An amplified AC signal is subject to clipping if it exceeds some specific levels. These levels are determined by the DC and AC load lines and the operation point Q. To explain why the output clipping occurs, we will first work with the simplest case in which the DC load line is the same as the AC load line. This happens (as we explained before) if the transistor output has no load, or if the load has very high resistance, about ten times higher than the collector's resistor RC. Suppose that we have the following DC load line:
Now suppose that an AC signal is applied to the input of the transistor. Since the AC load line is the same as the DC load line, the Q point will move up and down on the DC load line. The amplitude of this oscillation depends on the base current of the input signal:
The purple waveform shows the input base current change caused by the input signal. It oscillates from approximately 7uA to 17uA. This input causes an output signal from 3 to 11 Volts at VCE (orange waveform). Now suppose that the input signal amplitude is further increased:
A B-Class amplifier clips one complete semi-period of the input signal
As you see, the right side of the output waveform is clipped. We call this "distortion" because the output signal is distorted. In some situations, this distortion is legitimate. A B-Class audio amplifier for example has the Q point very close to one end of the load line, and it amplifies only one side of the input waveform. But there are many situations where the signal must be amplified without any distortion. So, extra care must be taken to avoid clipping.
It is obvious that the maximum output voltage depends on the position of the operating point Q and the maximum VCE. The maximum total VCE oscillation cannot exceed the maximum VCE as defined from the load line. But this is not enough. As you can see, the output waveform could be clipped only in one side. Therefore, we need to define two different maximum levels. Since the output waveform oscillates around the VCEQ point, we divide it into the left portion and the right portion relative to the VCEQ point:
VMax-Left = VCEQ VMax-Right = VCE(cut) - VCEQ
It is obvious that the maximum output can be achieved if the operating point is placed in the middle of VCE. As a matter of fact, the maximum output is achieved if the Q point is little above the middle of VCE, due to the saturation region.
Output Signal Clipping under Load
In the previous pages we explained how the load line is affected when load is connected at the transistor output. Moreover, we explained how to draw the AC load line along with the DC load line. I will use the drawing from the final AC-DC load lines from the example in page Drawing the DC and AC load line example.
The red line is the DC load line and the green is the AC load line. Both lines intersect at the Q point. The VCE oscillation will still take place around the VCEQ point. It is obvious that the maximum left change is the same like before, without any load being connected at the output. But the maximum right change is now different. Since the cutoff VCE of the AC load line occurs before the cutoff VCE of the DC line, the output signal will be clipped at the AC VCE(cut).
VMax-Left = VCEQ VMax-Right = VCE(cut)AC - VCEQ
There is a simpler way to calculate the VMax-Right. As we know, the VCE(cut)AC is:
VCE(cut)AC = VCEQ + ICQ x rc
If we replace this equation to the previous, the result is the following:
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?
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