Explain to us what you already know about a colpitts design so this can be a useful discussion.View attachment 347760
Hello everyone!
I'm trying to understand how the Colpitts oscillator works.
Could someone explain the roles of the capacitors labeled C and C3 in the circuit? What are they used for?
Thanks in advance!
Well, I know that capacitor C is supposed to configure the transistor in a common-base setup. But I'm not completely sure about that, because many sources say that this capacitor blocks the AC signal, and that the circuit wouldn't work without it.Explain to us what you already know about a colpitts design so this can be a useful discussion.
What do you think C and C3 are for?
It provides a alternative path that might be equal to a short at AC but not DC. So, in a way it would block the AC by removing it from the other signal paths.Well, I know that capacitor C is supposed to configure the transistor in a common-base setup. But I'm not completely sure about that, because many sources say that this capacitor blocks the AC signal, and that the circuit wouldn't work without it.
As for capacitor C3 — I have no idea what it's used for.
Why is it necessary to remove the AC component? What problem does it cause?It provides a alternative path that might be equal to a short at AC but not DC. So, in a way it would block the AC by removing it from the other signal paths.
C3, ok tell us what R3 does and think about what effect a AC alternative path would have to what it does.
R1 provides bias for the base along with R3, a steady DC current. Blocks is a misnomer here, bypass ( alternative path) would be better.Why is it necessary to remove the AC component? What problem does it cause?
Resistor R3 sets the operation mode of transistor T1 to enable oscillation.
I think if capacitor C blocks the AC component, it will pass through resistor R3, because it has nowhere else to go. Is that correct?
Great, yours then. Was trying to push the OP into using what they know to move forward but I guess just the answer is ok today.To understand why and how a harmonc oscillator works, one must remember the oscillation condition as formulated by Barkhausen:
"An active element (amplifier) with positive feedback must have a lopp gain which is (slightly) larger than unity at one single frequency only."
In the example under discussion we have a common base amplification stage with positive gain (between emitter and collector).
This gain is developped across an L-C resonant circuit. The resistor R3 sets a suitable DC bias point.
In order to allow a loop gain larger than unity (but not too large with respect to heavy distortions due to overdrive effects) the feedbacl loop is closed using a capacitive signal voltage divider C2-C3.
The values of these capacitors must be carefully selected in order to fulfill the above mentioned requirements and restrictions (loop gain somewhat larger than unity).
PS (comment): I cannot see how a capacitor coudld "block" any ac component.
nsaspook, I'm still not completely clear on everything. I know that R1 and R3 set the bias point of the transistor. Now I understand that capacitor C prevents the signal from going to the transistor's base, thanks to you. Could you please explain how the feedback works with some guiding questions? Thanks!R1 provides bias for the base along with R3, a steady DC current. Blocks is a misnomer here, bypass ( alternative path) would be better.
This is a common base configuration. Common means a reference point like circuit ground. The base needs the DC bias but but it also needs the AC signal to be at or near our common reference. C does that.
Short on time here, need to leave. Ask @LvWnsaspook, I'm still not completely clear on everything. I know that R1 and R3 set the bias point of the transistor. Now I understand that capacitor C prevents the signal from going to the transistor's base, thanks to you. Could you please explain how the feedback works with some guiding questions? Thanks!
As I have tried to explain, for a "good" oscillator which@LvW, i understand that the capacitive divider forms the feedback and causes the transistor to turn on at the right time to maintain oscillations. But why is a divider specifically used here? Is it possible to use only C2?