a switch and visa versa.

retrolefty: 40ma is absolute maximum, but 20-30 ma is a better maximum limit standard to stick to. Keep in mind due to internal resistance of the output pins the output voltage will sag as current draw from the output pin is increased. Just shows that there is a difference from fundamental principal and electrical reality of real world components. That is why the datasheet for devices are so important to study and understand a chip limits and capabilities.

Ok.

Because there is no exact answer. If the board is being powered via USB then there is a 500ma thermofuse that sets a maximum limit for the board and anything you wire up to it. But if powered from an external DC voltage source then there might be somewhat more but it depends on the exact DC voltage input as there is a heat dissipation limit for the on-board 5vdc voltage regulator. So hard answers are desirable but often allusive in the real world.

Hmmm... Ok. Where do I find the information for the on-board 5vdc regulator?

No, totally different components, and a switch doesn't share all the characteristic of a switch and visa versa.

:-\ Just when I thought I found a shortcut... ;) Ok.

When performing Ohm's Law calculations, what specifies the values for V and I? V, to me, should be the supply voltage (3.3V or 5V would be what I'm using in this case).

No, V is actually the voltage "drop" across the resistor. Eg, if 1K is tied between 5V and 3.3V, the current I = Vdrop / R = (5V - 3.3V) / 1K = 1.7 mA, and not 5 mA or 3.3 mA.

oric_dan(333):

When performing Ohm's Law calculations, what specifies the values for V and I? V, to me, should be the supply voltage (3.3V or 5V would be what I'm using in this case).

No, V is actually the voltage "drop" across the resistor. Eg, if 1K is tied between 5V and 3.3V, the current I = Vdrop / R = (5V - 3.3V) / 1K = 1.7 mA, and not 5 mA or 3.3 mA.

In your example, you mean a 1K(ohm) resistor, right? (just making sure--I'm really new, if you haven't figured it out yet. :blush:)

So, in this LED (for example), it says "forward voltage drop" is 1.8V-2.2V (DC). Let's say I have a circuit that has 5V---LED---1K(ohm)---Gnd. I want to find the current. I = Vdrop / R = 2.2V / 1K = 2.2mA.

Right? Or am I still confused... :/

No with a forward voltage drop on the LED of 2.2V the drop across the resistor is 5 - 2.2 = 2.7 V You can not use ohms law for an LED it does not work because it is a non linear device. Only the resistor.

32Mhz (forgive the capitalized M if it is incorrect please)

The M's right, but the h's not.... Hertz's symbol is Hz, so you should write MHz

My tuppence worth…

I think what might not be becoming clear to messrs anorton and macweenie yet, is the notion that all of the voltage across the LED and its resistor has to be accounted for: the 5v in the 5V—LED—1K(ohm)—Gnd scenario has to disappear by the time we get to the right hand end.

So here’s how it all hangs together…

1. What’s the input voltage? We’re saying 5v, so we have that.
2. What’s the voltage we must account for over the LED? Manufacturer will tell us, let’s go with the 2.2V mentioned earlier
3. So what voltage must we disappear over the resistor? We started with 5, accounted for 2.2, so we have 2.8 left to drop over the resistor
4. How are we doing with Ohm’s Law for the resistor so far? Well we don’t know R, that’s what we need to calculate; we do know the voltage (2.8v); we don’t know the current…
5. But we do know that in a series circuit like this, the current through the LED has to go through the resistor: there’s no other path. And we do know (or manufacturer will tell us) that it’s 20mA, or 0.020A… this is the value to which we need to limit the current, the object of the exercise.
6. So now we can apply Herr Ohm to the resistor with R=V/I giving R=2.8/0.02 = 140 Ohm

EDIT:

This is all a consequence of Kirchoff’s Voltage Law which Wikipedia describes thusly:

The algebraic sum of the products of the resistances of the conductors and the currents in them in a closed loop is equal to the total emf available in that loop.

The products of the resistances of the conductors and the currents is simply the voltages (Ohm’s Law).

Do we have a loop though? The 5V—LED—1K(ohm)—Gnd doesn’t look like a loop… but it is if you think of the 5v and Gnd being eg the two terminals on a battery… then it would look like a loop: 5V—LED—1K(ohm)—Gnd—Battery—5v

So now we can see Kirchoff’s Voltage Law at work in our simple circuit…

1. Total EMF available in loop: 5V
2. Sum of the voltages in the loop: Vled + Vresistor = 2.2 + Vresistor
3. So 5 = 2.2 + Vresistor or Vresistor = 5 - 2.2

Does that help?

No, V is actually the voltage "drop" across the resistor. Eg, if 1K is tied between 5V and 3.3V, the current I = Vdrop / R = (5V - 3.3V) / 1K = 1.7 mA, and not 5 mA or 3.3 mA. In your example, you mean a 1K(ohm) resistor, right? (just making sure--I'm really new, if you haven't figured it out yet. )

So, in this LED (for example), it says "forward voltage drop" is 1.8V-2.2V (DC). Let's say I have a circuit that has 5V---LED---1K(ohm)---Gnd. I want to find the current. I = Vdrop / R = 2.2V / 1K = 2.2mA.

Right? Or am I still confused... :/

Yeah, the other guys are correct, and as I said you use "the voltage "drop" across the resistor" to get the current through it. Ohm's Law is for resistors.

JimboZA's post really cleared it up for me--I needed that background information that is probably really basic to a lot of people here... :)

Thanks, everyone, for all your help!

//Andrew

Thanks JimboZA,

Although I got a great deal of information before your post, you gave me the piece of the puzzle I was lacking. I hadn't understood just WHERE the current I needed to power, in this case, an LED was coming from. Having looked briefly at the datasheets for different chips and sensors I wasn't making the now obvious connection that the items current requirements was in there. I was under the assumption that 15mA (right?) was the rule of thumb for ALL LEDs, but now it's so obvious I could face-palm.

So using the following datasheet:

I would subtract 4V (maximum forward voltage) from the 5V the Arduino is putting out to get a drop voltage of 1V which using V = r * I would be 1V = r * 20mA or 1 / .2 = a 5 ? resistor? I combed the datasheet for any other values that looked right but theses were the only ones that seemed to fit the bill.

or 1 / .2 = a 5 ? resistor?

No .2 is 200 ma, you need to use .02 in your calculation.

Lefty

The moment I hit post I knew I screwed up, but the dog asked to go out and you caught it before I could fix it :blush:.

so it should be a 50? resistor... Do you know HOW I know I want to learn this? Because I keep on trying even though I have been wrong 100% of the time!

macweenie: The moment I hit post I knew I screwed up, but the dog asked to go out and you caught it before I could fix it :blush:.

Oh, the old "the dog eat my correct posting" excuse, go to the corner. ;)

so it should be a 50? resistor... Do you know HOW I know I want to learn this? Because I keep on trying even though I have been wrong 100% of the time!

Grumpy_Mike:
No with a forward voltage drop on the LED of 2.2V the drop across the resistor is 5 - 2.2 = 2.7 V

Are you sure?

macweenie:
Thanks JimboZA,

Although I got a great deal of information before your post, you gave me the piece of the puzzle I was lacking. I hadn’t understood just WHERE the current I needed to power, in this case, an LED was coming from.