Basic Circuit: Why don't these two LEDs light up?

[bremenfallturm]

TL;DR Summary

Basic electronic question from an absolute beginner: I would assume that all LEDs are connected in parallel, and given the circuit in my image, why don't the two leftmost LEDs light up?

Hello!
Forgive me for this absolute beginner question.

I was trying to experiment with Charlieplexing a couple of LEDs. Charlieplexing is a technique that utilizes the fact that many microcontrollers attend a high impedance state when set to input and of course that diodes only pass current one way. Configuring it right, you can individually address $P(n,n-1)=n(n-1)$ LEDs (in this case) given $n$ pins.

I built the most basic Charlieplexing circuit in the Falstad Circuit Simulator. Below I am using the input state that would light up the second LED *from the right*

And indeed it does, according to the simulator:

However, I would assume that all LEDs are connected in parallel, and given the circuit in my image, why don't the two leftmost LEDs light up as well? This is not the expected behavior of Charlieplexing but I'm trying to understand why it works. The two leftmost LEDs are oriented the same as the one that lights up (i.e. the anode is oriented towards +). Why don't they light up?

I'm an absolute beginner in electronics so I may not have even understood serial and parallel correctly. Forgive me for the beginner question:)

 

[Baluncore]

... , why don't the two leftmost LEDs light up as well?

Because a bandgap voltage must be dropped across the LED.
The voltage is set low by the LED second from the right, there is insufficient voltage present to light the two series LEDs on the left.

 

[DaveE]

 

[Baluncore]

Red, orange, yellow, green, blue, violet.
How come Y and G have swapped places?

Voltage needed to start conducting = Energy in eV = 1239.84 / λ nm

 

[DaveE]

Red, orange, yellow, green, blue, violet.
How come Y and G have swapped places?

Voltage needed to start conducting = Energy in eV = 1239.84 / λ nm

Good catch! They're just special, I guess.
You just can't trust anything you steal these days!

 

[bremenfallturm]

Because a bandgap voltage must be dropped across the LED.
The voltage is set low by the LED second from the right, there is insufficient voltage present to light the two series LEDs on the left.

Ah ok, sounds very reasonable. Thank you for the clarification. And it is never possible to get "sufficient voltage present" to light up both the two leftmost series LEDs and the second LED to the right? Just trying to wrap my hand around why we don't get sufficient voltage on those two.

 

[Baluncore]

And it is never possible to get "sufficient voltage present" to light up both the two leftmost series LEDs and the second LED to the right?

It is not possible, because a LED drops a fixed forward voltage when on. Half that forward voltage is insufficient to turn on one of the two series LEDs that are in parallel.

If you want to discuss the circuit further, you will need to specify the colour of the LEDs, and label the components D1, D2, D ... and R1, R2, R ...

 

[DaveE]

A simple model for diodes is a battery in series with a resistor*. You could try substituting that into your circuit in place of the diodes to see the problem.

Then, theoretically with enough current in the single shunt diode you could make enough voltage to get some current flow in the two series diodes. But IRL, you'll break the single diode first.

OTOH, if your single diode was UV and your two series diodes were both IR you could illuminate the IR diodes with the UV staying off.

* Edit: oops! You also need a perfect rectifier in series, since current can only flow in one direction.

 

[phyzguy]

If you added another resistor between the top of the 2 LEDs and the top of the 4 LEDs, then you could get all three to light, because the voltage across the 4 LEDs would be higher than across the 2 LEDs.

 

[sophiecentaur]

TL;DR Summary: Basic electronic question from an absolute beginner: I would assume that all LEDs are connected in parallel, and given the circuit in my image, why don't the two leftmost LEDs light up?

Why don't they light up?

Sorry if the following is too elementary but no matter:
Diodes don't behave as you might expect. LEDs only light up when 'forward biased' (i.e. when the 'arrow' points from the + supply and the 'earth' supply that explains why the right hand single diode (which is reverse biased) doesn't pass any current and also the two diodes in series, left of centre.

Also, LEDs are not like resistors; current is not proportional to volts with them. Those two 200Ω resistors limit the current that can pass through. Many LED circuits have resistors in series to limit the current flowing through.
All the other three diodes are forward biased so why don't they all light up? The single diode will have about 2.5V across it (from the red curve on the graph). The remain two diodes are in series but they 'share' the volts across the single LED so they get half the volts each. Very little current will get through the two LEDs so you won't see any light from them. You only see the LED with the full 2.5 V across it.
PS It's always helpful to have labels on components in diagrams, eg D1 D2 D3, R1 R2 R3