Quote from: PRR on Today at 05:23:48 PM

Quote from: Steben on Today at 12:31:45 PM

Quote from: Vivek on Today at 12:17:47 AM

...if we place many diodes in series, the clipping threshold changes but the total effective resistance of multiple series diodes when conducting is same as one diode?

Mmmm, yes and no.... normally overdrive circuits tend to have several tens of kohms at max gain on the feedback loop and a couple of kohms in the leg to ground. The internal resistance of diodes is couple of ohms max. ...

The slope does add-up with series diodes. As you say, with limited voltage the current is less and this too increases diode impedance. Diode impedance can be many-K, higher than typical circuit impedance. Shockley's Law predicts 26r@1mA and 26k@1uA.

I was expecting higher slopes

The slope Vd vs Id for a single diode is rd = n VT / Id,

where VT = is the thermal voltage which is about 26mV at room temperature, Id is the diode current, and n is the ideality factor.

The slope is derived from the exponential diode relationship.

iD = I0 exp ( vD / (nVT) )

So,

vD = n.VT ln(iD / I0) = nVT ( ln(iD) - ln(I0))

Take the derivative,

rd = d vD / d iD = nVT/iD

The result is independent of vD, so if you put three diode in series to make a compound diode with voltage uD = 3 *vD, the VI equation is now

iD = I0 exp ( uD / (3 nVT) )

If you derive the slope it's the same maths but now get rd_3diodes = 3 * rd_1diode.

The difference between diodes is in the ideality factor n. A BJT junction has an n just above 1 whereas a 1N4148 signal diode tends to be around 1.8, +/- 0.2 depending on how it's derived.

EDIT: Dumb arsed mistakes fixed.