Finding Vout from circuit contain diodes.

[dla]

Homework Statement

I am suppose to find an R value so that Vo=80mV given that both diodes are identical, conducting 10mA at 0.7 V and 100 mA at 0.8 V

Homework Equations

$I = I_{s} e^{v/{V_{t}n}}$

The Attempt at a Solution

I get that if the diodes are the same n, V_t and I_s are the same. So first I need to solve for n by using the equation and information above to create 2 equations assuming room temperature. Then I can make two equations from D1 and D2 and combine them to solve for V.

My main question is how did they know that Vo=V_2-V_1, voltage drop of diode 2 - voltage drop of diode 1. Why not the other way around? It might be simple but I am having a hard time seeing it any help will be much appreciated.

The circuit is attached. Thanks.

 

[marcusl]

No, n is the ideality factor of the diode. It is a "fudge factor" that accounts for some real-life departures from the ideal Schockley equation, and is usually set to one in the absence of specific information about the device being used.

The voltage at the common anode connection is pinned by the voltage drop of diode 1--there is no choice, because its drop sits between the source and ground--so it is V1. The drop V2 is determined by the current through the right branch, which depends partly on R, so Vo=V1-V2. The trick is to find how the current splits. You'll use Kirchoff's laws, Ohm's law and the Schockley equation to solve your problem.

 

[CWatters]

My main question is how did they know that Vo=V_2-V_1

By applying KVL around the loop... Earth -> resistor -> D1 -> D2 -> Earth

+Vo + V_1 + (-V_2) = 0

Rearrange to give

Vo = V_2 - V_1

 

[rude man]

Homework Statement

I am suppose to find an R value so that Vo=80mV given that both diodes are identical, conducting 10mA at 0.7 V and 100 mA at 0.8 V

Homework Equations

$I = I_{s} e^{v/{V_{t}n}}$

The Attempt at a Solution

I get that if the diodes are the same n, V_t and I_s are the same. So first I need to solve for n by using the equation and information above to create 2 equations assuming room temperature. Then I can make two equations from D1 and D2 and combine them to solve for V.

This is exctly the right approach. Use the given diode voltages at 10 mA and 100 mA to solve for n.

Then the rest is just summing currents to zero at the anodes node.

 

[marcusl]

Sorry, looks like I misread the problem. I took it as D1 operated at 10 mA and .7V, D2 at 100 mA and .8V. Apologies to the dla and rude man. Those two sets of values are sufficient to solve for Is and n, after which the problem can be solved as the other respondents said.

 

[dla]

Thanks for the replies everyone. And I see it now CWatters thank you.

Actually marcusl your reply made me wonder how did we know that n is not 1?

When I made the two equations, I thought I had three unknowns at first: n, V_T and I_s. I didn't know if I should have assumed n=1 or assumed V_T was at room temperature and use V_T=25 mV. I ended up doing the latter because the solution showed it.

But if anyone can explain that would be great.

 

[gneill]

Thanks for the replies everyone. And I see it now CWatters thank you.

Actually marcusl your reply made me wonder how did we know that n is not 1?

When I made the two equations, I thought I had three unknowns at first: n, V_T and I_s. I didn't know if I should have assumed n=1 or assumed V_T was at room temperature and use V_T=25 mV. I ended up doing the latter because the solution showed it.

But if anyone can explain that would be great.

n and V_T are both constants for a diode that is operating at a fixed temperature. That means that $nV_T$ is also a constant. So even if you assumed the wrong value for V_T, the "variable" n that pops out of your solving for n from the given potential and current values would compensate. No problem!

I find it convenient to replace n and V_T with a single variable, V_x, when the particulars of n and V_T separately are not relevant.

 

[dla]

Oh that makes a lot more sense! Very much appreciated, thanks gneill!