# LED circuit question

#### buzaiandras

Joined Jul 18, 2011
54
Hi all,

I have some questions about LEDs connected in parallel.
I know I should not connect them in parallel (with only one limiting resistor; and I am not doing that in practice ).
I am just trying to correctly understand some behavior.

I uploaded two images of the circuit to be able to descirbe my issue.
In both images there is the same circuit; the only difference is that in one of them the S1 switch is turned on.
The specified Vf of both LEDs is ~1.7V and we assume they are identical in characteristics.
The current is limited throungh the 1K resistor.
The multimeter shows the voltage drop accros the LEDs, the ammeters show the current running through the LEDs and the total current (the bottom one).
When S1 is off the current flows through one LED and when S1 is turned on the current flows through both LEDs (in parallel).
I understand that when S1 is turned on, and the diodes are in parallel the Vf will decrease because now the current running throught the LEDs is lower (LEDs have identical characterstics) as shown by the meters.
What I do not understand is why , when S1 is turned on the total current is a little higher compared to the total current of when S1 is turned off.
Shouldn't the total current be the same in both cases (and when in parallel it will split between the LEDs), since it is limited only by the 1K resistor? Why does it behave this way?
What causes the total current to increase a little?

The circuit is simulated in Circuit Maker but the same behavior can be tested with the segments of a multiplexed 2 digit 7 segment LED display.
In this example the difference is low but when tested with a multidigit LED display the difference was a couple of milliamperes (depending on how many digits were turned on at the same time).
I am trying to understand what is the cause of this.

Thank you ,

Buzai

#### SgtWookie

Joined Jul 17, 2007
22,230
In the left schematic (S1 open), the voltage at point A is 1.678v, and the circuit current is 3.322mA.
Supply voltage is 5v; 5v-1.678v = 3.322; 3.322/1k Ohms = 3.322mA

In the right schematic (S1 closed) the voltage at point A is 1.613v, and the circuit current is 3.387mA
Supply voltage is still 5v, 5v-1.613v = 3.387; 3.387/1k Ohms = 3.387mA
The voltage across R8 increased, so the current increased. I=E/R.

This is why you need to use individual current limiting resistors for 7-segment LED displays.

#### buzaiandras

Joined Jul 18, 2011
54
Thank you for your response but unfortunately it does not clarify my issue.
Don't get me wrong. I don't want to be rude or anything like that.
The calculations you made are correct but that is not the point.

The resistors in series with each LED should be used so that one of the LEDs won't 'handle' more current than the other (caused by the difference in the LEDs characteristics - ex: small difference in Vf). I am aware of that.
But that is not the case since in our case the LEDs have identical/exactly the same properties (as I said in the first post) so the current will split in half between the two LEDs (as it can be seen in the simulation screenshot).

What I do not understand is why the total current is different between the two cases since the current is limited only by the 1K resistor (and the LEDs are identical).

Thank you ,

Buzai

#### Audioguru

Joined Dec 20, 2007
11,248
Nobody sells LEDs with identical forward voltage drops. You are gambling when you connect LEDs in parallel because one might use all the current and the others will not light.

#### bountyhunter

Joined Sep 7, 2009
2,512
The current changed because the load changed. Two "on" LEDs in parallel is not the same as one LED on. Even if the diodes are identical, one diode is not the same load as two in parallel (diodes are non linear). Different load means differnt current through the resistor.

Joined Dec 26, 2010
2,148
To make this quite clear to you, the constancy of LED forward voltage is only an approximation. In practice, the forward voltage of an LED is linked to the current through it, albeit in the rather weak way common to other diodes.

A rough approximation to this can be made by considering the diode to behave like a voltage slightly less than the normal forward voltage, in series with a small resistance. More accurate models, as found in some simulators, incorporate a modified form of Schockley's diode equation to give a better prediction of the diode voltage.

N.B. Individual diode voltages still vary from the typical. The fuller diode model is only relatively more accurate.

#### Wendy

Joined Mar 24, 2008
23,273
What you are not considering is what happens with failures. While LEDs are reliable, they do occasionally die. If you make a design that a single point failure can cascade then it is a failure.

I have a tutorial on LEDs that grew beyond the original goal.

LEDs, 555s, Flashers, and Light Chasers

Chapters 1 and the 1st ½ of chapter 2.

It has been mentioned, but it is true. LEDs are less likely to be identical than most parts, variation is the norm.

#### buzaiandras

Joined Jul 18, 2011
54
Thank you all for your responses #### JMac3108

Joined Aug 16, 2010
348
Buz,

I think most posters misunderstood the intent of your question. You are assuming the LEDs are identical (and they are in the simulator) and asking why the total current changes when the two LEDs are in parallel. Correct?

Re-read SgtWookie's post. He understood and answered your question correctly. In words, the reason the current changes when the two LEDs are in parallel is as follows...

- The LEDs split the current so there is less current through each one.
- Less current thru the LED means lower voltage drop across the LED
- The current limit is set by the voltage across the resistor
- The resistor voltage is Vcc - Vdiode
- Since Vdiode changed, the current changes

This is what SgtWookie explained in his post

#### buzaiandras

Joined Jul 18, 2011
54
Hello,

Sorry for my delayed response.
@JMac3108: Yes that is correct. I understand it now .

Thank you,

Buzai