Again, there are situations in which the signal clipping is eligible, for example in B and C class amplifiers. But in many application the signal must be amplified without any distortion or clipping whatsoever. Therefore, we must be able to design amplifiers with maximum unclipped amplification gain.
There are 2 steps to design an amplifier with maximum unclipped output. The first step is to determine the maximum output peak to peak voltage (Vp-p Max), and the second is to set the operation point at the half of the max Vp-p Max.
Let's first see the case that the load resistance is very high or no load is connected. In that case, the AC load line is almost the same as the DC load line, so we can safely work only with the DC load line. The maximum oscillation output can be from 0.5V to VCE(cut). We avoid operating the transistor near the saturation area because in that area the output signal is highly distorted, therefore we use the number 0.5 for our calculations. So, to set the Q point, all we have to do is to find the middle. There is a simple formula which does exactly this:
VCEQ = (0.5 + VCE(cut)) / 2
Let's see an example. Suppose that we calculated the IC(sat)=4mA and VCE(cut)=12V. From these points we draw the DC load line as shown bellow. To set the optimum Q point, we need to know only the VCE(cut) which is 12V:
VCEQ = (0.5 + 12) / 2 = 12.5 / 2 = 6.25V
This way we achieve maximum oscillation within the complete linear area of the transistor, without exceeding the VCE(cut) value (12V).
Maximum Unclipped Oscillation under Load
Suppose now that the load resistor is not that big, and the AC load line has different slope than the DC load line. First of all, let's make something clear: Since the load is connected in parallel (AC analysis) to the collector or emitter resistor, this means that the resulting AC resistor (rc or re) can only be smaller than the collector or emitter DC resistor (RC or RE). Therefore, it is easy to understand that the Ic(sat) current of the AC load line can only be higher than the IC(sat) current of the DC load line. And since the AC and DC load lines intersect at the Q point, it is absolutely certain that the Vce(cut)-AC voltage of the AC load line can only be less than the VCE(cut)-DC voltage of the DC load line.
The previous statement makes clear that, if the AC load line is not the same as the DC load line, the output oscillation without clipping becomes smaller. As a matter of fact, the new oscillation range will be 0,5V up to Vce(cut-off)-ac. To calculate the Q point we use the same formula as before, but we replace the VCE(cut)-DC term with the Vce(cut)-AC:
VCEQ = (0.5 + Vce(cut)-ac) / 2
This formula tells us that, in order to achieve the maximum unclipped output, we need to set the Q point in the middle (approximately) of the AC load line. Here is an example:
How to set the Optimum Q Point
There are many ways to change the Q point, since any change on the DC bias will also change the Q point. The designer may choose to go with the trial and error method, or by solving the mathematics equations. But no matter which method is used, the designer must be able to locate the biasing part that needs to be changed, so that this change will have big effect on the Q point and small or no effect on the rest of the circuit and its characteristics.
Changing the Q point in Common Emitter connection
As always, we suppose that the transistor is biased with a voltage divider, and the emitter has also a small feedback resistor. Let's remember how VCE is calculated:
VCC = IC x RC + VCE + IC x RE = > VCE = VCC - IC x RC - IC x RE
So, by changing either RC or RE, we can change the VCE, thus we change the VCEQ of the Q point. But which one to choose? The answer is simple. The capacitor CE acts as a bridge in AC signal, which means that that the resistor RE does not have any affect on the AC signal, and therefore has no affect on the AC load line. Therefore, we prefer to change the emitter resistor RE, since it affects only the DC load line. By increasing the IE (=IC) current the VCEQ point shifts rights. If IE is decreased the VCEQ point shifts left.
Changing the Q point in Common Collector connection
Here is an example of a common collector connection:
It is obvious that if RE is changed, it will have an affect on both AC and DC load lines, since there is no bypassing capacitor across this resistor. As we know, this type of connection is also called "emitter follower", because the emitter voltage follows the base voltage:
VE = VB - VBE (1)
Moreover, from the schematic we can calculate the VCE:
VCC = VCE + IE x RE = VCE + VE => VCE = VCC - VE (2)
We replace the term VE in the second formula (2) from the first formula (1):
VCE = VCC - (VB - VBE) => VCE = VCC - VB + VBE
This equation tells us that we can change the VCE and thus the VCEQ of the Q point by changing the base voltage VB. So, we can simply change the RB1 or RB2 resistor to achieve the optimum Q point. There is something that we need to take into account here. First, let's see the AC equivalent of the circuit:
As you see, RB1 and RB2 are still active components in the AC equivalent. These components have an affect at the input signal, since RB1 and RB2 will eventually define the input stage impedance. A large change on either resistor may require to re-design the circuit.
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.
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.
@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:
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?