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.
My, that's a lot of questions.
Presumably the diode model you instructor wants you to use is an ideal diode with 0.7V drop in series with a 4 ohm resistor. So, when the diode is conducting, it behaves like a voltage source with a resistor in series. When the forward bias is less than 0.7V it does not conduct.
Is the 4 ohms per diode negligible in comparison to 1.5K? Well, it's more than 0.5% for two diodes so it will drop tens of mV. That might be negligible or it might not be, depending on the application. Since the instructor gives you the value, I suggest it might not be negligible in terms of getting the correct answer.
Two such diodes in series behave like one 1.4V diode with 8 ohms in series.
Best Answer
Reverse bias puts a photodiode into photoconductive mode, as opposed to photovoltaic mode. This results in increased sensitivity, decreased capacitance, and increased speed. This is at the expense of increased dark current.
Good review at https://www.thorlabs.com/tutorials.cfm?tabID=31760
The best review I know of is the Sharp Application Note on Optoelectronics. I'm having trouble finding it on the Shape site, but one link is http://denethor.wlu.ca/pc300/projects/sensors/photdiod.pdf
In particular, that Sharp AN shows the voltage/current relationships as a function of the reverse bias.
FWIW, I'd recommend an active amplifier circuit for your photodiode just on general principles, but I have no idea what your application is.