--- title: "Barkhausen criteria and 50% duty cycle from 555" subject: "Electronics Devices and Circuit" chapter: 5 key: "semester_3/electronics_devices_and_circuit:barkhausen-555-duty-cycle" source: https://www.bctengineer.com/q/semester_3/electronics_devices_and_circuit/barkhausen-555-duty-cycle site: BCT Engineering --- # Barkhausen criteria and 50% duty cycle from 555 Electronics Devices and Circuit · Chapter 5: Signal Generator and Waveform-Shaping Circuits - Asked in: 2079 Baisakh, 2+3 marks - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_3/electronics_devices_and_circuit/barkhausen-555-duty-cycle - Whole chapter: https://www.bctengineer.com/md/semester_3/electronics_devices_and_circuit/chapter-5 ## Question State Barkhausen Criteria for sinusoidal oscillation. Is it possible to obtain 50% duty cycle square wave from 555 timer Astable Multivibrator? How? ## Short answer Barkhausen criteria: loop gain magnitude 1 and phase shift 0° or 360°. Basic 555 astable duty cycle always exceeds 50%, but 50% can be achieved by making RA much smaller than RB, adding a diode across RB, or dividing output frequency by two using a flip-flop. ## Answer **Barkhausen criteria:** An amplifier with feedback produces sustained sinusoidal oscillation at the frequency for which the loop gain Aβ satisfies both: 1. **Magnitude condition:** |Aβ| = 1 (loop gain is exactly unity) 2. **Phase condition:** the total phase shift around the loop is 0° or 360° (positive feedback), that is, ∠Aβ = 0°, 360°, ... Here A is the amplifier gain and β is the feedback factor. #### Can a 555 astable give a 50 % duty cycle? For the 555 astable circuit, tH = 0.693 (RA + RB) C and tL = 0.693 RB C, so the duty cycle is D = (RA + RB)/(RA + 2RB). This is always **more than 50 %** because the capacitor charges through RA + RB but discharges only through RB. So an exact 50 % is not possible with the basic circuit. It can be obtained in these ways: 1. **Make RA very small compared with RB** (for example RA = 1 kΩ, RB = 100 kΩ). Then D ≈ 50 %. However, RA cannot be made too small: during the LOW time a current VCC/RA flows from VCC through RA into the discharge transistor (pin 7), which wastes power and can overload it. 2. **Add a diode across RB** (diode D1 from pin 7 to the capacitor node, with its cathode towards the capacitor). The capacitor then charges through RA and D1 only, and discharges through RB only: ```text VCC --[RA]--+---|>|---+--- pin 6,2 --[C]-- GND | | pin 7 ---[RB]+ ``` tH = 0.693 RA C, tL = 0.693 RB C. Choose **RA = RB** to get D = 50 % (ignoring the diode drop), and f = 1/(0.693 (RA + RB) C) = 0.72/(RA C). 3. **Divide the frequency by 2.** Run the 555 at twice the required frequency and feed its output into a T flip-flop. The flip-flop output is an exact 50 % square wave. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What is the Barkhausen phase condition? A. Total loop phase shift of 0° or 360° B. Total loop phase of 90° C. Loop phase of 180° exactly D. No phase restriction Answer: A. Positive feedback requires zero (or 360°) loop phase. (Key point: Phase condition) 2. Why can the basic 555 astable not give exactly 50 % duty cycle? A. Charging is via RA + RB but discharging only via RB, so tH > tL B. The threshold is VCC/2 C. Pin 5 is grounded D. The discharge is faster than charge by 2x because of the output stage Answer: A. D = (RA + RB)/(RA + 2RB) > 50 %. (Key point: Why not 50 percent) 3. How does a diode across RB give 50 %? A. Charging bypasses RB through the diode so tH = 0.693 RA C; with RA = RB, tH = tL B. It doubles the supply C. It removes the capacitor D. It sets threshold to VCC/2 Answer: A. With the diode, tH = 0.693 RA C and tL = 0.693 RB C, equal when RA = RB. (Key point: Diode method) 4. How does the T flip-flop method give a 50 % square wave? A. The 555 runs at twice the frequency and the flip-flop halves it, giving equal high and low times B. It averages the output C. It adds a diode D. It lowers VCC Answer: A. A T flip-flop toggles on each pulse so its output is symmetrical. 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