Why there is reduction in voltage after diode?

Question

I have used simple resistor divider with the diode whose forward voltage drop is 0.617V as shown in figure. I am reading voltage in voltmeter as seen figure. Can anyone explain me why it is like this?

I am expecting no drop across diode.

Answer 1

What

The voltage drop is a fundamental characteristic of forward-biased doped-silicon semiconductor junctions. There are diodes with a smaller voltage drop but all diodes will have a voltage drop.

You can look at a graph of diode characteristics but all this tells you is that Vd exists, not why

Why

To understand this in detail, you need to read about semiconductor physics. See PN junction voltage drop

The way I think of it is as follows:

Because of the doping of the silicon, "free" electrons diffuse from the n-type region adjacent to the junction. These electrons combine with holes in the p-type region. This leaves a "depletion-zone" that lacks carriers and is therefore insulating. To drive current through that region you have to apply an electric field to create a force that moves carriers into the depletion region. This electric field across the junction is what we measure as a voltage across the junction.

^{From Wikipedia}

Answer 2

Can anyone explain me why it is like this?

That is how a diode behaves in forward direction, the forward voltage drop depends on the current through the diode and can be seen in the following graph in the datasheet

In your right side image there is no voltage output because the diode is connected in the reverse direction so there is no current through it (except from a negligible leakage current)

Answer 3

You can read the datasheet, or understand the physics, but here's the intuitive explanation: a diode provides a barrier to charge flow in one direction. The diode must somehow "know" that the flow is going in the right direction, so it must somehow observe that flow. The only way the diode can observe the flow is to let the flow do some work on the diode to see that the flow is pushing the right way. Since work was done, some of the energy is converted to heat in the diode, and this manifests as reduced voltage at the output.

An analogy: you can measure the direction of air flow past a car by putting your hand out the window. However, doing so necessarily means you will introduce some additional drag, either slowing the car or burning more gas.

Another analogy: check valves perform a similar function to diodes for fluid flows. A simple design is a ball pushed against a gasket by a spring:

A flow in the "wrong" direction merely pushes the ball against the gasket more strongly, but a flow in the "right" direction opposes the spring and pushes the ball away from the gasket, allowing fluid to flow:

However, holding the ball open against the force of the spring requires constant energy input, and this is manifest as reduced pressure on the output side of the check valve, represented here as a lighter blue color.

Although the physics of a diode are different, the concept is the same: there's a barrier, pushing it in the wrong direction holds it shut, pushing it sufficiently hard in the other direction opens it, and some energy is used in the process.

If you introduce active electronics, it's possible to arbitrarily reduce the voltage drop of a diode. This is called a precision rectifier, and is commonly realized with an op-amp:

Answer 4

To understand this properly, you have to include your multimeter in the schematic. Let us say that your multimeter has a resistance of 10 million ohms. (The exact figure doesn't matter, but it's usually very large).

^{simulate this circuit – Schematic created using CircuitLab}

So we have a complete circuit consisting of the voltage source, the diode and the internal resistance of the multimeter. (We will assume that the actual meter VM1 has a near infinite resistance, and so R1 determines the meter's resistance.)

Anyway, when the diode is forward-based, then current flows through D1 and R1. In accordance with diode behavior, D1 has a forward drop, which for a silicon diode may be in the neighborhood of around 0.6V. And so the measurement registered by the meter VM1 is about 5V minus that.

If we reverse-bias the diode, then almost no current flows. The diode looks almost like an open circuit, except for leakage through the diode which is measured in nanoamperes (or less). In other words, only a tiny trickle of current flows through D1 and R1; next to nothing.

The voltage measured at the top of R1 comes from Ohm's Law: V = IR: the current through R1 times the resistance of R1.

If the current through R1 is 1 nA (nano-Ampere), the voltage is:

$$\left(1\times 10^{-9}\text{A}\right)\times\left(1\times10^{7}\Omega\right) = 1\times 10^{-2}\text{V} = 10\text{mV}$$

With these assumptions (that I pulled out of thin air), there is a tiny voltage.

When you have the diode reverse-biased, try using the smaller voltage scales available in your meter: get the voltage with fraction of a millivolt precision. You may find that it's not exactly zero, due to the tiny leakage current trickling through your meter's internal resistance.