Graded diode labs: forward drop, an LED resistor, half-wave and bridge rectifiers, a smoothing capacitor for a ripple limit and a Zener regulator.
On the syllabus
The diode forward characteristic and forward voltage
LED current limiting
Half-wave and full-wave (bridge) rectifiers
Smoothing capacitors and ripple, ΔV ≈ I / (fC)
The Zener diode shunt regulator
You will be able to
Measure a diode’s forward drop and current.
Size a current-limiting resistor for an LED.
Predict the peak and average output of half-wave and bridge rectifiers.
Choose a smoothing capacitor for a ripple specification.
Design a Zener regulator that holds its voltage under load.
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. Measure the forward drop
A diode conducts one way only. Once it conducts, the voltage across it stays nearly the same whatever the current, so the resistor in series sets the current.
FormulaI = (Vs − Vf) / R
Worked example9 V through 2.2 kΩ into a silicon diode with Vf = 0.7 V: I = 8.3 / 2200 = 3.77 mA.
On the scope or meterLess than a volt across the diode, and the rest across the resistor. The current is the same in both.
Your task
A 5 V supply drives a 1N4148 diode, D1, through R1 (1 kΩ).
Measure the voltage across the diode (anode minus cathode) and the current through it.
2. Size an LED resistor
An LED drops a fixed voltage that depends on its color: about 1.8 V for red, 3 V for blue. The series resistor takes the rest of the supply and sets the current.
FormulaR = (Vs − Vf) / I
Worked exampleA blue LED with Vf = 3.0 V from 9 V at 15 mA: R = 6 / 0.015 = 400 Ω. The E24 value 390 Ω gives 15.4 mA.
On the scope or meterThe LED’s brightness follows its current. A meter in series reads the current; one across the LED reads Vf.
Your task
A red LED runs from 5 V through R1. Choose R1 for 10 mA through the LED, within 10%.
Take the LED’s forward voltage into account: measure it once, then calculate.
3. Half-wave rectifier: peak and average
A half-wave rectifier passes only the positive half of each cycle. The output peak is the input peak less one diode drop, and the average of a half sine is its peak over π.
FormulaVpeak out = Vpeak in − Vf, Vdc = Vpeak out / π
Worked example24 V peak in: about 23.3 V peak out and an average of 23.3 / π = 7.4 V.
On the scope or meterOn the scope, humps for the positive halves and flat gaps at 0 V between them.
Your task
A 50 Hz sine of 10 V peak feeds a 1N4007 diode into a 1 kΩ load, RL. No smoothing.
Put the scope on the load. Measure the peak output voltage and the average (DC) output voltage.
4. Smooth it: pick a reservoir capacitor
A reservoir capacitor charges at each peak and feeds the load in between. The output then sags a little before the next peak: that sag is the ripple.
FormulaΔV ≈ I / (f C), so C ≥ I / (f ΔV)
Worked exampleA full-wave rectifier (100 pulses a second) at 50 mA with 2 V of ripple: C ≥ 0.05 / (100 × 2) = 250 µF. 330 µF leaves about 1.5 V.
On the scope or meterOn the scope, a nearly flat line near the peak with a small sawtooth dip between charging pulses.
Your task
The same half-wave rectifier, now with a reservoir capacitor C1 across the 1 kΩ load.
Choose C1 so the ripple is at most 1 V peak to peak and the output never drops below 8 V.
Ripple ≈ I / (f C), with f the number of charging pulses a second.
5. Full-wave bridge rectifier
A bridge of four diodes steers both halves of the wave through the load the same way. Two diodes are in the path each half cycle, so it loses two drops.
FormulaVpeak out = Vpeak in − 2 Vf, Vdc ≈ 2 Vpeak out / π
Worked example24 V peak in: about 22.6 V peak out and an average of 2 × 22.6 / π = 14.4 V.
On the scope or meterOn the scope, a hump every half cycle with no gaps: twice the half-wave’s rate.
Your task
Four 1N4007 diodes make a bridge from a floating 50 Hz, 10 V peak source into a 1 kΩ load. No smoothing.
Measure the peak output and the average output. Compare them with the half-wave circuit.
6. Design a Zener regulator
In a Zener shunt regulator, R1 drops the difference between the supply and the Zener voltage. Its current splits between the load and the Zener, and the Zener needs a minimum current to hold its voltage.
FormulaR1 ≤ (Vin − Vz) / (IL + IZ min), P = I² R1
Worked example9 V to a 3.3 V Zener, a 2 mA load and 5 mA minimum: R1 ≤ 5.7 / 0.007 = 814 Ω. 750 Ω passes 7.6 mA and dissipates 43 mW.
On the scope or meterThe output stays close to the Zener voltage while the Zener carries current. With R1 too large it falls away.
Your task
A 1N4733A Zener (5.1 V nominal) regulates a 12 V supply for a 1 kΩ load. Choose the series resistor R1.
With the load on, the output must sit between 4.7 V and 5.3 V, the Zener must carry at least 5 mA so it stays in regulation, and R1 must dissipate no more than 250 mW.
Viva questions
The questions an examiner asks at the bench. 75% passes; every answer comes with why.