To analyze the AC transistor operation, we will use the T and Π models. These models are used to replace the transistor in the circuit. The transistor is replaced by the emitter resistor and a current source.
The transistor T model
Suppose that we have a Common Emitter transistor amplifier from which we've make the AC equivalent circuit as follows:
We can replace the transistor above with the T model. The circuit is then changed as follows:
It is called "T model" because the transistor is replaced by a T-shaped circuitry. In our example you can locate this model if you search for this T-shaped circuitry rotated 90 degrees clockwise. The top side of the T has the collector current source, and the bottom side of the T has the internal AC emitter resistor. This resistor is marked with the symbol "r'_{e}". The small "r" means that we are referring to an ac resistance, the "e" pointer means that we are referring to the emitter, and the prime symbol (') means that we are referring to an internal size of the transistor.
As you see, the base AC voltage is directly applied across the internal base-emitter resistance. Therefore we can extract the following equation:
i_{e} = u_{b} / r'_{e}
The input impedance of the base is this:
Z_{in(base)} = u_{b} / i_{b}
Finally, from the collector's side, the ac collector voltage is calculated with the following formula:
u_{c} = i_{c} x r_{c}
The symbol r_{c} is the total AC collector resistance. The collector's resistance in DC operation is different than the AC resistance. That is because, in AC operation, the coupling capacitor adds the load resistance R_{L} in parallel with the DC collector resistance R_{C}. Therefore, we use the symbol r_{c} in short for the total resistance R_{L}//R_{C}.
The transistor Π (II) model
Let's replace the transistor from the previous CE amplifier with the Π model:
It is obvious why this is called Π model. The letter Π comes from the Greek alphabet and is spelled like the letter "P". A double "I" letter can be used instead (II). From the T model, we know that:
Z_{in(base)} = u_{b} / i_{b} (1)
And we also know that:
u_{b} = i_{e} x r'_{e} (2)
(1)(2)=> Z_{in(base)} = r'_{e} x i_{e} / i_{b}
But from the theory we know that i_{e} / i_{b} is the current gain β. Therefore:
Z_{in(base)} = β x r'_{e}
Both models can be used for the AC transistor analysis with the same results. If you happen to know the AC base voltage u_{b} and the AC current i_{b}, the T model can be then used to analyze the circuit, without needing to know the β value. On the other hand, if the current gain β is known, then you can use the Π model for the analysis.
The Base-Emitter AC internal resistance of the transistor (r'_{e})
So far, we have seen how to do the DC analysis and the AC analysis separately, but we still do not know how the AC and the DC voltages are connected. The internal Base-Emitter AC resistance does exactly this: connects the emitter DC current with the base AC current. We use the symbol r'_{e} which is different from the symbol r_{e}. The prime symbol shows that we refer to an internal size. The r_{e} is used for the AC external emitter resistance.
The Base-Emitter internal AC resistance of the transistor depends on the DC current of the emitter. The equation which connects these two is this:
r'_{e} = 25mV / I_{E}
You may wonder what these 25mV are. The story goes back in 1947, when William Shockley invented the first transistor. Shockley used the diode current to determine the resistance:
I_{E} = I_{S} (e^{Vg/kT} - 1)
I_{S} is the reverse saturation current and V is the voltage across the diode. At 25 ^{o}C, the above equation can be rewritten like this:
I_{E} = I_{S} (e^{40V} - 1)
After some calculations, the equation becomes like this:
r'_{e} = 25mV / (I_{E} + I_{S})
And since I_{E} is many times greater than I_{S}:
r'_{e} = 25mV / I_{E}
The above equation is valid for operation at room temperature (25 ^{o}C). For an accurate calculation at different temperatures, the following equation can be used:
r'_{e} = (25mV x T+273) / (I_{E} x 298)
T is the contact temperature in degrees Celsius. Let's see now what this equation means. Suppose that we have a common emitter amplifier like the one we saw in previous pages, and we want to use the Π model to calculate the transistor input impedance. Suppose also that we did the DC analysis with the help of the DC equivalent, and found that the emitter current is 1.1mA. From this DC current, we can calculate the AC base-emitter resistance:
Hi,
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.
At 9 January 2014, 14:04:31 user Nash wrote: [reply @ Nash]
@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:
Question 1:
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
Question 2:
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.
Hey!
This is a good tutorial, however I am a bit lost in the math by the way you write it.
Would it be possible to rearrange it and have the formula go down the page e.g.:
5*(2 2)
=5*(4)
=5*4
=20
Also can you please explain to me the arrow that you use in the formulas, what does it mean? e.g.: IE = %u03B2 x IB => IB = IE / %u03B2 = 1mA / 30 => IB = 33uA
@john I know what you mean, but i do not show the conventional current flow but the electron flow (which is the reverse). I think i have to explain this somehow in a short paragraph.
At 14 April 2012, 9:19:47 user john wrote: [reply @ john]
I think the current flow in your explanation is for PNP transistor, because for NPN transistor current flow going in to basis not going out.
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.
At 16 January 2012, 11:07:08 user almalo wrote: [reply @ almalo]
Sorry I'am a bit confused:
First you wrote: hfc>hfe>hfb.
2nd: hfe = hfc + 1
So what is the truth?
@Russ i took a quick search around but i could not find one with more details about the biasing history.
At 20 October 2011, 1:01:32 user Russ wrote: [reply @ Russ]
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.
Thanks
At 13 June 2011, 15:40:12 user Fung wrote: [reply @ Fung]
The L7805 1A voltage regulator, there are 2 situations when a switch is open and close.
When open, only an LED and a 74HC00 IC are working, the total current may be about 30mA.
When closed, at least 5 ICs will be in operation, with an 7-segment dual display, the total current may increase to about 150mA. Then the regulator heats up in a short time.
The transistor is also heat up because of the voltage regulator.
Will the problem be solved by moving it away from the voltage regulator?
@Fung yet is varies. but the situation you describe does not sound normal to me. which voltage regulator you use, and how much current this is supposed to provide?
At 21 April 2011, 7:46:10 user Fung wrote: [reply @ Fung]
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?