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24 February 2009
Author: Giorgos Lazaridis
BJT Transistor theory




A Common Base amplifier example

We want to design a small signal common base amplifier. The input is 10mV at 1KHz. The voltage gain must be around 10 (Av=10). The transistor is a BC548 NPN with β=100. The DC supply (VDD) is 12V.

Here is the schematic diagram of the amplifier:





We extract and calculate the DC equivalent: We start with an estimation of RE. RE = 10K

VCE = VDD / 2 => VCE = 12 / 2 = 6V

VC = Av x 25 mV => VC = 10 x 25 mV = 250 mV

VE = VDD - VC - VCE = 12 - 0.25 - 6 = 5.75 V

IE = VB / RE = 5.75 / 10000 = 0.575 mA

VB = VBE + VE => VB = 0.7 + 5.75 => VB = 6.45 V

RB2 = (VB x 0.1 x β x RE) / VDD = (6.45 x 0.1 x 100 x 10000) / 12 => RB2 = 53750 Ohms

So we choose a 56K resistor for RB2.

RB1 = RB2 x (VDD - VB) / VB => RB1 = 56000 x (12 - 6.45) / 6.45 = 48186 Ohms

So we choose 47K for RB1.

RC = VC / IC = 0.25 / 0.575 = 434 Ohm

We choose 470 Ohms for RC.



Here is this circuit prototyped on a breadboard:





And here is a closer look at the transistor and it's biasing components:





The results can be seen in the oscilloscope. A 10 mVp-p signal is applied at the input. The first channel of the oscilloscope (yellow channel) is connected at the input, and the second channel (green channel) is connected at the output:





It is obvious that the output is 10 times bigger than the input (Av=100 / 10 = 10). The amplifier operates as expected. With the same methodology, we can design a common base amplifier with voltage gain 100 (Av=100). From the calculations we get the part values:

RB1 = 65000 Ohms [ We will use 68 K ]

RB2 = 35000 Ohms [ We will use 33 K ]

RC = 7142 Ohms [ We will use 8.2 K ]





And here are the results:





The input is still 10 mV, but the output is now 2 divisions multiplied by 500 mV, which makes 1000 mV (1 Volt). The voltage gain is now 100.









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    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:
      http://www.pcbheaven.com/forum/index.php?board=30.0


  • 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:
      http://www.pcbheaven.com/forum/index.php?topic=1773.0


  • 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

      regards


  • 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)
      =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

      - 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.
      Thanks


  • 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..!! ^_^



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