Calculating Q points for diodes

[whatphysics]

Homework Statement

In Figure Q22, the empirical junction diode equation is : (eqn is in the picture attached)
for both diodes D1 and D2, where VT = 26 mV at room temperature, n = 1, IS = 3.5 x 10-14 A, R1 = 600 Ω, R2 = 300 Ω, R3 = 200 Ω, VS = 9 V. Calculate the Q-points ID = _____ mA and VD = _____ V of diode D1.

Homework Equations

Shockley Diode Equation?

The Attempt at a Solution

I tried to take Id=Vs-Vd/R3 and Vd=0.026ln(Id/Is)
I would try to sub in value of Vd=0.7V into Id eqn to get Id value and then beginning the iteration process and continue this until I get a non changing Id and Vd values to 3 d.p. However, I didn't get the right answers.
Was I wrong to take R3 only for the Id equation?

The right answers are 4.18mA and 0.66V
The answers I got are way off 41.39mA and 0.72V.

Any help is greatly appreciated, thank you!

 

[CWatters]

I tried to take Id=Vs-Vd/R3

That's not correct. There will be a voltage drop across R1.

 

[whatphysics]

That's not correct. There will be a voltage drop across R1.

So should it be Id=Vs-Vd/(R3+R1) since R3 and R1 are in series?

 

[cnh1995]

since R3 and R1 are in series?

They are not.

Use node voltage analysis. You will get an equation of a straight line in terms of Vd and Id, and an exponential (or logarithmic) equation in terms of Vd and Id. You need to solve these two equations to find Vd and Id.

How do you solve this transcendental equation?

 

[gneill]

Hint: Thevenin to reduce the source and resistor network first. Then it's a classic load-line situation.

 

[CWatters]

+1

So should it be Id=Vs-Vd/(R3+R1) since R3 and R1 are in series?

Unfortunately it's more complicated than that because of R2.

See the replies above for the way forward.