Half-Wave Rectifier Using a PN Junction Diode

Experiment: Half-Wave Rectifier Using a PN Junction Diode

1. Aim

To study the rectifying action of a PN junction diode by constructing a half-wave rectifier, to observe the difference between the ideal textbook waveform and the actual output, to investigate the effect of junction capacitance on the reverse-half output, and to study the smoothing effect of a filter capacitor on the rectified waveform.

2. Apparatus / Components Required

• SEELab3 or ExpEYES-17 unit
• PN junction diode — 1N4148 (low junction capacitance, $C_j \approx 4\text{ pF}$)
• PN junction diode — 1N4007 (higher junction capacitance, $C_j \approx 20\text{ pF}$) for comparison
• Load resistor $R_L = 1\text{ k}\Omega$
• Filter capacitor $C = 1\text{ }\mu F$
• Breadboard and connecting wires
• PC or Smartphone with SEELab3 / ExpEYES software

3. Theory & Principle

3.1 The PN Junction as a Rectifier

A PN junction diode conducts current freely when forward biased (anode positive with respect to cathode) and blocks current when reverse biased. When an AC sinusoidal voltage is applied to a series circuit of a diode and a load resistor $R_L$:

• Positive half-cycle: The diode is forward biased and conducts. Current flows through $R_L$ and a positive output voltage appears.
• Negative half-cycle: The diode is reverse biased and ideally blocks all current. The output voltage across $R_L$ is zero.

Since only one half of the AC cycle appears at the output, this is called half-wave rectification.

3.2 Deviation from the Ideal Textbook Waveform

Many textbooks depict the output as a perfect half-sinusoid that begins exactly at $0\text{ V}$ and reaches the same peak as the input. The actual output differs in two important ways:

Forward voltage drop ($V_f$): A real silicon diode requires a minimum forward voltage of approximately $0.6$–$0.7\text{ V}$ before it begins to conduct significantly. The output therefore:

• Starts conducting only when $V_{in} > V_f \approx 0.6\text{ V}$ (the threshold or cut-in voltage)
• Has a peak amplitude reduced by $V_f$ compared to the input peak:

For $V_{in,\text{peak}} = 3\text{ V}$ and $V_f \approx 0.6\text{ V}$, the output peak is approximately $2.4\text{ V}$.

3.3 Junction Capacitance and Reverse-Half Leakage

Every PN junction has a small junction capacitance $C_j$ in parallel with the ideal diode. Its value depends on the diode type:

During the reverse half-cycle, this capacitance couples a small fraction of the input signal to the output through capacitive displacement current. The effect is:

• Visible at high frequencies (e.g., 1000 Hz) because $Z_C = 1/(2\pi f C_j)$ becomes small enough to pass a measurable signal.
• Not visible at low frequencies (e.g., 50 Hz) because $Z_C$ is very large.
• Suppressed by a load resistor: When $R_L = 1\text{ k}\Omega$ is connected, the tiny charge coupled through $C_j$ is quickly discharged through $R_L$, producing negligible voltage. Without $R_L$, the $1\text{ M}\Omega$ input impedance of A2 allows the capacitively-coupled charge to build up to a visible level.

3.4 The RC Filter (Smoothing)

The pulsating DC output of a half-wave rectifier contains a large ripple — the output alternates between the rectified peaks and zero. A capacitor $C$ connected in parallel with $R_L$ reduces this ripple:

• On the conducting half-cycle, $C$ charges rapidly to the peak output voltage.
• On the non-conducting half-cycle, $C$ discharges slowly through $R_L$, maintaining the output voltage at a nearly constant level instead of falling to zero.

The ripple voltage (peak-to-peak) is approximately:

where $f$ is the AC frequency. A larger $RC$ product means slower discharge and lower ripple. For $f = 1000\text{ Hz}$, $R_L = 1\text{ k}\Omega$, $C = 1\text{ }\mu F$: $RC = 1\text{ ms} = 1$ time period — moderate ripple. Increasing $C$ to $10\text{ }\mu F$ gives $RC = 10\text{ ms}$, reducing ripple significantly.

4. Circuit Diagram / Setup

Part A — Basic half-wave rectifier (no filter capacitor):

1. Connect the anode of the diode to WG.
2. Connect the cathode of the diode to one end of $R_L$.
3. Connect the other end of $R_L$ to GND.
4. Connect WG to A1 (monitors $V_{in}$).
5. Connect the junction of the diode cathode and $R_L$ to A2 (monitors $V_{out}$).

Part B — With RC filter:

1. Connect the filter capacitor $C = 1\text{ }\mu F$ in parallel with $R_L$ (between A2 node and GND). Do not add this capacitor until Part A observations are complete.

Note on junction capacitance test: To observe the $C_j$ effect clearly, temporarily remove $R_L$ (open-circuit load) and use a 1N4007 diode at 1000 Hz. The $1\text{ M}\Omega$ input impedance of A2 will allow the reverse-half leakage to become visible.

5. Procedure

Part A — Basic Rectifier

1. Open the SEELab3 / ExpEYES app and select the “Half-Wave Rectifier” experiment.
2. Set the WG to a sinusoidal output at $f = 1000\text{ Hz}$, amplitude $\approx 3\text{ V}$ peak.
3. Click “Start”. Observe both traces:
   • A1 — the input sinusoidal voltage
   • A2 — the rectified output across $R_L$
4. Note the following from the display:
   • The peak amplitude of A2 compared to A1 — the difference is the diode forward voltage drop $V_f$.
   • The input voltage at which the output just begins to appear — this is the cut-in voltage $V_{th}$.
   • The shape of the output during the negative half-cycle — it should be near zero (with $R_L$ connected).
5. Sketch both waveforms and compare with the ideal textbook picture.

Part B — Junction Capacitance Effect

1. Remove $R_L$ from the circuit (open-circuit output). Keep A2 connected directly at the diode cathode.
2. Replace the 1N4148 with a 1N4007 diode.
3. Observe the A2 trace during the negative half-cycle — a small sinusoidal signal should now be visible, caused by capacitive coupling through $C_j$.
4. Restore $R_L = 1\text{ k}\Omega$ and observe that the reverse-half signal disappears.

Part C — RC Filter

1. Reconnect the 1N4148 and $R_L = 1\text{ k}\Omega$.
2. Connect the filter capacitor $C = 1\text{ }\mu F$ in parallel with $R_L$.
3. Observe the A2 trace — the output should now be a near-DC level with a small ripple superimposed.
4. Note the ripple amplitude and the average DC level.

Halfwave (No Filter Capacitor. Notice voltage drop)

Half Wave With Filter

6. Observation Table

6a. Part A — Basic Rectifier

6b. Part B — Junction Capacitance (no $R_L$, 1N4007 at 1000 Hz)

6c. Part C — RC Filter

7. Results and Discussion

• The diode conducted only during the positive half-cycle, producing a half-wave rectified output, as expected.
• The output peak was ____ V, which is ____ V less than the input peak of ____ V, consistent with the silicon diode forward voltage drop of $\approx 0.6\text{ V}$.
• The output began appearing only when $V_{in}$ exceeded the cut-in voltage of ____ V, confirming the threshold behavior absent in the ideal textbook model.
• With the 1N4007 and no load resistor, a reverse-half signal of ____ V peak was observed, attributed to capacitive coupling through the $20\text{ pF}$ junction capacitance. Connecting $R_L$ reduced this to ____ V, demonstrating that the effect is suppressed by a low-impedance load.
• Adding a $1\text{ }\mu F$ filter capacitor reduced the ripple from ____ V (no filter) to ____ V, raising the average DC output to ____ V.

8. Precautions

1. Diode polarity: Verify the cathode (marked with a band) is connected toward the load ($R_L$) and the anode toward WG. Reversing the diode will block the positive half and pass the negative half instead.
2. WG amplitude: Keep the peak input voltage within $\pm 3.3\text{ V}$ to stay within the safe input range of A1 and A2.
3. Capacitor sequence: Observe the basic rectifier waveform thoroughly before connecting the filter capacitor — adding it too early masks the individual half-cycle shape.
4. Capacitor polarity: If using an electrolytic capacitor for $C$, connect the positive terminal to the A2 (output) node and negative to GND. Reversing it can damage the capacitor.
5. Frequency for junction capacitance test: The $C_j$ coupling effect is only visible at high frequencies (≥ 500 Hz). At 50 Hz the reactance of even 20 pF is $\approx 160\text{ M}\Omega$ — far too high to produce a measurable signal.

9. Troubleshooting