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.
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?
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.
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?
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
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.
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
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
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.
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.
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
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???
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.
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: ua-cam.com/video/hbH2OAdpdOk/v-deo.html Did this help? /Henrik
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
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?
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! 👍
Do you have any questions about this circuit? Ask below and I'll answer as many questions as I can :)
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.
Thank need oscilater circut and radio tuner curcuit your explainion easy to understand
How come there was no capacitor on the input between the base and emitter ?
I love your channel by the way.
@@adastra123 Thanks! I am not aware that adding a capacitor between the base and the emitter would improve the circuit in any way.
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?
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.
👏👏👏👏👏👏
This guy engineers.
How about you make a decent video
This design is ridiculous!
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?
Its ridiculous hard much ive had to search for a norkal breadboard tutorial lol. Atleast now im here so i appreciate you
Thank you! Everyone says "oh you can use transistors as amplifier" but I never got it to work and now I understand why XD
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
Interesting and informative. Hope you will do more about basic circuits. Thanks
Thank you ,this circuit is working well than i expected !!
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.
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
simple but awesome kinds.
nice video,
and i wait for next Vid.
I wish you made more electronics videos!
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
That's is exactly true
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.
Indeed i saw it too...
He has more than .18V at the base
Forgot a zero
You righy
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.
What type of mic can be used insted of that circuit
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
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???
What is the output signal voltage of your oscillator?
I would like to see the signal path in red , , so it goes to the base, but also through the transistor?
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.
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: ua-cam.com/video/hbH2OAdpdOk/v-deo.html Did this help? /Henrik
@@HenriksHobbyElectronics Thank you so much Henrik! This is extremely useful. Looking forward to more videos.
what kind of wave did you generate in the other sound circuit?
Thank you
say you want stereo audio though...
would you just mirror this circuit and have them both share the negative/ground rail?
common ground only.......just build it twice
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
What happens,
When you connect a microphone directly to the amplifier (omitting the oscillator circuit.)
Will distorting of the sound output happens if a 5V supply is used instead of 9V?
the dynamic range would b much reduced
Hi Henrik, hoy could you adjust values for a 24V circuit?
you should try iirst then ask
Can you aldo explain the source circuit ehich you used ?
Check it out here: ua-cam.com/video/hbH2OAdpdOk/v-deo.html
Where or how do you get the values of resistors , capacitors and Vcc ?? And how much source voltage should we use ?
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?
How to gain the circuit transistor?
Thanks
Can you make a single transistor bc547 oscillator
Yes, I already made a video about that: ua-cam.com/video/hbH2OAdpdOk/v-deo.html
The BC547 is very reliable.
What do you mean ?
Changing the capacitors to at least 1000uF will help with frequencies below 1kHz.
Can u draw the load line? (DC & AC)
can we replace BC547 with 2n222a?
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! 👍
is the capacitor a must because I dont have one?.
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
Tr base bias voltage is too low.
❤
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
i have alot of those parts no0w
Bad design my dear. Single transistor cannot power speech coil
😂
👎