Confused with diodes and conventional vs real current flow

Question

What would happen in the following hypothetical situation?

pd : images below

A diode must actually receive the electron current and voltage on its negative side, i.e. the N-side.

And that is why we connect the positive side of the battery to the anode and the negative side of the battery to the cathode of the diode, because in reality, a diode receives voltage and electrons from the negative side of the battery, because in truth, the current and voltage flows from negative to positive.

And I know that we, like everywhere else, we use conventional current but...

What if, for example...

We have two diodes in one circuit.(light light-emitting diodes)

One diode A is located above the circuit, and it needs a bias forward of 0.7 V to turn on, if it does not receive that amount, it will not let any electrons pass through it.

The other diode, called B, is on the right side of the circuit, and needs a 0.4V bias forward to light up.(let current flow)

And so, with the battery on the left side of the circuit, the battery is 0.6V. So this battery would let us turn on the B diode but not the A diode.

The battery has the symbol + up and - down, like all the others.

So...

in this case, wouldn't the diode led B light up without the need of the diode A to pass current?

Because according to the conventional current model: the voltage and the current would come out from the + side of the battery, it would go through Led A, and Led A would not receive enough forward bias, so the electrons would be stuck, and the voltage would reach Led B, but the current would not, so Led B would not light up.

Although according to the model of real current direction: the voltage and current of the battery would come out from the - side of the battery, it would go through the cathode of led B, with enough forward bias (+0.4 needed for led B) and the current and voltage would come out from the anode of led B, this current and voltage would reach the cathode of led A, and because it does not have the necessary forward bias, it would not turn on ...

So, based on the actual current direction, wouldn't diode B light up, even though no current passes through diode A?

Thanks :-)

images here

Answer 1

Look, you are just going to have to learn to deal with Ben Franklin's mistake (see hackastical's xkcd cartoon). Most of us have been there in our early days. pipe's comment is spot on. If you're doing solid state physics and talking about n-type semiconductors, then you need to think about electron flow. If you're building an LED circuit, forget electron flow and go with current. Current is produced (in conductors) by electrons, true enough, but electrons flow in the opposite direction as current. At some point you'll understand that it's just a convention, and has no deeper significance, and you can get on with your life.

Now. Your LED issue arises from your forgetting Kirchoff's Voltage and Current Laws: the sum of voltages around a loop is zero, and every connection point has zero net current. So, with 0.6 volts across the two LEDs, each LED must have at most 0.6 volts across it (KVL), and will in practice have less, with the two summing to 0.6. So the 0.7 volt LED has zero current through it, which means the 0.4 will too (KCL), and with no current through either LED, neither will light up.

And yes, this means that the actual distribution of voltages across the two LEDs is undefined. When you make a more reasonable (realistic) model of the LEDs, leakage effects will allow a more reasonable voltage distribution. But the voltage across the 0.7 volt LED will not reach 0.7 volts and the LED will not have much light output. Depending on the exact materials used, it's possible that the 0.4 volt LED will light up slightly.

Answer 2

1. It’s Ben Franklin’s fault. Deal with it.

From here: https://xkcd.com/567/

2. Your diagram shows a battery voltage of 0.6V. Not tall enough for your electrons to ride the Vf quantum-mechanical ride and light those LEDs, which have Vf (forward voltage) ranging from 1.1 for an infrared type to over 3V for ultraviolet.

Answer 3

[A] diode must actually receive the electron current and voltage on its negative side ...

Electron drift will be inbound on the cathode (if you want the LED to light). But note that the diode symbol contains an arrow that denotes the direct of conventional current flow.

And that is why we connect the positive side of the battery to the anode and the negative side of the battery to the cathode of the diode, because in reality, a diode receives voltage and electrons from the negative side of the battery, because in truth, the current and voltage flows from negative to positive.

Or, "We connect the anode to positive because, in truth, conventional current flows from positive to negative". It's the same from a circuit analysis point of view.

What if, for example ... We have two diodes in one circuit.(light light-emitting diodes). One diode A is located above the circuit, and it needs a bias forward of 0.7 V to turn on, if it does not receive that amount, it will not let any electrons pass through it.

It won't let conventional current pass through either.

The other diode, called B, is on the right side of the circuit, and needs a 0.4V bias forward to light up (let current flow). And so, with the battery on the left side of the circuit, the battery is 0.6V. So this battery would let us turn on the B diode but not the A diode.

Incorrect. They voltage will be spread across them and neither will turn on.

The battery has the symbol + up and - down, like all the others. So in this case, wouldn't the diode led B light up without the need of the diode A to pass current?

Figure 1. Typical I-V curves for a range of LEDs. Source: LEDnique.com.

No. Current flows in a loop. If any device in a series circuit blocks the current then no current will flow. Incidentally, the lowest LED forward voltage, V_f, is for infrared LEDs and is in the range of 1.2 to 1.4 V.

Because according to the conventional current model: the voltage and the current would come out from the + side of the battery, it would go through Led A, and Led A would not receive enough forward bias, so the electrons would be stuck, and the voltage would reach Led B, but the current would not, so Led B would not light up.

Fairly muddy thinking here.

Although according to the model of real current direction: the voltage and current of the battery would come out from the - side of the battery, ...

You are confusing current flow and mobile charge flow. We all just stick with conventional current flows from positive to negative while understanding that, at least in metallic conductors in particular, charge is carried by mobile electrons.

... it would go through the cathode of led B, with enough forward bias (+0.4 needed for led B) and the current and voltage would come out from the anode of led B, this current and voltage would reach the cathode of led A, and because it does not have the necessary forward bias, it would not turn on ...

As explained already, current can not flow through one of your LEDs and not through the other.

Figure 2. The diode check-valve analogy from What is an LED?

LEDs are diodes (that emit light). Diodes are electrical non-return valv

If you look at the check-valve in the figure above, it should be clear that the spring normally keeps the ball in position and prevents back-flow. When “forward-biased” the ball shut-off can be moved against the spring but it will take some initial pressure to move the ball. This results in a pressure drop across the valve: the pressure downstream will be less than the inlet pressure.

In a similar manner the PN junction causes a voltage drop. For silicon it is about 0.7 V. For LEDs it will be higher and depend on the dopants used to generate the wavelength or colour of the emitted light.

Pushing the analogy a little further, we can also see that further pressure drop will occur due to the constriction of the valve. The more water we push through the valve the more the pressure will drop. This will be added to the initial pressure drop required to open the valve in the first place. The resultant pressure drop graph will look remarkably like one of the I vs V curves in Figure 1.

It should also be clear that connecting two of these in series will require the pressure (voltage) to rise to double the value of one valve before current will flow.

So, based on the actual current direction, wouldn't diode B light up, even though no current passes through diode A?

No.