# I don't understand transistor amplifying current

#### JoonDong

Joined Aug 20, 2017
22
Is not current determined by voltage and resistance passively?

e.g. How can the gain factor be determined collectively because the current between the collector and emitter varies depending on the resistance of the load at below circuit?

#### Attachments

• 10 KB Views: 22

#### BobTPH

Joined Jun 5, 2013
9,265
Suppose you take out the transistor and replace it with a resistor. You have a load of 100 Ohms and a power supply of 10V.

First use a 1 Ohm resistor. How much current flows?
Now change it to a 100 Ohm resistor. How much current flows now?
Now change it to a 1 MegOhm resistor. How much current flows?

Now think of a resistor that has a third lead coming out of it. If you put a current into this third lead, the resistance changes such that the current flowing through the resistor is 10 times higher than the current you input.

That is how the transistor works.

Bob

#### MrAl

Joined Jun 17, 2014
11,704
Hi,

It's actually calculated a few different ways. The simplest is the view of what Beta is.
Beta is the ratio of collector current to base current, and it is almost always greater than 1.
Therefore we have something we can call 'current gain'.

If you need a better answer then you might have to be a little more descriptive about what it is that you dont understand. For example, there is also the voltage controlled current source view.

#### ArakelTheDragon

Joined Nov 18, 2016
1,365
Its "Om's law". Even if you have a coeficient of amplification of "80", you still have to follow :"Om's law", the voltage in the emitter is stable and the current through the emitter depends only on the resistance in it. If you have a load in the emitter of "100Ohm" that means that the current will be:
Vemitter/Remitter. The current through the collector is equal the current in the emitter.

#### WBahn

Joined Mar 31, 2012
30,294
Is not current determined by voltage and resistance passively?

e.g. How can the gain factor be determined collectively because the current between the collector and emitter varies depending on the resistance of the load at below circuit?
I'm not quite sure I understand your question, so keep that in mind when reading my response.

If the transistor is conducting (either in the active region or the saturation region) then the base-emitter junction is going to have roughly one diode voltage drop, or about 0.7 V across it.

That means that the 2 kΩ resistor will have about 0.7 V across it and will have about 350 μA flowing in it.

The 10 kΩ resistor has about 4.3 V across it and so it will have about 430 μA flowing in it.

The difference between these two currents, or about 80 μA, flows into the base.

If the transistor is in the active region, then it will respond by adjusting the voltage difference between the collector and the emitter until a multiple of the base current if flowing in the collector. The multiple is call the current gain, or beta, of the transistor and it is a pretty poorly known value. Figure for a 2n2222 that it is something in the range of 100 to 200 in the active region. So the transistor will TRY to set the collector current in the range of about 8 mA to 16 mA. If that current through the load results in a large enough voltage drop to require that the collector-emitter voltage of the transistor be less than the saturation voltage (about 200 mV to 300 mV, roughly), then the transistor will hold the voltage there and whatever current flows under those conditions is what will flow (which would be less than the 8 mA to 16 mA active region current).

The amount of current flowing in the active region is dictated primarily by the transistor. The amount of current flowing in the saturation region is dictated primarily by the load.

For the circuit you've shown, a load resistance below about 300 Ω to 600 Ω should place it comfortably in the active region. As the load resistance increases, the transistor will start to enter saturation and will be heavily saturated by the time the load resistance reaches 6 kΩ.

#### jayanthd

Joined Jul 4, 2015
945