I am wondering if the energy loss in a voltage drop over an entire circuit is equal to the energy transferred from the circuit (e.g. light/sound/mechanical energy/heat).
Yes. Energy is conserved. Any electrical energy that is lost by the circuit is transformed into some other kind of energy.
It is often heat, for example in a resistor or a diode.
It could be something like chemical energy, in a battery charger.
If this is true, wouldn't that make it so that the voltage drop in a short circuit would be 100% since all of the electrical energy must be converted into something else?
Yes, the energy lost from the circuit will be converted to heat. The wire forming the short circuit will heat up (because of its resistance, even though that resistance is too small to matter in more useful circuits), as will the battery (because of its internal resistance).
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
The forward biased semiconductor junction takes a voltage level to be able to push the electronic charges over the P-N zone. Think of it as similar to how you have to "lift" each marble up to the table top from on the floor. In addition to the energy level difference needed to transport charge across the junction there is also a resistive part of the diode that drops some voltage as well. The resistive drop in the diode will be dependent upon the amount of current flow allowed through the junction.