555 astable reads 1.25 kHz on the CRO but 1.00 kHz on the counter
A trainee checks a 555 astable board built to a lab sheet: R1 = 10 kΩ, R2 = 68 kΩ, C = 10 nF, Vcc = 9 V, pin 5 decoupled with 10 nF, output on pin 3. The calculated frequency is about 1 kHz. Using an analog CRO with a ×1 probe, the trainee counts 10 complete cycles across 8.0 horizontal divisions with the timebase set to 1 ms/div, and reads a peak amplitude of 0.42 V. The lab record therefore says 1.25 kHz and 0.42 V, a 25% error, and the trainee plans to change R2 to 82 kΩ to correct it. A handheld DMM in frequency mode on the same output reads 1.002 kHz. The scope's VAR/HOLD knob is pulled out, and the probe switch is at ×1 while the scope input menu shows ×10. Vcc measures 9.02 V; R1 = 9.98 kΩ, R2 = 67.9 kΩ, C = 9.9 nF.
1. From the CRO data, compute the measured period and frequency, and compare with the calculated value. Should the board be modified?
Read the worked answer
10 cycles in 8.0 divisions at 1 ms/div = 8.0 ms total, so one period = 0.8 ms and f = 1/0.8 ms = 1250 Hz.
Calculated: T = 0.693 × (R1 + 2R2) × C = 0.693 × (9.98 kΩ + 135.8 kΩ) × 9.9 nF ≈ 1.00 ms, f = 1.44/((R1 + 2R2)C) ≈ 996 Hz.
All three components are within 1%, so the board cannot be 25% high; component tolerance cannot explain it. The DMM counter reading of 1.002 kHz agrees with the calculation, so the board is correct and the CRO reading is wrong. Changing R2 would damage a working circuit - do not modify it.
2. Identify the measurement errors and state the correct amplitude of the output.
Read the worked answer
Two errors. First, the VAR/HOLD knob is pulled out, so the horizontal timebase is not calibrated: the scale is no longer 1 ms/div, which is exactly the kind of 20–30% error seen here. Push the knob to the CAL detent and repeat.
Second, the probe is on ×1 while the scope menu is set to ×10, so the scope multiplies the displayed value by 10. The 0.42 V reading therefore corresponds to about 4.2 V peak, i.e. ≈8.4 V peak-to-peak, which is the expected swing for a 9 V supply.
Correct procedure: set the probe and the scope input to the same attenuation, confirm the CAL detent, use the finest timebase that gives several cycles, and cross-check with the frequency counter. Then record ≈1 kHz and ≈4 V peak.
3. If 1.25 kHz had been the true frequency, what R2 would be needed, and what would fitting 82 kΩ actually do?
Read the worked answer
From f = 1.44/((R1 + 2R2)C), the required total is R1 + 2R2 = 1.44/(f × C) = 1.44/(1250 × 10 nF) = 115.2 kΩ. With R1 = 10 kΩ, 2R2 = 105.2 kΩ, so R2 ≈ 52.6 kΩ - a standard 51 kΩ or 56 kΩ part, not 82 kΩ.
Fitting 82 kΩ would give R1 + 2R2 = 174 kΩ and f = 1.44/(174 kΩ × 10 nF) ≈ 828 Hz, moving the frequency further from 1 kHz and wasting a component. The correct decision is to trust the counter, fix the scope settings, and leave the board as built; verifying with a second instrument before changing hardware is the key judgement here.