How to wire a transistor with separate power source for base

Question

I'm using a separate low-voltage source to apply a small current to a transistor base which will then turn on a higher-voltage current to power an LED. I guess this is more like two separate circuits instead of a parallel circuit powered by one battery. I wasn't sure how to connect the negative terminal of the low-voltage source, so I tried three different ways, illustrated below. All three worked, lighting the LED. Does it matter which I use and if so why?

Setup 1 - negative terminal to higher-voltage battery's ground:

^{simulate this circuit – Schematic created using CircuitLab}

Setup 2 - negative terminal to transistor emitter:

^{simulate this circuit}

Setup 3 - negative terminal to transistor collector:

^{simulate this circuit}

Answer 1

It is normal to use ideal symbols in logic diagrams yet the real circuits have equiv. series R or ESR in the bulk resistance of every junction, caps, batteries, and diodes (even current meters). Most of the time, these (Level 0 logic schematics) with RLC parasitics are inserted mentally by those with experience to see the effects.

Let me show you a level 1 example. Imagine each level going to higher complexity with Miller Capacitance and trace inductance with complex impedance and variable Ic vs Vbe curves. The ESR values are just ballpark estimates on my side, but when a current flows thru a diode, you expect the exponential behaviour for voltage, it will be also increased by the series resistance and voltage drop V/R=I internal to each part. On batteries it is relatively easy to measure this and on transistors it can be looked up in datasheets as Vce(sat)/Ic {= Rce which is not shown}.. There are also thermal effects, so all specs in tables state the standard test temp of 25'C which is easy to control above room temp of say 22'C.

Now my schematic was not simulated so the values are wrong, but used your results. So with Vbat- Vce - Vf(LED) = error, do some KVL to verify the results. What if the 2.5V battery was ESR=10 ohms ? what would you expect?

Notice the ratio Ic/Ib = 22/1.2 is about 20:1 which is a reasonable value for a saturated switch, but which part would you compute is really limiting the current? What is a closer value for ESR1?