Graded 555 labs: set an astable to 1 kHz, measure period and duty cycle, time a 100 ms monostable pulse, and get below 50% duty with a diode.
On the syllabus
The 555 inside: two comparators at 1/3 and 2/3 of VCC, a flip-flop and a discharge transistor
Astable frequency: f = 1.44 / ((R1 + 2R2) C)
Duty cycle: (R1 + R2) / (R1 + 2R2)
Monostable pulse width: t = 1.1 RC
Duty below 50% with a steering diode
You will be able to
Choose the timing parts of an astable for a given frequency.
Measure period and duty cycle on a scope and match them to the formulas.
Design a monostable for a given pulse width.
Explain why a plain astable cannot go below 50% duty, and fix it.
The first challenge is free. The rest, the viva and the certificate are part of Pro, and everything is here to read.
Graded challenges
Values like 4.7k, 100n or 2.2u are fine. Readings are the number only, in the unit shown. Every check measures the simulation, not a bench: ideal parts within their models, no noise.
1. Set an astable to 1 kHz
In an astable the capacitor charges through R1 and R2, then discharges through R2 alone, swinging between 1/3 and 2/3 of VCC. Each charge and discharge is one cycle of the output.
Formulaf = 1.44 / ((R1 + 2R2) C)
Worked exampleR1 = 2 kΩ, R2 = 10 kΩ, C = 47 nF: R1 + 2R2 = 22 kΩ, so f = 1.44 / (22 000 × 47 nF) = 1.39 kHz, a 0.72 ms period.
On the scope or meterA square wave on OUT. The period is the time from one rising edge to the next.
Your task
An NE555 runs as an astable from 5 V: R1 = 1 kΩ, R2 from DIS to THR, and C1 = 100 nF.
Choose R2 so the output runs at 1 kHz: one cycle every 1.0 ms, within 5%.
2. Measure the period and the duty cycle
The output is high while the capacitor charges through R1 + R2, and low while it discharges through R2 only. So the high time is longer than the low time, and the duty cycle is above 50%.
Worked exampleR1 = 4.7 kΩ, R2 = 10 kΩ, C = 1 µF: tH = 10.2 ms, tL = 6.9 ms, a 17.1 ms period at 60% duty.
On the scope or meterA square wave that is high for longer than it is low. Measure one full cycle, then the high part of it.
Your task
An astable with R1 = 10 kΩ, R2 = 4.7 kΩ and C1 = 1 µF.
Put the scope on OUT. Measure one period in milliseconds and the duty cycle: the share of each cycle the output is high, in percent.
Then check both against the formulas.
3. Time a 100 ms monostable pulse
A monostable sits with its output low. A trigger below 1/3 VCC sets the output high and lets C charge through R. When C reaches 2/3 VCC the output goes low again. One trigger, one pulse of fixed length.
Formulat = 1.1 R C
Worked exampleR = 47 kΩ and C = 4.7 µF give t = 1.1 × 47 000 × 0.0000047 = 243 ms.
On the scope or meterOUT jumps high at the trigger and drops back after the pulse width, whatever the trigger does next, as long as the trigger is shorter.
Your task
An NE555 runs as a monostable: R1 from VCC to DIS and THR, C1 = 10 µF. A trigger pulls TRIG low for 50 ms at 450 ms.
Choose R1 so the output pulse lasts 100 ms, within 5%.
4. Get below 50% duty with a diode
A diode across R2 gives the charging current a shortcut, so C charges through R1 only and discharges through R2 only. Now the high time and the low time are set separately, and the duty can go below 50%.
FormulaIdeal duty ≈ R1 / (R1 + R2)
Worked exampleFor 40% duty with R1 = 10 kΩ: 10 / (10 + R2) = 0.4, so R2 = 15 kΩ in theory. The diode drop slows charging, so the real duty comes out a few points higher.
On the scope or meterA square wave that is high for a shorter time than it is low.
Your task
A plain astable charges through R1 and R2 but discharges through R2 alone, so it is always high for longer than it is low.
Here a 1N4148 across R2 lets C1 charge through R1 only. With R1 = 10 kΩ and C1 = 1 µF, choose R2 for a duty cycle of 30%, within 4 percentage points.
Start from the ideal formula, R1 / (R1 + R2), then measure: the diode’s drop makes charging slower than the formula says, so adjust.
Viva questions
The questions an examiner asks at the bench. 75% passes; every answer comes with why.