RC Phase Shift Oscillator Circuit Using a NPN Transistor
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- เผยแพร่เมื่อ 19 ส.ค. 2024
- This electronics video tutorial explains how to design the RC Phase Shift Oscillator Circuit using a single npn transistor. This circuit generates a sine wave at the ouput. The frequency of the oscillations depends on the resistor-capacitor network that is coupled to the base of the transistor.
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In these crazy ass times it’s nice to know that The Organic Chemistry Tutor is still trying to teach
I’m really liking this electronics playlist, but some of these videos leave me wanting to know more. Or more specifically, *why*. I looked up the theory behind this circuit online and I enjoyed the discussion about how the output is successively shifted in each stage to arrive back at the input in-phase. I feel like mentioning that detail might have been helpful to people like me who may have been wondering why certain values were chosen for the components or why things work the way they do. I’m all for actually getting out a breadboard and building something, but I like to know why I’m doing things the way I am.
Either way, just a suggestion, but thanks again so much for all of these videos!
Have u found the answers?
is there anything that Organic Chemistry Tutor does not know?🤣👍🏻
Computer Science: he has videos on boolean logic, but I really hope he posts videos on Java/C++ and Data Structures.
@@joeyGalileoHotto i believe his major is either Physics or Organic Chemist
@@ohmslaw6856 a few sources says he has a chemistry degree from UCLA
Ignorance
❤
You're the best!
Big mistake: The leftmost R doesn't count as part of the phase shift line, 'cause it causes shifts just in connection with the capacitor CB, whose value is large in comparison with the three C's. Instead, the base voltage divider is part of the phase shift line.
Thankyou for your fantastic video. I was able to make your circuit work with power coming from the 5v power pin of an Arduino Uno
I'm sorry, I couldn't focus on the guy with all these beautiful electronics...
you could use and old air coil from a tube radio to tune the capacitors.
I see a mistake in the calculation of the base bias. You must consider the collector voltage not the supply voltage in the potential divider formula. But practically, because the collector resistor is well less than series connected RB resistor, the omission of the collector resistor causes no problem in this particular case.
Nice job! Thank you.
I would like this to use as a practice CW oscillator with a 9V battery and put a little speaker in the line to the collector.
I think I'll still need a kind of a resistor in series with it ....
is there a way to calculate the output amplitude
sorry for the question that can be use on audio amplifiers too?
I want to why you use voltage divider here? Please explain this...
The voltage in point B is determined by the transistor's base to emitter diode which is normally around 0.7 volts. we cannot use the voltage division of RA and RB. the base-emitter diode creates a reference voltage at point B
OCT, please do a couple videos on biology
You are just keep on draw and draw the same circuits i saw in google without explaining how EXACTLY it works because i can also draw those circuits but i am not able to exactly know how it works the same as you do
I have a query...in the circuit you drew on the screen you drew no emitter resistor. Is cool. Then you calculated a voltage at the base using a voltage divider with R(A) and R(B). Is also cool, but then you told us the voltage at the base would concur with the voltage you calculated for the voltage divider. Isn't the voltage drop across the Base-Emitter just the 0.6 volts of the PN junction? You told us it should be 7.42 Volts. I've never seen a 7.42 V semiconductor PN junction forward bias voltage but there's always a first time I suppose. I'm just confused.
How did you cause the voltage at the base to rise to 7.42 volts without an emitter resistor again? I've no doubt you caused your oscillator to work with the values you used, it's just...well I'm confused. Seems like 7.42 volts across a PN junction would require a reverse bias. Wouldn't that turn the transistor off? I don't get it.
the 7.42 volts calculated is a voltage divider being fed into the base of the transistor. He is saying we need to make sure the voltage divider is high enough to turn on the transistor because if the BE .7v drop. So he was just clarifying that the voltage divider he used is in fact large enough to turn on the transistor.
@@marisavittorio7347 Uhm...ok. Is the emitter of the transistor on ground? What is the voltage across a forward biased PN junction? Just how does the voltage at the base rise above ground potential +0.6 or 0.7 volts? It's a lovely voltage divider...that isn't doing what we think.
in what region is the transistor operating?
Why are you focusing so much on the amplifier when the 3 piece RC oscillator is the highlight of the circuit?
The schematic is wrong. The author does not understand how it works. From collector towards base, C must go first, then R, not vice versa. Only this way the first RC couple will create any phase shift. Of course, then you do not need Cb. Ra constitutes a negative feedback from collector to base, so it can prevent oscillation if Ra/Rc is too small (100:1 should be OK). Ra parallel to Rb acts as R, so if you chose 100K||470K, R would be around 82K to roughly meet the formula (dynamic base resistance also plays a role in this). As for the bias: for BC549 and 2 mA collector current Vbe would be between 0.5 and 0.6 V. Ra and Rb acts as voltage divider, so for Ra:Rb = 1:4.7 as in this example the collector voltage would be almost equal to Vbe: Vce = Vbe * ( Ra+Rb / Rb). Therefore Vce would be 0.66 volt, pretty close to saturation. Author's claim the Vbe would be 7.42 volts is a complete bullshit - first, the source voltage for the Ra/Rb is not 9 volts, but Vce, second, if you put 7.4 volts hard to base-emitter junction, you would blow it because the base current would exceed the maximum allowed current by several orders of magnitude. If you want a decent working point, Ra should be at least 3 times the Rb (then you would get Vce = 1.5) but it can be as high as 7 times Rb - then you get Vce around 3-5 volts, about half the supply voltage. So in fact, swap Ra and Rb, make Ra = 470K and Rb = 100K, then you should get nicer sine wave without distortion caused by transistor saturation.
Basically, you could replicate this circuit and it may even oscillate, but the author has no clue what he's talking about.
kindly,explain me in more depth please
@@mpkdon let me think please, it was a year ago :-) I have to recall what i thought then
@@mpkdon OK, I got it. In the feedback circuit: the big Cb capacitor (100 uF) and the resistor R connected between the Cb and ground, has very little effect and contributes almost zero phase shift. So, if you remove both of them and connect the first C directly to collector, you will get a circuit with the same behavior, but save two useless parts. This is because the phase shift is created by differences between voltage and current across a resistor vs capacitor. Voltage phase at the first R does not depend on anything (because Cb is big) and the first phase shift occurs only after the first C.The phase-shift network hence constitutes of C-R, C-R and C-Ra||Rb.
As for the DC bias: Vbe of the transistor is approximately 0.6 volts. Rc, Rb and Ra form a feedback network - the greater collector current Ic, the lower will be collector voltage Vce. Therefore the collector voltage will settle at Vbe * (Ra + Rb) / Rb. In this case, Rb / (Ra + Rb) is approximately 5/6, therefore Vce will be approximately Vbe * 6 / 5, which is about 0.72 volts. If you imagine that the circuit produces sinewave oscillations, it is clear that the output swing must be limited but the lowest possible Vce, which is the transistor's saturation voltage. For BC549 type of transistor this saturation voltage could be about 0.2 volt. Therefore, negative halfwave output amplitude (half amplitude) can be maximum ~0.5 V, but most likely it will be distorted. Positive halfwave would be approximately the same as the negative.
But how does this circuit produce a 180 degree phase shift?
In the 3 RC circuits in a chain, each stage makes an 60 degree phase shift. 3*60=180.
This circuit as shown is really not very good but it will work. The NPN transistor has to provide a gain greater than 29 for it to oscillate. His circuit does a rather poor job of setting the gain.
Is this oscillator could reach mhz?
Anything built on a breadboard is limited to
I can't get it to oscillate. Varied the voltage on the base from 0-1.5V and I know the transistor was on because I was measuring Ic. Ic would vary from 0-15mA with Rc=1K.
Even measured Ib to make sure. Just don't get it. I built it on a breadboard so maybe parasitic capacitances. I don't know.
I found this th-cam.com/video/PN8iwDNj658/w-d-xo.html video that talks about the problems with this kind of oscillator.
Hi...c1+c2+c3 total capacitance are incorrect
why are you scrolling to the comments ???
Will you say no to this comment???
(look in the replies)
You looking the replies thank you!
If you want to know how this sounds if you connect it to a small loudspeaker, then check it out here: th-cam.com/video/hbH2OAdpdOk/w-d-xo.html :)
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Unfortunately, your practical observations do not agree with the theory. As you decrease the resistance almost to a half, the frequency does not double. And when you increase the resistance almost 5 times, the frequency decrease merely by a factor of 2.5.