As we said previously, a transistor amplifier usually operates both with AC and DC voltages, the DC voltage is used to bias the transistor and the AC voltage is the signal that will be amplified. To analyze a transistor circuit, both voltages must be analyzed. But the transistor itself as well as the biasing components react differently in AC and DC signals. Obviously, there must be a method to analyze each signal separately. The simplest and most widely used method is using the DC and AC equivalents. According to this method, two equivalent circuits are extracted from the original circuit, the DC and the AC equivalent. The currents and voltages for each circuit are calculated separately, and then, using the superposition theorem we can calculate the final values. For your information, the superposition theorem states that:
"The response -voltage or current- in any branch of a bilateral linear circuit having more than one independent source, equals the algebraic sum of the responses caused by each independent source acting alone, while all other independent sources are replaced by their internal impedances."
Making the DC equivalent
To make the DC equivalent circuit, the following steps must be taken:
I] All AC sources become zero.
II] All capacitor are replaced with an open circuit
Suppose for example that we have the following circuit from which we want to make the DC equivalent:
According to the first rule, all AC sources (if any) must become zero. This applies for the "Input" AC source that we have. When a power source becomes zero, it means that the output voltage will always have the same potential as the grounding signal - which is zero. Therefore we replace it with a grounding signal. According to the second rule, all capacitors must be replaced with an open circuit. CIN must be replaced with an open circuit, thus the Input AC supply can be omitted. COUT is also replaced with an open circuit, thus RL can be omitted. Here is the resulting DC equivalent circuit:
The above circuit can be simplified:
Having this circuit, it is very simple to make the DC analysis, since there is no AC source whatsoever. From this analysis, the designer is able to calculate all the DC biasing values, draw the DC load line and set the quiescence point.
Making the AC equivalent
To analyze the AC signals, we need to make the AC equivalent circuit. The following steps must be taken:
I] All DC sources become zero
II] All capacitors are replaced with a bridge
When we designed the DC equivalent, we simple removed the AC sources. That is because the AC sources were coupled through decoupling capacitors, and due to the fact that all capacitors were replaced with open circuits, we simple removed the AC sources. But its not the same for the DC equivalent. The DC sources are directly coupled to the circuit so we cannot remove them. Therefore, according to the first step, all DC sources become zero. In other words, every component that is connected to the positive DC supply must be grounded. Then, we replace the capacitors with bridges:
The above circuit can be simplified:
In many cases (like this one) the resulting circuit can be further simplified. The two resistors of the voltage divider are now connected in parallel, since the top side of RB1 is now grounded. Moreover, the emitter resistor and the load resistor (RE and RL) are also connected in parallel. We can calculate the equivalent resistors for RB1//RB1 and RE//RL and replace them in the circuit:
The above circuit is ready for the AC analysis since there is no more a DC source.
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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?