The common emitter connection is a very commonly used connection, mainly due to the high voltage gain that can be achieved. Therefore, we will examine many different common emitter circuits.
Common Emitter with fixed bias
This is the simplest common emitter amplifier and this is the reason why this connection is important to know. We will discuss this connection again when we talk about the transistor as switching devices. In the meanwhile, it's good to know how to design a common emitter amplifier with fixed bias. This type can achieve the maximum voltage amplification possible. Here is the schematic:
To design such a circuit, it is very important to know the precise h_{FE} and hfe values (DC and AC current gains), otherwise the calculations will be wrong. This is not simple though, because even two similar transistors from the same batch of the same manufacturer might have different gain, plus this value is sensitive to temperature. So we expect to have some difference between the calculated and the measured values.
One good way to start the design is from the output impedance. The output impedance of the amplifier depends on the circuit impedance that will be connected after the Cout capacitor. If the transistor must provide high current (for a low impedance circuit), then the amplifier must also have low impedance, otherwise we prefer to have high impedance. The amplifier's output impedance is defined by the R_{C} resistor. Therefore, we can begin by selecting an R_{C} value. We can begin with the DC analysis of the DC equivalent:
As we know, if no load is connected or if the load's resistance is high enough, we can safely select a V_{CE} value V_{CE} = V_{CC} / 2. We will see later how to calculate V_{CE} if load is connected. In the meanwhile, let's calculate the appropriate R_{C} value to achieve the required V_{CE}. We know that:
From the above equation we get an I_{C} value. What we need to do now is set an appropriate I_{B} value to achieve this I_{C}:
I_{C} = h_{FE} x I_{B} => I_{B} = I_{C} / h_{FE}
The I_{B} can be calculated with the ohm's law:
I_{B} = V_{CC} / R_{B} => R_{B} = V_{CC} / I_{B}
Now we can proceed with the AC analysis. As we saw before, we can calculate the internal emitter resistance by the emitter current (which is equal to the collector current):
r'_{e} = 25mV / I_{C}
And from the Î model we can define the base input resistance:
Zin(base) = hfe x r'_{e}
We proceed with the AC analysis by drawing the AC equivalent:
R_{B} is parallel to the Zin(base) of the transistor. So we can calculate the amplifier's input impedance by calculating the total resistance of R_{B} // Zin(base):
Zin = (R_{B} x Zin(base)) / (R_{B} + Zin(base))
Now we can calculate the base AC voltage from the AC input voltage. Rg and Zin are connected as a voltage divider, so the base AC voltage is calculated like this:
u_{B} = ACVp-p x Zin / (Rg + Zin)
Now we can calculate the AC base current:
i_{B} = u_{B} / Zin(base)
The AC collector current depends on the hfe value:
i_{C} = hfe x i_{B}
Finally, we calculate the AC voltage across R_{C}:
u_{Rc} = i_{C} x Rc
We have everything we need to determine the amplifier's characteristics. First, we can determine the voltage gain (Av):
Av = u_{Rc} / ACVp-p
We can also determine if the amplifier operates as a small signal amplifier. As we've already said, a small signal amplifier has ac peak to peak collector current at least 10 times smaller than the DC collector current. We have already calculated both currents, so we can compare them and determine if the amplifier is a small signal amplifier. This way we can determine if the output will have significant distortion or not.
This methodology may be rather stiff for a beginner, but it is also quite accurate. An experienced designer can calculate such an amplifier within a few minutes by heart.
to get more information about <a href="http://911electronic.com/tunnel-diode-characteristic-symbol-definition/">tunnel diode</a> click hotlink. I found this site yesterday and i think there is a lot information about diodes.
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?