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

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Henrik's Hobby Electronics

Henrik's Hobby Electronics

2 жыл бұрын

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

Пікірлер: 73
@HenriksHobbyElectronics
@HenriksHobbyElectronics 2 жыл бұрын
Do you have any questions about this circuit? Ask below and I'll answer as many questions as I can :)
@HenriksHobbyElectronics
@HenriksHobbyElectronics 2 жыл бұрын
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 жыл бұрын
Thank need oscilater circut and radio tuner curcuit your explainion easy to understand
@adastra123
@adastra123 2 жыл бұрын
How come there was no capacitor on the input between the base and emitter ? I love your channel by the way.
@HenriksHobbyElectronics
@HenriksHobbyElectronics 2 жыл бұрын
@@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 жыл бұрын
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?
@RexxSchneider
@RexxSchneider 2 жыл бұрын
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 жыл бұрын
👏👏👏👏👏👏
@flapjack9495
@flapjack9495 Жыл бұрын
This guy engineers.
@SheikhN-bible-syndrome
@SheikhN-bible-syndrome Жыл бұрын
How about you make a decent video
@SuperFredAZ
@SuperFredAZ 10 ай бұрын
This design is ridiculous!
@Craftinges
@Craftinges 2 ай бұрын
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?
@potatogamer6239
@potatogamer6239 8 ай бұрын
Thank you ,this circuit is working well than i expected !!
@hortchester1255
@hortchester1255 8 ай бұрын
Its ridiculous hard much ive had to search for a norkal breadboard tutorial lol. Atleast now im here so i appreciate you
@Enderbro3300
@Enderbro3300 Жыл бұрын
Thank you! Everyone says "oh you can use transistors as amplifier" but I never got it to work and now I understand why XD
@Mangputt
@Mangputt 2 жыл бұрын
simple but awesome kinds. nice video, and i wait for next Vid.
@spacecraftbuildingservices
@spacecraftbuildingservices 6 ай бұрын
Interesting and informative. Hope you will do more about basic circuits. Thanks
@mufeedco
@mufeedco 2 жыл бұрын
Thank you
@salimkumar9748
@salimkumar9748 2 жыл бұрын
Thanks
@arunkurma25
@arunkurma25 2 жыл бұрын
What type of mic can be used insted of that circuit
@sosscs
@sosscs Жыл бұрын
what kind of wave did you generate in the other sound circuit?
@batata_bajji
@batata_bajji 8 ай бұрын
I wish you made more electronics videos!
@Damian_Is_Epic
@Damian_Is_Epic 5 ай бұрын
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 8 ай бұрын
What is the output signal voltage of your oscillator?
@ltlt6117
@ltlt6117 Жыл бұрын
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???
@Nrhrdi
@Nrhrdi Жыл бұрын
How to gain the circuit transistor?
@jw228w
@jw228w 5 ай бұрын
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
@JohnnieCool
@JohnnieCool 5 ай бұрын
I would like to see the signal path in red , , so it goes to the base, but also through the transistor?
@gino3286
@gino3286 4 ай бұрын
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
@edengetachew1056
@edengetachew1056 2 жыл бұрын
Where or how do you get the values of resistors , capacitors and Vcc ?? And how much source voltage should we use ?
@HenriksHobbyElectronics
@HenriksHobbyElectronics 2 жыл бұрын
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 жыл бұрын
Changing the capacitors to at least 1000uF will help with frequencies below 1kHz.
@uni-byte
@uni-byte 9 ай бұрын
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 9 ай бұрын
Indeed i saw it too...
@marckman1411
@marckman1411 4 ай бұрын
He has more than .18V at the base
@marckman1411
@marckman1411 4 ай бұрын
Forgot a zero
@marckman1411
@marckman1411 4 ай бұрын
You righy
@Danielhernandez-vu9fg
@Danielhernandez-vu9fg Жыл бұрын
Hi Henrik, hoy could you adjust values for a 24V circuit?
@kennmossman8701
@kennmossman8701 4 ай бұрын
you should try iirst then ask
@logangraham2956
@logangraham2956 Жыл бұрын
say you want stereo audio though... would you just mirror this circuit and have them both share the negative/ground rail?
@kennmossman8701
@kennmossman8701 4 ай бұрын
common ground only.......just build it twice
@unebonnevie
@unebonnevie 7 ай бұрын
Will distorting of the sound output happens if a 5V supply is used instead of 9V?
@kennmossman8701
@kennmossman8701 4 ай бұрын
the dynamic range would b much reduced
@DwiNoviantoUntidar
@DwiNoviantoUntidar 2 жыл бұрын
Can u draw the load line? (DC & AC)
@pachi06
@pachi06 2 жыл бұрын
Tr base bias voltage is too low.
@RahulSharma-oi4es
@RahulSharma-oi4es 2 жыл бұрын
Can you aldo explain the source circuit ehich you used ?
@HenriksHobbyElectronics
@HenriksHobbyElectronics 2 жыл бұрын
Check it out here: kzbin.info/www/bejne/npOrY4J3mdWXhc0
@priyanshugoswami241
@priyanshugoswami241 2 жыл бұрын
Can you make a single transistor bc547 oscillator
@HenriksHobbyElectronics
@HenriksHobbyElectronics 2 жыл бұрын
Yes, I already made a video about that: kzbin.info/www/bejne/npOrY4J3mdWXhc0
@drekkerscythe4723
@drekkerscythe4723 Жыл бұрын
can we replace BC547 with 2n222a?
@HenriksHobbyElectronics
@HenriksHobbyElectronics Жыл бұрын
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! 👍
@sam-kx3ty
@sam-kx3ty 2 жыл бұрын
is the capacitor a must because I dont have one?.
@Avidiy
@Avidiy 2 жыл бұрын
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
@alicanturk7713
@alicanturk7713 Жыл бұрын
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
@OmarTravelAdventures
@OmarTravelAdventures 2 жыл бұрын
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 жыл бұрын
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: kzbin.info/www/bejne/npOrY4J3mdWXhc0 Did this help? /Henrik
@OmarTravelAdventures
@OmarTravelAdventures 2 жыл бұрын
@@HenriksHobbyElectronics Thank you so much Henrik! This is extremely useful. Looking forward to more videos.
@mdzisanislam4991
@mdzisanislam4991 10 ай бұрын
@elektron2kim666
@elektron2kim666 Жыл бұрын
The BC547 is very reliable.
@kampa3531
@kampa3531 5 ай бұрын
What do you mean ?
@Madmaxkkb
@Madmaxkkb Жыл бұрын
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
@larryshaver3568
@larryshaver3568 3 ай бұрын
i have alot of those parts no0w
@alexanderzohrens2945
@alexanderzohrens2945 7 ай бұрын
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.
@mtryambon
@mtryambon Ай бұрын
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
@johanwahyudi8508
@johanwahyudi8508 5 ай бұрын
😂
@alexisrodme
@alexisrodme Жыл бұрын
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