Home     Contact     Projects     Experiments     Circuits     Theory     BLOG     PIC Tutorials     Time for Science     RSS     Terms of services     Privacy policy  
 Home      Projects     Experiments     Circuits     Theory     BLOG     PIC Tutorials     Time for Science   

24 February 2009
Author: Giorgos Lazaridis
BJT Transistor theory

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:

Relative pages
  • Basic transistor circuits
  • 555 theory of operation
  • Learn how to interface ICs
  • Learn how dimmers work
  • The TRIAC theory
  • Learn about the most popular PC Cooling methods
  • Dr.Calculus: Checking transistor functionality
  • Op-Amp IC Pinouts

  • Comments


      Email (shall not be published)


    Notify me of new posts via email

    Write your comments below:
    BEFORE you post a comment:You are welcome to comment for corrections and suggestions on this page. But if you have questions please use the forum instead to post it. Thank you.


  • At 25 October 2015, 17:44:11 user Mohammad Irfan wrote:   [reply @ Mohammad Irfan]
    • Hey keep it up! nice work

  • At 28 August 2015, 21:23:03 user tomaeh wrote:   [reply @ tomaeh]
    • 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.

  • At 17 June 2015, 15:13:52 user Vahhab wrote:   [reply @ Vahhab]
    • 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 6 April 2015, 23:20:04 user Billy Keith wrote:   [reply @ Billy Keith]
    • These are perfect explanations. Thank you for a great reference source.

  • At 19 February 2015, 8:23:36 user Giorgos Lazaridis wrote:   [reply @ Giorgos Lazaridis]
    • @lee smith Sure. It is here:

  • At 18 February 2015, 16:20:16 user lee smith wrote:   [reply @ lee smith]
    • Thanks George!

      Just went through your transistor theory pages. Do you have a pdf/doc package with all the pages? I have some serious studying to do.

  • At 5 January 2015, 16:56:28 user sadusaisaandeep wrote:   [reply @ sadusaisaandeep]
    • how to dc power changing ac power?

  • At 13 October 2014, 18:29:02 user pranav wrote:   [reply @ pranav]
    • wow you have explained all in very detailed form . thank you very very much.

  • At 26 June 2014, 5:39:22 user tharaka wrote:   [reply @ tharaka]
    • what is the anode voltage if the cathode is grounded in a diode?

      A-|>---(grnd) what is the voltage value of A if the diode is silicon ?

  • At 25 May 2014, 19:48:48 user Giorgos Lazaridis wrote:   [reply @ Giorgos Lazaridis]
    • @Nash Its in this link:

  • At 25 May 2014, 15:51:22 user Niyas wrote:   [reply @ Niyas]
    • where can I find pdf version of this article?

  • 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

  • At 7 May 2013, 22:45:13 user mangyi wrote:   [reply @ mangyi]
    • your explanation is far better than my course book, thank you very much

  • At 15 April 2013, 20:29:39 user Giorgos Lazaridis wrote:   [reply @ Giorgos Lazaridis]
    • @Ganesh Nadar No, the biasing must be DC

  • At 15 April 2013, 14:16:25 user Ganesh Nadar wrote:   [reply @ Ganesh Nadar]
    • Can Vcc of transistor be AC 12V???

  • At 9 December 2012, 10:05:26 user Giorgos Lazaridis wrote:   [reply @ Giorgos Lazaridis]
    • @peter boltink in this example, VC is Vre because only Vre is there to change Vc

  • At 22 November 2012, 4:12:45 user peter boltink wrote:   [reply @ peter boltink]
    • I think there is a mistake
      VC is Vce VE = 7.76V
      and the voltage at the collector resistor is (1.93 x 2200) = 4.24v

      Vc is not the same as Vre


  • At 30 October 2012, 14:21:18 user ravi shankar wrote:   [reply @ ravi shankar]
    • working of bjt tansistor {clearly}

  • At 16 July 2012, 17:59:37 user Giorgos Lazaridis wrote:   [reply @ Giorgos Lazaridis]
    • @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.

  • At 9 July 2012, 1:58:04 user Mint Electronics wrote:   [reply @ Mint Electronics]
    • Can somebody please explain to me how the following formulas were constructed?

      IC = (VCC - VBE) / (RE (RB / hfe)) (pg. 3)

      How is the voltage divider current bigger than the base current? Isn't it the other way around? (pg. 4)

      VCE = VCC - IC x RC - IE x RE = VCC - IC (RC RE) (pg. 4)

      IVD > 10 x IB => RVD < 0.1 x %u03B2dc x RE (pg. 4)

  • At 3 July 2012, 8:18:54 user Giorgos Lazaridis wrote:   [reply @ Giorgos Lazaridis]
    • @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.

  • At 3 July 2012, 2:26:34 user Mint Electronics wrote:   [reply @ Mint Electronics]
    • 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)

      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

      - Cheers,
      Mint Electronics.

  • At 17 April 2012, 7:40:29 user Giorgos Lazaridis wrote:   [reply @ Giorgos Lazaridis]
    • @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.

  • At 23 February 2012, 4:14:02 user fulton g.w. wrote:   [reply @ fulton g.w.]
    • 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

  • At 16 January 2012, 20:09:23 user Kammenos wrote:   [reply @ Kammenos]
    • @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?

  • At 21 November 2011, 9:18:21 user Kammenos wrote:   [reply @ Kammenos]
    • @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.

  • At 16 July 2011, 0:55:40 user Juan Carlos wrote:   [reply @ Juan Carlos]
    • The two best invention of mankind are the transistor and the Condon

  • At 12 July 2011, 13:51:05 user ramzal wrote:   [reply @ ramzal]
    • it is very effective to understand basics of transistors!

  • At 13 June 2011, 17:09:57 user Kammenos wrote:   [reply @ Kammenos]
    • @Fung which circuit are you referring to?

  • 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?

  • At 22 April 2011, 5:29:31 user Kammenos wrote:   [reply @ Kammenos]
    • @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?

  • At 25 September 2010, 4:28:00 user pradeep wrote:   [reply @ pradeep]
    • motor invetor

  • At 2 July 2010, 16:52:58 user suguna wrote:   [reply @ suguna]
    • why is the fixed bias designed so?

  • At 18 May 2010, 23:49:41 user Raldey wrote:   [reply @ Raldey]
    • Thankz..it really helps me a lot.. now i understand my lesson..!! ^_^

    reddit this Reddit this

     HOT in heaven!

    NEW in heaven!

    New Theory: AC electric motor working principle

     Contact     Forum     Projects     Experiments     Circuits     Theory     BLOG     PIC Tutorials     Time for Science     RSS   

    Site design: Giorgos Lazaridis
    © Copyright 2008
    Please read the Terms of services and the Privacy policy