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Need help understanding how an R-C oscillator works

by gyre
Tags: oscillator
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gyre
#1
Apr7-12, 08:08 AM
P: 3
Hoping someone can help me with some basic questions about resistor-capacitor oscillator (timer) circuits. I can put them together, I see them working, but I don't understand why!

In the circuit in the attached drawing, if one installs the correct values of resistor and capacitor, I know that the transistor should repeatedly switch on and off as the capacitor repeatedly charges up then discharges through the base-emitter junction. But what I don't understand are the following:

- At time=0, I understand that voltage and current flow right through the un-charged capacitor. But I do not understand why voltage/current do not instantly also flow right through the base-emitter junction of the transistor, switching it on -- after all, the base-emitter junction is basically just wired in parallel with the capacitor, no? Does the base-emitter junction have a very high resistance or something?

- Why does a resistor-capacitor circuit oscillate at all, as opposed to reaching an equilibrium in which the battery and the capacitor both discharge to ground simultaneously via the base-emitter junction of the transistor, with the battery constantly keeping the capacitor at some charge level?

- Why does the capacitor take a while to charge up to the voltage at which it "opens" the base-emitter junction then, when it reaches that voltage, completely empty itself through this junction as opposed to just bringing itself back down to just below the "transistor switch-on" voltage? Does the base-emitter junction take a while to turn itself off or something?
Attached Thumbnails
2012 04 07 R-C circuit.jpg  
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Jony130
#2
Apr7-12, 08:24 AM
P: 410
The circuit you post in the diagram is not a oscillator it is delay-on time circuit.
Also to open BJT you need 0.6V across base-emitter junction.
So capacitor need time to charge up to the 0.6V. After this happen, the BJT start to conduct (transistor switch-on ) and capacitor stop charging and we reach an equilibrium.
And this circuit is also potentially dangerous, because the collector current is not limited by any collector resistor. The only limitation is base resistor and hfe. And this can cause the BJT to blow.
gyre
#3
Apr7-12, 12:18 PM
P: 3
Jony, thanks for the quick reply. It was reassuring: I could see this working as a time delay, but I thought there was also some way to make it oscillate that I just didn't get. Glad to hear that is not the case!

Still, I don't understand why at t=0 there would not be a voltage difference across the base of the transistor. At t=0 the battery pumps out 9v (for example); at t=0 the capacitor sees this 9v difference; so why does transistor's base, which is in parallel with the capacitor, also not see this 9v difference?

So if I add a second BJT triggered by the first and a loudspeaker, as in the new diagram attached, does that change it from a time-delay circuit into an oscillator circuit? If so, I still have the same question: why would the current oscillate instead of just settling into an equilibrium?

(also threw in a resistor to protect both BJTs... was just keeping it simple at first).

Thanks very much for the help.
Attached Thumbnails
2012 04 07 R-C circuit #2.jpg  

Jony130
#4
Apr7-12, 04:35 PM
P: 410
Need help understanding how an R-C oscillator works

Quote Quote by gyre View Post
Still, I don't understand why at t=0 there would not be a voltage difference across the base of the transistor. At t=0 the battery pumps out 9v (for example); at t=0 the capacitor sees this 9v difference; so why does transistor's base, which is in parallel with the capacitor, also not see this 9v difference?
Your main problem is that you don't understand how capacitor work.
At t=0 the capacitor is empty, 0V across capacitor terminals.
But also we have a resistor between a 9V voltage source and capacitor. And this resistor limit the charging current.
At t=0 empty capacitor act like short-circuit (0V). And this means that the BJT base is also short to the gnd.
The voltage on capacitor cannot change suddenly form 0V to 0.6V. We need time to voltage on capacitor to grow. This is because the resistor limits the the charging current.
And after the time

[tex]T = R *C* ln(\frac{Vcc}{Vcc-Vbe})[/tex]

The transistor will be switch-on.


Quote Quote by gyre View Post
So if I add a second BJT triggered by the first and a loudspeaker, as in the new diagram attached, does that change it from a time-delay circuit into an oscillator circuit? If so, I still have the same question: why would the current oscillate instead of just settling into an equilibrium?
Your new diagram indeed show a oscillator circuit.
And on this page you have explained why this circuit is oscillated.
http://www.talkingelectronics.com/pr...ml#OSCILLATORS
gyre
#5
Apr8-12, 06:25 AM
P: 3
Thanks for the info, think I understand it now. That link is great.

As for the time-delay circuit, believe it or not it was the charge on the base I wasn't understanding rather than the capacitor. I was sort of thinking of voltage as water pressure and envisioning it "pushing" against the BJT's base while "filling up" the capacitor. There truly are limits to that analogy!

Much appreciated. I'll probably be writing back with some other basic issues that stump me.


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