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Currents overview

Topic 7 of 7

Half-wave rectification

A single diode can allow load current during one half of an alternating input and block it during the other. The result is current in one direction, with zero-current intervals.

Use an ideal diode: zero forward voltage drop while conducting and no reverse current. Its cathode is identified by the bar in the circuit symbol. This model gives a simple rule for the two circuit states; a real diode's forward drop can reduce the measured output peak.

Fix the circuit and voltage references

The source is between A and return C, with input vin = VA - VC. The diode connects A, its anode, to B, its cathode. A 6.00 Ω resistor connects B to C. Output is measured across that resistor as vout = VB - VC; positive load current goes B to C.

Take vin = 24.0 sin(100πt) V, with t in seconds and the initial upward zero crossing at t = 0. Choosing C as zero potential makes it easier to compare the other two nodes.

Keep the circuit and voltage references fixed

The ideal diode has its anode at A and cathode at B. The source is between A and C, and the 6.00 Ω load is between B and C. Input is vin = VA - VC; output is vout = VB - VC. The load's + and - marks define the output reference in both panels.

Positive input: the diode conducts

The forward-biased ideal diode completes the source and load current pathA single closed circuit connects source node A to the diode anode, the diode cathode bar to node B, B through the resistor to node C, and C through the a.c. source back to A. No wire bypasses either the diode or resistor. In the positive input half-cycle A is at a higher potential than C. The ideal forward diode has zero drop, so B is at A's potential. Brown arrows trace conventional current from A through the diode to B, down through the load to C and along the return to the source. Output follows the positive input. The plus sign at the upper load terminal and minus at the lower are the fixed voltage-reference labels, not a claim that the instantaneous output is nonzero. This is an ideal low-voltage model with no smoothing capacitor or bridge.VA> VCa.c.sourceAnodeCathodeABC6.00 Ωload+-B is at A's potential.vout = vin > 0

Negative input: the diode blocks

The reverse-biased ideal diode blocks current while input voltage is negativeA single closed circuit connects source node A to the diode anode, the diode cathode bar to node B, B through the resistor to node C, and C through the a.c. source back to A. No wire bypasses either the diode or resistor. In the negative input half-cycle A is below C. The diode is reverse biased. There are no current arrows because the ideal circuit current is zero. There is then zero potential difference across the resistor, so B is at C's potential even though A is below both. Input remains negative while output is zero. The plus sign at the upper load terminal and minus at the lower are the fixed voltage-reference labels, not a claim that the instantaneous output is nonzero. This is an ideal low-voltage model with no smoothing capacitor or bridge.VA< VCa.c.sourceAnodeCathodeABC6.00 Ωload+-B is at C's potential.i = 0 and vout = 0

Conventional current goes B to C through the load during conduction and returns through the source. The ideal diode blocks the other half-cycle; it does not consume charge. A real diode's forward voltage drop can reduce the measured output peak.

Both states keep the same source, anode/cathode orientation, resistor and A/B/C references. The positive half-cycle has a complete conducting path; the negative half-cycle blocks current. B is at C's potential in the blocked state.
  • Positive input: A is above C. The diode conducts, so B has A's potential in the zero-drop model. Current passes from A through the diode to B, then through the load to C and back through the source. The resistor output follows the positive input.
  • Negative input: A is below C. The diode is reverse biased and blocks current. With zero current through the resistor, its p.d. is zero and B is at C's potential. The source input can be negative while the load output is zero.
Positive half: vout = vin,   iload = vout/R
Negative half: vout = 0,   iload = 0

The negative input half remains visible; the load output is zero

These exact ideal-model traces share the 0-20 ms time scale. Both voltage graphs also share the same vertical scale. Load current has its own ampere scale and positive direction B to C. Blue-grey shading marks the blocked half-cycle, not transferred energy.

Input: the complete sinusoid

Signed input voltage has both half-cyclesElapsed time is horizontal from zero to twenty milliseconds, with the time axis at the quantity's zero. Input voltage is measured as V A minus V C, in volts. It rises to positive twenty-four volts at five milliseconds, crosses zero at ten and reaches negative twenty-four volts at fifteen before returning to zero at twenty. The shaded second half-cycle identifies the blocked interval, not an energy area. The output and current are unidirectional pulses, with a twenty-millisecond repetition period, rather than steady d.c. values.05101520-240+24vin / VBlocked intervalt / ms

Output: positive voltage pulses

Half-wave output voltage follows only the positive inputElapsed time is horizontal from zero to twenty milliseconds, with the time axis at the quantity's zero. Output is V B minus V C in volts, on exactly the same voltage scale as input. It follows the positive input half, peaking at twenty-four volts at five milliseconds, and is identically zero from ten to twenty milliseconds. The shaded second half-cycle identifies the blocked interval, not an energy area. The output and current are unidirectional pulses, with a twenty-millisecond repetition period, rather than steady d.c. values.05101520-240+24vout / VBlocked intervalt / ms

Load current: one direction, with gaps

Load current consists of positive pulses separated by zero intervalsElapsed time is horizontal from zero to twenty milliseconds, with the time axis at the quantity's zero. Current in the six-ohm load is positive from B to C. It reaches four amperes at five milliseconds and is zero throughout ten to twenty milliseconds. The vertical axis is amperes, not volts. The shaded second half-cycle identifies the blocked interval, not an energy area. The output and current are unidirectional pulses, with a twenty-millisecond repetition period, rather than steady d.c. values.05101520-40+4iload / ABlocked intervalt / ms

When comparing heating, use the full 20.0 ms input period, including the zero interval. The retained squared-current area gives 0.480 J per period, half the unrectified 0.960 J. Its rms current is 2.00 A; the sinusoid's divide-by-√2 rule does not apply to this half-wave shape.

The three traces use the same time reference. Input voltage keeps its negative half-cycle, while output voltage and load current are zero during that interval. The voltage axes share a scale; current has its own units.

The ideal positive output peak is still 24.0 V, giving peak load current 24.0/6.00 = 4.00 A. Positive pulses repeat every 20.0 ms, so their repetition frequency is 50.0 Hz. This is one pulse per input cycle. It differs from the unrectified resistor's two power peaks per current cycle.

The rectified current is unidirectional but not steady. A single diode does not hold the last peak across the resistor during a blocked half-cycle. No energy-storage or smoothing component is included in this circuit.

Reversing the physical diode produces negative output pulses under the same VB - VC reference. Reversing only a voltmeter's leads changes the displayed sign, not which half-cycle the diode conducts. Keep the circuit orientation and measurement reference separate.

Optional check An ideal diode connects source node A (anode) to B (cathode), with a resistor from B to return node C. Input is V_A - V_C and output is V_B - V_C. What happens during a negative input half-cycle?
An ideal diode connects source node A (anode) to B (cathode), with a resistor from B to return node C. Input is V_A - V_C and output is V_B - V_C. What happens during a negative input half-cycle?

For measured waveforms, preserve the time-base, voltage scale and diode/load details. A real forward drop and instrument response can explain differences from the ideal peak; an a.c. meter calibrated for a sinusoid may not correctly report this rectified waveform's rms value.