Why is the current in the main circuit lower with Transistor vs without it?
I know that you must connect everything up correctly in a circuit in order to make things work. But I want you to realize that there are two main classifications of schematics.


A schematic designed to show all of the physical connections so that you know how to construct it; and,
A schematic designed to communicate how it works and functions.


These are very different goals. The first one helps a builder of circuits who doesn't really care to understand them, but just needs to know the exact details required in putting it together correctly so that it operates as designed. The second one helps a designer communicate to other designers how the design functions. In this latter case, it's not so important to show all the connections.

Since you need to understand what you did and do not really need to know how to build another one, let's look at a schematic designed to communicate to designers.







simulate this circuit &ndash; Schematic created using CircuitLab

The above two schematics are equivalent. Notice that I've remove all the wiring to voltage sources and simply placed the right voltage at the right place. This helps avoid distracting wiring and makes you focus on what the circuit actually looks like to a designer.

(I've added a little dashed box around an NPN to highlight the fact that there are actually two diodes in the NPN transistor. These diodes aren't exactly mirror images of each other: one is designed differently than the other one. But this lets you see that if you wire it "upside down" so to speak, it can still operate. Just... differently.)

The left side schematic "looks weird" to a designer. The NPN seems wired up backwards. It looks like the LED current would have to flow against the diode direction indicated in the standard BJT diagram. But the right side diagram shows the arrangement more clearly. And if you look over the dashed box illustration of the NPN in the upper left corner, you can see that the base is indeed forward-biased -- except that it is forward-biased with respect to the collector, not the emitter.

This isn't a total disaster. The NPN will work like that, too. It will just have a much worse active mode \$\beta\$.

So looking at the right side schematic, you can see that you've turned the emitter into the collector and the collector into the emitter. (Which is fine, just not often done.)

The real problem here is that your base is set to a hard \$1.25\:\text{V}\$ across the "new" base-emitter junction. This means there is no current limiting taking place. So lots of base current will flow. As a result, the "new" collector will be very close to the "new" emitter voltage (ground, basically.) This "new" collector will also be forward biased relative to the base.

So everything kind of works. Except for the one huge problem that you are missing any current limit to the forward-biased base-emitter and base-collector junctions. As a result, very very large currents will flow from the base. And this means very, very big dissipation in the BJT. So it should get VERY HOT.



Put a resistor in the base-emitter loop.







simulate this circuit

Either one of the above two will work. Try them both out.

The left side one works because the emitter current is \$I_\text{LED}=\frac{1.25\:\text{V}-700\:\text{mV}}{27\:\Omega}\approx 20\:\text{mA}\$. The BJT is not saturated in that case. The right side one works because the BJT is in fact saturated and the LED current is about what you figured out in the first place, limited by the value of \$R_1\$.