# Relation between voltage drop across incremental resistance and VD of diode

In small signal analysis of diode (silicon) we take voltage drop across diode (VD ) is 0.7 volt. using other data we calculate current through diode " I " and then incremental resistance $r_d = \frac{nV_t}{I}$ . As it is a resistance there must be a voltage drop across it which is $nV_t$ (constant for a certain temperature). What is the significance of this voltage drop ? Does it add up with 0.7 voltage drop ? Is there any other noticeable effect of this voltage drop ?

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.

The problem is that you are mixing two models.

The first one is known as the ideal diode in series with voltage source, and that's your $V_\gamma=0.7V$. Remember that $V_\gamma$ has no physical meaning, it is just chosen so that the average diode characteristics looks a bit like the ideal one.

The second one is the Shockley model, that is more precise and allow you to compute things like the small signal resistance $r_d$.

Is there a mid point? Actually, yes, it's called piecewise linear model. What you basically do is fit the exponential with two or three segments to simplify calculations and avoid the huge errors you might encounter with the first model.

• But in my book many circuit is analysed by using Vγ=0.7V to find the current I and then use this I to find rd.Is't it using two model for one circuit? – Anklon Apr 20 '14 at 18:22
• Well, let's say that using $V_\gamma$ is just convenient. You can either solve the circuit using the Shockley model or say "ok let's assume this voltage drop", that is a lot easier. The trick is that $V_\gamma$ is correct enough in most of the cases, for most of the diodes, so you are speeding up your calculations without affecting precision. $V_\gamma$ belongs to all the diode models in some ways, it's just an agreement. – Vladimir Cravero Apr 20 '14 at 18:26

The voltage-plus-ideal-diode model is a large-signal model. It's perfectly valid to use the large signal model to find the operating point, then use the small signal model. For example, if you were asked to find the AC voltage here:

simulate this circuit – Schematic created using CircuitLab

The DC current through D1 is about 4.3V/10K or 430uA. Say n = 1.836 for the 1N4148

So, $r_d$ = 111 ohms. The output voltage (assuming the capacitors do not affect the AC voltage) is going to be $V_{OUT} = \frac {r_d \cdot 100mV}{(r_d + 10K)}$ (ignoring R1) or approximately $V_{OUT} = \frac {r_d \cdot 100mV}{10K}$

We'd expect output voltage of 1.11mV, and indeed when we simulate it in PSPICE, the result is 1.06mV, with an average diode current of about 443uA, accounting for the slightly lower than expected output voltage.