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Consider this circuit which consists of a DC bias and an AC source connected with a diode and resistor:

Boylestad and Nashley book 11th ed. Source: Boylestad and Nashley book 11th ed.

Using superposition theorem the circuit was solved.

In the book,the DC resistance across the diode can be found to be around 150 Ohms, whereas the AC resistance (dynamic resistance) is only 8 Ohms which was found using the formula:

$$r=\frac{26\mathrm{mV}}{i_d}$$

where \$i_d\$ is the current through the diode.

Similarly, the DC voltage across the diode was assumed to be 0.7 volts which is the barrier potential, while the AC voltage across the diode was only 0.01 volts!!

What is the reason behind this weird behaviour? If the DC voltage across the diode is 0.7V, then to my thought it appears that AC voltage must also be around 0.7V.

Questions

  1. Why is the DC resistance high in the circuit whereas the AC resistance is low?
  2. Why does the diode have such a low (0.01 volts) AC voltage across it? or why does the diode have higher impact on DC voltage while lower impact on AC voltage?

I also say that the above behavior is almost true in all the p-n junction diode circuits which I've seen.

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  • \$\begingroup\$ I am actually typing an answer when I think to myself... Where do these questions come from? Because it sounds a little like home-work, since the book should actually explain this near the schematic if it isn't... \$\endgroup\$ – Asmyldof Jul 21 '15 at 20:28
  • \$\begingroup\$ 2Vp-p (peak to peak) is 0.707Vrms for sinusoidal signals. Look for datasheet of any rectifier silicon diode and find forward current vs. forward voltage graph - for such a small currents forward voltage is also very very small, around a few milivolts not 0.7V! Remember that dynamic resistance is relation of forward voltage to forward current and it varies with them! \$\endgroup\$ – Jakub Rakus Jul 21 '15 at 20:45
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    \$\begingroup\$ @lustful-rat In the question the diode is sufficiently forward biased due to the DC offset. So the AC analysis should be done around the operational point. \$\endgroup\$ – Eugene Sh. Jul 21 '15 at 20:48
  • \$\begingroup\$ @Asmyldof, this sounds like a conceptual question, not a homework question. The homework part would be calculating the voltages and currents, and that part is already mostly done. \$\endgroup\$ – Adam Haun Jul 21 '15 at 21:21
  • \$\begingroup\$ @Adam That what I actually meant, I want to understand this conceptually \$\endgroup\$ – Andrew Flemming Jul 21 '15 at 22:23
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1) The DC "resistance" is symply \$V/I\$, which in your case is \$0.7\cdot2k / (10 - 0.7) = 150\Omega\$. The dynamic resistance is \$\frac{dV}{dI}\$ of the diode, which can be easily found on the diode datasheet. Like this, for example:

http://www.davidbridgen.com/images/diode.gif

Here \$\frac{dV}{dI}\$ is the inverse of the slope of the line around the \$0.7V\$ point. As you can see the line is almost vertical, so the slope is very big. So the inverse is very small.

2) The voltage across the diode is not \$0.01V\$. It is \$0.7\pm 0.01V\$. But since the dynamic resistance is the derivative, the DC component is going away.

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  • \$\begingroup\$ +1 for your great explanation. Learn't some things and also found if we use only ac sources the results would change a bit.Anyway thanks a lot dude. \$\endgroup\$ – Andrew Flemming Jul 21 '15 at 22:58
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This isn't really superposition. Superposition only works in linear circuits, and diodes are nonlinear. It takes about 0.7 volts to "turn on" the diode, but after that a small change in voltage gets you a large change in current. Dividing the DC voltage by the DC current doesn't really tell you anything. To get a linear circuit, you have to make an approximation like this:

schematic

simulate this circuit – Schematic created using CircuitLab

Now superposition will work. In the AC analysis, both the 10V DC source and the 0.7V drop in the diode disappear, and you're left with only the 8 \$\Omega\$ equivalent resistance.

This is a simplified version of a common technique for dealing with nonlinear things:

  1. Use the full nonlinear behavior ("large-signal model") to find the operating conditions at for a constant (DC) input. Here, that's your 10V DC offset.

  2. Make a linearizing approximation ("small-signal model") around the DC operating point.

  3. Calculate the effect of a small input variation (AC) using the linear approximation.

  4. Add the DC and AC results together.

You'll be doing this a lot with transistors, so you might as well get comfortable with it now. :-)

In a diode, the nonlinear current/voltage relationship is given by the Shockley equation:

$$I_D = I_S (e^{\frac{V_D}{nV_T}} - 1) \approx I_S e^{\frac{V_D}{nV_T}}$$

If \$n \approx 1\$, the linear approximation is:

$$\frac{dI_D}{dV_D} = \frac{1}{V_T} I_S e^{\frac{V_D}{V_T}} = \frac{I_D}{V_T}$$

We want a resistance, which is voltage divided by current, so we flip the fraction upside-down. \$V_T\$ is about 26 mV at room temperature, so:

$$r_d \approx \frac {V_T} {I_D} \approx \frac{26\ \mathrm{mV}} {I_D}$$

Here's a graph showing what I'm talking about. The numbers don't match your problem, but the behavior is the same.

Diode I-V curve with bias point and linear approximation shown

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  • \$\begingroup\$ Just a small - but in my view: important - comment: The above derivation of a dynamic (differential) resistance applies for small signals only. And for this reason we should use a small-letter symbol only (rd instead of RD). According to my experience with students it is very important to use different symbols for static and dynamic resistances. All electronic circuits are non-linear and contain external (ohmic) resistors. Hence, it can cause confusions if we do not distinguish between static and dynamic (small-signal) resistances. \$\endgroup\$ – LvW Jul 22 '15 at 7:31
  • \$\begingroup\$ Good point. I changed RD to rd in my answer. \$\endgroup\$ – Adam Haun Jul 22 '15 at 12:56
  • \$\begingroup\$ (+1) Even just for pointing out that that is not superimposition (I'd like to click +10)!!! I've seen tons of crappy books and articles (and blog/forum posts) explaining linearization techniques as application of "superimposition". \$\endgroup\$ – Lorenzo Donati Apr 21 '18 at 15:00
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The AC resistance in the total circuit is not very much different than the DC resistance, as it is drawn.

The AC resistance is low measured over the Diode. The 2k resistor is 2k. Point. At least, in the theoretic domain. Real world always has some tiny gotcha's and such, but I'm leaving that out for the remainder.

Why is the AC effect or AC resistance of the diode low?

Simple! The diode is in forward conduction for all voltages that the AC source generates. If the AC source is 2Vpp, it goes from +1V on the top compared to the bottom to -1V on the top versus the bottom. So at it's most positive peak the total voltage is 11V, but at its lowest low peak, the total voltage is 9V.

In both cases the current flows clockwise through the circuit. So the only effect the AC voltage has in the Diode is a small change (9V to 11V isn't a huge jump at all!) in the forward current. The forward current through the diode does change its forward voltage, but not by a huge degree.

So apparently they calculated the forward voltage at 9V and the forward voltage at 11V from some data not known to me at this time, and the difference is 26mV between them. So the only effect the diode has on the AC component is 26mV, which gives the resistance, which makes that difference 26mVpp, which is on average a 13mV deviation from the DC component, which is 0.7V.

So the DC voltage drop of the diode then is 0.7VDC and the AC voltage drop effect is 13mVAC.

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