# Voltage across the diode

What will be the voltage across the terminals of the diode if its directly connected (forward biased) to a variable voltage source that varies from 0.7 to 5 volts?

Will it remain 0.7 volt at the output of the diode terminal or the output of the diode will vary with the voltage to which the diode is directly connected. I know it's a basic question but I am confused.

On thing that comes in my mind is that the resistance of the diode will decrease as the voltage/current increases across the diode and thus the voltage across the diode will remain a constant 0.7 across the diode but I am confused. Can anyone please clear this confusion?

• In case of suppling with voltage source the diode will be destroyed. Commented Jan 12 at 4:59
• There are at least two ways to interpret your question. Please edit your question with a sketch of the connections and the measurement points. Commented Jan 12 at 5:02
• Is this the same as the Voltage across the diode? question posted on the Physics stack exchange? Commented Jan 12 at 19:21

What will be the voltage across the terminals of the diode if its directly connected (forward biased) to a variable voltage source that varies from 0.7 to 5 volts?

When a (5 V) voltage much higher than usual (0.7 V) is applied to the diode, an extremely low-resistance voltage divider is formed from the source and diode resistances. So depending on the ratio between the two resistances, the voltage is somewhere between the extreme values ​​(0.7 V and 5 V).

If the source is a very good constant voltage source, the current and power dissipation will be significant and any of the devices may be damaged.

# CircuitLab experiments

## Real diode

Let's select three typical points of the diode IV curve obtained by the arrangement below.

726.5 mV, 20 mA: Let's start with a 20 mA current value. For this purpose, we can take an empirical approach by carefully changing the voltage in the Vin parameters window until the current becomes 20 mA.

simulate this circuit – Schematic created using CircuitLab

798.7 mV, 60 mA: Then we can apply the same procedure to set IF = 60 mA...

simulate this circuit

844.4 mV, 100 mA: ... and finally, to set 100 mA.

simulate this circuit

5 V, 7.079 A: Now is the time to set the abnormally high 5 V input voltage. This procedure is different from the previous ones because we directly set the voltage without caring about the current. And the result is there - abnormally high (7 A) current!

simulate this circuit

Figuratively speaking, there is a conflict like a tug of war between two constant-voltage devices - a constant voltage source and a diode, because each of them tries to set its own voltage.

Rin = 908 mΩ inserted: To mitigate the conflict, we need to insert a small resistance in the circuit. We can get crafty and do it via the ammeter by giving it a resistance in the parameters window. But how much should this resistance be?

simulate this circuit

Let's approach it like this - to give equal chances to the source and the diode in this "fight". In other words, adjust Rin so that the voltage drop across the ammeter becomes equal to the voltage drop across the diode (half of the whole, i.e. 2.5 V).

This gives us an idea - that the diode behaves like the same 908 mΩ resistance. Then let's develop it below.

## Dynamic resistor

On thing that comes in my mind is that the resistance of the diode will decrease as the voltage/current increases across the diode and thus the voltage across the diode will remain a constant 0.7 across the diode...

Yes, that is a very good (my favorite) way to functionally imagine how the diode does this magic of keeping the voltage across itself constant. Now we need to repeat the three experiments above with a real diode replacing it with an equivalent variable resistor.

726.5 mV, 20 mA: To set 20mA current at this voltage, the diode in Schematic 1.1 behaved as a 36.33 Ω resistor; so now we replace it with such a resistor. We determine this value experimentally by adjusting the resistance so that the current is 20 mA.

simulate this circuit

798.7 mV, 60 mA: When the input voltage increases, RF decreases to set 60 mA current (we adjust it so that the current is like so)...

simulate this circuit

844.4 mV, 100 mA: ... and finally, to set 100 mA.

simulate this circuit

5 V, 7.079 A: When Vin increases abnormally to 5 V, the dynamic resistance becomes only 706.3 mΩ, and the current is 7.079 A!

simulate this circuit

Rin = 908 mΩ inserted: Like above, to mitigate the conflict, we insert a small resistance Rin (set the ammeter resistance in its parameters window). We choose it equal to the RF resistance. Figuratively said, two voltage sources - Vin and ground, "pull" the midpoint through equal 908 mΩ resistances; Vin "pulls" it up and the ground "pulls" it down. Because they are equally strong, the point stands in the middle.

simulate this circuit

## Other diode models

The dynamic resistor diode model is very good for intuitive understanding of diode operation but not useful for calculation. That is why, other models are used for this purpose. Let's consider them.

"Ideal" diode + resistor: The idea behind this simplified model is to represent the real diode as an "ideal" diode and resistor connected in series.

simulate this circuit

Accordingly, the diode IV curve is a sum of the two IV curves.

Voltage source + resistor: The problem of the "ideal" diode is that there are not such diodes:-) But it can be replaced by a voltage source. The "ideal" diode D is auxiliary; its role is to stop the reverse current when Vin < VF.

simulate this circuit

"Ideal" diode: The simplest diode model consists of only "ideal" diode. To get its IV curve, we need to pass a current through it and read the voltage across it.

simulate this circuit

Note that the current is plotted along the abscissa and the voltage along the ordinate.

So can I say that as voltage increases across the diode then the resistance of the diode decreases and because of it the output voltage across the diode remains constant at approximately 0.7 volts?

Yes, with the proviso that the voltage across the diode is set by a real voltage source with some internal resistance.

So it's the resistivity of the diode that changes with voltage? More voltage in diode will cause more current in diode so the resistivity decreases and thus resistance decreases.

Yes, with the proviso that "resistivity" and "resistance" is figuratively speaking (in quotation marks).

So I can say that increasing the voltage across the diode decreases it's resistivity?

Yes, you can figuratively say that the diode decreases its “resistivity".

In my answer above, I explained this phenomenon through many experiments to be more convincing, and here I will do it in a more human-friendly verbal way.

Generally speaking, a diode has the property of making it difficult for current to pass through it under the influence of a voltage. We can call this property "static resistance", "DC resistance"... well, if you so desire, "resistivity" too... but do not advertise it much because it is not accepted in semiconductor theory; use it only for "your own use". In life (at school, university...) we often have to "play a double game" when we express ourselves according to the canons of the paradigm but explain it to ourselves in our own way :-) Basically, "resistivity" is reserved for conductive materials with uniformly distributed resistance (metals, graphite, etc.)

Things (circuits) are understood through concepts, and the concept here is called voltage divider. It is formed by the two "resistances" - Rd of the diode and Rin of the voltage source. There is always both, though either may be very little. Figuratively speaking, they "fight" each other to "move" the midpoint (change its voltage) - Rin "pulls" it up and Rd "pulls" it down. For example, if the two resistances are the same, the voltage will be half of the input. More precisely speaking, the input source tries to change (increase) the midpoint voltage by passing current through Rd but the latter opposes it by changing (decreasing) its resistance in the opposite direction.

However, this divider is a bit more special because only one of its resistances (Rin) is constant; the other (Rd) is "self-changing", "dynamic", "non-linear". So when the input voltage changes, at the same time the divider changes its transfer ratio Rd/(Rd + Rin) in the opposite direction.

If Rin = 0, i.e. you connect an "ideal" voltage source to the diode, the latter will only change the current but will not be able to change the voltage. A transistor in a CE stage will do the same if there is no collector resistor, also R2 in a voltage divider if R1 = 0, etc. In all these cases it is desirable to have some resistance in series with the "ideal" source.

• So can I say that as voltage increases across the diode then the resistance of the diode decreases and because of it the output voltage across the diode remains constant at approximately 0.7 volts?
– Alex
Commented Jan 14 at 4:27
• So it's the resistivity of the diode that changes with voltage? More voltage in diode will cause more current in diode so the resistivity decreases and thus resistance decreases.
– Alex
Commented Jan 14 at 6:43
• So I can say that increasing the voltage across the diode decreases it's resistivity?
– Alex
Commented Jan 14 at 6:45
• @Alex, I will answer you today more systematically at the end of my answer. Commented Jan 14 at 7:12
• Thanks alot sir. I will wait. So much grateful to you.
– Alex
Commented Jan 14 at 7:23

If your voltage source is ‘ideal’ (that is, it has no internal resistance), then the voltage will be whatever you set it to.

The current is a different matter. The diode current will be small until the voltage reaches the diode’s forward threshold, after which it will rise exponentially. Eventually the diode will heat up and burn, at which point its current will drop back to zero.

Here’s an example diode I-V curve:

The diode I-V curve has four regions, as voltage goes from negative to positive, from left to right:

• Reverse (‘Zener’) breakdown (lots of current, continued rise with more-negative voltage)
• Reverse bias (little or no current, just leakage)
• Forward subthreshold (little current, rising exponentially as voltage approaches Vf)
• Forward bias (lots of current, continued rise with more-positive voltage)

After the ‘knee’ of the threshold, about 0.7V for a silicon diode, the diode forward voltage rises with current. It behaves roughly like a low resistance.

Try this out for yourself (simulate it here):

As you move the slider from left to right you can see the behavior in each state as the voltage swings from -100 to 25V.

The diode being modeled here is a 1N4148 type, a medium-current silicon switching diode. You can see that once it's in forward bias the current rises linearly with voltage. And, it Zeners out at -75V.

Since there is both this internal resistance and the diode forward drop, as diode current increases so does its power dissipation, heating it to the point where eventually turns into a ‘fry-ode’, pitching its last in a fiery heat death.

That said, there are diodes out there designed to handle very large currents. A common place you’ll find these in switching power supplies, usually as Schottky type owing to its lower forward bias voltage (about 0.3V) than a regular silicon type. Given a strong enough voltage source, these too can be made to blow up. Sometimes they will save you the trouble and blow up all on their own if the power supply they’re installed in is overstressed.

One more thing. The diode’s exponential V-I behavior in the ‘knee’ area has another use: a logarithmic amplifier can be constructed from a diode and an op-amp.

• What if the diode doesn't burn or fry even when the voltage is increased beyond 0.7? What will be the terminal output voltage in this case?
– Alex
Commented Jan 12 at 7:26
• Like I said, the diode behaves like a resistor. So the voltage will rise. Commented Jan 12 at 16:11
• @Alex - If you try to increase the supply voltage across the diode the forward voltage will rise slowly but the current will rise by a large amount, (per the right side of the curve). The V/I slope of the curve (right side) is the small equivalent resistance of the diode at that point. Depending on the power supply current rating vs the diode's current rating one or the other will eventually fail, the diode will burn out or the power supply fuse will burn.
– Nedd
Commented Jan 12 at 16:23

Only one winner here:

1. The diode lowers the voltage source voltage, because the source's own internal resistance is more significant than the diode's, or

2. The voltage source's internal resistance is low enough that the diode passes many hundreds of amps.

Maybe you get something between the two, but this is just another irresistible force vs. immovable object thing. Neither the voltage source nor the diode has zero impedance, so pitting one against the other is just going to explode the least perfect one.

Copied

It conducts at

.7

Once it conducts its like a piece of wire

And theres no.potential.across it

Cause there's no

Across

Any more…

Before Its open

You had something to measure across

Uts how you check a fuse

If its open

If its good its like a piece of wire

Sometimes looking at the other way and its converse explains it

With it closed

The test leads arr basically touching

By Michael Paglia

• Where did that come from? Is this an answer, or is it the reason you asked the question?
– JRE
Commented Jan 12 at 13:36
– Alex
Commented Jan 12 at 13:39
• Is that the reason you asked the question in the first place? If so, a link and a quote belong in the question.
– JRE
Commented Jan 12 at 13:41