When we discussed about the transistor operation in AC, i mentioned the term "Small Signal". It is important to know what we call "Small Signal", and why it is necessary to know the difference. The following characteristic is a typical IC to VBE input characteristic. It shows the increment of IC current in relation to the VBE base voltage.
The collector current is zero as long as the VBE voltage is less than about 0.65 volts. This is something that we've talked before many times. The VBE voltage has to do with the material that the transistor is made of. Above this voltage level, the collector current (along with the emitter current of course) climbs up rapidly. This is the typical transistor operation. What you need to notice here is the region of the characteristic around the 0.7 volts. The line seems to be curved at that point. This is a typical problem that designers face, if they want to have an undistorted signal amplification. The curve becomes more intense as temperature increases. At sub-zero temperatures things are usually much better and the curve is not so intense. The above characteristic corresponds to a temperature of around 150oC. I chose this high temperature because the distortion is more obvious.
So, let's take a closer look at the region that the transistor will work at. That's usually above 0.68V for VBE. The following diagram is a portion from the input characteristic shown above, but only for a VBE range from 0.68 to 0.72 Volts. The transistor is biased with DC and the Q is set. At that point, the VBE is stable at around 0.7 volts. Then we apply a large AC signal at the base. This signal causes the Q point of VBE to oscillate:
Although the input AC signal is symmetrical, due to the curvature of the input characteristic, the output current change is not symmetrical. The result is a distorted amplified signal which in certain applications it is totally unwanted.
Now, suppose that we apply a smaller input signal:
The difference is obvious. Although the output signal is much smaller in terms of amplitude, it seems to have almost no distortion even at that high temperature. This is normal because now we used a much smaller portion of the characteristic, and this portion can be considered as a straight line. As a conclusion we can say that if the AC input signal is small, the AC current change at the collector is proportional to the AC voltage change at the base.
But, how do we define a signal as "small"? There is a general rule to define the small signal which states that:
The AC peak to peak current of the emitter must be smaller than 10% of the DC current of the emitter.
Although the distortion will not be eliminated, it will be radically limited. The amplifiers that satisfy this 10% rule are called small signal amplifiers. They are usually used to amplify small signals such as the TV or radio signals.
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Would you please tell that why the mentioned curvature happens? I have this problem in my TFT (Thin Flim transistor) and in the low voltage of D-S, D-S current does not like a diode curve and it has a curvature like you mentioned.
@Giorgos Lazaridis Your Diagrams for the current flow for a NPN connection are wrong. BIG MISTAKE Fix it please http://pcbheaven.com/wikipages/images/trans_theory_1317761009.png & http://pcbheaven.com/wikipages/images/trans_theory_1317761285.png
@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?