# Colpitts Crystal Oscillator Design

A qualitative and quantitative analysis of the Colpitts crystal oscillator circuit, oscillation requirements, and practical circuit design considerations.

Description Reference
Quantitative analysis of Colpitts crystal oscillators Crystal Oscillator Design and Temperature Compensation, Marvin E. Frerking, Chapter 7.3, Appendix F
Oscillator phase vs frequency, and crystal loaded Q analysis Crystal Oscillator Circuits, Robert J. Matthys, Chapters 6, 7
Colpitts crystal oscillator phase vs gain analysis Oscillator Design and Computer Simulation, 2nd Edition, Randall W. Rhea, Chapter 11.2
Crystal drive level equations Intel application note AP-155 (Oscillators for Microcontrollers), Appendix A
Common collector gain equations Common collector, Wikipedia
Transistor biasing RF Circuit Design, Chris Bowick, Chapter 6

## 3 thoughts on “Colpitts Crystal Oscillator Design”

1. Sree says:

I Have few doubts in designing a colpitts crystal oscillator Can i get your personal mail id to contact you Please

2. Dan says:

Hi Craig,
I have a question about your “Colpitts Crystal Oscillator Fundamentals” video on YouTube. At 4:35, you give example values for the feedback network. If e1 is considered a voltage source, then its impedance will ideally be 0 ohms. Therefore, C2’s capacitive reactance (-jX2) will not have any effect on the rest of the tank circuit since it is in parallel with 0 ohms. Resonance will then be solely determined by C1’s capacitive reactance (-jX1) and the crystal’s inductive reactance (+jXe), and therefore X1 should equal Xe.
In the actual circuit, e1 is the emitter, which will have a very low source impedance in parallel with C2, reducing C2’s effect on the rest of the tank considerably. Again, resonance would be determined primarily by C1 and Le.
Can you please resolve this discrepancy?
Thanks!

• Craig says:

Hi Dan, if I understand you correctly I believe your confusion is the result of perspective. The e1 voltage source would ideally be a 0-ohm impedance from the perspective of the load that it is driving (in this case, the feedback network of C1/C2/X1). You are including the impedance of the e1 voltage source when analyzing the circuit from the perspective of the e1 voltage source, which is invalid; otherwise the 0-ohm ideal voltage source would short-circuit the entire feedback network and no current would flow through any part of the feedback network at all!

In fact, applying this same logic to any other circuit would suggest that an ideal voltage source (or any power source that approached the 0-ohm output of an ideal voltage source) would never be able to provide power to anything; it would not be able to light an LED or light bulb, power an op amp or CPU, etc.

Consider the simple example of a 1k-ohm resistor connected across a 9v battery (let’s just assume that the 9v battery is an ideal voltage source with 0-ohm impedance). Ohm’s law tells us that the current through the 1k-ohm resistor will be I = 9/1000 = 9mA, which can be easily verified in the real world with a current meter. However, using the same logic you applied to the oscillator, the battery is a perfect 0-ohm short-circuit across the 1k-ohm resistor, so NO current will flow through the 1k-ohm resistor, and all the current will flow through the battery. This of course is impossible, as the 1k-ohm resistor provides the only path for electrons to flow out of one battery terminal and into the other. If there was an actual short-circuit internal to the battery, it would get really hot really fast just sitting on the shelf! The output resistance of a voltage source represents its inherent internal losses (nothing is a perfect conductor).

Modeling sources and loads as lumped impedances or as ideal components can be very useful, but don’t confuse what the load “sees” with what the source “sees”. If you and your friend are facing each other, your visual perspectives would necessarily be different (assuming that you aren’t identical twins :); it would not make sense to use your friends perspective to describe what you are seeing, and vice versa.

Thanks for the comment and hope this helps!