I'm experimenting with the Colpitts oscillator, configured in the common base topology. Here is a picture:

enter image description here

In my own circuit, the coil is approximately 20 microhenry, and I've experimented with several values of capacitors ranging from around 10pF to 1nF.

I understand that the coil and the capacitors form a resonant circuit which has the output fed to an amplifier (here a common base amplifier). I also understand how common base amplifier works.

From oscilloscope measurements I've noticed that changing the resistor between the emitter of the transistor and ground (R1 in the picture) to a smaller value dramatically increases the amplitude of the oscillation. But I can't understand the reason for this!

Even though I understand resonant circuits and transistor amplifiers separately, I somehow fail to see exactly how the resistor affects the amplitude of the oscillation. In the common base configuration, the emitter is the input and the collector is the output. If we lower the emitter resistor resistance, more current would (according to my reasoning) somehow bypass the second capacitor.. And higher the voltage across the resistor, higher the input to the amplifier and therefore higher amplified output. But how does this all work? This could be a very simple thing but I've been trying to make sense of this all day but I still don't really understand why lowering the emitter resistor increases the output amplitude. I've also looked at the common base amplifier gain formulae but they all just show a collector resistor (which is not even present at this circuit).


4 Answers 4


Although you are correct that the emitter resistor will bypass some of the signal current the more important factor is that it affects the biasing of the transistor.

For an amplifier to become an oscillator one of the requirements is that there is feedback and that the overall loop gain is more than unity to account for the losses in various parts of the circuit.

Once oscillation is started the amplitude will keep increasing until something causes the gain to decrease so that it is exactly unity. In this type of amplifier it can be caused by saturation of the amplifier where it is giving as much output as it can - this is usually related to the current flowing through the active device.

At that point the amplitude will be determined by the effective loss resistance of the resonant circuit and the output current of the amplifier.

By reducing the emitter resistor you are making the loss resistance worse by bypassing part of the signal current, but you are also increasing the quiescent current in the transistor and increasing its output current capability. In your case the latter effect is larger than the former so the output amplitude is increased.

The transconductance of the transistor will also increase with current that will also change things and probably increase the gain.

You can calculate the DC conditions by ignoring the feedback and treating it as a simple amplifier.

With oscillators it may not be as simple as I have just described as there are also non-linear effects. The AC signal at the emitter may get rectified by the base-emitter junction and affect the voltage at the base and so change the DC bias conditions - in many cases this may be the dominant factor, it can also cause the oscillations to be modulated at a low frequency where the oscillations will repeatedly stop and start, referred to as squegging. This behavior is often exploited in super regenerative receivers.

  • \$\begingroup\$ What do you mean by output current capability? And how does increasing the quiescent current increase the gain? Does it have something to do with the transconductance of the transistor? Perhaps my understanding of the amplifier is not as good as I thought.. \$\endgroup\$
    – S. Rotos
    May 20, 2017 at 21:48
  • \$\begingroup\$ On a class A amplifier the AC output current cannot exceed the quiescent current. The transconductance of a BJT increases with current, although it is not necessary for the gain to increase for the output current capability. \$\endgroup\$ May 20, 2017 at 22:00
  • \$\begingroup\$ Ah, I think I understand now, thank you. So it's just about increasing the quiescent point to allow more current "swing". \$\endgroup\$
    – S. Rotos
    May 20, 2017 at 22:39

Old style sine wave oscillator is an amp with a feedback circuit that results infinite total amplification at the certain frequency(see NOTE 1). Colpitts oscillator was one such construction that could provide reasonable operation with low gain triode tubes. It's advantage over other common circuits was no need of multiple or tapped coils.

The circuit is inherited to transistor radios. But it's today not at all critical (=no need for careful calculations) due the high available gain in modern transistors.

But back to theory: If the amp were ideal, the oscillation amplitude would grow infinitely. In practice there exist 2 clipping mechanisms that limit the output amplitude:

  • there's not enough Vce, transistor saturates
  • Ic cannot be negative.

If you make R1 smaller, the operating point shifts to larger rest Ic value. It has more room to swing before clipping occurs. That means higher AC voltages are possible in reactive parts. That shows as higher output voltage amplitude if the clipping was caused by Icmin=0.

NOTE1: The LC ciruit really is a frequency dependent feedback circuit for a common base (=input in emitter) amplifier. Oscillation happens at the frequency, where

  • the amplifier amplifies more than the feedback circuit attenuates
  • the total phase shift, when the signal travels through the amplifier and returns to amp input, is 0 degrees or a multiple of 360 degrees.

If the amp happens to be inverting, then the feedback circuit need to invert the phase again. It's not the case here.

LC resonant circuits have a steep curve for phase shift vs frequency near the resonant frequency. That's why the oscillating frequency generally is near the resonant frequency.

The infinite gain: Imagine a noise circulating in the loop. It is amplified in every turn. At the oscillation frequency it doesn't fight with the accumulated oscillation in the LC circuit, but increases its amplitude until the physical amplitude maximum (=clip limit) is reached.

In the university level system theory they prove all this with complex variable transfer functions.That theory wasn't specially for electronics, it's essential for the control systems, too.

  • \$\begingroup\$ I'm sorry but you have made a mistake. The colpitts oscillator does not have infinite amplification at a certain frequency because the tank circuit can never run at magnitude resonance - it has to run off-resonance to achieve the correct phase shift to meet the barkhausen critieria for oscillator. The tank has theoretically 6 dB gain at the oscillating frequency if equal value capacitors are used. \$\endgroup\$
    – Andy aka
    Jul 9, 2017 at 9:01
  • \$\begingroup\$ @Andyaka the loop gain > 1, no matter how little above 1, is needed for starting the oscillation. Of course the phase condition also must be fulfilled and that generally specifies the frequency. But loop gain >1 implies an infinite gain to the noise that circulates in the loop. But I rewrite the answer to be more compliant with the theory, because shortening causes harm. Thanks. \$\endgroup\$
    – user136077
    Jul 9, 2017 at 9:11
  • \$\begingroup\$ OK I see what you mean - an accumulated amplification that is infinite due to reinforcement of the signal because it is in phase - I shall withdraw my d/v. OK it won't let me upvote unless you make some minor edit - maybe just do that and I can upvote. \$\endgroup\$
    – Andy aka
    Jul 9, 2017 at 9:14
  • \$\begingroup\$ @Andyaka rewritten. \$\endgroup\$
    – user136077
    Jul 9, 2017 at 10:25

Inside the emitter (and well beyond reach) is what is known as \$r_e\$. Often it is referred to as "little r e". That resistor becomes in series with the signal that comes from the junction of the two capacitors hence it limits the gain of the transistor circuit.

\$r_e\$ has a value of 26 mV divided by emitter current so, for an emitter current of (say) 1 mA it has a value of 26 ohms. This internal resistor dominates the external emitter resistor in your circuit (2.2 kohm) and so much so that you can virtually ignore the 2.2 kohm.

However, because the 2.2 kohm sets the emitter current, if you halved it to 1.1 kohm, emitter current will double but, significantly, \$r_e\$ now drops to 13 ohms.

Can you see that whatever you do (within reason) to the external emitter resistor it makes no difference? Any increase in attenuation caused by the external resistor due to it halving or quartering is more than countered by the same proportional drop in \$r_e\$.

Behind \$r_e\$ is what could be called the "true emitter" and this, in effect, is a constant voltage set by the base voltage minus the forward diode drop of the base-emitter. It's pretty much like the virtual earth in an op-amp configuration - the "true emitter" voltage is fixed and \$r_e\$ is very much akin to the input resistor in the inverting op-amp circuit - the lower its value, the more the circuit gain increases.

So if \$r_e\$ is reduced by 2 you double the effective potential gain of this type of colpitts oscillator and get a bigger signal at the collector.


Lower resistor R1 in emitter produces lower Rin[1/gm, or 0.026/Ie_amps] at the emitter, which extracts more energy from the capacitive divider.

However, with resonant circuits, energy transfer sometimes needs attention to matching. Read up on that.

Also, LC circuits only store energy if there is a closed-loop path for energy to circulate. Please add a large Cbypass from Top of Inductor to Bottom of Emitter Resistor (bottom of C2). ESR of that Cbypass.....will affect the Q.


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