Relationship between Zener diode voltage and knee voltage in forward bias? Also the resistance in forward bias
The 'knee' voltage of a Zener diode refers to the point where voltage drops more rapidly as current is lowered. 

In reverse bias (ie. Zener or avalanche mode) the knee generally gets sharper and occurs at smaller current as the Zener's nominal breakdown voltage increases, but the relationship is not linear and may be different for different devices with the same nominal voltage. 

Here are the V/I curves for NXP BZX84 series Zeners rated from 2.4 V to 6.8 V. Below 5 V the knee becomes very 'soft' and occurs at higher current. The transition to a sharper knee above 5 V is due to the voltage regulating mechanism changing from the Zener effect to avalanche breakdown.

 

At higher voltages the knee becomes sharper as it gets closer to zero, and the breakdown voltage becomes more 'ideal'. Here are the curves for BZX84 C7V4 to C24:-



In forward bias a Zener has an I/V curve similar to a normal silicon diode, which can be approximated with the Shockley diode equation:-

$$I = I_S e^\frac{V_D}{nV_T}$$

where 
    \$I_S\$ is the scale current,
    \$V_D\$ is the voltage across the diode,
    \$V_T\$ is the thermal voltage, and
    \$n\$ is the ideality factor.

But is it right to call any place on this curve the knee? Here is the I/V curve for BZX84 series Zeners:-



This curve is obviously exponentional (as expected due to the exponential term in the equation), so if a different current scale was used the 'knee' would appear to occur at lower or higher current. However if plotted on a logarithmic current scale the curve would be almost a straight sloping line, not at all like the definite knee seen in reverse biased Zeners > 5 V.