It is not clear why the experimenter expects \$I_B\$ to be constant in the face of changing transistors, including substitutions of Darlingtons for regular NPN's.
We can assume that the voltage source is ideal (valid: since these results are from a simulation), then the magnitude of \$I_C\$ does not disturb the voltage. (Indeed in the circuit on the right, we have the simulated transistor cheerfully passing 39000 Amperes, yet the source delivers!)
Even two different diodes will not pass exactly the same current if they are hooked up to the same voltage source and same resistor, because they have different curves.
In the case of a Darlington versus NPN base junction, we are looking at two diodes versus one. Plus, the Darlington may incorporate an internal bypass resistor.
The base current can be approximated using simple diode arithmetic: subtract the forward \$V_{BE}\$ drop from 5V, and divide that by the 100 ohm base resistance:
- Regular NPN: \$(5 - 0.7)/100 = 0.043\$
- Darlington: \$(5 - 1.4)/100 = 0.036\$.
If we use the exact \$V_{BE}\$ figures given in the diagram, we get the exact \$I_B\$ current values:
- Regular NPN: \$(5 - 0.72945)/100 = 0.0427055\approx 0.0427\$
- Darlington: \$(5 - 1.01)/100 = 0.0399\$.
The base current is simply that: application of Ohm's Law to the base resistor, subject to to voltage which remains when \$V_{BE}\$ is subtracted from 5V.
It is not a variation of base current with \$beta\$. That is to say, of course it varies with beta, but beta is perhaps not the relevant parameter to choose as an independent variable for understanding the variation. \$beta\$ is a high level summary of the characteristics of a transistor, connected with the simplified model.
If you wish to hold \$I_B\$ absolutely constant, then you have to drive the base with a current source. (Your circuit simulation software surely has an ideal current source component that you can immediately plant into the circuit.)
Take the most sensitive transistor that you want to measure, and choose the base current so that this transistor is just barely saturated. Less sensitive transistors will then derate the collector current from there. Include a collector resistor to protect the transistors, and as a basis for measuring current.
Rule 1 isn't a "good idea", it isn't a "guideline", it is a fundamental tenet of transistor physics. If for any reason (during normal usage) it is unable to hold then the circuit will not operate.
As for the lamp, it is a purely resistive element. It should have 10V across it, but thanks to the transistor it won't. So the transistor gets 0.2V and the lamp gets 9.8V and reality is saved.
Best Answer
Unfortunately you do not describe in which context this statement is made.
I have seen such statements before in the context of analog IC design. On a chip we can make transistors that are "identical" in the sense that they have very similar transistor parameters like \$\beta\$ and what \$V_{BE}\$ you get under certain circumstances.
This has nothing to do with "external parameters" like \$V_{CE}\$ because these are mostly defined by the circuit in which a transistor is used.
On an IC transistors can be identical as they're all made at the same time.
Discrete transistors can be identical but you'll have to do your best to find a pair that is identical enough. You might never find a pair like that.
The most basic application of "identical" transistors is a current mirror:
Where TR1 converts I1 into a (\$V_{BE}\$) voltage which is then applied to an identical transistor TR2 which then makes a copy of I1 flow, which is I2.
For identical transistors I1 = I2 (ignoring some 2nd order effects).