# Diode Questions - (I just don't get it)

1. Sep 23, 2006

Q:
In the circut below, $I$ is a dc current and $v_s$ is a sinusoidal signal. capacitors $C_1$ and $C_2$ are very large; their function is to couple the signal to and from the diode, but block the dc current from flowing into the signal source or the load (not shown). Use the diode small-signal model to show that the signal component of hte output voltage is:
$$v_0 = v_s \frac{nV_T}{nV_T + IR_s}$$

I really do not know what to do here. I need a hint of some sort, cause I'm lost. I'm thinking that the capacitors will go away, but I don't know if that's right... and I don't know why that is the case.

How are the capacitors blocking the dc current from flowing into the signal source?

I don't understand this circuit!

Last edited: Sep 23, 2006
2. Sep 23, 2006

### Gokul43201

Staff Emeritus
What is the impedance of a capacitor at frequency f=0?

PS: What do the terms n and V_T refer to? There's no mention of them in the body of the question.

3. Sep 24, 2006

Impedence of a capacitor is $\frac{1}{j\omega C}$, as the frequency drops to zero, the capacitor becomes an open circuit.

Sorry for not defining n, and V_T. I wasn't sure if it was a standard or not. They are characteristics of the books model for a semiconductor junction diode.

$$V_T$$ is the thermal voltage
$$V_T = \frac{kT}{q}$$
k, boltzmann's constant
T, absolute temperature
q, magnitude of electronic charge

$$n$$ has a value between 1 and 2.
Diodes using the standard integratedcircuit fabrication process exhibit n=1 when operated under normal conditions. Diodes available as discrete two-terminal components generally exhibit n=2.

4. Sep 24, 2006

### doodle

Begin with the dc analysis. The dc current I is blocked from flowing back to the source or to the output due to the capacitors. Thus its only path is to flow through the diode. This gives, rd = n*VT/I as the small-signal resistance of the diode.

Then, proceed with the small-signal (ac) analysis. In this part, you can assume that the capacitors are shorted and the dc current source opened (deactivated). It should then be easy to derive the output (ac) voltage with respect to the supply voltage.

5. Sep 24, 2006

Let me run this by you, to see if I'm understanding this properly.

When we first find $I_D$, with the DC analysis, we set the frequency to 0. This opens the capacitors, and the DC current flows fully across the diode, thus giving $I_D$ as you said. This makes sense :) thank you.

Now, we replace the diode with rd right? So why is it that we can assume that the capacitors are shorted? I don't undrstand this. Why is the DC current source also deactivated?

I see if I short the capacitors, and deactive the current how to get the output voltage. I don't understand why this is happening.

6. Sep 24, 2006

### doodle

The general approach to solving circuits having both dc and ac components is to separate the analysis for the dc and ac parts. Thereafter, you will use superposition to sum up the dc and ac responses. Separating the dc and ac analyses invariably requires all ac sources to be deactivated in dc analysis, and vice versa. This explains...
As for...
If the capacitance of capacitors is big enough, they will act like shorts. Recall that the impedance of capacitors is, after all, inversely proportional to its capacitance. Furthermore, as the capacitors used in the circuit are explicitly labelled as coupling capacitors, it should be safe to regard them as shorts in the ac analysis.

7. Sep 24, 2006

thank you so much man!! It's friggen nice to actually understand something. Well, while I'm at it. What are coupling capacitors?

8. Sep 25, 2006

### doodle

Looking up in Wikipedia, I found this:
http://en.wikipedia.org/wiki/Coupling_capacitor" [Broken]

Last edited by a moderator: May 2, 2017
9. Sep 25, 2006

wow I feel like a jack a