Output characteristic, Load Lines and Quiescence point
There is a very interesting methodology to graphically analyze the operation of a transistor amplifier using the output characteristics and the load lines. By properly setting the quiescence point along the load line, one can select how the transistor amplifier will operate. For example, the same transistor can implement a Hi-Fi audio amplifier, B-class amplifier or a load switch, simply by setting the quiescence point onto different positions.
The DC Load Line
If you open a transistor datasheet, you will probably find a set of diagrams and characteristics. One of these is the Common Emitter output characteristic or IC to VCE characteristic and looks like this:
This is the IC to VCE characteristic of a BC547 transistor. The horizontal axis (x) has the VCE voltage in volts, and the vertical axis has the IC current, usually in milliamperes. Between them, there are several different curves. Each one of these lines corresponds to a different base current, usually measured in microamperes. From now on, we will work extensively with these characteristics, so it is important for you to understand how to read them and how to use them to determine the operation of the amplifier.
The DC Load Line is a line that we draw on these characteristics, which eventually determines all the points that the transistor will operate at. In other words, the operation point (usually called Q from the word "Quiescence") will be somewhere on the DC load line. We use the term "DC" because -as we will see later on- there is also an AC load line. Many times, when we talk about the DC load line, we omit the term "DC" and we write only "Load Line" meaning the "DC Load Line". To draw this load line, we need to know the collector current and the collector-emitter voltage. Suppose for example that IC=40mA and VCE=12V. The load line is drawn with red color:
We will explain how to draw the load line, but before we do, we must first discuss about the 4 basic regions of this characteristic, the saturation area, the cut-off area, the linear area and the breakdown point.
Region 1: The Saturation Area
The saturation area is the area where the load line intersects with the saturation point of the characteristics. In the following drawing, we've marked the saturation area with a red transparent filter:
Due to the fact that the VCE potential is low at the saturation area, sometimes we consider that the saturation point is at the top of the load line (VCE=0) with a small error. Normally, the VCE voltage required to achieve saturation current is a few tens of a volt. This voltage is called "Saturation Voltage" and we use the symbol VCES.
If the collector current is very high, the collector contact of the transistor is overheated and eventually the transistor is destroyed. Therefore, if the transistor is planned to operate at the saturation area -usually for switching applications-, caution must be taken to maintain the collector current bellow harmful levels.
If the transistor operates as an amplifier and it is driven in the saturation area, then the output signal is distorted. That is because the transistor does not operate in it's linear areas. To avoid this, the transistor amplifier must be calculated in a way that the collector-emitter voltage will not fall bellow the saturation voltage. Generally:
VCE > 0.5V => IC < VCC / RC
Region 2: The Cut-Off area
The cut-off area is the area in which the collector current becomes zero. In the following drawing, the cut-off area is marked with yellow mask:
Generally, we can say that in order for a transistor to work in the cut-off area, the base current IB must become zero. This comes out of the IC to IB equation:
IC = β x IB
The precise equation to calculate the cut-off point is this:
IB = (ICO / (1-α)) + (β x IB)
ICO is the reverse saturation current. Since it is very small, usually around 10 to 50nA, we can safely remove it from the previous equation.
When a transistor operates in the cut-off area, no current flows within the collector. Usually we drive the transistor in this area when we want it to operate as a switch. If the transistor operates as an amplifier, then the output signal will be clipped.
Region 3: The Linear Area
The linear area is the area between the cutoff and the saturation area of the transistor, as shown bellow with a green mask:
It is called "linear area" because in this area the transistor has the most linear operation. This area represents the normal transistor operation. To successfully design a transistor amplifier, the designer must be able to put the transistor to work within this area, otherwise the output signal will be either clipped or distorted. There are though special occasions where an amplifier operates beyond the linear area, such as a class-B or class-C amplifier.
On the other hand, if the transistor operates as a switch, it must not operate within this area. A switch must be either ON or OFF, and this can only be achieved if the transistor operates in saturation of cutoff areas.
Region 4: The Breakdown Point
The breakdown point is the point on the VCE axis above which the collector current increases rapidly and the transistor is destroyed. This area is marked with a purple mask in the following drawing:
@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?