# Practice Diode Problem From Old Test

Discussion in 'Homework Help' started by Sleepcakez, Jun 26, 2010.

1. ### Sleepcakez Thread Starter New Member

Jun 26, 2010
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So the circuit has an input Vi = 1.4 + 0.1cos(wt) V.
Both Diodes have a turn on value of .7V
There is an initial current of 17mA running through Diode 1.
n = 1.2

I am supposed to find the DC current through Diode 2 and then I have to find the small signal resistance r1 and r2.

I do not have the answer to the problem, and I do not believe I'm going quite in the right direction on this. At first I went and solved out that the current running through the 1k resistor was .7mA, and the current running through D2 was 16.3mA. I don't have my work in front of me right now but I know I used the genero rd equation to try and solve for r1 and r2.

Thanks for any help.

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2. ### t_n_k AAC Fanatic!

Mar 6, 2009
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If the DC current in D1 is 17mA then what will be the DC current in D2?

Note that the 1k resistor across D2 will take 0.7mA - so how much is left from 17mA to flow in D2?

Once you know the DC currents in each diode you calculate their effective dynamic resistances.

$r_{dynamic}=\frac{nV_T}{I_{diode$$DC$$}}$

where

$V_T=\frac{kT}{q}=0.026 \ at \ room \ temp$

3. ### Sleepcakez Thread Starter New Member

Jun 26, 2010
13
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Where did you get the equation Rdynamic? Did you manipulate some equations to obtain that or what? I was unable to find that in my book. Because I knew the question wanted me to use (n), I figured I needed to use the ideal diode equation since I saw no other equations in my book that involved (n). That being said:

So the dc current running through D2 is 16.3mA.
Using that equation to solve for Rd1 and Rd2 I obtain the following:
Rd1 = (1.2)(.026v)/(.017A)
Rd2 = (1.2)(.026v)/(.0163A)

Solve for each of those and I'm done?
I really didn't take this problem for being this easy after working on it for so long, although in my first few attempts I did get pretty close to going in that direction.

4. ### Georacer Moderator

Nov 25, 2009
5,151
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The formula for the diode dynamic resistance is pretty much given in any textbook. The steps for getting to it are the following:
We know that $i_{\small{D}}(t)=I_{\small{D}}e^{\small{v_{\tiny{d}}/nV_{\tiny{T}}}}$
The small signal approach supposes that $\frac{v_{\small{d}}}{n{\cdot}V_{\small{T}}}<<1$
and the above equations can be rewritten as its Laplace Transform as $
i_{\small{D}}(t)=I_{\small{D}}\left(1+\frac{v_{\small{d}}}{n{\cdot}V_{\small{T}}}\right)$

wich can be groupped as follows:$i_{\small{D}}(t)=I_{\small{D}}+\frac{I_{\small{D}}}{n{\cdot}V_{\small{T}}}{\cdot}v_{\small{d}}$
Now we can see that the total current is composed by a DC current and an AC component wich has the following formula: $i_{\small{d}}=\frac{I_{\small{D}}}{n{\cdot}V_{\small{T}}}{\cdot}v_{\small{d}}$ (note: in the denominator it says Vt, I don't know why TEX keeps skewing it)
As a result, the dynamic resistance is $\frac{v_{\small{d}}}{i_{\small{d}}}=r_{\small{d}}=\frac{n{\cdot}V_{\small{T}}}{I_{\small{D}}}$

I hope this is clear enough.

5. ### Sleepcakez Thread Starter New Member

Jun 26, 2010
13
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Thank you very much for the reply. It's about 5:30 a.m. here right now, so when I wake up I'll check out all that derivation.

6. ### Ron H AAC Fanatic!

Apr 14, 2005
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What does the (t) refer to in $i_{\small{D}}(t)$? The equation I am familiar with is $i_{\small{d}}=I_{\small{s}}e^{\small{v_{\tiny{d}}/nV_{\tiny{T}}}}$

I would have just solved for Vd and taken the derivative with respect to Id.

7. ### Georacer Moderator

Nov 25, 2009
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1,266
I' m glad you asked! This is a point that really got me thinking when I started studying circuits. What I 'm about to say is an exact transfer of what Sedra/Smith teaches:

The (t) is just meant to show that the diode current is a function of time and not a constant. But the real juice of the analysis is coming up next:

When describing a quantity (voltage or current) we will be using the following assumptions (d refers to diode and can be substituted with any letter):
$x_{\small{d}}$ refers to a time-variable quantity (AC signal)
$X_{\small{D}}$ refers to a constant quantity (DC signal)
$x_{\small{D}}$ refers to the sum of the above, the overlay of the AC and the DC signal.

For our example, the AC signal is the small signal applied to the diode and the DC is the bias signal.

The equation you are familiar with is written, with respect of the above:
$i_{\small{D}} =I_{\small{S}}{\cdot}e^{v_{\tiny{D}}/n{\cdot}V_{\tiny{T}}$
wich can be rewritten by distinguishing the DC and the AC components:
$i_{\small{D}}=I_{\small{S}}{\cdot}e^{(V_{\tiny{D}}+v_{\tiny{d}})/n{\cdot}V_{\tiny{T}}} {\Leftrightarrow}$
$i_{\small{D}}=I_{\small{S}}{\cdot}e^{V_{\tiny{D}}/n{\cdot}V_{\tiny{T}}}\ {\cdot}\ e^{v_{\tiny{d}}/n{\cdot}V_{\tiny{T}}}$
and replacing $I_{\small{D}}=I_{\small{S}}{\cdot}e^{V_{\tiny{D}}/n{\cdot}V_{\tiny{T}}}$ in the above equation, we get:
$i_{\small{D}}=I_{\small{D}}{\cdot}e^{v_{\tiny{d}}/n{\cdot}V_{\tiny{T}}}$

This equation conveniently relates the dynamic response of a diode in a small signal, when the diode is biased by a given (or known) dc current.

Last edited: Jun 27, 2010
8. ### Ron H AAC Fanatic!

Apr 14, 2005
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I still think solving for Vd and differentiating WRT Id is simpler.

Nov 25, 2009
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10. ### t_n_k AAC Fanatic!

Mar 6, 2009
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Hopefully that question is resolved in your mind.

Seems that was all that the question was asking. So yes you are probably done.

However, I'm not sure why in the posted circuit, the source has a small signal ac component superimposed on the DC. Was there no other question in relation to the ac voltage across the diode or something of that nature?

11. ### Georacer Moderator

Nov 25, 2009
5,151
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Maybe the teacher has my formula in mind, wich can only be applied if the AC component of voltage on the diode is under 10mV (sufficiently small).

12. ### Sleepcakez Thread Starter New Member

Jun 26, 2010
13
0
I have no clue as to why the circuit was presented in that form. Literally all that was asked was to find the current through Diode 2 and the small signal resistance of Diode 1 and 2.

Edit: I used LTSpice to draw that circuit up. On the problem itself, it was shown as a single sinusoidal source *1.4 + 0.1cos(wt)*

Last edited: Jun 28, 2010