Voltage drop across a diode
Let me explain a mathematical origin of the magical number 700 mV. This may help you to understand what's wrong with your question.

Let \$V_T = \frac{kT}{q} \approx 26\,\text{mV}\$ at room temperature. For \$V_d &gt; V_T\$ we can rewrite the diode equation this way:
$$I = I_s \left[\exp\left(\frac{V_d}{V_T}\right) - 1\right]
\approx I_s \exp\left(\frac{V_d}{V_T}\right)
= 1\,\text{A}\cdot\exp\left(\frac{V_d + V_T\log(I_s)}{V_T}\right)$$

A typical value of \$I_s\$ for a silicon p-n diode is around \$1\cdot10^{-12}\,\text{A}\$, so \$\log(I_s) \approx -28\$.

The numerator becomes positive if \$V_d &gt; -V_T\log(I_s) \approx 28\cdot26\,\text{mV} = 728\,\text{mV}\$. Nothing abrupt happens at this point; just a border between negative and positive values of \$x\$ in \$\exp(x)\$, where \$x = \frac{V_d + V_T\log(I_s)}{V_T}\$.

Define \$V_d = 728\,\text{mV}\$ as "the threshold voltage", and the magical number is ready.