Practical case: Simple half-wave rectification

Simple half-wave rectification prototype (Maker Style)

Level: Basic. Visualizing how a diode converts AC to pulsating DC by removing the negative half-cycle.

Objective and use case

You will build a fundamental analog circuit that uses a single semiconductor diode to block the negative portion of an alternating current (AC) signal, passing only the positive portion to a resistive load.

Why it is useful:
* Power conversion: It represents the first stage in converting AC mains power to DC for electronic devices.
* Signal demodulation: Used in AM radios to extract audio signals from radio frequency carriers (envelope detector).
* Polarity protection: Similar logic prevents damage to DC circuits if batteries are inserted backward.

Expected outcome:
* Input Signal: A complete sine wave swinging between positive and negative voltages (e.g., +10 V to -10 V).
* Output Signal: A pulsating waveform showing only the positive «humps» of the sine wave; the voltage sits at 0 V during the negative cycle.
* Voltage Drop: The peak output voltage will be approximately 0.7 V lower than the input peak due to the silicon diode forward voltage drop.
* Frequency: The output frequency remains identical to the input frequency.

Target audience and level: Students and hobbyists learning basic analog components.

Materials

  • V1: 10 V (peak), 60 Hz AC voltage source (sine wave), function: main power input.
  • D1: 1N4007 (or 1N4148), function: rectifier diode.
  • R1: 1 kΩ resistor, function: resistive load.

Wiring guide

This guide defines the connections between components using specific node names (VIN, VOUT, 0).

  • V1 (Source): Connect the positive terminal to node VIN and the negative terminal to node 0 (GND).
  • D1 (Diode): Connect the Anode to node VIN and the Cathode (marked with a stripe) to node VOUT.
  • R1 (Load): Connect one terminal to node VOUT and the other terminal to node 0 (GND).

Conceptual block diagram

Conceptual block diagram — Half-Wave Rectification
Quick read: inputs → main block → output (actuator or measurement). This summarizes the ASCII schematic below.

Schematic

[ SOURCE / INPUT ]             [ RECTIFICATION ]               [ LOAD / OUTPUT ]

[ V1: AC Source    ]           +----------------------+           [ R1: Resistor   ]
[ 10 V Peak, 60Hz   ] --(VIN)-->| Anode (A) -> Cathode | --(VOUT)--> [ 1 kΩ         ] --> GND
                               | D1: 1N4007           |
                               +----------------------+
Schematic (ASCII)

Measurements and tests

To validate the circuit, you will need a dual-channel oscilloscope or a simulation tool.

  1. Setup Probes:
    • Connect Channel A (Yellow) to VIN to monitor the source.
    • Connect Channel B (Blue) to VOUT to monitor the voltage across the resistor.
    • Ensure the ground clips of both probes are connected to node 0 (GND).
  2. Visual Inspection:
    • Observe that VIN is a full sine wave centered at 0 V.
    • Observe that VOUT follows VIN during the positive cycle but stays flat at 0 V during the negative cycle.
  3. Cursor Measurement:
    • Measure the peak voltage of VIN (e.g., 10.0 V).
    • Measure the peak voltage of VOUT. It should be approximately 9.3 V.
    • Calculate the difference (Vin – Vout). This confirms the roughly 0.7 V forward voltage drop of the silicon diode.

SPICE netlist and simulation

Reference SPICE Netlist (ngspice) — excerptFull SPICE netlist (ngspice)

* Practical case: Simple half-wave rectification

* --- Circuit Description ---
* V1 (Source): 10V Peak, 60Hz Sine Wave
* D1 (Diode): 1N4007 Rectifier
* R1 (Load): 1k Ohm Resistor

* --- Components ---

* V1: Main power input
* Connected: Positive -> VIN, Negative -> 0 (GND)
* Syntax: SIN(Voffset Vamp Freq)
V1 VIN 0 SIN(0 10 60)

* D1: Rectifier diode (1N4007)
* Connected: Anode -> VIN, Cathode -> VOUT
D1 VIN VOUT 1N4007

* R1: Resistive load
* Connected: VOUT -> 0 (GND)
* ... (truncated in public view) ...

Copy this content into a .cir file and run with ngspice.

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* Practical case: Simple half-wave rectification

* --- Circuit Description ---
* V1 (Source): 10V Peak, 60Hz Sine Wave
* D1 (Diode): 1N4007 Rectifier
* R1 (Load): 1k Ohm Resistor

* --- Components ---

* V1: Main power input
* Connected: Positive -> VIN, Negative -> 0 (GND)
* Syntax: SIN(Voffset Vamp Freq)
V1 VIN 0 SIN(0 10 60)

* D1: Rectifier diode (1N4007)
* Connected: Anode -> VIN, Cathode -> VOUT
D1 VIN VOUT 1N4007

* R1: Resistive load
* Connected: VOUT -> 0 (GND)
R1 VOUT 0 1k

* --- Models ---
* Standard model for 1N4007 Diode
.model 1N4007 D (IS=7.69n RS=0.042 N=1.45 BV=1000 IBV=5u CJO=14.2p VJ=0.5 M=0.333 TT=4.32u)

* --- Analysis Commands ---
* Transient analysis
* Frequency is 60Hz (Period ~16.67ms).
* Simulate for 50ms to capture approximately 3 full cycles.
.tran 0.1ms 50ms

* Operating Point for initial check
.op

* --- Output Directives ---
* Print input voltage and rectified output voltage
.print tran V(VIN) V(VOUT)

.end

Simulation Results (Transient Analysis)

Simulation Results (Transient Analysis)
Show raw data table (515 rows) 
Index   time            v(vin)          v(vout)
0	0.000000e+00	0.000000e+00	-2.01593e-21
1	1.000000e-06	3.769911e-03	5.704546e-05
2	2.000000e-06	7.539822e-03	5.927562e-05
3	4.000000e-06	1.507964e-02	6.305993e-05
4	8.000000e-06	3.015924e-02	7.111847e-05
5	1.600000e-05	6.031821e-02	1.021853e-04
6	3.200000e-05	1.206342e-01	3.070797e-04
7	5.378437e-05	2.027484e-01	2.167324e-03
8	7.424258e-05	2.798514e-01	1.250260e-02
9	9.741093e-05	3.671480e-01	4.715921e-02
10	1.262516e-04	4.757778e-01	1.182339e-01
11	1.839330e-04	6.928557e-01	2.983890e-01
12	2.467131e-04	9.287461e-01	5.130162e-01
13	3.467131e-04	1.303359e+00	8.676123e-01
14	4.467131e-04	1.676120e+00	1.226655e+00
15	5.467131e-04	2.046499e+00	1.587509e+00
16	6.467131e-04	2.413969e+00	1.947514e+00
17	7.467131e-04	2.778010e+00	2.305173e+00
18	8.467131e-04	3.138102e+00	2.659882e+00
19	9.467131e-04	3.493735e+00	3.010809e+00
20	1.046713e-03	3.844404e+00	3.357375e+00
21	1.146713e-03	4.189609e+00	3.698904e+00
22	1.246713e-03	4.528861e+00	4.034877e+00
23	1.346713e-03	4.861677e+00	4.364712e+00
... (491 more rows) ...

Common mistakes and how to avoid them

  1. Reversing the diode:
    • Error: The diode is installed with the cathode pointing toward the source.
    • Result: The circuit produces negative voltage pulses instead of positive ones.
    • Solution: Verify the stripe (cathode) points toward the load resistor.
  2. Ignoring power ratings:
    • Error: Using a very small resistor (e.g., 10 Ω) with a standard 1/4W resistor.
    • Result: The resistor overheats and burns.
    • Solution: Calculate power (P = V^2 / R) or use a resistor value like 1 kΩ or higher for demonstration purposes.
  3. Floating Ground:
    • Error: Measuring VOUT without a common ground reference between the source and the oscilloscope.
    • Result: Noisy or floating signals on the screen.
    • Solution: Ensure all grounds (Source, Resistor, Oscilloscope) are tied to node 0.

Troubleshooting

  • Symptom: No output voltage (0 V flatline).
    • Cause: Diode is open (blown) or disconnected.
    • Fix: Check continuity with a multimeter; replace the diode.
  • Symptom: Output is identical to Input (full sine wave).
    • Cause: Diode is shorted internally.
    • Fix: Replace the diode; a shorted diode acts like a wire.
  • Symptom: Output peak is significantly lower than expected (e.g., 5 V drop).
    • Cause: High internal resistance of the source or an incorrect component (e.g., using a Zener diode in reverse breakdown).
    • Fix: Verify the diode part number is a standard rectifier (1N400x series).

Possible improvements and extensions

  1. Filter Capacitor: Connect a capacitor (e.g., 10 µF) in parallel with R1 to observe how the capacitor fills in the gaps between pulses, smoothing the DC output.
  2. Full-Wave Bridge: Replace the single diode with four diodes (bridge configuration) to utilize both the positive and negative halves of the AC cycle, improving efficiency.

More Practical Cases on Prometeo.blog

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Quick Quiz

Question 1: What is the primary function of the semiconductor diode in this circuit?




Question 2: What is the expected shape of the output signal?




Question 3: How does the output frequency compare to the input frequency in this half-wave rectifier?




Question 4: Why is the peak output voltage slightly lower than the input peak voltage?




Question 5: Approximately how much voltage is typically dropped across a standard silicon diode?




Question 6: In a standard schematic for this circuit, which component typically acts as the load?




Question 7: What happens to the voltage at the output during the negative cycle of the input?




Question 8: Which of the following is a practical use case for this type of circuit mentioned in the text?




Question 9: To allow positive current to flow from the source to the load, how should the diode be oriented?




Question 10: If the input signal swings between +10 V and -10 V, what is the approximate peak output voltage?




Carlos Núñez Zorrilla
Carlos Núñez Zorrilla
Electronics & Computer Engineer

Telecommunications Electronics Engineer and Computer Engineer (official degrees in Spain).

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