Electronic – 2N2222 transistor datasheet saturation current

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.

It may be helpful to separate two usages of the term power.

The first is the technical usage. According to Ohm's Law, this is current multiplied by voltage: \$IE\$. (Or \$I^2R\$, or \$\frac{E^2}{R}\$.)

The second is the common misuse of the word in everyday speech. People say things like, "It's not plugged in, I need power!" or "Power up that amplifier." It's a convenient expression compared to "It's not plugged in, I need electrical energy."

Power is a rate: how quickly something uses energy. In electrical terms, it's measured in watts. Watts, in turn, are Joules (energy) per second. If a circuit uses 2 watts, it's using 2 joules per second. If the circuit requires 1 ampere of current, we can know (using Ohm's law) that the voltage must be (E = P / I) 2 volts. Another way to look at it might be to say that it is using 1 ampere at 2 volts every second, but that is cumbersome and more easily stated using watts.

Energy is a quantity of something. In electronics, a joule is a the energy required to produce one watt of power for one second. More specifically, it's how much work is required to move one coulomb (a lot of electrons) of electric charge at one volt.

Energy doesn't necessarily have to be electric in nature. It can also be thermal, gravitational, kinetic, acoustic, etc. For example, you could move an object on a table by using sound waves to vibrate it.

You can have some amount of available energy that, when delivered (or consumed) in a short duration, produces a high amount of power. Delivered or consumed slowly, produces a small amount of power.

A battery stores energy, in chemical form, which can be used at different rates. A remote control for a television, for example, draws very little current. Batteries in such a device last a very long time. You could say that a remote is a low power device. By comparison, a small electric motor using the same battery would be a high power device, because it consumes energy more quickly.

Keep in mind, power is still watts, and watts are a defined unit (joules per second). So when we say something is high or low power, we are speaking relatively. That same electric motor would be considered a low power device, if you were comparing it to a much larger motor.

A 100 watt light bulb requires more power than a 50 watt bulb. It doesn't really matter whether it is low voltage and high current, or high current and low voltage. The wattage is a product of the two, however it is arrived at.

So, circling back to the first part of your question: You're never supplying a load with power, technically. You supply energy and it consumes it at some rate, which in turn is power. But as previously mentioned, people tend to use power as a substitute for other concepts.

So what is the difference between an amplifier and a power amplifier?

Amplifier is a very general term. It is simply a category of circuit or device that increases the magnitude of a signal. It might be an operational (differential) amplifier, an audio amplifier (microphone, instrument, headphones, etc.), radio frequency amplifier, and so on. (To say nothing of amplifiers non-electrical in nature, like fluid or mechanical. A jack to raise your vehicle could be considered a type of amplifier).

The answer is "it depends." Remember how the term "power" gets thrown around inappropriately?

If you're talking about an operational amplifier in terms of the portion of a circuit, it only amplifies voltage. It's definitely not a power amplifier.

Any amplifier that amplifies audio, is technically an audio power amplifier. The output from such an amplifier has both voltage (amplitude) and current. Since P = IE, increasing either voltage or current will, by definition, increase power.

Class A amplifiers, are always power amplifiers. If you encounter one labeled only as an "amplifier," it is just someone taking a shortcut and not calling it an "audio power amplifier." (Similar to how we might say "fill up the car with gas" instead of "fill up the car with unleaded gasoline.") The specifics have simply been omitted.