adding a series resistor to clipping diodes

[m7b52000]

This is not a post about Si vs Ge at equivalent gain/volume levels!

Adding a series resistor appears to increase the forward voltages of a diode pair but my question is: does the voltage curve just get shifted to the right or is the shape/gradient of the curve altered as well. I can't find a clear answer in the forum.

cheers

[Rob Strand]

Adding a series resistor appears to increase the forward voltages of a diode pair but my question is: does the voltage curve just get shifted to the right or is the shape/gradient of the curve altered as well. I can't find a clear answer in the forum.

There's been quite a few posts over the years.

Adding a resistor isn't the same as adding more diode in series.

When you add a resistor is sort of adds the clean signal to the clipped diode signal.   That's pretty much the case for feedback diodes.  For post opamp clipping like, say like an MXR distortion or a RAT, the opamp is clipping so the series resistor adds the clipping signal to the diode signal; there is no clean.   If the opamp *is* clipping the series resistor does kind of tweak the level.  When you back things off so thing are *just* clipping you can get differences.   At the end of the day these things are what they are, try it, like it/hate it/indifferent.

[m7b52000]

Rob, thanks. I am just considering the case of post BJT clipping e.g Electra type circuit. I follow what you say but when I measure the Vf of a diode and resistor in series there is a measurable increase in the Vf. As R increases the Vf increases. In that sense it resembles adding a second series diode?

[R.G.]

Old question, hasn't been looked at live on the forum for a while.

Semiconductor diodes have an incremental forward resistance (that is, incremental change in forward voltage divided by the incremental forward current) that is quite low - from fractions of an ohm to several ohms, usually. What  diode does is change from a high incremental resistance at voltages too low to make it conduct, over to the low forward resistance as the voltage increases.
There is always some resistance/impedance inside the signal source driving the clipping diodes. At voltages too low to turn on the diodes, the source impedance and diode forward impedance form a resistive divider, but the diode's incremental resistance is too high to make any difference. When the signal voltage gets big enough to start turning on the diodes, the diodes' resistance starts flipping between off (high resistance) and on (low resistance) as the signal level varies. This changes the equivalent voltage divider of the source resistance and diode resistance between not much if any signal loss to nearly complete loss of any voltage higher than the diode forward conduction voltage. So the signal appears clipped at the forward conduction voltage of the diode. See footnote below about what happens in the actual "knee" of the diode's conduction.
If you insert a resistor in series with the diode: the diode does the same thing it always did, it is high resistance below its conduction voltage, and low resistance above its conduction knee. But in this case, the voltage divider consists of the source impedance and the sum of the resistor and the diode resistance. The diode has the same low resistance as always at any given conduction current, but the resistor in series with it means that the divider resistance of the diode+series resistor can't ever go lower than the series resistance. On a V-I curve basis, the diode by itself looks like a zero(ish) current right up to the conduction start, then a turn for higher and higher currents with a slope of 0.1 to a few amperes per volt. The series reistance makes that slope after the diode turns on be 1/the resistance amps per volt.
If what you meant was does the voltage at start of conduction change, the answer is no. The diode still does what it always did, so the start of conduction (and the start of change of voltage divider) happens at the same voltage. But the resistor in series limits how hard the diode can clip, so the signal above the diode conduction is slanted to the right; the signal above the diode conduction has a lower loss than just the diode itself.

Did that help clear it up, or muddy it up even more?

[m7b52000]

Thanks, R.G. So in very simple terms: adding a series diode will shift the curve to the right by 1 Vf whereas adding a series resistor will not change the voltage at start of conduction but decrease the gradient of the curve to the right of that point?

[R.G.]

Yes.
With the diode not conducting, the signal voltage all appears across the diode. When the diode starts conducting, the current in the diode+resistor goes up, and the resistor does V=I*R, while the diode stays stuck at whatever voltage its conduction curve says to be at that current.
I promised a footnote on the knee region and didn't type it in.
The knee of the diode is the region where the diode goes from substantially not conducting to a quite-low resistance. This region is small in most semiconductor diodes, about 25-50mV total. The diode current is actually an exponential function of the forward voltage, so the knee is rounded, not abrupt. Some diodes have a soft, sloppy, rounded knee, some have a sharper curve at the start of conduction. Some have higher internal wire resistances, and some have different bulk semiconduction resistance, These and a few other more abstract parameters are responsible for the wide and somewhat bewildering variety of comments on which diode sounds great, which doesn't.
In the knee, the diode's shifting resistance can be exploited. For signals under about 25mV, the distortion of a signal traversing that small part of the diode curve is under 1% or so. Diodes can be arranged in a bridge where signal is injected and received on the two sides of the bridge while current is fed through the other two junctions. Done well, it's a surprisingly good variable attenuator. Thomas Organ used this modulator as the basis for its tremolo in most of the USA Vox amplifier line. A cheaper version can be made with a single diode or two in series, but you lose the cancellation of the modulating LFO signal that the bridge approach gives you.

[antonis]

I can't find a clear answer in the forum.

Maybe 'cause resistor is considered linear item while diode isn't..
(Ohm's law contrary to Shockley equation..)

[m7b52000]

Thanks again. I am gaining a small amount of wisdom...