I've been reading up on clipping diodes and understand how they work when there is only one. What I don't get is how they work when set in an anti-parallel fashion.
If there IS a straight path at all times where the signal can go across the diodes, why is there clipping at all?
And by that I mean, why doesn't THIS happen (see attachment)
With ideal diodes it would indeed be a useless circuit, but real diodes are
not ideal and have a forward voltage that's logarithmically dependent on
the current - as the voltage rises at the input to the resistor its current
will rise roughly in proportion, but the voltage across the diodes will soft-clip
around 0.5 to 0.7V. For most silicon diodes there is only uA flowing below 0.4V
so signals in the +/- 0.4V range will pretty much be undistorted.
Forward voltage in silicon diodes at room temperature rises about 20mV per
doubling of current - a factor of 1000 in current gives about 0.2V increase
until parasitic resistances start to take over as the device overloads.
Mark, are you saying they behave more like zeners in this combination?
MarkT:
...but real diodes are not ideal and have a forward voltage
As I was reading your answer I realized how silly my question was and had one of those a-ha moments.
Thanks!
cjdelphi:
Mark, are you saying they behave more like zeners in this combination?
Not at all, the zener effect is a totally different phenomena
See : Zener diode - Wikipedia
I understand zeners... I'm confused as to the clipping in the 0.6v range
A Zener starts acting once a certain amount of current is flowing through it and will continu to do so (more or less) until it's max current ability is reached.
A Sicilium diode has a voltage drop across it of about 0.6-0.7V. So in your circuit with the AC apllied to it always one of the diodes will be conduction. On the positive halve of the wave, the right diode will conduct. Since there is a voltage drop over the diode of 0.7V the rest of the applied voltage is lost in the resistor (in the form of heat).
In the lower half of the wave, the diode does the same thing. Thus reducing the wave to a clipped version of +0.7V - -0.7V).
A germanium diode has a voltage drop of about 0.2V btw.
For more precise mathemathical treatment you'll have to look at the diode equation
(aka Shockley ideal diode equation)
ahh! - finally, understand it lol.
Thanks 
cjdelphi:
I understand zeners... I'm confused as to the clipping in the 0.6v range
The 'clipping' is the result of the applied voltage being at or above the Forward Voltage Drop (Vf )of the specific diode and will be defined in the datasheet for the diode. It's value can change by small amount because of temperature variations or forward current variations.
