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I am a student working on a project where we are using CC Colpitts oscillators as local oscillators.Image of circuit used

We've chosen the above values based on the condition gm > w02C1C2RL,series, which for the given image translates to gm > w02C1C2R6. This is a formula we worked with as a given result (assuming biasing resistors are large in comparison to impedances at the resonant freq), which I gather comes from the feedback network needing to have closed-loop gain > 1 and have the feedback signal be in phase with the input.

I wanted to arrive at this result myself but have been having some trouble. Here is how I have framed the problem and tried solving it:

  1. The feedback topology is shunt-shunt, i.e., I am sampling the voltage at the emitter and then feeding a current back in at the base.
  2. The feedback gain, B = j(1 / XC1), since the feedback input comes into one end of C1 and the other end is grounded for feedback analysis(shunt connection).
  3. Then for an open-loop gain A, the condition of A*B > 1 should allow me to solve for the previous result

I then tried finding the loaded forward open-loop gain (A), loading the emitter with C1||C2 and the base with L||C1. However, here I got lost in the math and haven't been able to reduce A*B > 1 to the above result after multiple attempts.

I would appreciate any input of mistakes I may have made in the problem setup or help on how to proceed from here if the rest is right.

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  • \$\begingroup\$ from practical point of view.. why not use a more classical topology with tank circuit between collector and emitter ? \$\endgroup\$
    – gino
    Oct 25, 2021 at 8:02
  • \$\begingroup\$ Related answer \$\endgroup\$
    – Andy aka
    Oct 25, 2021 at 8:19
  • \$\begingroup\$ @gino the CC Colpitts is a classical oscillator and is used in many circuits. It's no harder or easier to analyse than any other Colpitts type so, I'm not sure your recommendation makes any sense. \$\endgroup\$
    – Andy aka
    Oct 25, 2021 at 8:44
  • \$\begingroup\$ @LorenzoMarcantonio that paper is for a common emitter Colpitts oscillator and not common collector. It goes all over the place in my opinion and needn't be so long-winded. \$\endgroup\$
    – Andy aka
    Oct 25, 2021 at 8:49
  • \$\begingroup\$ @sarthak it does indeed work - could you clarify what makes the biasing wrong? \$\endgroup\$
    – kraken234
    Oct 25, 2021 at 18:02

2 Answers 2

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Consider the small signal model of the circuit (excluding the inductor) as shown below:

schematic

simulate this circuit – Schematic created using CircuitLab

Here, I am assuming that, the operating frequency is much higher than the cutoff frequency of the bias circuit, \$\omega >> \frac{1}{C_3(R_7||R_8)}\$, which can be ignored. Also,I assume that \$\beta>>1\$ so that the base current can be neglected and \$\alpha = 1\$. The output resistance and the degeneration are assumed to be \$r_o, R_9 >> \frac{1}{g_m}\$ and are consequently ignored.
With these simplifications, KCL at node \$v_o\$ gives: $$(v_x-v_o)sC_1=v_osC_2+(v_o-v_x)g_m$$ $$v_o = \frac{(g_m+sC_1)v_x}{s(C_1+C_2)+g_m}$$ $$v_x-v_o = \frac{sC_2v_x}{s(C_1+C_2)+g_m}$$ KCL at node \$v_x\$ gives: $$i_x = sC_1(v_x-v_o)$$ $$i_x = sC_1.\frac{sC_2v_x}{s(C_1+C_2)+g_m}$$ $$\frac{v_x}{i_x} = \frac{C_1+C_2}{sC_1C_2}+\frac{g_m}{s^2C_1C_2}$$ $$Z_{eq}= \frac{1}{j\omega C_{eq}}-\frac{g_m}{\omega^2C_1C_2}, C_{eq} = \frac{C_1C_2}{C_1+C_2}$$ This is the impedance looking into the transistor which is a series combination of capacitance and negative resistance. The total impedance including the inductor becomes: $$Z_T = j\omega L+ R_6 + \frac{1}{j\omega C_{eq}} - \frac{g_m}{\omega^2C_1C_2}$$ For sustained oscillations, $$X_T = 0 \implies \omega L = \frac{1}{\omega C_{eq}} \implies \omega = \frac{1}{\sqrt{LC_{eq}}}$$ $$R_T < 0 \implies \frac{g_m}{\omega^2C_1C_2} > R_6$$ This gives the condition you mentioned in the question.

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One can use 2 - port parameters as well. We will use the Barkhausen Criterion with current gains. It would seem natural to do this since the circuit is in common collector configuration. It is assumed that sinusoidal oscillation takes place. The Barkhausen Criterion in terms of current gain is:

$$A_i \beta_i = 1\ + \ \ j0$$

We briefly describe \$ \ \ A_i \ \ \$ and \$ \ \ \beta_i \ \ \$ from:

$$A_i = \frac {y_{fc}}{y_{ic} + \Delta_c^y z_{in \ \ \beta}}$$

$$\beta_i = \frac{y_{21\beta}}{y_{11\beta} + \Delta^y_{\beta}z_{in \ \ c}}$$

\$ \ \ A_i \ \ \$ is the active device current gain. Assume \$ \ \ \Delta^y_c \approx 0 \ \ \$. Then $$ \ \ A_i = \frac{y_{fc}}{y_{ic}} = h_{fc}. $$ where \$ \ \ y_{fc} \ \ \$ is the forward common-collector trans-admittance parameter and \$ \ \ y_{ic} \ \ \$ is the input admittance common-collector parameter. The subscript c refers to common-collector parameters. $$\ \ \beta_i = \frac{y_{21\beta}}{y_{11\beta}} = -\frac{C_1}{C_1 + C_2} \ \ (when \ \ \Delta^y_{\beta}=0) \ \ $$ This is the current gain of the feedback network or " the feedback fraction " of the oscillator . \$ \ \ y_{21\beta} \ \ \$ is the feedback network trans-admittance parameter and \$ \ \ y_{11\beta} \ \ \$ the feedback network input admittance parameter. The subscript \$ \ \ \beta \ \ \$ refers to the feedback network.

where:

$$z_{in \ \ c} \approx h_{ic}$$

and is the input impedance of the amplifier.

$$z_{in \ \ \beta} = \frac{1 + y_{22 \beta} h_{ic}} {y_{11\beta} + \Delta_{\beta}^{y} h_{ic}}$$

and is the input impedance of the feedback network.

The deltas \$ \ \ \Delta \ \ \$ are the determinants of the y matricies.

\$ C_1 \ \ and \ \ C_2 \$ make up the usual split capacitor arrangement in a Colpitts Oscillator. \$ \ \ C_2 \ \ \$ is connected to ground and \$ \ \ C_1 \ \$ goes to the base of the BJT. The junction of the two capacitors goes to the emitter of the BJT. The inductance L is in parallel with \$ C_1 \ \ and \ \ C_2 \$.

The feedback network y parameters are:

$$ y_{11\beta} = j\omega(C_1 + C_2)$$ $$ y_{22\beta} = j\omega C_1 - j\frac{1}{\omega L}$$ $$ y_{21\beta} = -j\omega C_1 $$ $$ y_{12\beta} = -j\omega C_1 $$ $$ \Delta^y_{\beta} = -\omega^2 C_1 C_2 + \frac{C_1 + C_2}{L}$$

Let \$\Delta_{c}^{y} z_{out\beta} \approx 0 \$

From these we obtain:

Equation 1: $$ h_{fc} y_{21\beta} = y_{11\beta} + \Delta_{\beta}^{y} h_{ic}$$

The real and imaginary parts must be separated and equated and we arrive at the real part corresponding from equation 1: This is equation 1a (criterion for oscillation):

$$h_{fc} y_{21\beta} = y_{11\beta}$$

$$ \omega \neq 0 $$

From the imaginary part of equation 1 we obtain equation 1b. (frequency of oscillation):

$$ \Delta_{\beta}^y \ \ h_{ic} = 0 $$

We now analyze equations 1a and 1b:

From equation 1a: $$h_{fc} = -\frac{C_1 + C_2}{C_1}$$

$$h_{fc} = -(1+h_{fe})$$

$$h_{fe} = \frac{C_1 + C_2}{C_1} - 1$$

$$h_{fe} = \frac{C_2}{C_1}$$

From equation 1b (or the imaginary part) :

$$ \Delta_{\beta}^y \ \ h_{ic} = 0 $$

the frequency of oscillation is obtained:

$$frequency = \frac{1}{2\pi \sqrt{L C_{eq}}}$$ where

$$C_{eq} = \frac{C_1 C_2}{C_1 + C_2}$$

We now wish to analyze this oscillator with negative resistance theory. The total (feedback network+amplifier) y parameters become:

$$y_{11 \beta + c} = \frac{1}{r_{\epsilon}} + j \omega C_1 $$ $$y_{22 \beta + c} = j\omega C_1$$ $$y_{21 \beta + c} = -(g_m + j \omega C_1) $$ $$y_{12 \beta + c} = -j\omega C_1$$ $$ \Delta{^y}_{\beta + c} = -\omega^2 C_1 C_2 $$

We use:

$$y_{out \beta + c} = y_{22\beta + c} - \frac{y_{12\beta + c} y_{21\beta + c}}{y_{11\beta + c} + y_{in}}$$

where: $$y_{in} = Re(y_{in}) + j Im(y_{in})$$

Let $$ Re(y_{in}) = \frac{1}{R} $$ $$ Im(y_{in}) = j \omega C_2 $$

where R generalizes and helps simplify \$ \ \ y_{out\beta + c} \ \$.

Also:

$$ R \neq 0 $$

Then:

$$y_{out\beta + c} = \frac{ -\omega^2 C_1 C_2 + j \omega C_1(\frac{1}{R} - g_m)}{\frac{1}{R} + j \omega( C_1 + C_2 )}$$

We wish to make the imaginary part of the numerator of \$ \ \ y_{out\beta + c} \ \ \$ vanish. This can be done with condition 1:

$$ 1 \ \ - \ \ g_m R = 0 $$

There is also condition 2:

$$R = r_{\epsilon}$$

We assume conditions 1 and 2 hold.

The governing equation is Equation 2:

$$ y_{out\beta + c} = -y_{coil \ \ circuit} $$

We let:

$$y_{coil \ \ circuit} = \frac{1}{R_p} - j \frac{1}{X_L}$$

where \$ \ \ R_p \ \$ is the resonator coil circuit parallel resistance and \$\ \ X_L \ \ \$ is approximately the inductive reactance.

Equations 2a and 2b, the next equations , are put together to make the governing equation which is equation 2:

$$y_{out\beta + c} = - y_{coil \ \ circuit}$$

Equation 2a:

$$ y_{out\beta + c} = \frac{-\omega^{2} C_1 C_2}{g_m + j\omega (C_1+C_2)}$$

Equation 2b is:

$$y_{coil \ \ circuit} = \frac{1}{R_p} - j \frac{1}{X_L} = \frac{1}{ R_{series coil} + R_6 + j \omega L} $$

where: $$ R_p = Q X_L + R_{series coil} + R_6 $$

$$ X_L \approx \omega L $$

$$ Q = \frac{X_L}{R_{series coil} + R_6}$$

From equation 2:

$$-\frac{\omega^2 C_1 C_2}{g_m + j\omega(C_1 + C_2)} = -\frac{1}{R_{series} + R_6 +j\omega L}$$

By multiplying both sides by their LCM we obtain:

$$ (\omega^2 C_1 C_2)(R_{series} + R_6) + j \omega^3 L C_1 C_2 = g_m + j\omega (C_1 + C_2)$$

The real parts are separated and equated from equation 2 (when conditions 1 and 2 are met):

$$ ( \omega^2 C_1 C_2 )( R_{series coil} + R_6) = g_m $$

Something that must be said:

$$ \omega \neq 0 $$

From this equation the criterion for oscillation is found:

$$ \frac{g_m}{\omega^2 C_1 C_2} = R_{series coil} + R_6 $$

The imaginary parts are separated and equated from equation 2 so that:

$$ \omega^3 L C_1 C_2 = \omega (C_1 + C_2) $$

again:

$$ \omega \neq 0 $$

This implies the frequency of oscillation:

$$ \omega^2 = \frac{1}{L C_{eq}}$$ where:

$$ C_{eq} = \frac{C_1 C_2}{C_1 + C_2}$$

Maybe this might help.

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