# Voltage divider equation tunnel diode - Art of Electronics

Early on in the book "The Art of Electronics" (p.15, 2nd edition) a voltage divider circuit is described: an input voltage $V_{in}$ over a tunnel diode $D$ and a resistor $R$, and an output voltage $V_{out}$ the voltage across the resistor.

It is then written that a change in $V_{in}$, denoted $v_{sig}$, results in an (amplified) change $v_{out}$ in the output:

$v_{out} = \frac{R}{R + r_{t}} \cdot v_{in}$

where $r_{t}$ is (negative) dynamic resistance of the diode $D$.

Unfortunately no derivation is given.

Now I have been trying for the last two hours to derive this form but keep getting stuck. Unfortunately there seem to be no leads on the internet. Could someone derive the formula (in a clean manner)?

• It appears you are having difficulties with the small signal model. I suggest you tune your interned search on that term. Basically, you set a quiescent point (Vq,Iq) on the diode V-I characteristic and then create a new v-i frame centered on that point. Then, if you linearize the characteristic there, you will have a straight line passing through the v-i origin - that's the characteristic of a resistance. It is dynamic since it refers to v (variation of voltage about Vq) and i (variation of current about Iq). Its value also depends on the quiescent point chosen - it's the slope of the char. – Sredni Vashtar May 11 '18 at 21:32

There will be a small signal AC current through the two devices

$$i = \frac{v_{in}}{R+r}$$

Therefore the output voltage (the voltage across the resistor) will be given by

$$v_{out} = iR = \frac{R}{R+r}v_{in}$$

• How did you derive/come up with the first equation? For a voltage divider with two resistors $R_1$ and $R_2$ it is clear to me that $i = \frac{v_{in}}{R_1 + R_2}$. However with the diodes resistance depending on the voltage, it is not clear to me how the dynamic resistance enters the equation. The second equation you state is clear. – balletpiraat May 11 '18 at 20:50
• "Small signal" means the amplitude of the signal is small enough that $r$ doesn't change (significantly) due to the signal itself. – The Photon May 11 '18 at 20:52
• Intuitively the change in output voltage follows from the change in the resistance of the diode AND the change in current. But how to derive the exact form? See my answer below. – balletpiraat May 11 '18 at 22:47
• In a small-signal model, it doesn't depend on changes in $r$. The whole point of a small-signal model is to ignore nonlinearity for small variations about a bias point. If you're coming from a physics background you might have seen the same thing under the name of "perturbation theory". I'm sure Horowitz and Hill didn't jump through all those hoops just to get back to the same result as the small signal model. – The Photon May 12 '18 at 0:32

I came up with the following solution which seems sound. Note, I am interested in a clear mathematical derivation from simple premises.

Consider the current through the diode and resistor

$I_i = \frac{V_i}{R+D_i}$

where $D_i$ is the diodes resistance at voltage $V_i$ and current $I_i$.

We define

$\Delta I := I_2 - I_1 = \frac{V_2}{R+D_2} -\frac{V_1}{R+D_1}$ where

$\Delta V := V_2 - V_1$

and make the linear (small signal) approximation for the diodes resistance:

$D_2 := D_1 + r$.

If (a) $D_1 \ll R$ and (b) $\| r \| \ll R$ it follows that

$\Delta I = \frac{V_2}{R+D_2} -\frac{V_1}{R+D_1} = \frac{V_1 + \Delta V}{R+D_1+r} -\frac{V_1}{R+D_1} = \frac{V_1}{R+D_1+r} -\frac{V_1}{R+D_1} + \frac{\Delta V}{R+D_1+r} \approx \frac{\Delta V}{R+r}$

In the textbook the I-V curve of the diode makes one believe it operates at around 0.2 volts and 7 mA with $r= -6 \Omega$ and $D \approx 7 \Omega$.

Given a mean input voltage of around 20 volts the total resistance must be around $3K \Omega$ s.t. premises (a) and (b) are satisfied!

Correction: I see that I completely mis-understood the concept of resistance stating $D_2 := D_1 + r$.