What if I switch the place of the load in a common emitter NPN transistor circuit?

Question

As you can see in the picture, there's an NPN transistor in a common emitter configuration. In the first configuration, the load is placed between the collector pin and Vcc, and it is the most commonly used configuration.

Then, in the second configuration, I moved and put the load between the emitter pin and GND. I have never seen this configuration where the load is at the emitter pin, but both configuration seems to work just the same. I have tested both configuration using some generic NPN BJT transistor as a switch to turn on a load (in this case, an LED). I used 2N2222A, BC547, and the higher power TIP3055).

Is there any advantages or disadvantages between the two configurations? In my opinion, it seems to work just the same way, but I'm not sure because I have never seen the second configuration applied in circuits (or there's any?) I don't know.

Answer 1

Then, in the second configuration, I moved and put the load between the emitter pin and GND. I have never seen this configuration where the load is at the emitter pin,

You've created what's called the common collector configuration. It's also called an emitter follower. It's very common and well-known.

Is there any advantages or disadvantages between the two configurations? In my opinion, it seems to work just the same way,

They really aren't the same at all. Where the common emitter has (potentially) voltage gain and high output impedance, the common collector has voltage gain very near 1, and low output impedance.

The common collector circuit is typically used as a unity gain buffer (no voltage gain by can produce more power output than is input). The common emitter is typically used as a voltage gain stage.

Answer 2

Please learn about the basics of transistors. Configuration 1 is called Common Emitter. A small Vin of 1 or 2 volts causes the collector to go almost to ground then the load gets almost the complete supply voltage.

Configuration 2 is called Emitter Follower or Common Collector. The emitter is always about 0.7V less than the base voltage so it is rarely used as a switch. Even if Vin is Vcc then the load does not get the entire supply voltage.

Answer 3

It's good that you specify what transistors you tried this with, but you also need to note Vcc value, Rl, Rb Vb.
These all affect the results.

In addition to what others say:

You need to measure more closely and/or also be aware of transistor saturation.
Apparently small differences can matter greatly.

In circuit 1. the base will be at ABOUT 0.6V and Vce can be lower - typically 0.1 - 0.3V for a saturated transistor.

In circuit 2. the base is about at Vcc (less Ib x Rb drop, and V load is 0.6V or more below Vb.

The dissipation in the transistor may be 5+ times as high in 2.

If you use FETs rather than bipolar transistors the differences become more pronounced.
For most FETS Vgs_on is in the order of 3 to 12 volts. (Some allow lower voltages and a few need higher).

Example:

With a low Rdson (on resistance FET) - say 10 milliOhms - and say 12V supply and 3A load and Vgson = 3V.

In cct 1:

Vds = Fet on voltage = I x R = 8A x 0.010 ohms = 0.08 V
FET dissipation = V x I = 0.08 x 8 = 0.64 watt.

In cct 2.

Vg = 12V
Vgs = 3V so Vs = 9V.
Vds = 3v
FEt dissipation = V x I = 24 watt

So, in fig 2 the FET has 24/0.64 ~= 37 x as much dissipation. Case 1 will need minimal heatsinking. Case 2 will need major heatsinking.

Case 1 will deliver a maximum of 12 - 0.08 = 11.92V to the load.
Case 2 will deliver a maximum of 12 - 3 = 9V to the load.
(For 8A load current the load resistor will need to be smaller in case 2.)