Solving the Diode Puzzle: Finding V with Diodes

[vermin]

Homework Statement

Hi so I have a problem that I don't exactly know how to approach, here's the diagram:
http://images.fr1ckfr4ck.fastmail.fm/probA.jpg
The diodes are supposed to have saturation current I_s = 10^(-15) A

Homework Equations

What is V?

The Attempt at a Solution

First I tried assuming that 2mA must be going through D_1, which means 8mA must be going through D_2, so I used the exponential model to solve for the voltage drop across each diode. But then what? Is that even what I should be trying to figure out here?
What that got me was that the voltage drop across diode 1 would be ~708 mV and ~742 mV across diode 2. But even if that's right I don't know what it means for the voltage at V.

If I assumed that both diodes should have a voltage drop of 0.7 volts (which I don't think is right) they would have a current of 1.45mA through them - which doesn't seem to work. So that seemed like a dead end also.

Any pointers would help, thanks.

 

[gneill]

Homework Statement

Hi so I have a problem that I don't exactly know how to approach, here's the diagram:
http://images.fr1ckfr4ck.fastmail.fm/probA.jpg
The diodes are supposed to have saturation current I_s = 10^(-15) A

Homework Equations

What is V?

The Attempt at a Solution

First I tried assuming that 2mA must be going through D_1, which means 8mA must be going through D_2, so I used the exponential model to solve for the voltage drop across each diode. But then what? Is that even what I should be trying to figure out here?
What that got me was that the voltage drop across diode 1 would be ~708 mV and ~742 mV across diode 2. But even if that's right I don't know what it means for the voltage at V.

If I assumed that both diodes should have a voltage drop of 0.7 volts (which I don't think is right) they would have a current of 1.45mA through them - which doesn't seem to work. So that seemed like a dead end also.

Any pointers would help, thanks.

If you have the potentials across the diodes thanks to the known currents flowing through them, then you can fix the potential with respect to ground at the top of D2. Then you're just one diode drop away from the potential at the bottom of D1...

Can you show how you computed your values of ~708 mV and ~742 mV? What value did you use for nV_T?

 

[vermin]

For nV_T I'm using 25mV, so to solve for the voltage drops across D1 and D2 I used;

D2:
8mA=(10^-15A)*e^(V₂/25mV)

solving for V₂ it came out to 742.8mV
I used the same equation for D1, V₁ except swapping 2mA for 8mA.
V₁ came out to 708.1mV

It does seem obvious now I could go up from 0V (below D2) to 742.8mV, then drop 708.1mV across D1 to give me V=34.7mV

I think that might actually be the right answer..

 

[gneill]

For nV_T I'm using 25mV, so to solve for the voltage drops across D1 and D2 I used;

D2:
8mA=(10^-15A)*e^(V₂/25mV)

solving for V₂ it came out to 742.8mV
I used the same equation for D1, V₁ except swapping 2mA for 8mA.
V₁ came out to 708.1mV

It does seem obvious now I could go up from 0V (below D2) to 742.8mV, then drop 708.1mV across D1 to give me V=34.7mV

I think that might actually be the right answer..

Yup, looks good if nV_T is taken to be 25 mV. I think that 26 mV is the usual assumed value for room temperature silicon, but the 'n' can vary depending upon the diode's construction. n between 1 and 2 is typical.

 

[rezeew]

As a scientist, my first approach to this problem would be to analyze the circuit using Kirchhoff's laws and Ohm's law. From the given information, we know that the saturation current of the diodes is 10^-15 A, which means that they are operating in the forward biased region.

Using Kirchhoff's laws, we can write the following equations:
1. For the loop containing the voltage source, resistor R, and diode D1:
V - Vd1 - IR = 0
where V is the voltage of the source, Vd1 is the voltage drop across diode D1, and I is the current flowing through the circuit.

2. For the loop containing the voltage source, resistor R, and diode D2:
V - Vd2 - IR = 0
where Vd2 is the voltage drop across diode D2.

3. Applying Ohm's law to the resistor R:
V = IR

We can now solve these equations simultaneously to find the value of V. Substituting V = IR into the first two equations, we get:
V - Vd1 - V = 0
V - Vd2 - V = 0

Simplifying these equations, we get:
Vd1 = 0.5V
Vd2 = 0.5V

This means that the voltage drop across each diode is 0.5V. We can now use the exponential model to find the current flowing through each diode:
I = Is*(e^(Vd/Vt) - 1)
where Is is the saturation current, Vt is the thermal voltage (kT/q), and Vd is the voltage drop across the diode.

Substituting the given values, we get:
I1 = (10^-15)*(e^(0.5/0.026) - 1) = 1.46 mA
I2 = (10^-15)*(e^(0.5/0.026) - 1) = 1.46 mA

Since the current flowing through each diode is the same, we can use Ohm's law again to find the value of V:
V = IR = (1.46 mA)*1000 ohms = 1.46 V

Therefore, the voltage at V is 1.46 V. This approach takes into account the properties of the diodes and