Until now, we've been talking only for transistors in DC operation. We've learned how to bias a transistor correctly, and we also saw a quick example on how to set the operation point. But many times, transistors are used to operate with AC signals. A transistor audio amplifier for example is an AC signal amplifier, since the microphone generally generates an AC output. And here is a point that many people confuse: Transistors are NOT AC components: Transistors can only operate with DC signals!
That sounds kinda paradox, since we all know that transistors are also used to amplify AC signals. But think about this: Suppose that we have an NPN common emitter transistor amplifier, and we feed an AC signal at its base. As long as the input signal is higher than 0.7 Volts, it will be amplified normally and it will appear at the output of the amplifier. But what happens when the signal becomes less than 0.7 Volts? And worst, what happens when the signal becomes negative? As we know, an AC signal has a positive and a negative period.
The answer is this: A negative signal at the base of an NPN transistor means that the base-emitter diode is reverse-biased. The diode acts like an open circuit (cut-off) and the amplifier does not work at all. There is also a tight limit: If the negative voltage become too high, the base-emitter diode will be destroyed. The maximum reverse voltage that this diode can handle is usually around 5 volts for small signal transistors. The exact value for each transistor can be found in the manufacturer's datasheet, usually with parameter name VEBO (Emitter-Base voltage). The same situation happens of course if we use a PNP transistor and we reverse-bias the base-emitter diode.
So, how is it possible to amplify an AC signal? The answer is by biasing the transistor with DC voltage. Suppose for example that we want to amplify an 1 Vp-p AC signal. This means that the signal has +0.5 Volts positive period and 0.5 Volts negative period. If we bias the input with let's say 1.5 volts DC, then the input will vary from 2 Volts (1.5 + 0.5) to 1 Volt (1.5 - 0.5). This is considered as a DC signal and the transistor can amplify the complete period normally.
An AC signal fed into the base of the transistor will not be amplified correctly since it has a negative period as well.
Biasing the transistor with DC voltage we cause the AC signal to be shifted upwards, therefore it has no more negative period.
Coupling and Bypassing Capacitors
As we said before, transistors are DC components. This means that the output will also be a DC voltage. But if we amplify an AC voltage, then we probably want to get an AC voltage at the output as well. How is this done? Simple, with a coupling capacitor. A capacitor operates as a resistor in AC and as an open circuit in DC. It is not the purpose of this theory to analyze in details the capacitor's behavior in AC and DC. Nevertheless, it is important to have some very basic knowledge about capacitors.
The Coupling Capacitor
A coupling capacitor is a capacitor connected in series with the circuit that we want to couple. The AC signal is free to go through the capacitor, while the same capacitor acts as an open circuit against any DC current. Let's see an example of a coupling capacitor:
Both Cin and Cout are coupling capacitors. Their job is to block any unwanted DC voltage from between the stages that they couple. This is the what would appear in the oscilloscope's screen if two channels were connected, one before the Cout coupling capacitor (Channel 1-Blue line) and one after the same capacitor (Channel 2-Red line):
Looking at channel 1, it is obvious that the amplified AC voltage has been shifted above the zero line, and has become a DC voltage. That is because the DC voltage from the emitter of the transistor has been added to the amplified ΑC signal. Looking at channel 2, any DC voltage has been removed due to the coupling capacitor. It is possible only for the AC voltage to cross, and therefore it has become an AC voltage again.
The Bypassing Capacitor
A bypassing capacitor is a capacitor connected in parallel with a circuit that we want to bypass. Unlike the coupling capacitor, the bypassing capacitor removes any unwanted AC signal from this circuit, since any AC current goes through the bypassing capacitor, leaving only the DC current to go through the parallel circuit. Let's see an example:
A bypass capacitor (Ce) is connected in parallel with a resistor (Re). What we want is to have only DC current flowing through the resistor, in order to maintain the current stable (Ie=Ve/Re). The actual problem is that when an AC signal is applied at the base of this circuit, this AC signal will also appear at the emitter of the transistor. This will change the emitter voltage which will eventually change the emitter current, and we do not want that. Therefore, we add the bypassing capacitor Ce. All the AC voltage will be grounded through this capacitor. Therefore, the voltage across the resistor Re will not change, and the current will remain stable. In the following graphs you can see what would appear in an oscilloscope's screen, if the probe was connected across Re. The left graph shows the output if Ce was NOT connected across Re, and the right graph shows the output with the bypassing capacitor Ce connected across Re:
@Mint Electronics Sorry for 2 reasons: first for the loooong delay (i thought that i had posted the answer immediately), and sorry for not explaining this in the article. I will re-read the whole theory when i finish it and fix some issues like that. I though that it was not so important to explain it, but maybe i will put some spoilers with the proof. Anyway:
Vcc = Vb Vbe Ve => Vcc - Vbe = Ve Vb
But we can approximate that Ve = Ic x Re (since Ic almost = Ie)
Also, Vb = Ib x Rb => Vb = (Ic x Rb) / hfe
From the above:
Vcc - Vbe = (Ic x Re) [(Ic x Rb) / hfe] => (We get Ic in common)
Vcc - Vbe = Ic x [Re (Rb / hfe)] => (divide both sides with term)
(Vcc - Vbe) / [Re (Rb / hfe)] = Ic
The voltage divider current will always be bigger than the base current, since it is composed by the voltage divider current PLUS the base current.
@Mint Electronics Although i try to keep the math as simple as possible, the way you want me to re-arrange it would be more like a math tutorial rather than a transistor tutorial. I keep it simple but not that simple, it would be tiring and confusing for those who want to learn transistors.
As for the "arrow", it is not an arrow, it is the Greek letter %u03B2 (Beta) which probably you cannot see due to your browser's encoding used. It is good to know that there are people who cannot follow this encoding. This %u03B2 letter is the same as the hfe. In formulas we use %u03B2 rather than hfe for short. I think i have to find something else to show this....
In the meanwhile try to change your encoding and the correct letter will reveal.
The use holes to explain any part of a transistor function is confusing, a positive charge does not move because of the mass of the positively charged nucleus.[ie; Protons, so called holes.]The flow of electrons is convincingly demonstrated by the cathode ray tube and other experiments carried out a hundred years ago. Electron deficiency and excess better explain the attraction or repulsion which is used for a transistor to function
@almalo you're right, the typo is obvious. I used the minus sign used to show the reverse current directin, as an algebric sign. I had to use ABS numbers for the comparison. Common collector has the maximum current amplification. hfc>hfe>hfb
I write -hfc = IE / IB but i should write instead |-hfc| = IE / IB. The result with this change is:
hfe = hfc - 1
I tripple check this editorial because i do not want to make such mistakes, sometimes i fail to locate them though. Thank you for noticing.
Can you recommend a book that goes into the history of transistor biasing and the development of other basic circuits? I am interested in how this developed. From hindsight it seems so clear and I wonder how difficult it really was.
The resistance of a transistor varies as its temperature changes, am I right?
I have a circuit which has a +5V voltage regulator, a transistor is used to amplify the signal of a PIEZO sounder and it is placed nearby the voltage regulator because of the routing of tracks. However, due to the quite-high current, the regulator heats up when it is in operation, heatsink is not setup yet, not only itself, the tracks under the board are also heated up.
Due to the case stated above, the transistor is heated up later on, it changes the frequency to the PIEZO sounder (ie unable to keep the original frequency) because of the change of its internal resistance. As I need to add the heatsink on the regulator, what things should I do in order to keep the frequency (that is to keep the internal resistance of the unit) of the amplifying circuit?