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### Topic: Current flow & Voltage Drops in series & parallel (Read 4817 times)previous topic - next topic

#### Andy_Cool

##### Jan 11, 2013, 10:38 am
Hi

Though my query is about LED resistance & Voltage drops it really breaks down to basic electronics in context to Arduino.

Lets assume:
Arduino output pin sources 40ma max current
LED has 3V Fwd voltage Drop
LED Max Current is 20 ma

1. In the following figure, what would be the forward voltage of the LED? (& why)

2. In the each of the following 3 figures
2.1  how much current would the pin source? (& why)
2.2  what would be the Vf on each LED? (& why)
2.3  will 20 ma flow through each LED? (& why)

Thank you.

#### Boffin1

#1
##### Jan 11, 2013, 11:46 am
First drawing forward drop of LED = 3v  ( you told us so )

second drawing, no current as you have 2 3v LEDs in series

third drawing   -  damage, 2  3v LEDs will try to drag the output of the Arduino down to 3v

third drawing,    5v  less the 3 volts accross the diode = 2v accross the 250R res, by ohms law  = 8 Ma each LED
45 years of editing projects with a knife and soldering iron, then I found Arduino !

#### Andy_Cool

#2
##### Jan 11, 2013, 01:04 pm
Thanks for replyin, Boffin1. Kindly explain this.

third drawing,    5v  less the 3 volts accross the diode = 2v accross the 250R res, by ohms law  = 8 Ma each LED

Also since each Atmega8 pin can source 40ma, in what configuration should the
above 2 LEDs be connected to a single pin so that each LED gets 20ma?
(My confusion stems more from the Voltage drop & how it adds/subtracts in all of this)
Thanks.

#### Grumpy_Mike

#3
##### Jan 11, 2013, 01:19 pm
Quote
Also since each Atmega8 pin can source 40ma,

No 40mA is the absolute maximum the data sheet says of this value:-
Quote
Stresses beyond those listed under "Absolute Maximum Ratings" may cause permanent damage to the device. This is a stress rating only and functional operation of the device at these or other conditions beyond those indicated in the operational sections of this specification is not implied. Exposure to absolute maximum rating conditions for extended periods may affect device reliability.

The LED is a non linear device so ohms law will not work, so you have to use ohms law on the resistor. As the LED will have 3V across it the rest of the voltage must be dropped across the resistor,
That is 5 - 3 = 2V
So if a 250R resistor (where do you get such a thing I don't know as it is not a standard value ) has 2V across it the current through the resistor has to be 2 / 250 = 0.008 Amps or 8mA. As it is a series circuit the current flowing through the resistor is the same as the current flowing through the LED.
The two circuits are in parallel so each will have 8mA flowing through each leg making the total current draw from the pin 8 + 8 = 16mA.

#### Boffin1

#4
##### Jan 11, 2013, 01:31 pm
You will find the LEDs are pretty bright at 15mA,, so you could use your bottom circuit with a 120 ohm in each leg, which will give a total current drawn from the pin of about 33 ma.

Do you want to use 2 LEDs  as they are in different locations, or was it to try and get more illumination ?
45 years of editing projects with a knife and soldering iron, then I found Arduino !

#### dhenry

#5
##### Jan 11, 2013, 01:34 pm
Quote
LED has 3V Fwd voltage Drop

That assumption is actually far more complicated than it is stated above. Most of the times, people make the highly simplified assumption that a led will drop 3v once it starts to conduct (aka as long as If > 0). A real led starts to conduct at Vf > 0 (when If is very small).

Quote
1. In the following figure, what would be the forward voltage of the LED? (& why)

Assuming that the led conducts, the voltage drop over the resistor is 5v - 3v = 2v. Its current is 2v/250ohm = 8ma. Which means the led is conducting. so we are good.

Quote
2. In the each of the following 3 figures

Assuming that the leds conduct. The voltage drop over the 2 leds is 6v, which means they cannot be conducting.

Assuming that the leds are not conducting. The voltage drop over the 2 leds is zero, which means there is current going through the resistor. which means the leds must be conducting.

We fall into a fallacy, cause by the assumption that the led only conducts when Vf = 3v.

The other cases are similar and fairly simple.

#### Boffin1

#6
##### Jan 11, 2013, 02:05 pm

Quote
Assuming that the leds are not conducting. The voltage drop over the 2 leds is zero, which means there is current going through the resistor. which means the leds must be conducting.

No,  there is 5v accross the LEDs and zero accross the resistor , so zero current .
45 years of editing projects with a knife and soldering iron, then I found Arduino !

#### Andy_Cool

#7
##### Jan 11, 2013, 02:22 pmLast Edit: Jan 11, 2013, 02:27 pm by Andy_Cool Reason: 1

@Boffin1

So this is how it works. The voltage drop (Vd) across the resistor is first calculated.
It is 2V (since 5v-3v). Now we decide on the value of resistor so that we get desired current.
Aha! I had tried to read through various posts, but was confused & had to post this noobish question.
(I remember the current/voltage bifurcation always used to confuse me )

Thanks all. This was an education.

Do you want to use 2 LEDs  as they are in different locations, or was it to try and get more illumination ?

For more illumination. Actually, I want to connect at least 2 RGB LEDs to 3 PWM Arduino pins.

Are there some tutorials to brush up the electronics basics (with respect to Arduino.)?
Thanks again.

#### dhenry

#8
##### Jan 11, 2013, 02:36 pm
Quote
so zero current .

I will make it simpler for you: what is Vfwd when If = 0, for any led / diode of your choice?

Quote
So this is how it works.

That's the 1st order approximation. Works well for just lighting up a led.

#### Boffin1

#9
##### Jan 11, 2013, 03:00 pm
@  Andy Cool    yes you have got it.

@ dhenry

Quote
I will make it simpler for you: what is Vfwd when If = 0, for any led / diode of your choice?

The Vfwd is what the manufacturer states

the forward voltage accross the LEDs  is 5v  ,    they have not reached their combined series Vf of 6 volts yet.
The voltage accross the resistor is zero, so no current.

Try it with a meter the voltage  accross the LEDs will be 5v.
45 years of editing projects with a knife and soldering iron, then I found Arduino !

#### dhenry

#10
##### Jan 11, 2013, 03:08 pm
OK, even simpler than that.

Here is the V-I curve for Osram LAE7F datasheet.

Question: what is the Vf when If = 0?

Quote
he Vfwd is what the manufacturer states

What most people fail to comprehend is that Vfwd differs at different current levels. Vf = 3v @ 20ma has no meaning when If = 0ma.

#### dhenry

#11
##### Jan 11, 2013, 03:13 pm
Here is a quick sim of your circuit, with two leds.

The X-axis is for the voltage on the whole circuit, and Y-axis is the current, plotted in a logarithmic scale.

Those two leds have Vfwd of roughly 3v. and the current going through them is just shy of 1ma.

#### Boffin1

#12
##### Jan 11, 2013, 03:25 pm
Once again you are getting into nit picking pedantics when the OP was asking the very basics.

Sorry Andy_Cool ,  this seems to be happening more, and doesn't help newbies who ask about the basics.

Good luck with your project, if you want more illumination, you can run several 20 mA LEDs ( with their resistors ) in parallel with a simple transistor driver.
45 years of editing projects with a knife and soldering iron, then I found Arduino !

#### dhenry

#13
##### Jan 11, 2013, 03:46 pm
To fully satisfy you, two blue leds (3v@20ma) under 5v, current is 0.25ma without any resistor and 0.11ma with a 1.5kohm resistor.

Quote
Once again you are getting into nit picking pedantics

#### Papa G

#14
##### Jan 11, 2013, 04:15 pm

OK, even simpler than that.

Here is the V-I curve for Osram LAE7F datasheet.

Question: what is the Vf when If = 0?

Quote
he Vfwd is what the manufacturer states

What most people fail to comprehend is that Vfwd differs at different current levels. Vf = 3v @ 20ma has no meaning when If = 0ma.

Hard to say since your example curve starts at If of 1.

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