In small signal analysis of diode (silicon) we take voltage drop
across diode (VD ) is 0.7 volt.
That's not quite correct. In small signal analysis, one linearizes about the operating point so, in fact, no assumption is made about the DC operating voltage - one should in fact solve for the operating point.
What is the significance of this voltage drop ? Does it add up with
0.7 voltage drop ?
No, to see the significance, let's review small-signal analysis. First write the total diode voltage as the sum of a constant and a time varying component:
$$v_D = V_D + v_d $$
where \$v_D\$ is the total voltage,\$V_D\$ is the DC (time average) voltage, and \$v_d\$ is the AC voltage.
Next we assume that the total voltage is at all times not very different from the time average which allows us to do small signal analysis in the first place.
The following is the justification for this approach.
The ideal diode equation is (assuming significant forward diode current)
$$i_D = I_S e^{\frac{v_D}{nV_T}}$$
Setting \$v_D = V_D + v_d\$ in the above yields
$$ i_D= I_S e^{\frac{V_D + v_d}{nV_T}} = I_S e^{\frac{V_D}{nV_T}}e^{\frac{v_d}{nV_T}} = I_De^{\frac{v_d}{nV_T}}$$
where \$I_D\$ is the DC diode current.
Expanding the exponential in a Taylor series yields
$$i_D = I_D (1 + \frac{v_d}{nV_T} + \frac{1}{2}(\frac{v_d}{nV_T})^2 + ... )$$
Now, here's the crucial move. If we assume \$v_d\$ is small enough, we can ignore the 2nd order and higher terms in the expansion yielding
$$i_D \approx I_D (1 + \frac{v_d}{nV_T}) = I_D + \frac{I_D}{nV_T}v_d = I_D + \frac{v_d}{r_d} = I_D + i_d$$
where
$$r_d = \frac{nV_T}{I_D}$$
Thus, assuming \$v_d\$ is small enough, this linear model gives good agreement and allows us to find the total diode current by superposition of the DC current and the small-signal current.
As it is a resistance there must be a voltage drop across it which is
nVt
It isn't a resistance. As shown above, \$r_d\$ is the ratio of the small-signal voltage \$v_d\$ to the small signal current \$i_d\$ which means
\$r_d\$ is the inverse slope of the diode IV curve at the operating point; it is the dynamic resistance at the operating point.
Best Answer
I pretty much agree with your approach. Setting \$I_\text{CAL}=1\:\text{mA}\$ and \$V_\text{CAL}=700\:\text{mV}\$ and given \$\eta=1\$, the complete equation (using only Shockley diode equation) I get it, is:
$$R=\frac{10\:\text{V}-3\:\text{V}}{I_\text{CAL}\cdot\frac{e^{V_D/V_T}-1}{e^{V_\text{CAL}/V_T}-1}}$$
(Noting that \$V_D=750\:\text{mV}\$.)
Discounting the \$-1\$ terms, this simplifies a little:
$$\begin{align*}R&\approx\frac{7\:\text{V}}{I_\text{CAL}\cdot e^{{\left(V_D-V_\text{CAL}\right)}/V_T}}\\\\&\approx \frac{7\:\text{V}}{1\:\text{mA}\cdot e^{{\left(750\:\text{mV}-700\:\text{mV}\right)}/V_T}}\\\\&\approx 7\:\text{k}\Omega \:\cdot e^{{-50\:\text{mV}}/V_T}\end{align*}$$
Depending on the room temperature value for \$V_T\$ that you use (Spice defaults to about \$25.876\:\text{mV}\$), you will roughly get \$950\:\Omega\lt R\lt 1020\:\Omega\$.
I've rounded your value, which I take to be at the low end of this range. But I think your approach is solid.