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24 February 2009
Author: Giorgos Lazaridis
BJT Transistor theory




The Transistor operation in AC

Until now, we've been talking only for transistors in DC operation. We've learned how to bias a transistor correctly, and we also saw a quick example on how to set the operation point. But many times, transistors are used to operate with AC signals. A transistor audio amplifier for example is an AC signal amplifier, since the microphone generally generates an AC output. And here is a point that many people confuse: Transistors are NOT AC components: Transistors can only operate with DC signals!

That sounds kinda paradox, since we all know that transistors are also used to amplify AC signals. But think about this: Suppose that we have an NPN common emitter transistor amplifier, and we feed an AC signal at its base. As long as the input signal is higher than 0.7 Volts, it will be amplified normally and it will appear at the output of the amplifier. But what happens when the signal becomes less than 0.7 Volts? And worst, what happens when the signal becomes negative? As we know, an AC signal has a positive and a negative period.

The answer is this: A negative signal at the base of an NPN transistor means that the base-emitter diode is reverse-biased. The diode acts like an open circuit (cut-off) and the amplifier does not work at all. There is also a tight limit: If the negative voltage become too high, the base-emitter diode will be destroyed. The maximum reverse voltage that this diode can handle is usually around 5 volts for small signal transistors. The exact value for each transistor can be found in the manufacturer's datasheet, usually with parameter name VEBO (Emitter-Base voltage). The same situation happens of course if we use a PNP transistor and we reverse-bias the base-emitter diode.

So, how is it possible to amplify an AC signal? The answer is by biasing the transistor with DC voltage. Suppose for example that we want to amplify an 1 Vp-p AC signal. This means that the signal has +0.5 Volts positive period and 0.5 Volts negative period. If we bias the input with let's say 1.5 volts DC, then the input will vary from 2 Volts (1.5 + 0.5) to 1 Volt (1.5 - 0.5). This is considered as a DC signal and the transistor can amplify the complete period normally.


An AC signal fed into the base of the transistor will not be amplified correctly since it has a negative period as well. Biasing the transistor with DC voltage we cause the AC signal to be shifted upwards, therefore it has no more negative period.




Coupling and Bypassing Capacitors

As we said before, transistors are DC components. This means that the output will also be a DC voltage. But if we amplify an AC voltage, then we probably want to get an AC voltage at the output as well. How is this done? Simple, with a coupling capacitor. A capacitor operates as a resistor in AC and as an open circuit in DC. It is not the purpose of this theory to analyze in details the capacitor's behavior in AC and DC. Nevertheless, it is important to have some very basic knowledge about capacitors.




The Coupling Capacitor

A coupling capacitor is a capacitor connected in series with the circuit that we want to couple. The AC signal is free to go through the capacitor, while the same capacitor acts as an open circuit against any DC current. Let's see an example of a coupling capacitor:





Both Cin and Cout are coupling capacitors. Their job is to block any unwanted DC voltage from between the stages that they couple. This is the what would appear in the oscilloscope's screen if two channels were connected, one before the Cout coupling capacitor (Channel 1-Blue line) and one after the same capacitor (Channel 2-Red line):





Looking at channel 1, it is obvious that the amplified AC voltage has been shifted above the zero line, and has become a DC voltage. That is because the DC voltage from the emitter of the transistor has been added to the amplified ΑC signal. Looking at channel 2, any DC voltage has been removed due to the coupling capacitor. It is possible only for the AC voltage to cross, and therefore it has become an AC voltage again.




The Bypassing Capacitor

A bypassing capacitor is a capacitor connected in parallel with a circuit that we want to bypass. Unlike the coupling capacitor, the bypassing capacitor removes any unwanted AC signal from this circuit, since any AC current goes through the bypassing capacitor, leaving only the DC current to go through the parallel circuit. Let's see an example:





A bypass capacitor (Ce) is connected in parallel with a resistor (Re). What we want is to have only DC current flowing through the resistor, in order to maintain the current stable (Ie=Ve/Re). The actual problem is that when an AC signal is applied at the base of this circuit, this AC signal will also appear at the emitter of the transistor. This will change the emitter voltage which will eventually change the emitter current, and we do not want that. Therefore, we add the bypassing capacitor Ce. All the AC voltage will be grounded through this capacitor. Therefore, the voltage across the resistor Re will not change, and the current will remain stable. In the following graphs you can see what would appear in an oscilloscope's screen, if the probe was connected across Re. The left graph shows the output if Ce was NOT connected across Re, and the right graph shows the output with the bypassing capacitor Ce connected across Re:













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  • Comments

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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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