Configuration of transistor for amplification

You have a crude Class A amplifier there now.
Input to base.
Output from collector.
Gain is about Rc/Re = 10k/1k = 10.

Brief answer re base input current appears at "cut to the chase" below, but ...

Close enough,

• Ib = (Vdd x Rbu/(Rbu+Rbl) - Vbe) / Re / Beta

Don't even start to try and wonder about it or which resistor is which.
By the end it should make sense.

Calculate voltage at base point with transistor removed.
Call 110k = Rbu= R_base_upper.
Call 10k connected to base Rbl = R_base_lower.
Call Voltage where base connects Vb.
Call 20 V supply Vdd

Vb = 20v x Rbl/(Rbu+rbl) = 20 x 10/120 = 1.666V.

V base to emitter = Vbe

Vbe for an operating silicon transistor is about 0.6V
Can be somewhat different but use 0.6V for now.

As Vb = 1.666V then
Ve = Vb - Vbe = 1.666 - 0.6 = 1.066V. Ve appears across Re (1K) so I_Re = 1.07/1000 = 1.07 mA.

We can call this 1 m or 1.1 mA close enough for this example. I'll use 1 mA for convenience.

Now "it happens" as a function of the formulae related to transistor action that the impedance of the emitter is 26/I for I in mA.
"Don't ask why for now" is good advice. The answer is - because as you will discover in due course, that's the way it is.

So at 1 mA Re =~~ 26 ohms. At 2 mA Re = ~= 13 ohms. At 0.5 mA Re ~= 52 ohms.
This is the effective resistance of the emitter junction to current flow. I'll call that Rqe rather than Re as I've already used Re as the external emitter resistor.

Call transistor current gain Beta, because that's what it is traditionally called for traditional reasons.

If you look into the base you effectively see Re multiplied by the current gain of the transistor. That's because for every mA that flows ij the emitter circuit you only need 1/Beta as much in the bases ciurcuit to control it so it APPEARS that the resistance is beta times as large.

Assume our example transistor has Beta = 100. This is well inside the range of normal for small signal transistors.

Looking into the base we see Beta x Resistance in base circuit =

• Rbase to signal = Beta x (Re + Rqb) = here about 100 x (1000 + 26) = 102600 ohms or ~= 100 k ohms.

Note I said "to signal" as DC will or may have its own rules.
(All obey the same rules but other factors affect what is seen -
eg if we put a 10 uF capacitor across Re it is approximately 0 ohns to AC at audio signals so "vanishes". I said before that gain was ~= Rc/Re = 10
That was before we allowed for Rqe and before we bypassed Re to remove it for AC.
If we do the above gain becomes about Rc/Rqe = 10000 / 26.4 =~ 385

Cut to the chase:

Now, during the hand waving and mirrors we hid something. I said Vb worked iut at 1.66V. The current down the Rbu + Rbl string to ground will be i=V/r = 29/(110k+10k).
This current is just enough to set Vb = 1.666V as we calculated BUT with 1.666v on Vb the same current will flow via Rbl to ground. ie no base current will flow. Your original questiion was "how much base current" and that seems to say "none". However, with no base current the transistor will turn off, Ic will drop, Vre will drop and so Ve will drop causing more than 06V to appear on Vne so the ransistor will turn on and restore. Vb will fall just enough to draw the extra current needed fro Rbu and to reduce the current in Rbl. It will do this automatically and it will draw "just the right amount".

JTRZ (h=just theright amount is enough such that Ib = Ie/Beta.

So we see that is more and less to what happens than appared. The correct example is dynamic and needs load lines on a graph. But "bood enoug" result goes. Based on above.

• Vb = 1.666V so

• Ve = !.066 V.

• I_re = 1.066/1000 = 1.066 mA. ~+ 1.1 MA as before.

• BUT beta = 100 so Ib = Ie/Beta = 1 mA/100 = 10 microamp.

Close enough,

• Ib = (V+ x Rbu/(Rbu+Rbl) - Vbe) / Re / Beta

After going through the above that should not be as scary as it would hev been previously.

E&OE - could easily have typo'd something there.
Please point out if errors seen.

I'll start first with definition of amplification. In the most general way amplification is just a ratio between two values. It does not imply that the output value is greater than the input value (although that's the way it's most commonly used). It is also not important if the current change is big or small.

Now let's move to some common amplification values used:

The most important (and the one your question talks about) is \$ \beta\$. It is defined as \$ \beta= \frac {I_c} {I_b} \$, where \$I_c\$ is the current going into the collector and \$I_b\$ is the current into the base. If we rearrange the formula a bit, we'll get \$I_c=\beta I_b\$ which is the most commonly used formula. Because of that formula, some people say that the transistor "amplifies" the base current.

Now how does that relate to the emitter current? Well we also have the formula \$I_c+I_b+I_e=0\$ When we combine that formula with the second formula, we get \$\beta I_b + I_b + I_e=0\$. From that we can get the emitter current as \$-I_e=\beta I_b + I_b= I_b (\beta + 1)\$ (note that \$ I_e\$ is current going into the emitter, so it's negative).

From that you can see that using the \$ \beta \$ as a handy tool in calculations, we can see the relationship between the base current of the transistor and the emitter current of the transistor. Since in practice the \$ \beta \$ is in the hundreds to thousands range, we can say that the "small" base current is "amplified" into "large" collector current (which in turn makes "large" emitter current). Note that I didn't speak about any deltas until now. That's because the transistor as an element does not require current to change. You can simply connect the base to a constant DC current and the transistor will work fine. If the change in current is required, it's not because of the transistor but because of the rest of the circuit which could be blocking the DC part of the input current.

There is another value also used and it's name is \$ \alpha\$. Here's what it is: \$ \alpha = \frac {I_c} {I_e} \$. When we rearrange that, we can see that \$I_c= \alpha I_e\$. So \$ \alpha\$ is the value by which the emitter current is amplified in order to produce collector current. In this case, the amplification actually gives us a smaller output (although in practice \$ \alpha \$ is close to 1, something like 0.98 or higher), because as we know, the emitter current going out of the transistor is the sum of the base current and collector current which are going into the transistor.

Now I'll talk a bit about how transistor amplifies the voltage and current. The secret is: It doesn't. The voltage or current amplifier does! The amplifier itself is a bit more complex circuit which is exploiting properties of a transistor. It also has input node and output node. The voltage amplification is the ratio of voltage between those nodes \$A_v = \frac {V_{out}}{V_{in}}\$. The current amplification is ratio of currents between those two nodes: \$ A_i=\frac {I_{out}}{I_{in}}\$. We also have power amplification which is the product of current and voltage amplification. Do note that the amplification can change depending on the nodes we chose to be input node and output node!

There are few more interesting values related to transistors which you can find here

So to sum this up: We have transistor which is doing something. In order to safely use transistor, we need to be able to represent what transistor is doing. One of the ways of representing processes happening in the transistor is to use the term "amplification". So using amplification, we can avoid actually understanding what is happening in transistor (if you have any semiconductor physics classes, you'll learn that there) and just have few equations which will be useful for a large number of practical problems.