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

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.

• 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 1. the base will be at ABOUT 0.6V and Vce can be lower - typically 0.1 - 0.3V for a saturated transistor. || In 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. || Jan 30 '20 at 2:56
• As a switch, when enough current flows through the base ( to saturate the transistor), in the first circuit, RL gets almost all of the supply voltage (e.g Vcc - .3V). In the second circuit, to get enough current through the base to saturate the transistor (and get almost the full Vcc across RL), you'd need a voltage higher than Vcc at the base (which is not always available/feasible). After all, when using a transistor as a switch, you want the full Vcc across the load and the second circuit makes it harder to achieve.
– Big6
Jan 30 '20 at 2:58
• Google high side and low side switch with BJTs.
– Big6
Jan 30 '20 at 3:00
• it seems to work just the same way ... try the two configurations in a scenario in which a microcontroler uses the transistor to drive a 12 V load ... make Vcc = 12 V .... then drive the base with a 5 V signal Jan 30 '20 at 3:00

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.

• I think I didn't see any difference because my test only using the transistors as a simple switch. Maybe the difference would be greater in high frequency switching or AC signal amplification? Jan 30 '20 at 7:26

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.

• That's why I'm asking the question here. Googling can be inconclusive. Jan 30 '20 at 3:53
• In Config 2, (the EF or CC), would a NPN Darlington mitigate the voltage drop and heat sink requirement substantially? I'm about to experiment with the so-called "Capacitance Multiplier" for SMPS filtering, where only a high value (the higher the better) of Rb makes it worthwhile. A 3v drop across the device is actually necessary in my case. Only the device, some resistors and caps will be used--no more complexity than that. Aug 13 '21 at 14:27
• A darlington has two transistors each with 0.7V base-emitter voltage loss. Then if the base voltage is 5V, the load gets only 5V - (0.7V x 2)= 3.6V. A power darlington works well as a capacitance multiplier. Aug 14 '21 at 15:38

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.)

• I don't see any difference because I'm testing the transistors by using them as a switch to turn on an LED. Maybe the difference would be visible in some high frequency switching stuff or audio frequency amplification? Jan 30 '20 at 7:24
• @YudhiG.NO. You do not see any differences because you are not measuring as accurately as you need to - or as you could. If you have a common DMM (digital multimeter) you will be able to see some difference. They will not be very significant in this case - but will be in others - even at low frequency. Also, read what others say about gain and impedance. Jan 30 '20 at 10:36