Output characteristic, Load Lines and Quiescence point
How to draw the Load Line (DC Load line)
To get a first clue about the Load Line, I will use an example with a Common Emitter amplifier with voltage divider biasing for the base. More details are yet to come for each connection type. Here is the circuit:
For an instance, let's forget about the input portion of the circuit (the voltage divider) and let's work only with the output portion. According to Kirchhoffs' law we have:
V_{CC} = I_{C} x R_{C} + V_{CE} + V_{E}
Let's now calculate the first point of the load line. This point (like any other point in a x-y Cartesian system) has 2 terms: an x and a y term. We need to find a V_{CE} and a I_{C} pair of values which corresponds to the x and y terms respectively. We can make this calculation much easier with a simple and common trick: We will calculate a V_{CE} value for I_{C}=0. This way, the first pair will be on the V_{CE} axis. Let's solve the previous equation for V_{CE}:
V_{CE} = V_{CC} - I_{C} x R_{C} - V_{E}
First thing that we notice is that I_{C} x R_{C} is zero, since I_{C} is zero. V_{E} is zero as well, because I_{E} is -almost- equal to I_{C}, and V_{E} is equal to I_{E} x R_{E}. The equation can be re-written as follows:
V_{CE} = V_{CC} - 0 - 0 => V_{CE} = V_{CC} = 10V
Now for the second pair. Similarly, we will calculate a I_{C} value for V_{CE} = 0. So, the second point will be on the I_{C} axis. Let's solve for I_{C}:
So, now we have the 2 points that we need to draw the load line. The points are:
Point A (10, 0) Point B (0, 2.17)
The green dots indicate the 2 points of the load line, and the red line is the DC load line itself.
The Operation Point, AKA Quiescence point - Q
We will continue the previous example and we will calculate the Q point. There is something that you need to make clear: The Q point is a point on the Load Line. The Load Line is calculated (as we saw before) by finding two points in the cut-off and saturation area. The Q point is usually placed within the linear area. The Q point is determined by the DC biasing of the transistor. The fact that the circuit uses VDB (Voltage divider Bias), allows us to neglect the base current in our calculations. So, the base voltage is:
V_{B} = V_{CC} x R2 / (R1 + R2) = 10 x 2200 / 12200 = 1.8 V
We can now calculate the emitter voltage as follows:
V_{E} = V_{B} - V_{BE} = 1.8 - 0.7 = 1.1 V
And the emitter current is:
I_{E} = V_{E} / R_{E} = 1.1 / 1000 = 1.1 mA
And since the collector current is almost equal to the emitter current, we can calculate the voltage drop across R_{C} as follows:
V_{RC} = I_{C} x R_{C} = I_{E} x R_{C} = 1.1 x 3600 = 3.96 V
Finally, the Collector-Emitter voltage is calculated as follows:
V_{CE} = V_{CC} - V_{C} - V_{E} = 4.94 V
Now we have everything we need to set the Q point:
I_{C} = 1.1 mA V_{CE} = 4.94 V
Let's analyze for one moment what we've done so far. First, we found two points to draw the load line. The first point was found on the V_{CE} axis by zeroing the I_{C} current, and the other point was found on the I_{C} axis by zeroing the V_{CE} voltage. Both points were calculated with the same equation:
V_{CC} = I_{C} x R_{C} + V_{CE} + V_{E}
For the first point we solved this equation for V_{CE}, and for the second we solved it for I_{C}. Then, we calculated the Q point. For the Q point, we need another pair of I_{C}-V_{CE} values. These values are calculated from the DC transistor bias. If the DC bias is not in the cut-off or in the saturation area, then it must be somewhere on the DC load line that we draw before. That is true since both the load line points and the quiescence point are calculated with the exact same equation. The only difference is that for the load line, we choose (for our own ease and only) to find points on the two axis, whilst for the quiescence point we solve the equation for the DC values defined by the biasing resistors.
Since the quiescence point is only one, and since the I_{C} and V_{CE} values of the quiescence point are critical, we usually use the pointer "q". So, for the quiescence V_{CE} voltage we use the symbol V_{CEq}, and for the quiescence I_{C} current we use the symbol I_{Cq}.
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?