Let's take a look at the DC and AC equivalents of a common emitter amplifier:
Let's remember the equation that we used before to draw the load line:
VCC = IC x RC + VCE + VE
It is obvious that if the resistor RC is changed, the slope of the load line will also change. As you see from the equivalent circuits above, the RC resistor of the DC equivalent is different from that of the AC equivalent. That is because the output of the amplifier has a load coupled through a coupling capacitor. This load (RL) takes no part on the DC equivalent since the capacitor acts as an open circuit in DC, but the same load is connected in parallel with the collector resistor RC on the AC equivalent. If the resistance of the load is many times higher than the collector's resistor, then the parallel total resistance (RC//RL) is almost equal with the collector's resistance RC. Otherwise, the resulting total parallel resistance is significantly smaller than RC. This means that the saturation current is increased and the VCE voltage is decreased. Let's see what changes in the scene. In the AC equivalent, we can add the voltages according to Kirchhoff's law:
uce + ic x rc = 0 => ic = - uce / rc (1)
The minus sign means that the current is reversed, but for now we can simply omit it. The AC collector current is given by the following equation:
ic = ΔIC = IC - ICQ (2)
And the AC collector voltage:
uce = ΔVCE = VCE - VCEQ (3)
We can replace the equations 2 and 3 to the equation 1 and extract the following equation:
IC = ICQ + (VCEQ - VCE) / rc
This is the new equation from which we get the 2 points for the AC load line. To find them, we do the same trick as we did for the DC load line: First we zero the IC current to extract the VCE voltage, and then we zero the VCE voltage to extract the IC current:
VCE(cut) = VCEQ + ICQ x rc (for IC=0)
And now let's reset the VCE to get the IC:
IC(sat) = ICQ + (VCEQ / rc)
Drawing the DC and AC load lines - An example
Here is a typical common emitter amplifier:
Rg is the AC generator's internal resistance and RL is the load resistance. Let's first draw the DC load line. For this, we will be needing the DC equivalent circuit. Very quickly we have:
VB = 12 x 2200 / 12200 = 2.16 V VE = 2.16 - 0.6 = 1.56 V IE = 1.56 / 800 = 1.95 mA IC = 0.99 x 1.95 = 1.93 mA VC = 1.93 x 2200 = 4.24 V VCE = 12 - 4.24 - 1.56 = 6.2 V
The Q point is at VCEQ = 6.2V and ICQ = 1.93mA. For the Load line, we have:
For IC=0: VCE = 12V
For VCE=0: IC = 12 /(2200 + 800) = 4 mA
Now we can draw the DC load line and set the Q point:
Now we will work on the AC equivalent to draw the AC load line:
The 1k8 resistor is the result of the 10k (R1) parallel to the 2k2 (R2), and the 1k4 is the result of the 2k2 collector resistor parallel to the 4k load resistance. First the VCE:
VCE(cut) = VCEQ + ICQ x rc = 6.2 + 1.93 x 10-3 x 1400 = 8.9 V
And for IC:
IC(sat) = ICQ + (VCEQ / rc) = 1.93 + (6.2 / 1400) x 103 = 6.35 mA
We will now draw the AC load line with green color on the same characteristic with the DC load line (red color):
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?