Did I calculate resistors correctly?

Other Hardware General Electronics
1

Hello,
are this calculations correct?
1st picture:

2nd picture:

In both pictures I forgot to add toggle switch. I want leds to physically(not digitaly) indicate when the power source is on.
I have specific power sources:
-9V battery
-5V 5A power from buck converter

In both scenarios I'm going to use green leds, which input voltage should be around 2.2V, but I don't have proper resistors, so I need to work with what I have.

My main question is that, wouldn't this 5A power in the 2nd picture be harmfull for the led? Or led draw this amount of current that it needs (0.02 to be right)?

2

Current is not pushed from a power supply it is drawn by a component like an LED which takes what it needs.

In your 2nd picture the fact that LED may be drawing 0.02A doesn't mean that something else must be taking 4.98A. If the LED is the only thing in the circuit then the supply will only be delivering 0.02A. 5A is the maximum it can deliver not what it is always delivering.

You haven't shown any calculations for the resistors. And you say the LED voltage is 2.2V then on the pictures you use 2.4V and 2.6V so I have a feeling you're getting confused about something but 'm not sure exactly what.

Steve

3

I've read on the charts that green LED needs around 2.2V. The Voltage that I can achieve using the resistors that I have can be 2.4V adn 2.6V, I know that it's more than 2.2 and it will be the main cause of the shorter lifespan of the LED.
To calculate the value of the resistors I used the online calculator. I might look for different resistors in the future.

slipstick:
Current is not pushed from a power supply it is drawn by a component like an LED which takes what it needs.

In your 2nd picture the fact that LED may be drawing 0.02A doesn't mean that something else must be taking 4.98A. If the LED is the only thing in the circuit then the supply will only be delivering 0.02A. 5A is the maximum it can deliver not what it is always delivering.

Thank you, I was concerned about that.

4

This is quite elementary in order to use any online calculator...

Green LED usually need 1.8V, 5 or 20mA at maximum brigthness. By basic laws, total current usage of the circuit is sum of all component usage, in this case 5 or 20mA max, depending of LED.

However, LED cannot stand more voltage than specified without damage, then if use 9V power source and nothing else than one LED, (9-1.8V) = 7.2V need to be handled by resistor. For maximum brightness and 20mA LED, that would be 7.2V/0.02A = 360 Ohms. However, for a 5mA LED that would be 7.2/0.005 = 1.44K Ohms minimum.

And that is, resistor just limit needed current and total delivered power in this case is up to 5A is too much here. For example, you would be theoretically able to use 5A/0.02A = 250 LEDs rated 20mA in series with your 9V, 5A power supply with maximum brightness.

5

Use LED Resistor Calculator

You have your LEDs draw 20 mA. That's unnecessarily high. 5 mA is usually more than enough to light it up pretty brightly, for indicator LEDs 1-3 mA is often bright enough.

6

Lets see here. The LED determines the voltage drop, not you. You're probably calculating it in the wrong way.

If you use a LED that drops 2.2v, that's the voltage drop across it, not 2.4 or 2.6.

In your example you used 5v, with the 2.2 voltage drop of the LED, the resistor will drop 2.8v

Now, as you have determined the voltage drop across the resistor, you determine how much current you want to pass on the led.
Suppose you want to pass 0.01A, so, now you calculate with U=R.I
U = 2.8 (voltage drop across the resistor)
I = 0.01 A
So, you have R = 280 ohms, and it will draw 0.01A no matter what's the current limit of your source, as long as it's 5v.

7

Yes I knew you were confused about something. The LED voltage is fixed for a particular LED. You calculate the resistor to provide the amount of CURRENT you need (usually somewhere between 5mA and 20mA - 5mA is usually enough for most purposes).

If you use a resistor that is smaller than the calculator says you don't get more voltage across the LED (because that's fixed) what you get is more CURRENT through the LED than expected and that extra current is what will reduce the life of the LED. So when you do your calculation, if you don't have the correct size resistor use the next size UP not down. E.g. if calculator says 140, use 150 NOT 120. That gives a little less current and that's a lot safer.

Steve

8

Thank all of you for a quick and best replies.
Now I know that I choose the resistor to provide how much current will be placed on LED. The LED's drop is fixed depending on the type of it.

Both leds will be working as an indicators. Let's assume that the LED drop is 2.2V. So for the:
-1st circuit the resistor have to drop 9V-2.2V=6.8V
-2nd circuit the resistor have to drop 5V-2.2V=2.8V

Current that will pass the LED is 10mA maximum.

Using R=U/I
For the:
-1st R=680 Ohm
-2st R=280 Ohm

Also, I know that It's safer to use higher values of the calculated resistors. I guess I was sleeping on the physics classes :smiley:

9

Yes, you got it.
Nah, it's normal, we are used to think of electronics as fixed resistance, thus we provide different voltages.
LEDs are diodes, and they work in a different way.

10

Because of what I think is a bit of a curve for LED's - my calcs never quite came out as advertised - close but - so my solution for some of my projects was to put a meter in the circuit and adjust the resistance as needed - a note on the "adjustment" was to get around 18 to 19 ma for some projects as I wanted as bright as I could since the projects were used outside some times.

11

Whereever you have a circuit like diagram 2, with two circuits sharing a voltage source, you can treat each half of the circuit independently (assuming the voltage source isn't overloaded). Its just like all the lights in the house stay the same brightness when others are switched on and off.

12

saildude:
Because of what I think is a bit of a curve for LED's - my calcs never quite came out as advertised - close but

Of course. That's the difference between the theory and the real world.
The forward voltage of an LED is considered a fixed value, but in fact does increase slightly with increasing current.
There are slight difference between individual LEDs - especially so between production batches.
The resistor value is never exact - 5% tolerance is still common, though nowadays it's more and more 1%.
Your power supply voltage is not exact, there's a tolerance as well.

13

@tadashimori

Lets see here. The LED determines the voltage drop, not you. You're probably calculating it in the wrong way.

If you use a LED that draws 2.2v, that's the voltage drop across it, not 2.4 or 2.6.

In your example you used 5v, with the 2.2 voltage drop of the LED, the resistor will drop 2.8v

Now, as you have determined the voltage drop across the resistor, you determine how much current you want to pass on the led.
Suppose you want to pass 0.01A, so, now you calculate with U=R.I
U = 2.8 (voltage drop across the resistor)
I = 0.01 A
So, you have R = 280 ohms, and it will draw 0.01A no matter what's the current limit of your source, as long as it's 5v.

When discussing leds, you don't talk about a "voltage drop across the led".
Leds are NONLINEAR.
It is NOT a voltage drop per se , it a FORWARD VOLTAGE (Vf) , found on a datasheet.
"Voltage drop across ..." is an Ohm's Law expression used to describe LINEAR resistance, NOT
NON-LINEAR FORWARD VOLTAGE.
Technically , yes the voltage is "dropped" across the led but it is not described that way.
One would say "the forward voltage for the led is ____"
When you start talking about "voltage drop across" any newbie is going to think they can apply Ohm's
Law to a led. "
If they then see the schematic with 9V and a 330 ohm resistor going to a 2.4V led, they might think
"Well, let's see, the resistor has (9V-2.4V) = 6.6V across it so it is drawing 20mA , and the led has
2.4V across it and is drawing 20mA, so it must have a resistance of 2.4V/0.02A = 120 ohms.
But when they try to measure the led resistance they are not going to see 120 ohms because it is
NONLINEAR.
This is why we try to avoid using the term "voltage drop across" when discussing leds. Instead we say
"the led forward voltage is 2.4V".

If you use a LED that draws 2.2v

Ok, if we're trying to learn (or teach) electronics, we need to get our nomenclature correct.

The above statement is incorrect because you don't say a led "draws 2.2V", if you WERE going to say
something like that you would say a led "DROPS 2.2V" and even this is really not the correct way to
state this information.
The correct way to state this is :
"If a led has a forward voltage drop (Vf) of 2.2V..."

"draws .." is a term used to refer to CURRENT , NOT VOLTAGE.

When referring to voltage, one would say "DROPS" (NOT "DRAWS")

(If you're going to try to be a forum electronics consultant, at LEAST learn to "talk the talk".)

14

Sure, learn to talk the talk and let someone who is confusing voltage supply on led without understanding what I'm saying.

Meh, Thanks for informing all this, as English is not my native language it's hard to get the terminologies right, but I gotta say, you guys are way too agressive all the time. No need for that at all.
Didn't he understand what he wanted to learn?

raschemmel:
Leds are NONLINEAR.

I KNOW they are not linear, so what? Can't he just approximate to 2.2v? Or just tell him to see the datasheet and figure this?

raschemmel:
When you start talking about "voltage drop across" any newbie is going to think they can apply Ohm's
Law to a led. "

While I especifically said that he can't, nope, a newbie would not think that, only people who are really concerned about terminologies would. In fact, the author of the post thought he could and after reading my answer he figured out he can't.

raschemmel:
(If you're going to try to be a forum electronics consultant, at LEAST learn to "talk the talk".)

Not trying to be a consultant or anything. Never claimed to be. This is a forum full of hobbists. I started learning this about 2 months ago, and already faced several issues, so I can help as another learner as most of the people here. Why would you stop a student from trying to teach something to another student? That's how we learn.
You guys are so focused on terminologies and conventions that you don't focus on the problem at all.
A simple "Hey, just learn that the correct terminology is not voltage drop, but forward voltage" or "The forward voltage is not always the same, but you can approximate it ok?" would be way more helpful than these rants you guys give.

Something like wvmarle said:
"That's the difference between the theory and the real world.
The forward voltage of an LED is considered a fixed value, but in fact does increase slightly with increasing current."
See? Easy.

15

tadashimori:
you guys are way too agressive all the time.
...
these rants you guys give.

Raschemmel is by far the most aggressive and rude current poster. I say current: there was one worse, he hasn't posted in 4 months.

For all that though sometimes Raschemmel sneaks in a gem of info which turns out quite useful. Whether or not it's worse oops worth putting up with the attitude is of course, up to each individual.

16

Hey less of the "you guys". Trust me there's only raschemmel. He seems to have elected himself the resident pedant. Any use of a term other than the way he thinks it should used gets jumped on. It's a bit wearing but he does more less know what he's talking about.

He just seems to have no understanding of the idea that it often helps to simplify things a bit for people who are just starting to learn this stuff.

Steve

17

he could assume 2.2v.

You don't ASSUME anything. You look up the Forward Voltage on the datasheet. If you don't have a datasheet you can find the forward voltage empirically by adjusting the resistor value until you get a current of 20mA which is a typical spec for the forward voltage :

"Vf = 2.2V@20mA"

LED DATASHEET

18

Some posts DELETED and somebody is now on a 7 day holiday.

Keep it civil !

Bob.

19

All help is appreciated, but please don't fight.

Take a look at my project. I'm soo happy with it.

The voltmeter is upside down, actually the whole construction is. There are few missing parts but you'll never gonna guess what it's going to be.

20

Looks nice, but indeed there seem to be bits and pieces missing: that Nano is not connected at all, and that PWM board only has power connected. No outputs.

If that's upside-down you will have problems with that solderless breadboard. They're already known for unreliable connections, and if used upside-down parts may simply fall out..