In the previous section we learned how can someone analyze a transistor circuit. But if someone wants to design an amplifier, the methodology is different. In this section we will discuss the design methodology for the three different transistor connections.
Common Base Transistor Design Methodology
A common base amplifier is usually used as an input amplification stage for high frequency applications (small signal amplifier). We will not discuss much about this type of connection, since it is very rarely used. Here is a typical Common Base transistor amplifier schematic with an NPN transistor:
There will be a lot of mathematical functions and equations that have to be solved to analyze the circuit, and therefore i made an on-line calculator. You can use it to speed up your work. Here is the link:
The resistor Rg is the resistance of the AC signal generator. The amplifier is biased with Voltage Divider (R_{B1}-R_{B2}). The output is taken from the collector and the input signal is applied at the emitter.
The designer must know the input signal size (amplitude and frequency range) and the required amplification factor. Also, the designer must know the DC voltage that will be used for the transistor bias (V_{DD}). A common base amplifier has no current amplification, therefore we only need to calculate the voltage amplification. The design can begin with the DC equivalent:
First thing we can define with an educated guess is the R_{E} resistor. We know that the AC input must be able to provide enough current for the emitter and the collector circuit. Therefore, we want to maintain the emitter current as low as possible. We can safely choose a large resistor such as 10 Kohms for R_{E}.
Usually, we place the Q point in the middle of V_{CE}. This means that:
V_{CEQ} = V_{DD} / 2
It can be proved that the voltage gain is calculate from this formula:
Av = R_{C} / r_{i}
To calculate r_{i} we must first calculate the emitter resistor r'_{e}. But we cannot calculate r'_{e} since we do not know the emitter current. So we must find a way to calculate the emitter voltage first. Let's remember the formula for the internal emitter resistor:
r'_{e} = 25mV / I_{C} = 25mV / I_{E}
We replace this formula to the gain amplification formula:
Av = R_{C} / ( 25mV / I_{E} ) => Av = I_{E} x R_{C} / 25mV = I_{C} x R_{C} / 25mV = V_{C} / 25mV => V_{C} = Av x 25mV
From the above formula we can directly calculate the collector voltage V_{C}. Now we can use the Kirchhoff's law to calculate the emitter voltage:
From the voltage divider formula, we can now calculate the resistor ratio needed to get this voltage:
V_{B} = V_{DD} x R_{B2} / ( R_{B1} + R_{B2} )
Let's assume that we choose to design a firm voltage divider. This means that the current through the voltage divider (I_{VD}) must be at least 10 times greater than the base current. In other words:
I_{VD} = 10 x I_{B} => Rvd = 0.1 x Î² x R_{E} => R_{B1} + R_{B2} = 0.1 x Î² x R_{E}
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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?