Simple Diode Analysis Problem - Calculating Current/Voltage in a diode:

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aly34

Joined Jan 29, 2012
20
1. The problem statement, all variables and given/known data

Refer to the figure attached. Let IS = 25 fA and VCC = 10V. What is the total current, the current in the diode, and the current in the 5KΩ resistor? What is VD? Use the iteration technique.


2. Relevant equations

Ideal Diode Equation: I = Is(e^(Vd/Vt) - 1)
Ohm's Law: V= IR
Thermal Voltage: V = kt/q = 25.9mV



3. The attempt at a solution

Total Current in the circuit = 10 V / 10kΩ = 1mA

We need to find the Vd and Id using the constant voltage model and then use the iteration method using the ideal diode equation to get closer and closer to Vd and Id using iterations.

I am not sure how to analyze the circuit using the constant diode equation. I know how to use the iteration method.

Current through the diode = Current through the 5kΩ resistor.


Please help - Any suggestions will be helpful.

Thank you!
 

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t_n_k

Joined Mar 6, 2009
5,455
Consider firstly determining the Thevenin equivalent looking from the diode terminals.

The problem is the logarithmic function - there isn't a closed solution - hence the need to use an iterative solution method.
 
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Thread Starter

aly34

Joined Jan 29, 2012
20
Can we not use the constant voltage model?

That is how we learned about this in class?

Not sure how to calculate the Thevenin's equivalent to this circuit?
 

t_n_k

Joined Mar 6, 2009
5,455
Can we not use the constant voltage model?

That is how we learned about this in class?

Not sure how to calculate the Thevenin's equivalent to this circuit?
Sure you can use the constant voltage model for the diode.

So replace the diode with a battery (Vd=0.7V say) and solve for the circuit currents.
 

Thread Starter

aly34

Joined Jan 29, 2012
20
Sure you can use the constant voltage model for the diode.

So replace the diode with a battery (Vd=0.7V say) and solve for the circuit currents.

Ok - So when I do that I get a KVL loop like this:

10 V. = (10k)(I) + 0.7 V.

And I = 0.93mA

That is the current going around in the circuit - Correct?

So Vd = 0.7 V.? How do I find the current through the diode? Is it just 0.93mA?

And the current through the 5k resistor? That is also 0.93mA?

And then I would use the ideal diode equation to do iterations and get the current close to 0.93mA by changing the Vd value in the equation?


Thanks for your help so far! <3
 

t_n_k

Joined Mar 6, 2009
5,455
Yes - now find the diode current.

From there you would plug that current into the ideal diode equation and get new estimate for Vdiode.

With the updated value of Vd repeat the process to improve the estimate for I_diode & hence an updated Vd.

Keep going until you get a stable result for Vd & hence Id.
 
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Thread Starter

aly34

Joined Jan 29, 2012
20
Interestingly I found I needed only one iteration to obtain stable value.
Ok - So:

I = Idiode + Iload

0.93mA = Idiode + 0.14mA

So, Idiode = 0.73mA

And then plug 0.73mA into ideal diode and solve for Vd?

0.73mA = (25x10^-15)*(e^(Vd/25.9mV) - 1)

And solving for Vd we get 0.624 V?


How is the iterations done after this?

Thank YOU!
 
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Thread Starter

aly34

Joined Jan 29, 2012
20
Check your diode current value.
Oh! Sorry! Punched it in wrong! :]

I = Idiode + Iload

0.93mA = Idiode + 0.14mA

Idiode = 0.79mA

And then plugging it in the ideal diode equation and solving for Vd I got 0.626 V?
 
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panic mode

Joined Oct 10, 2011
1,865
new iteration is simply repeating the process but using more precise data (result of previous iteration).

for example in first iteration you assumed that Vd=0.7V, then did bunch of calculations and found it is not really 0.7 but more like 0.626V

so you repeat the whole thing using Vd=0.626V and you will get another result.

this can be repeated until calculated Vd closley matches assumed Vd.

This iterating process can take multiple attempts (5-15). Getting stable result (one that matches assumption) in second iteration is very good.
 

t_n_k

Joined Mar 6, 2009
5,455
OK now instead of starting the process by using Vd=0.7V you now repeat the whole process with Vd=0.626V in the circuit and re-estimate the three currents for the updated value of Vd=0.626V.

You then end up with a new estimate for the diode current. Again, plug that value into the ideal diode equation and come up with a revised estimate for Vd. If Vd changes only by a few mV then it's probably a reasonably stable estimate for the value.

If not, repeat the process one or more times and hopefully the difference in the estimated Vd decreases with each successive approximation. Decide on when near enough is good enough (within a few mV variation) - end of task.
 

Thread Starter

aly34

Joined Jan 29, 2012
20
new iteration is simply repeating the process but using more precise data (result of previous iteration).

for example in first iteration you assumed that Vd=0.7V, then did bunch of calculations and found it is not really 0.7 but more like 0.626V

so you repeat the whole thing using Vd=0.626V and you will get another result.

this can be repeated until calculated Vd closley matches assumed Vd.

This iterating process can take multiple attempts (5-15). Getting stable result (one that matches assumption) in second iteration is very good.

So for this problem - Getting Id = 0.79mA and then solving for Vd I got 0.626V. , is this not the most precise result I can get for this problem?

I'm not sure how to do the next iteration?


To answer the problem, i got the following values:

Total Current = 0.93mA
Diode Current = 0.79mA
Current through 5k (load) = 0.14mA
Vd value = 0.626 Volts


Is this not correct?
 
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t_n_k

Joined Mar 6, 2009
5,455
You'll find Vd=0.626V a pretty good estimate anyway once you repeat the process. I think at the next iteration it comes out with I_diode=0.812mA and hence Vd closer to 0.627V which seems stable enough to me.

Yep: Just noted you have it correct.
 

Thread Starter

aly34

Joined Jan 29, 2012
20
You'll find Vd=0.626V a pretty good estimate anyway once you repeat the process. I think at the next iteration it comes out with I_diode=0.812mA and hence Vd closer to 0.627V which seems stable enough to me.

Yep: Just noted you have it correct.


Awesome!

Thank you so much for your help!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
 

t_n_k

Joined Mar 6, 2009
5,455
You are welcome.

You may be able to get rid of those 'strange' posts back there. I think it depends on your post count as to whether you have the delete post permission or not.
 
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