How do diodes work, physically?
What is the source of the forward voltage drop across a diode, and why is this current dependent?
Do diodes act instantaneously to prevent current flow, or is there a short “ramp up” interval during which the resistance effectively increases to infinity?
-FK
Here’s a neat page that answers part of your question: How a Diode Works
The voltage drop is a function of the bandgap energy needed to turn the diode “on”. I’m not sure what you mean by “current dependent” in this case. The voltage drop is fixed by the diode chemistry, and is constant after saturation has ocurred.
Diodes, like any real-world devices, don’t act instantaneously, but require some time to turn “on”. This is why some types of diodes, like silicon, aren’t good for very high-frequency applications like cell phones.
Just to be sure I understand this:
barrier voltage = forward voltage drop at saturation?
If there is a minimum voltage of 0.6 V to produce current flow, if I am using a digital signal (0 or +5 V), then a single diode in the circuit will act as a noise filter for the low state?
Looking at the label on a Radio Shack package of diodes, it indicates a specific voltage drop (0.6 V) at a specific amperage (1.5 A). I took this to mean than the voltage drop would be a different value if the current is not 1.5 A. I am mistaken?
-FK
Ah, I see the source of the confusion. No, the current rating of a diode is the max. I[sub]F[/sub], or forward current. The forward voltage drop is independent of the forward current, after the conduction voltage has been exceeded.
Yes, I believe the barrier voltage is the same as the forward voltage drop, but I’m not 100% sure.
If by a “noise filter” you mean will a diode prevent, say, a < 1 V signal from being read as “low”, then yes, I suppose you could use them in this manner. I’m not really clear on what you mean, though.
This is a first-order approximation. In reality, the forward voltage drop is a function of current and temperature. As the current goes up, so does the voltage drop.
This is true for the “garden variety silicon diode,” which has a forward voltage drop of around 0.6 to 0.7 V. But there are other families that have lower voltage drops. Hot carrier diodes (Schottky diodes) often have forward voltage drops < 0.3 V, and the voltage drop of back diodes (tunneling diodes used in the nontunneling direction) can approach 0 V. Of course, these families have disadvantages, but that’s probably beyond the scope of this thread.
To answer the first two questions you really must take an EE course called “solid state devices.” I took it last quarter as part of my EE masters program. There’s not enough space in this thread to do it justice.
As far as the third question goes… yes, there’s a finite “switching speed” for every diode. It’s dependent on a number of factors, the primary one being junction capacitance.
What I was asking was, with a standard silicon diode, if I have a signal in the low state (nominally 0 V) which is a bit noisy (e.g. varies randomly between 0 and 0.2 V), my understanding is that a diode will filter this, so a voltmeter would indicate exactly zero volts until 0.6 V is reached, at which point it will read whatever the actual voltage is minus 0.6 volts. Correct?
-FK
There wouldn’t really be any need to do that, since most, if not all, logic chips already have circuitry to “clean up” the input signal (this is called signal conditioning).
No. A diode doesn’t suddenly “turn on” at 0.6 V; there’s a knee in the I-V curve, and often it’s not very sharp. Even if it were very sharp, I probably wouldn’t use a diode to do the job of detecting signals above 0.6 V. There are better ways of doing it, e.g. using a comparator circuit. Besides, as Q.E.D. stated, logic gates already use predetermined threshold voltages for defining logic states, and they often do a pretty good job of filtering noisy signals. Some even use hysteresis to help reduce false signals.
What, exactly, are you trying to do?