Single-Transistor Audio Amplifier - How the Common Emitter Amplifier Works

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  • เผยแพร่เมื่อ 28 ก.ย. 2024
  • I demonstrate how to make an audio amplifier with a single transistor on a breadboard, which is capable of running a 8 Ohm loudspeaker. The amplifier is a common emitter amplifier, which is a class A amplifier. I will show the circuit diagram, explain how it works, and show how to build it on a solderless breadboard. In the end of the video, we will listen to the tone from an audio oscillator, amplified using this audio amplifier. The NPN-transistor used is BC547.
    BC547 Audio Amplifier
    BC547b Audio Amplifier
    Simple Amplifier
    Useful references, showing how to calculate the optimal resistance values:
    • EECE 251 - BJT Design ...
    • EECE 251 - BJT Tutoria...
    #miniaudioamplifier #amplifier #amplifiermini #amplifiermaking

ความคิดเห็น • 78

  • @HenriksHobbyElectronics
    @HenriksHobbyElectronics  2 ปีที่แล้ว +9

    Do you have any questions about this circuit? Ask below and I'll answer as many questions as I can :)

    • @HenriksHobbyElectronics
      @HenriksHobbyElectronics  2 ปีที่แล้ว +1

      There is a typo in the circuit diagram: the 470 kOhm resistor should actually be 47 kOhm. Both values will create a decent sound in this experiment though.

    • @addkas470
      @addkas470 2 ปีที่แล้ว

      Thank need oscilater circut and radio tuner curcuit your explainion easy to understand

    • @adastra123
      @adastra123 2 ปีที่แล้ว +2

      How come there was no capacitor on the input between the base and emitter ?
      I love your channel by the way.

    • @HenriksHobbyElectronics
      @HenriksHobbyElectronics  2 ปีที่แล้ว +1

      @@adastra123 Thanks! I am not aware that adding a capacitor between the base and the emitter would improve the circuit in any way.

    • @StrsAmbrg
      @StrsAmbrg 2 ปีที่แล้ว

      I have two questions:
      1). What is the explanation of why the output capacitor is non-polar while the by-pass capacitor is polar?
      2). If you put capacitor in input side, what should it be? Polar or not-polar?

  • @hortchester1255
    @hortchester1255 10 หลายเดือนก่อน +4

    Its ridiculous hard much ive had to search for a norkal breadboard tutorial lol. Atleast now im here so i appreciate you

  • @spacecraftbuildingservices
    @spacecraftbuildingservices 8 หลายเดือนก่อน +2

    Interesting and informative. Hope you will do more about basic circuits. Thanks

  • @RexxSchneider
    @RexxSchneider 2 ปีที่แล้ว +44

    Taking your modification of 47K instead of 470K, here's the DC analysis:
    The collector current (Ic) can't exceed 9V / (2K + 680R) = 3.4mA and the BC547 has a minimum Hfe (β) of 110 at those sort of currents (typically around 300).
    The base current can't be more than 30μA and will almost certainly be much less, probably around 10μA.
    The current through the base bias resistors is around 9V / (47K + 10K) = 158μA, which is significantly greater than Ib, so we can ignore Ib to a first approximation.
    The voltage at the base (Vb) is 158μA x 10K = 1.58V.
    The base-emitter voltage (Vbe) is around 0.65V, so the voltage at the emitter (Ve) is 1.58V - 0.65V = 0.93V.
    The current through the emitter resistor (Re) is the same as the emitter current (Ie) and almost the the same as the collector current (Ic). It is 0.93V / 680R = 1.37mA.
    The quiescent Ic is 1.34mA, so the actual Ib = 1.34mA / 110 = 12μA, confirming the earlier estimate and justifying ignoring it compared to the 158μA in the base bias resistors.
    The voltage across the 2K collector resistor (Rc) is 2K x 1.34mA = 2.68V, so the collector voltage (Vc) is 9V - 2.68V = 6.36V.
    The voltage across the BC547 (Vce) is Vc - Ve = 6.36V - 2.68V = 3.68V.
    All of that is very reasonable design.
    However, the output impedance is 2K, so you will potentially loose 99.6% of your signal when driving an 8 ohm load like the loudspeaker.
    The AC analysis shows serious design flaws, unfortunately.
    The intrinsic emitter resistance (re) is 26mV / Ic = 26mV / 1.34mA = 19R at the quiescent point. That will change as the collector current changes so is non-linear.
    The same current flows through the collector resistor (Rc) as does through the total emitter impedance. Since the change in voltage across the total emitter impedance is the same as the change in voltage at the base (i.e. the input signal), the ratio of the the change in voltage across the collector (i.e. the ac output) to the change in voltage at the input is simply the ratio between the collector resistor and the total emitter impedance. That is the voltage gain (Av) of the circuit.
    Av = Rc / (re + (Re || Xc) where Xc is the reactance of the bypass capacitor and (Re || Xc) is the impedance of Re in parallel with Xc. Xc = 1/2πCf where f is the frequency.
    When Xc is much larger than Re (low frequencies), then Av = Rc / (re + Re) and no bypass occurs.
    In this case, Xc is greater than Re when f < 1/2πC.Re = 1/2π.10μF.680R = 23Hz, and the gain below that is 2K / (680R + 19R) = 2.86 (i.e. the DC gain).
    When Xc is much smaller than re (high frequencies), then Av = Rc / re and the emitter resistor (Re) is bypassed to ac.
    In this case, Xc is less than re when f < 1/2πC.re = 1/2π.10μF.19R = 840Hz, and the gain above that is 2K / 19R = 105, but that is non-linear and will change as the collector current changes, producing distortion for large signals.
    Between 23Hz and 840Hz, the gain will increase from x2.86 to x105 giving a very "tinny" sound. The circuit needs either no bypass capacitor for a gain of 2.86, or a bypass capacitor of 470μF to give a (non-linear) gain of around 105 from 20Hz upwards (the usual audio range).
    The output capacitor is also far too small for use with an 8R speaker. You generally need 1000μF for audio use, although in this case it won't matter because the high output impedance attenuates the signal so much anyway.
    Finally, the input impedance is the parallel combination of 47K, 10K and the impedance of the transistor looking into the base, which is β times the emitter impedance. At high frequencies, that will be 47K || 10K || (110.19R) at the worst case, which equals around 1.7K, but the dominant factor is the transistor impedance, so the circuit input impedance will depend strongly on the transistor β. That's not a desirable design.

    • @attilarivera
      @attilarivera 2 ปีที่แล้ว +2

      👏👏👏👏👏👏

    • @flapjack9495
      @flapjack9495 2 ปีที่แล้ว +7

      This guy engineers.

    • @SheikhN-bible-syndrome
      @SheikhN-bible-syndrome ปีที่แล้ว +8

      How about you make a decent video

    • @SuperFredAZ
      @SuperFredAZ ปีที่แล้ว +2

      This design is ridiculous!

    • @Craftinges
      @Craftinges 4 หลายเดือนก่อน

      So I need to increase collector current? to decrease re below 8 ohm to have a positif gain? cuz when you connect load its always parallel with the RC so load impedence is always around 8ohm. Rc/re =8 /19 always smaller than 1. if there was a 50 mA collector current then 26mV/50mA re would be 0.5 ohm? then 8/0.5= gain is 16. Would this work?

  • @arunkurma25
    @arunkurma25 2 ปีที่แล้ว +2

    What type of mic can be used insted of that circuit

  • @alicanturk7713
    @alicanturk7713 ปีที่แล้ว +3

    It would be better if you explain why you use components in that ways and why do select that specific values? After all it is better to understand what is happening in the circuit. Otherwise it doesn't make any sense to someone like me who is just a beginner. Thanks

    • @soorajsuresh636
      @soorajsuresh636 16 วันที่ผ่านมา

      That's is exactly true

  • @alexanderzohrens2945
    @alexanderzohrens2945 9 หลายเดือนก่อน +2

    The 2 kiloohm resistor IS to high
    IT will Limit the Maximum current to 4.5 mA.
    The Transistor can Drive more than 100 mA. So the Power of the amplifier IS Limited unnecessarily.

  • @JohnnieCool
    @JohnnieCool 8 หลายเดือนก่อน

    I would like to see the signal path in red , , so it goes to the base, but also through the transistor?

  • @edengetachew1056
    @edengetachew1056 2 ปีที่แล้ว

    Where or how do you get the values of resistors , capacitors and Vcc ?? And how much source voltage should we use ?

    • @HenriksHobbyElectronics
      @HenriksHobbyElectronics  2 ปีที่แล้ว +3

      Thank you for your question! The 10 uF capacitance is a typical value which blocks DC while allowing audio signals to pass through. (If you want to verify this, you can calculate the cutoff frequency for the capacitor.) You can calculate the resistance values using the method described in the videos linked in the video description. I used 9 V source voltage. Did this answer your questions?

  • @SilentWeeb
    @SilentWeeb 2 ปีที่แล้ว

    Changing the capacitors to at least 1000uF will help with frequencies below 1kHz.

  • @pachi06
    @pachi06 2 ปีที่แล้ว +1

    Tr base bias voltage is too low.

  • @DwiNoviantoUntidar
    @DwiNoviantoUntidar 2 ปีที่แล้ว

    Can u draw the load line? (DC & AC)

  • @sam-kx3ty
    @sam-kx3ty 2 ปีที่แล้ว

    is the capacitor a must because I dont have one?.

    • @Avidiy
      @Avidiy 2 ปีที่แล้ว

      If it is the capacitor speaker 🔊 you can remove it but there will be distortions but if it is the emitter to ground then it is a big nono

  • @elektron2kim666
    @elektron2kim666 ปีที่แล้ว

    The BC547 is very reliable.

    • @kampa3531
      @kampa3531 7 หลายเดือนก่อน

      What do you mean ?

  • @priyanshugoswami241
    @priyanshugoswami241 2 ปีที่แล้ว

    Can you make a single transistor bc547 oscillator

    • @HenriksHobbyElectronics
      @HenriksHobbyElectronics  2 ปีที่แล้ว

      Yes, I already made a video about that: th-cam.com/video/hbH2OAdpdOk/w-d-xo.html

  • @awaismushtaq5719
    @awaismushtaq5719 6 วันที่ผ่านมา

    Bad design my dear. Single transistor cannot power speech coil

  • @johanwahyudi8508
    @johanwahyudi8508 7 หลายเดือนก่อน

    😂

  • @uni-byte
    @uni-byte 11 หลายเดือนก่อน +8

    The biasing is all wrong. The quiescent DC voltage at the collector should be 4.5V for best voltage swing. To get that you would need 2.25mA through the collector resistor. This would imply 1.53V at the emitter and 2.23V on the base. But you have only .18V at the base, so the transistor is cut off. ??? This thing will make a better guitar "fuzz" than an audio amplifier. The distortion would be off the scale, so no wonder it sounds like a (distorted) square wave.

    • @fursino
      @fursino 11 หลายเดือนก่อน +1

      Indeed i saw it too...

    • @marckman1411
      @marckman1411 6 หลายเดือนก่อน

      He has more than .18V at the base

    • @marckman1411
      @marckman1411 6 หลายเดือนก่อน

      Forgot a zero

    • @marckman1411
      @marckman1411 6 หลายเดือนก่อน

      You righy

  • @Madmaxkkb
    @Madmaxkkb ปีที่แล้ว +2

    I have checked your design on oscilloscope and thats sad that its not a desirable design for audio amplification u need to adjust the values of R and C properly

  • @Enderbro3300
    @Enderbro3300 ปีที่แล้ว +2

    Thank you! Everyone says "oh you can use transistors as amplifier" but I never got it to work and now I understand why XD

  • @jw228w
    @jw228w 7 หลายเดือนก่อน +1

    enjoyed your video on the one transistor amplifier,so much that I see you do but not sure why you pick the components and where they go in the circuit.can you point to a basic video that will help explain the reasons for the components you use? thanks John

  • @mtryambon
    @mtryambon 3 หลายเดือนก่อน

    How is it amplified? You have 9v so why not just connect to the 9 v. You show how the circuit is wired it doesn’t explain what is happening

  • @potato_sirius
    @potato_sirius 10 หลายเดือนก่อน +1

    Thank you ,this circuit is working well than i expected !!

  • @larryshaver3568
    @larryshaver3568 5 หลายเดือนก่อน

    i have alot of those parts no0w

  • @gino3286
    @gino3286 6 หลายเดือนก่อน

    hi thank you sincerely for this excellent video
    I'm trying to find a minimalist circuit to build a line preamp that should have low voltage gain (2 or 3 times) low distortion and not too high output impedance
    Unfortunately a single transistor circuit may have, from what i understand, too much distortion and high output impedance
    Could you suggest a two-transistor scheme that might work?
    I'm very interested in minimalist circuits because many say they could sound very musical
    I would really like to have your opinion
    Kind regards Gino

  • @djdayvi961
    @djdayvi961 23 วันที่ผ่านมา

    Thank you very much for sharing your knowledge and especially make an amplifier with the BC547 using a capacitor on the speaker. That was what I was looking for

  • @ltlt6117
    @ltlt6117 ปีที่แล้ว

    Hi i made a common emmiter amplifier for 200mhz signal with 2× gain
    I can amplify a 50 mv p-p to 100mv p-p but I can't do it for 2v to 4v p-p
    Why what values should be change???

  • @Damian_Is_Epic
    @Damian_Is_Epic 7 หลายเดือนก่อน

    Thank you! I was struggling to understand why the components are laid out the way they are. I've watched a bunch of other videos but couldn't grasp the concept.

  • @forbeginnersandbeyond6089
    @forbeginnersandbeyond6089 10 หลายเดือนก่อน

    What is the output signal voltage of your oscillator?

  • @sosscs
    @sosscs ปีที่แล้ว

    what kind of wave did you generate in the other sound circuit?

  • @batata_bajji
    @batata_bajji 10 หลายเดือนก่อน

    I wish you made more electronics videos!

  • @logangraham2956
    @logangraham2956 ปีที่แล้ว

    say you want stereo audio though...
    would you just mirror this circuit and have them both share the negative/ground rail?

    • @kennmossman8701
      @kennmossman8701 6 หลายเดือนก่อน +1

      common ground only.......just build it twice

  • @Mangputt
    @Mangputt 2 ปีที่แล้ว

    simple but awesome kinds.
    nice video,
    and i wait for next Vid.

  • @Nrhrdi
    @Nrhrdi ปีที่แล้ว

    How to gain the circuit transistor?

  • @unebonnevie
    @unebonnevie 9 หลายเดือนก่อน

    Will distorting of the sound output happens if a 5V supply is used instead of 9V?

    • @kennmossman8701
      @kennmossman8701 6 หลายเดือนก่อน

      the dynamic range would b much reduced

  • @alexisrodme
    @alexisrodme ปีที่แล้ว

    👎

  • @mdzisanislam4991
    @mdzisanislam4991 ปีที่แล้ว

  • @mufeedco
    @mufeedco 2 ปีที่แล้ว

    Thank you

  • @salimkumar9748
    @salimkumar9748 2 ปีที่แล้ว

    Thanks

  • @drekkerscythe4723
    @drekkerscythe4723 ปีที่แล้ว

    can we replace BC547 with 2n222a?

    • @HenriksHobbyElectronics
      @HenriksHobbyElectronics  ปีที่แล้ว

      Both are NPN transistors, so probably yes! Do an experiment, and let me know how it works! If it doesn't work immediately, then check the data sheet for clues on how to optimize the circuit. Good luck! 👍

  • @RahulSharma-oi4es
    @RahulSharma-oi4es 2 ปีที่แล้ว

    Can you aldo explain the source circuit ehich you used ?

  • @Danielhernandez-vu9fg
    @Danielhernandez-vu9fg 2 ปีที่แล้ว

    Hi Henrik, hoy could you adjust values for a 24V circuit?

    • @kennmossman8701
      @kennmossman8701 6 หลายเดือนก่อน

      you should try iirst then ask

  • @OmarTravelAdventures
    @OmarTravelAdventures 2 ปีที่แล้ว

    I am totally impressed by your explanation. Thank you Henrik. I have wanted such explanation since I was 13 years old, now I found it. I am a novice to analyzing circuits like this, so I am so happy as a hobbiest to listen to you. I am looking forward to your next video! May I ask you to answer a few questions, either here in writing or in another video (if you want to make another video :) 1. I am confused why in the circuit diagrams that uses DC battery, instead of showing -ve and +ve leads, you draw an earth, can you explain that a bit more. Why isn’t the negative and positive leads shown to make it easier to understand the flow of current? 2. Probably related to the previous question, at 1:19 you say you want to limit the DC current but not limit the AC current. Is the DC current the one coming from the battery and the AC current coming from the oscillator? 3. At 3:14, when you say you apply an input signal, what does that mean? I have a difficult time conceptualizing what an input signal means in terms of electronic circuit. Is the oscillator circuit a closed circuit that Takes an DC current and changes it to an AC that oscillates? I think I might be getting it or, on the other hand, I am missing a fundamental concept that I to figure out. Did I get it right Henrik? Greetings from San Diego, California.

    • @HenriksHobbyElectronics
      @HenriksHobbyElectronics  2 ปีที่แล้ว +1

      Hey Omar! Thank you for your questions. (1) In this circuit, the "ground" is not actually grounded. We call this "ground" only in the sense that we use the battery's negative lead as the 0 V reference voltage. Basically, in the electrical engineering "lingo", we often call the 0 V reference ground. (2) Yes! (3) To start with, an amplifier is a device which takes an input signal and produces an output signal, which has been amplified by taking energy from the battery. An oscillator on the other hand produces an output signal without needing an input signal (except for the battery). For more information about that, check out my other video on oscillators: th-cam.com/video/hbH2OAdpdOk/w-d-xo.html Did this help? /Henrik

    • @OmarTravelAdventures
      @OmarTravelAdventures 2 ปีที่แล้ว +1

      @@HenriksHobbyElectronics Thank you so much Henrik! This is extremely useful. Looking forward to more videos.