A resonant circuit This circuit is intended to produce frequencies between 3Hz and 1.5GHz, given the response frequency of the P2N2222A, which is 350MHz.

Here's how I think this circuit works:

  1. While the capacitor C6 is charging through Q1's base, the transistor is conductive; when the capacitor reach its fill of charge, current can no more flow from the base of the transistor Q1.

  2. After Q1 is turned off, a momentary AC signal, which lasts for a fraction of a second, is induced in both capacitors C1, which is connected to the output, and C4, connected base of Q2. The signal from C1 goes right to the common output not, labeled Output, and that from C4 goes to the base of Q2, thus turning it on accordingly.

  3. Q2 turns off, and the same process is repeated with C2 and C5.

  4. Q3 turns on and off according to the signal from C5, and lastly, C3 dumps its output signal to common output.

This process can be made to repeat itself by means of the variable resistor R2, and hence the frequency can be controlled. the 100K resistor is there for the capacitor to discharge. the 1mF capacitors are intended to pass the AC signal with little, if no, impedance.

But my problem is with the 1µF cap, the 100K resistor, and the variable resistor placed between C6's negative and Q1's base.

Is my arrangement of those three afore mentioned components correct for the whole circuit to oscillate?

  • \$\begingroup\$ The circuit shown on the diagram will never work because Q2 and Q3 will never open (cut-off all the time). \$\endgroup\$ – G36 Jun 29 '18 at 16:42
  • 2
    \$\begingroup\$ I think you will need to add at least some resistance or such at the emitter. With it, I think this can act as a third-harmonic oscillator, where the third harmonic of the three transistors will constructively interfere. \$\endgroup\$ – Joren Vaes Jun 29 '18 at 16:48
  • \$\begingroup\$ Also, notice that when C6 stop charging the Q1 will still conduct current. Because the base current will flow via R3 resistor. \$\endgroup\$ – G36 Jun 29 '18 at 16:53
  • \$\begingroup\$ This circuit will definitely not oscillate with GHz -> the transit frequency of the N2222 ~ 250MHz! Q4, Q5 are ac short-ciruited to VEE. So I see now way that this works. \$\endgroup\$ – abu_bua Jun 29 '18 at 17:41
  • 1
    \$\begingroup\$ Upvote for the origilality. Even the pot tolerance is added. Now seriously , I think 1uF is to small for 1.5GHz. \$\endgroup\$ – Dorian Jun 29 '18 at 18:30

No, the circuit will not oscillate at any frequency.

For one thing, Q2 and Q3 do absolutely nothing, because one end of C4 and C5 is connected directly to ground.

Q1 has negative feedback (R2, R3 and C6 connected from collector to base). In order to get oscillation, you need positive feedback at the frequency of interest.

Note also that C1, C2 and C3 are simply connected in parallel.

At UHF frequencies, parasitic effects (wire inductance, junction capacitance) matter A LOT. You would do well to study some existing UHF oscillator circuits before trying to come up with your own.

  • \$\begingroup\$ Regarding C1, C2 and C3 being wired in parallel, I think this may be desired to generate a third harmonic oscillator. The fundamental of each of the signal at each of the emitters will be 120 degrees out of phase, which for the third harmonic translates to 360 degrees out of phase (ie, In phase). This is a technique used a lot in CMOS oscillators that operate past the \$f_t\$ and \$f_{max}\$ of a technology (hundreds of GHz). \$\endgroup\$ – Joren Vaes Jun 30 '18 at 4:39
  • \$\begingroup\$ @JorenVaes: No. The three collectors and the three capacitors are simply wired in parallel, period. Twisting the schematic around in that funky way does not alter this fundamental fact. \$\endgroup\$ – Dave Tweed Jun 30 '18 at 13:09
  • \$\begingroup\$ I meant "if you turn it into an actual oscillator by putting some resistance between the emitters and ground, then the capacitors can make sense". Indeed, as shown now it will not do any oscillating. \$\endgroup\$ – Joren Vaes Jul 1 '18 at 9:08

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.