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Engineering Voltage drop across a diode

Homework Statement
What is the voltages across each of these two diodes.
Homework Equations
none
2oppositediodes.png


So I have this circuit up above and I need to find the voltages across each of the diodes.
The only info given is that they are identical silicon diodes at T = 300K.
My first thought was that since the diodes are opposite, D2 would be in reverse bias and would act as an open. However, I realized that they are not ideal diodes, so there may be current passing through the diode and therefore a voltage drop.
I guess my question is am I overthinking the question? Is the voltage drop across each of the diodes just the standard 0.7 V drop for silicon diodes in general?
 

BvU

Science Advisor
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Hello marc, :welcome: !

0.7 wouldn't work: doesn't add up to 5 V !!

I don't understand what you mean with 'open' (D2).

What are the I-V characteristics you want to base your answer on ?
 
304
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Ireverse=Iforward.
Ireverse~Isaturation (Because 5V>>0.7V)
Dividing Shockley equation by Isaturation
exp(Vforward/(Vt*n))-1=1
Vforward/(Vt*n)=ln2
 

BvU

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Let me ask in another way: without D1 you would get 5 V over D2, right ? With a pretty small current (I hope).
What does the insertion of D1 change ?
 
Let me ask in another way: without D1 you would get 5 V over D2, right ? With a pretty small current (I hope).
What does the insertion of D1 change ?
In that case, the voltage across D1 is 0.7 and the voltage across D2 would then be 4.3 V?
 

BvU

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What current would that correspond to according to your expressions ?
 
What current would that correspond to according to your expressions ?
In this case, I think I would get current using the Shockley diode equation I = I_s(e^(V_d/V_t)-1). I would then plug in the voltages 0.7 and 4.3 for the diode voltage. Although if I do this, each diode would have different currents, and I'm not sure if this is possible as I'm used to loops having a single current value.
 

BvU

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There definitely is only one current in this circuit, so you can equate the I in the expressions for D1 and D2.
Out comes V over each of the diodes (VD1 and 5V - VD1 for example).

I don't know if the composer of your exercise intends that you go into this depth, or perhaps wants you to round off the current to zero.
 
There definitely is only one current in this circuit, so you can equate the I in the expressions for D1 and D2.
Out comes V over each of the diodes (VD1 and 5V - VD1 for example).

I don't know if the composer of your exercise intends that you go into this depth, or perhaps wants you to round off the current to zero.
My understanding is that it should be a simple exercise to prove that we comprehend the idea of diodes. The final question posed in this two-part exercise was to actually find the current in the circuit if the diodes break down at 4.5V. Given that I had to find voltage in the first part, and all the other variables in the diode equation are constants, I think my final answer would be the result of plugging my numbers into the diode equation?
 

BvU

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:smile: Too lazy to do the math. Does something really small come out for the current ?
 
I'm actually getting a relatively high number when I plug in the numbers. I = 10^-9 (e^(.7/.026)-1).
The constants I used were I_s = 1 nA and kT = 0.026 V which were given by my professor.
I'm getting around 493 A, which seems wrong.
 
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I'm actually getting a relatively high number when I plug in the numbers. I = 10^-9 (e^(.7/.026)-1).
The constants I used were I_s = 1 nA and kT = 0.026 V which were given by my professor.
I'm getting around 493 A, which seems wrong.
Incremental hint:
Vforward=Vt*n*ln2 = 18 mV for typical n=1
 
With this diode voltage, the current would then be 9.98 *10^-10 A
However, does this mean that the voltage drop for the diodes is 18mV? I'm not sure I understand since in class, we use the constant voltage drop model of the diode where there is a drop of 0.7 volts for a diode in forward bias. But according to that equation, the voltage drop is 18 mV.
 
My understanding of those data sheets is that voltage and current are dependent on each other. However, why was I taught in class to treat the voltage drop of a diode as a constant 0.7V (for silicon)?
 

berkeman

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Only at around room temperature and at a few mA of forward current. That's a fine way of thinking for the first introductory class in circuits. But that's not how diodes work in the real world. For real world circuits (or when you have a problem posed with nose-to-nose diodes), you will need to use the diode equation with reasonable values for Is from datasheets to help you figure the circuit out, IMO.
 

berkeman

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Homework Statement: What is the voltages across each of these two diodes.
Homework Equations: none

My first thought was that since the diodes are opposite, D2 would be in reverse bias and would act as an open. However, I realized that they are not ideal diodes, so there may be current passing through the diode and therefore a voltage drop.
Sorry that I'm late to the thread, but are you supposed to assume ideal or real diodes for this schoolwork problem? That is something that needs to 1N4148be part of the problem statement, since the answers will be different.

If you need to use real diode models (like the small 1N4148 diode) then use the datasheet to find a realistic Is value for the diode equation. If the diodes are supposed to be ideal, what is your answer then?
 
So then my final answer would be V(D1) = 18mV, V(D2) = 4.974V and I = 0.998 nA.
Would these be reasonable values for my final answer?
 

berkeman

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So then my final answer would be V(D1) = 18mV, V(D2) = 4.974V and I = 0.998 nA.
Would these be reasonable values for my final answer?
Looks pretty close to me, but I'd add +/- signs to those values. Typically Id is measured in the positive direction (anode to cathode), and Vd is measured likewise. :smile:
 

berkeman

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And sorry, did you verify that they don't want a simplistic answer assuming ideal diodes? In that case the Id total would be zero, right?
 
And sorry, did you verify that they don't want a simplistic answer assuming ideal diodes? In that case the Id total would be zero, right?
They are not ideal diodes, and saturation current was given, which I assumed was for calculating using the diode equation.
 

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