Did you calculate it differently than how I did it?Well I get 68.41978355514435mA for I2 current
If my equation for i2 is okay why are we calculating different values?If we want 1.5V across the the two series-connected diodes we need Id = 68.4197mA of current. (Your equation for I2 looks ok)
Load current is equal 1.5V/150 = 10mA
So the resistor current is equal ID+IL
R = (5V - 1.5V) / ( Id + Iload) = 3.5V/78.41mA = 44.63 = 47Ω.
Ah that's what it was, we have been using VT = 0.025V. This must be the reason for the difference.Using VT=.026 I obtain the same values as Jony130.
If the load is disconnected I have the diode current increasing to ~78.27mA which gives a new regulated voltage of 1.507V, using the 'exact' value of R=44.63Ω.
Here's my crack at it. (see figure attached)Your prof's method works OK. It produces a result very close to the 'exact' method which involves the numerical solution of two equations - one of them being the Shockley diode equation.
For instance, the 'exact' solution I got for the 50Ω load case was Vout=1.4828V.
And the Prof's method would be something like ...
Taking the 1.5V as the first pass value for Vout would give Itot=78.423mA with R=44.63Ω.
The load would take 1.5/50 or 30mA leaving 48.423mA in the diodes.
This would give Vd=0.741V with VT=.026 & Is=2.02981e-14
Hence, Vout=2*Vd=1.482V
Which is very close to the 'exact' value of 1.4828V.
In the end this is all very academic given the temperature sensitivity of VT.
But as an approximation, is my method of solution demonstrated in my previous post correct for RL's of 100, 75, and 50 ohms?Picking up on Adjuster's comments, I checked what happens when the load is reduced to 15Ω (10% of 150Ω), for instance.
Vout falls to around 1.25V (a decrease of 250mV) and the diode current falls to ~573uA. Itot is ~84mA.
So as Adjuster correctly points out, the assumption of stability over a wider range of loads is not valid at the increasing load current end of range. One needs to apply an 'exact' or more suitable method of solution.
Ahhh now I've seen what I've done wrong, this makes much more sense!Also keep in mind the assumed change in diode current is the new load current minus the load current at 150Ω.
So the load current at 150Ω would be 10mA.
With a 50Ω load the new load current is notionally 30mA, which equates to a 20mA change in current.
So at RL=50Ω the Vd change would be rd*20mA = 0.338*20e-3 = 6.76 mV.
This would give a total Vo change of 2*6.76mV=13.52mV.
An exact solution yields 15.2mV change based on your VT=0.025 and R=41.72Ω values.