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.
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\$\begingroup\$ This is exactly as expected. See any diode datasheet. \$\endgroup\$ – Olin Lathrop Dec 20 '13 at 13:17
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\$\begingroup\$ You say in the first sentence "...diode whose forward voltage drop is 0.617V..." and then in the last line, "I am expecting no drop across diode." These seem to be contradicting statements. Can you explain why you are expecting no drop? \$\endgroup\$ – JYelton Dec 20 '13 at 17:33
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\$\begingroup\$ The answer is the basic working of a diode at atomic level and could easily fill a 20 page chapter in a textbook. \$\endgroup\$ – flup Dec 20 '13 at 17:45
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\$\begingroup\$ There is no resistor, and thus no resistor divider, in your figure. \$\endgroup\$ – Phil Frost Dec 20 '13 at 18:34
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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
answered Dec 20 '13 at 10:39
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\$\begingroup\$ IIRC the carriers are depleted at the junction due to diffusion, not attraction. \$\endgroup\$ – apalopohapa Dec 20 '13 at 11:55
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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)
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\$\begingroup\$ But in the left i have not provide path from cathode of diode to ground, therefore there should be no voltage drop across diode. But I am seeing drop across it \$\endgroup\$ – bharath think Dec 20 '13 at 11:14
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\$\begingroup\$ Yes you have: through the voltmeter. It's not a very high current path, but it is a path. \$\endgroup\$ – pjc50 Dec 20 '13 at 11:19
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2\$\begingroup\$ The voltmeter has an input resistance that is usually in the 10M Ohm range, that essentially closes the circuit and lets current flow through the diode. \$\endgroup\$ – alexan_e Dec 20 '13 at 11:27
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\$\begingroup\$ These graphs always bother me; If the diode Vf depends on the current through the diode, should it not be on the dependent axis in the graph? \$\endgroup\$ – dext0rb Dec 20 '13 at 17:57
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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:
answered Dec 20 '13 at 18:52
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\$\begingroup\$ Just making sure I follow...the spring in the check-valve could be though of as the forward voltage drop, yes? You need a certain voltage to overcome the spring in the valve, and then your charge can flow. Once things are flowing, if you mess with the flow rate, it will put more/less pressure on the ball/spring. So if you increase the flow rate, the force on the spring will increase (pushes it in further, increase Vf). If you pull the spring back out (to lower the Vf, like in the precision rectifier) you will have to increase the flow rate to achieve the same output, right? \$\endgroup\$ – dext0rb Dec 20 '13 at 20:01
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\$\begingroup\$ @dext0rb Yes, that sounds about right. \$\endgroup\$ – Phil Frost Dec 20 '13 at 23:59
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\$\begingroup\$ @ Phil Frost, Exactly... we human beings understand the unknown new by the means of the well-known old... i.e. by analogies... BTW what is the idea behind the "precision rectifier" above? How is the voltage drop across the diode reduced? \$\endgroup\$ – Circuit fantasist Jan 2 '20 at 18:09
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1\$\begingroup\$ @Circuitfantasist that sounds like a good new question. \$\endgroup\$ – Phil Frost Jan 2 '20 at 18:12
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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.
