Continue to Site

Welcome to EDAboard.com

Welcome to our site! EDAboard.com is an international Electronics Discussion Forum focused on EDA software, circuits, schematics, books, theory, papers, asic, pld, 8051, DSP, Network, RF, Analog Design, PCB, Service Manuals... and a whole lot more! To participate you need to register. Registration is free. Click here to register now.

Register Log in

Crystal oscillator circuit based on BJTs in collector feedback configuration questions

N

NigthMoth

Member level 1
Member level 1
Joined
Aug 28, 2023
Messages
34
Helped
0
Reputation
0
Reaction score
6
Trophy points
8
Activity points
691
Hello all

I built and tested a 32.768kHz crystal oscillator circuit based on BJTs in collector feedback bias configuration.
Now I’m trying to understand how it works.
I calculated quiescent voltages for BJTs, some AC parameters and drown equivalent (hope so) model for AC analysis as on pictures below and stucked here.
I'll be very appreciate if someone finds time to check and answer my questions below (sorry if too many).

Circuit:
Circuit.png
AC model
AC model.png
Breadboard
LF_Crystal_Oscillator_10kHz_to_150kHz_BB_01.png
Real circuit
LF_Osc_Pic1.jpg
Output
LF_Osc_WF.jpg

Question 1:
Does my AC model correct?

Question 2:
Am I correct in understanding that feedback voltage gain is ½ ?

Question 3:
How to calculate Vin1?
Should it be
Vin1 = Vfb*Zin1/(Zout_fb + Rcryst + Zin1) voltage divider calculation?

Question 4:
How to calculate Vin2?
Should it be
Vin2 = Vout1*(Zin2||Xc3)/[(Zin2||Xc3) + Xc2 + Zout1] voltage divider with Xc3 parallel with Zin2?

Question 5:
Where to start analysis?
My guess is:
I should postulate, for example: “Let Vout2 oscillates between 0V and 5.2V”.
Then Vfb oscillates between +2.6V and +5.2V. Taking into account Vfb(quiescent) = 3.26V, it dives -0.66V below and +1.94V above quiescent point.
So AC voltage is -0.66V…+1.94V and I need to feed this AC voltage value into my AC model?

Question 6:
Why there is C2-C3 capacitor voltage divider? I tried to make calculations using formula as in my Question 3, and get Vin2 = 0.035*Vout1 with -60O
phase shift. As I understand, for oscillator one need 360O phase shift, But Q1 and Q2 already provide 180O + 180O shift.
Or C2 and C3 here for other purposes?

Question 7:
According to my calculations, for BJT Collector feedback bias configuration, Vout2 sits at 1.3V and adds 3.9V (up to 5.2Vcc) for High output or drops 1.3V (down to 0V) for low output. So I expect that after C4 (DC blocking capacitor) V(OUT) waveform should altering between +3.9V and -1.3V. But in reality it is between +1.25V and -3.54V.
It looks like test result is mirrored image of my calculations, but I can’t find where I’m wrong (maybe I misunderstood how DC blocking capacitor works?).

Question 8:
What is possible reason I got such a large difference in real and calculated values for Vout1(quiescent) and Vout2(quiescent)?


******
For my analysis I used these formulas and values:

For BJT DC analysis:
Ib = (Vcc - Vbe)/(Rf + hfe*Rc);
Ic = hfe*Ib;
Vout(quiescent) = Vc = Vce = Vcc - Ic*Rc;

For Vfb q-point
Vfb(quiescent) = Ic*Rdown + Vc;

For BJT AC analysis:
re = 26mV/Ic;
ro = 1/hoe = 1/9 uMho = 111k for 2N3904 @IC = 1mA;
Zi = re/(1/hfe + Rc/(Rc+Rf)) (applicable when Rc>>re);
Zo = Rc||Rf (applicable when ro>10*Rc);
Av = -Rc/re (applicable when Rf >> Rc) (“-“ for -180O phase shift);

For crystal + CL
Rcryst ~= Rs*(1+Co/CL)^2 @resonance frequency (from this techical note);

For feedback:
Zo_fb = Rup||(Rdown+Rf);
Av_fb = -Rup/re, or ½ of Av2, so Av_fb = 1/2 is for Vout2;

Values from datasheets:
2N3904 BJT:
- Minimum hfe (@IC = 1mA, Vce = 1V) = 70 (Use it for worst case scenario);
- Vce (@IC = 1mA) = 0.05V;
- Vbe (@IC = 1mA) = 0.7V; [Oscilloscope Q1 = 0.636V, Q2 = 0.636V];
- Rise time (@IC = 1mA) >500 nS (~1uS?);
- Fall time (@IC = 1mA) = 400 nS;
- hoe (@IC = 1mA) ~= 9 uMho;
CFS145 32.768kHz tuning fork crystal:
-Co = 1pF;
-Rs = 40k;
-CL = 12.5p;
 

Attachments

  • Circuit.png
    Circuit.png
    118 KB · Views: 42
Sort by date Sort by votes
I recognize your excellent documentation and effort to validate your assumptions.

The only fault I see is the assumption for hFE using only the hFE min =70 instead an approx. upper limit of 70% of 300 or 270. Although no max. was given, higher hFE’s are possible due to mfg. Tolerances and preferences.

If you revise your assumption and compare, you may be satisfied for Vce(Q).

You may then estimate the effect on % d.f. from 70 to say 300. The choice for negative feedback of Rbc/Rc =10 for -ve f.b. reduces the hFE sensitivity on DC bias.

The Rbc value also provides adequate series loop gain based on about 50% of the ratio for feedback R to source R impedance @ fo to saturate the oscillation.

I did not evaluate if your design was optimized for yields since there was no criteria given up front. (Specs or acceptance criteria)
--- Updated ---

Next to reply to your queries.
Question 1:
Is my AC model correct?
Not quite.
Av & Zo in each closed loop depends on feedback ratio lowers Zo output and Av but it is about 50% if the Rf/Rs ratio depending on hFE.

Question 2:
Am I correct in understanding that feedback voltage gain is ½ ?

1/2 of what? What is the FB/Source impedance ratio of each stage and passive attenuation? This FB also reduces Vbe variation and impedance

Question 3:
How to calculate Vin1?
Should it be
Vin1 = Vfb*Zin1/(Zout_fb + Rcryst + Zin1) voltage divider calculation?

The -fb/source ratio stabilizes everything which makes the math more complex. e.g. lower Rbe(f), Vin2, Zo2(f)

Question 4:
How to calculate Vin2?
Should it be
Vin2 = Vout1*(Zin2||Xc3)/[(Zin2||Xc3) + Xc2 + Zout1] voltage divider with Xc3 parallel with Zin2?

Vin2 sums Vout2 & Vout1 with divider ratios.

Question 5:
Where to start analysis?
My guess is:
I should postulate, for example: “Let Vout2 oscillates between 0V and 5.2V”.
Then Vfb oscillates between +2.6V and +5.2V. Taking into account Vfb(quiescent) = 3.26V, it dives -0.66V below and +1.94V above quiescent point.
So AC voltage is -0.66V…+1.94V and I need to feed this AC voltage value into my AC model?

The Vpk when AC coupled is asymmetric due to duty factor, d.f. from better hFE assumptions. Then impedance divider ratios depend on open loop gains * forward + fb ratios.

Question 6:
Why there is C2-C3 capacitor voltage divider? I tried to make calculations using formula as in my Question 3, and get Vin2 = 0.035*Vout1 with -60O
phase shift. As I understand, for oscillator one need 360O phase shift, But Q1 and Q2 already provide 180O + 180O shift.
Or C2 and C3 here for other purposes?

C3 attenuates forward gain only 50% while attenuating AC feedback 99% thus the AC gain is now increased to open loop to the point of non linearity with high current gain saturation.
(all for now)
Question 7:
According to my calculations, for BJT Collector feedback bias configuration, Vout2 sits at 1.3V and adds 3.9V (up to 5.2Vcc) for High output or drops 1.3V (down to 0V) for low output. So I expect that after C4 (DC blocking capacitor) V(OUT) waveform should altering between +3.9V and -1.3V. But in reality it is between +1.25V and -3.54V.
It looks like test result is mirrored image of my calculations, but I can’t find where I’m wrong (maybe I misunderstood how DC blocking capacitor works?).

Question 8:
What is possible reason I got such a large difference in real and calculated values for Vout1(quiescent) and Vout2(quiescent)
 
Last edited:
Upvote 0 Downvote

LaTeX Commands Quick-Menu:

Similar threads

Part and Inventory Search

Welcome to EDABoard.com

Sponsor

Back
Top