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POWER ELECTRONICS / INTERACTIVE LABUniversity of Vigo

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Rectifier laboratory

Single-phase half-wave uncontrolled rectifier

Change the circuit. Follow the current. Understand the output.

CSV: comma-separated fields; decimal point (.) in all three languages. Displacement power factor key: dpf (formerly df). Technical provenance is appended as metadata columns (waveforms) or rows (metrics); numerical fields are unchanged.

Learning level

Start with ideal diodes and an R or RL load: the circuit, voltage and current waveforms, and mean/RMS output values. Switching modes keeps your circuit settings.

Colored edge = level where a block is first introduced.

Choose how much detail to show

The current path

Periodic steady state
Single-phase half-wave rectifier circuitA sinusoidal source drives D1 and a series R or RL load, optionally including an opposing DC voltage E. Its positive terminal is above its negative terminal. The optional freewheeling diode is connected across the complete load, including E, with its cathode at the positive output terminal. Alternatively, the load can be R with a filter capacitor C in parallel. vs+−D1is = iD1 Rio DFiDF +−vo

Scroll the circuit horizontally if needed.Desplaza el circuito horizontalmente si es necesario.Despraza o circuíto horizontalmente se é necesario.

At the linked probe:90.0°

Playback is slowed to one electrical cycle every eight seconds.

Mean vs RMS: when can voltage be divided by R?

Here vo is the voltage across the complete load, not just the resistor. All averages use one complete source period in periodic steady state. Bypassed L or E count as zero.

With the capacitor filter, iR = vo/R, IR,dc = Vo,dc/R, and IR,rms = Vo,rms/R refer to the resistor branch. The complete rectifier output current is io = iD1 = iR + iC, whose peaks can be much higher. The following equations for io are scoped to the series load types.

Mean current — valid for R, RL, RE and RLE loads. The inductor voltage has zero mean over a complete period, so:

Io,dc = (Vo,dc − E)/R

This holds with or without L and E, for uninterrupted, discontinuous or zero current, with either diode model and with or without DF. It does not mean that L cannot change the mean current: L may change Vo,dc as well.

RMS shortcut — use it only for a purely resistive load. When L = 0 and E = 0, current is proportional to the complete-load voltage at every instant:

Io,rms = Vo,rms/R   (L = 0, E = 0)

With inductance or a nonzero E, use the voltage across the resistor alone instead. The following relation is valid for the series load types R, RL, RE, and RLE:

Io,rms = VR,rms/R
vR = vo − vL − E

VR,rms is the RMS voltage across the resistor. RMS means square, average, then take the square root; it is not a linear operation. A zero mean vL does not imply zero RMS vL. With L = 0 but E > 0, calculate the RMS of the waveform vo − E first; do not simply subtract E from Vo,rms.

Steady-state waveforms

Move across either plot to follow the current path. Dashed horizontal guides show the averages.

Time in ms · two cycles

01 / Source & output voltage

V

02 / Load & diode currents

A
What to notice now

Experiments to try

Predict first, change the circuit, then check the waveforms and reveal the explanation.

Inductive load: why can D1 stay on?

Use the load-type button to compare noninductive and inductive loads, keeping the other settings fixed. With inductance selected, increase L. Predict what happens when the source voltage becomes negative.

Reveal the explanation

For the ideal circuit without inductance, io = vo/R. With L > 0, current cannot change abruptly. Without DF, D1 remains on until its current reaches zero; if current persists after the source voltage becomes negative, the output voltage is also negative.

With E or VF: without inductance, io = (vo − E)/R. Turn-on from zero current requires Vm > E + VF, and the opposing voltage and diode drop can shorten or remove the current tail. With no current, vo = E, not necessarily zero.

Further explanation · Extended

Without DF, D1 turns off at the first subsequent current zero β. While it conducts after 180°, the inductor releases energy to R, to the DC load (Eio), to any diode loss, and back to the AC source. For E = VF = 0 and L > 0, β > 180°. A positive E + VF can move β before 180°; there is then no energy-return interval.

Adding DF: where does the current go?

Use the load-type button to select an inductive load with L > 0, then add and remove DF, keeping the other settings fixed. Predict the current path during the negative source half-cycle; compare the path, output voltage, source current and load current.

Reveal the explanation

During the negative source half-cycle, any remaining load current has two possible paths. Without DF, it continues through the source and D1. With DF, it circulates through the load and DF instead: the source current is then zero.

With ideal diodes, vo = 0 while DF conducts. With no effective inductance, or if current is already zero at 180°, adding DF creates no freewheeling current.

With VF > 0, vo = −VF while DF conducts. When the current extinguishes, vo returns to E.

Further explanation · Extended

The freewheeling loop comprises resistor R, inductor L, the DC source of voltage E, and diode DF. Thus vL = −Rio − E − VF. Stored magnetic energy supplies resistive heating, the DC load and any diode loss. Finite-time extinction is possible whenever E + VF > 0, even with ideal diodes if E > 0. Only when E = VF = 0 does ideal RL freewheeling decay without a finite-time zero.

Mean current: maximize it or smooth it?

Use an RL load with ideal diodes and no opposing voltage E. Keep Vm, f and R fixed. Without DF, compare L = 0 with larger L. Then add DF and vary L again. Predict which change increases the mean load current and which mainly reduces its ripple. Compare with the R load.

Preparation selects RL and disables E and VF; Vm, f, R, L and DF are retained.

Reveal the explanation

In this ideal circuit, the maximum mean current for fixed Vm and R is Io,dc = Vm/(πR). A purely resistive load reaches it. With L > 0 and no DF, conduction extends into the negative source half-cycle and lowers the mean output voltage and current. Adding DF removes that negative-voltage interval and restores the same maximum mean current. With DF present, increasing L smooths the current but does not raise its mean: the inductor voltage has zero mean in periodic steady state.

Beyond the ideal circuit: VF and E

From Intermediate on, the controls include the diode forward drop VF and the opposing voltage E. Change one of them at a time and compare with the ideal circuit.

VF: what changes?

Use the load-type button to select a noninductive load, then enable and disable VF. Keep Vm, f, R and E fixed. Predict the output voltage, load current and conduction interval. Compare with E disabled too.

Reveal the explanation

Without inductance, current flows only when vs > E + VF: io = max(vs − E − VF, 0)/R. During conduction, vo = vs − VF; otherwise vo = E. Increasing VF narrows the current pulse and reduces its mean/RMS values. The complete positive half-cycle is conducted only when E = VF = 0. If Vm ≤ E + VF, no current flows.

Further explanation · Extended

A finite VF produces diode conduction loss while current flows. For nonzero transferred power, converter efficiency is below 100% with VF > 0 and equals 100% with ideal diodes. When input and output powers are both zero, the ratio is undefined, not 100%. Power absorbed by E is part of the load power, not a diode loss.

Mean current with E and VF: what is attainable?

Use the load-type button if needed. For fixed Vm and R, first find the unrestricted maximum mean output current, allowing the load, DF, VF and E to vary. The unrestricted reference has no opposing voltage (E = 0) and ideal diodes. If E represents a battery voltage or motor back-EMF that must stay fixed and nonzero, keep E fixed and optimize only the remaining choices.

Reveal the explanation

First set E = 0 and use ideal diodes. If E is allowed to vary, the unrestricted maximum is Io,dc = Vm/(πR): a purely resistive load reaches it, and an inductive load with DF reaches the same value. With E = VF = 0 and DF present, increasing L reduces current ripple but does not increase its mean. If E instead represents a real battery voltage or motor back-EMF that must remain fixed at E > 0, then Vm/(πR) is not the attainable target: keep E fixed, vary only the other choices, and the maximum achievable mean current is lower because E opposes the current.

Further explanation · Extended

In general, Vo,dc = RIo,dc + E. With uninterrupted freewheeling operation, Io,dc = (Vm/π − VF − E)/R. This last expression is valid only if current remains positive throughout the period. If it would predict a negative current, or if current reaches zero at any phase, use the interval solution instead.

Battery or back-EMF: what does E change?

Enable E, then vary it while keeping the other settings fixed. Compare a noninductive and an inductive load, with and without DF. Predict the current-start threshold, the zero-current output voltage, and what happens when E + VF reaches the source peak.

Reveal the explanation

Increasing E raises the restart threshold vs > E + VF and lowers current for otherwise fixed settings. With no current, vo = E; a nonzero terminal voltage does not imply power transfer. If E + VF ≥ Vm, all currents and powers are zero in periodic steady state, while the output remains at E. With L > 0, E can cause finite-time freewheel-current extinction even when VF = 0.

The DC load absorbs pE = Eio, with mean PE = EIo,dc: idealized battery charging or electromechanical conversion, not diode loss. PE need not decrease monotonically as E increases, because voltage rises while current falls. E is held constant; state of charge, speed dynamics and a practical charging controller are not simulated.

One step further: load voltages & conductionOne step further: element currents & conduction

This plot shares the probe with the two plots above. Decompose the complete load voltage into the resistor, inductor and optional DC-source contributions.

This plot shares the probe with the two plots above. With the capacitor filter, it splits the diode current into its resistor and capacitor parts: iD1 = iR + iC.

03 / Load-element voltages

03 / Load-element currents

V

Connect the waveforms

Why can the inductor voltage be negative?

iD1 = iR + iCiR = vo/R;   iC = C dvo/dt
vo = vR + vL + EvR = Rio;   vL = L dio/dt
D1 turn-on angle α
D1 turn-off angle β
Time constant τ = L/R
Time constant τ = RC

From waveforms to power & ripple

Use the four core values above. All averages below use one complete source period.

How much power reaches the load?

What do these power values mean?

Pout is the full-period mean of output instantaneous power: Pout = ⟨voio⟩ = RIo,rms² + EIo,dc. The inductor absorbs zero average power. The RMS product is apparent power, not generally active power. PE = EIo,dc is power absorbed by the battery/back-EMF; it is not a converter loss.

Pdc = Vo,dcIo,dc is the DC-component power, including EIo,dc. The rectification ratio η = Pdc/Pout is not converter efficiency and is not a battery-charging efficiency. Converter efficiency Pout/Pin counts the complete load power, including resistive heating and the DC load.

How smooth is the output?

Read and calculate the ripple factors

Subtract the mean from the waveform to obtain its zero-mean ripple. Its RMS value, divided by the DC value, is the ripple factor. A smaller percentage means a smoother waveform relative to its mean; it is not a peak-to-peak percentage.

Vo,ac = Vo,rms2−Vo,dc2
Io,ac = Io,rms2−Io,dc2
For E = 0, η = 1/(1 + RFi²). In general, η = (RIo,dc² + EIo,dc)/(RIo,rms² + EIo,dc); RFi is a ratio, not a percentage. Voltage and current ripple factors need not be equal, even without inductance when E > 0.

More experiments: ripple

Does smooth current mean smooth voltage?

Load the smooth-current example; it retains the selected E and VF. Compare voltage/current waveforms and ripple factors above, then change f with the other settings fixed. For the familiar 121% voltage-ripple benchmark, disable both E and VF.

Reveal the explanation

Smooth current does not imply smooth voltage. With E = VF = 0 and DF, vo is the fixed half-wave sine: RFv ≈ 121.14%. Increasing ωL/R reduces current ripple. With nonzero E or VF, gaps can appear; during each gap vo = E. With uninterrupted current, vo = max(vs, 0) − VF regardless of E, while E lowers the mean current. With the other parameters fixed, multiplying f or L by the same factor has the same effect on the phase-domain waveforms.

Capacitor filter: ripple and diode-current peaks

Load the capacitor-filter example (R = 100 Ω, C = 1000 µF). Predict the voltage ripple and diode-current waveform. Keep Vm, f and the selected VF fixed. First compare C = 500, 1000 and 2000 µF at R = 100 Ω. Then restore C = 1000 µF and compare R = 100 and 200 Ω. Compare ΔVo, the D1 conduction interval and its peak current. Does reducing ripple always increase the absolute current peak?

Reveal the explanation

While D1 is off, the capacitor supplies R and vo decays with time constant RC. Increasing C with R fixed reduces the ripple and, in this comparison, increases the peak diode current. Increasing R with C fixed also reduces the ripple, but lightens the load: the absolute peak current decreases in this comparison. Do not infer the peak current from ripple alone. For ωRC ≫ 1, ΔVo ≈ (Vm − VF)/(fRC). The diode-current pulse lies around the voltage maximum, ending after it; over one period the charge supplied by D1 equals the charge drawn by R.

With small ripple, C holds vo close to Vm − VF while the source reaches −Vm. D1 then blocks almost 2Vm − VF (almost 2Vm with an ideal diode). DF does not conduct because vo never becomes negative.

One guided comparison

Predict → change → explain

Choose an inductive load with L > 0. Keep Vm, f, R, E and the diode settings fixed. Compare DF absent/present, then increase L with DF present. Check whether current remains at 180°. Predict the output and inductor voltages, mean current and both ripple factors.

Check your explanation

Without DF: D1 carries current until it reaches zero at β. Any conduction after 180° returns energy to the source and has negative output voltage. Only for E = VF = 0 is β necessarily after 180°.
With DF: current surviving at 180° transfers to it because both diode drops are equal. During freewheeling, vo = −VF and vL = −Rio − E − VF. When current extinguishes, vo = E.
Ripple factors: for E = VF = 0, adding DF lowers RFv to ≈ 121.14% and lowers RFi. With a nonzero opposing term, compare the actual values; the duration of the gaps and the voltage plateau matter. If current already reaches zero by 180°, DF never conducts and neither waveform nor ripple factor changes.

Increase L with DF present: with uninterrupted current, vo = max(vs, 0) − VF, so RFv is independent of L and Io,dc = (Vm/π − VF − E)/R. Increasing L reduces the current harmonics and RFi. With E = VF = 0, this includes all finite L > 0 and RFv ≈ 121.14%. If current extinguishes, vo = E in the gaps; changing L can then change the mean and ripple of voltage too. A positive E + VF permits finite-time extinction; a negative vL means positive current is decreasing, not reversing.

Continue to Extended for instantaneous power, energy flow and diode electrical stresses. Your circuit settings stay the same.

Energy flow & diode electrical stresses

Keep everything above. Now connect instantaneous power to stored energy, and examine the current and reverse-voltage stresses on each diode.

All five waveform plots share the same probe, time/angle axis and cycle count. Shaded bands identify the conduction intervals.

04 / Component voltages

V
Diode voltage = anode − cathode; negative means reverse bias.

05 / Instantaneous power

W

Where does the stored energy go?

ps = po = pR + pLpR = Rio²;   pL = dWL/dt

Positive ps means AC-source delivery; negative means energy return. Positive pL stores magnetic energy; negative releases it. The DC load absorbs pE = Eio ≥ 0. Output active power is Pout = ⟨po⟩ = RIo,rms² + EIo,dc, not just resistive heating.

Positive ps means AC-source delivery; with this capacitor filter it is never negative. Positive pC stores electric energy; negative pC releases it to R, including the part after the voltage maximum while D1 still conducts. Output active power is Pout = ⟨po⟩ = Vo,rms²/R because ⟨pC⟩ = 0.

At the linked probe:

Power is not energy

Power is an instantaneous rate (W); energy is its time integral (J). WL = ½Lio². In periodic steady state the inductor ends each cycle with the same energy, so ⟨pL⟩ = 0 even though it stores and releases energy within the cycle.

Returned energy integrates only ps < 0 over a source period. It is zero with DF, or without inductance. Positive E absorbs energy from the load current; it does not cause regeneration in this fixed-polarity model. Any energy returned to the AC source comes from the inductor, not from reverse load current.

Power is an instantaneous rate (W); energy is its time integral (J). WC = ½Cvo². In periodic steady state the capacitor ends each cycle with the same energy, so ⟨pC⟩ = 0 and ⟨iC⟩ = 0, even though it stores and releases energy within the cycle.

What must the diodes withstand?

Read current and reverse-voltage peaks from the selected operating waveforms. These are not component ratings.

Diode current and reverse-voltage peaks for the selected circuit
Diode Peak current Peak reverse voltage

Read voltage and current together

Diode voltage is anode minus cathode. A conducting diode has vD = VF, which is zero in the ideal model. A negative voltage indicates reverse bias; with a finite drop, an off diode can also have a small positive voltage below VF. Voltage alone does not establish conduction: check the current trace and highlighted circuit path.

The reverse-voltage value is max(−vD, 0) over a full period. Device selection also requires ratings, margins and nonideal effects not modeled here, including switching overshoot and reverse recovery. Full-period diode average and RMS currents are available in Advanced.

Diode losses & conversion efficiency

Diode conduction loss is PD = VFID,avg, using the full source period. The same power balance applies to both diode models: ideal diodes have zero loss; a finite drop gives the calculated conduction losses. Conversion efficiency compares load active power with source active power, not DC-component power. It is 100% in the ideal case with nonzero power flow, and undefined (—) when both powers are zero. Switching and reverse-recovery losses are not modeled.

One energy-flow check

Predict the power signs

Use L > 0 without DF. Locate an actual source-conduction interval after 180°; reduce E or VF if none exists. Predict the signs of ps, pR, pL, pE and diode loss. Add DF and examine its actual conduction interval. Which power vanishes and which diode conducts?

Check the energy balance

Without DF: while D1 conducts after 180°, io > 0, ps < 0, pR > 0, pL < 0 and pE ≥ 0. The balance is −pL = pR + pE + pD1 − ps. If β ≤ 180°, this interval does not exist.

With DF: during freewheeling, source current and ps are zero. The inductor supplies −pL = pR + pE + pDF. Output power po = −VFio is zero for an ideal diode and negative for a finite drop, even though the battery/back-EMF absorbs pE. If current extinguishes, all powers are zero but the output voltage is E.

Continue to Advanced for source-current harmonics, the full metric tables, equations and exports. All Extended explanations remain available there.

Advanced analysis — building on Extended

Earlier material remains available above. Continue with the full performance tables, source-current harmonics, equations and exports.

Advanced interpretation

Performance metrics

Always averaged over one complete source period, not just the conduction interval.

For loads with inductance or E, active power is ⟨voio⟩, not generally Vo,rmsIo,rms. Source-current DC is reported separately from harmonic distortion. Read the definition notes.

DC output, RMS values & power

Common alternative notation · Vo,cc ↔ Vo,dc; Vo,Ef ↔ Vo,rms

Output and power metrics

η = Pdc/Pout is the rectification ratio: the fraction of active output power associated with DC. Conversion efficiency is ηconv = Pout/Pin. R is the only dissipative element within the load; any diode losses belong to the converter. Ideal diodes have zero loss and 100% conversion efficiency for nonzero power flow.

Output ripple & waveform shape

Voltage and current ripple metrics, including the RL load

Ripple metrics

“AC” means the complete zero-mean ripple, including all its harmonics. Ripple factors are RMS ratios, not peak-to-peak ratios.

Source current & utilization

All current-distortion metrics here refer to is

Input-side metrics

THDi,DC excl.=Irms2−Idc2−I1,rms2I1,rms
THDi,DC incl.=Irms2−I1,rms2I1,rms

TUF is defined here as Pdc/(Vs,rmsIs,rms). No transformer is simulated; this number does not account for magnetic bias from the DC source current.

Conduction, devices & energy

Additional operating-stress and timing diagnostics

Conduction and device metrics

Why the source power factor is not just cos φ1

Bars show DC and harmonics 1–15. The THD metrics use the full waveform's mean square and its fundamental, not just these 15 bars.

With a purely resistive load (L = 0, E = 0) and ideal diodes, the source current has the shape of the ideal half-wave reference in the Fourier laboratory (investigation 07, «1φ · half-wave»). A nonzero diode drop changes that shape. The rectifier metrics use the complete waveform; the Fourier metrics use the active truncated series.

A nonzero half-wave source current has a nonzero mean, so the two source-current THD values differ. Both divide by the fundamental RMS; DC is excluded only from the numerator of the DC-excluded definition. With zero fundamental, THD is undefined. With zero source current, current ratios and PF are also undefined.

Definitions, equations & model scope
Metric and terminology conventions

Subscripts s and o identify the source and the complete output load; dc denotes the mean, rms the total RMS, and ac the RMS of the zero-mean ripple. Load types are R, RL, RE, RLE, and R‖C. In R‖C, the resistor and filter capacitor are in parallel; the opposing series voltage is not included. E is the effective series DC voltage (zero when disabled); vE = E, pE = Eio, PE = EIo,dc. The symbol Ereturned elsewhere denotes returned energy in joules, not the DC voltage E in volts. Unless explicitly labeled as a reference, readouts use the selected circuit.

Reference directions and model scope

vs = Vm sin θ, θ = 2πft. The amplitude is peak, not RMS. vo spans the complete series load, top minus bottom: vo = Rio + vL + E. Positive io flows down and enters the + terminal of E. DF is antiparallel to the complete load including E. Diode voltages are anode minus cathode. The ideal AC source can accept returned instantaneous power.

R > 0, L ≥ 0 and E ≥ 0. Selecting the noninductive load bypasses the saved L; enabling E changes R/RL into RE/RLE. Disabling E bypasses it without deleting its saved setting. Both diodes have the same constant VF. With fixed magnetic flux and motor speed, E approximates back-EMF and Eio electromechanical conversion power; as a battery it is an ideal constant-voltage charging model. Speed, torque dynamics, state of charge, battery chemistry, magnetic saturation, switching losses, reverse recovery and source impedance are not modeled. Negative E and battery discharge are excluded. Each edit selects periodic steady state; it is not a live switching event.

R/RL reference: exact current solution (E = VF = 0)

Reference only: E = VF = 0. For L > 0, define q = ωL/R, τ = L/R, A = (Vm/R)/(1 + q²), B = Aq. All angles below are radians. Use the general solution in the next subsection when either opposing term is nonzero.

g(θ) = A sin θ − B cos θ + B e−θ/q

Without DF: io = g(θ) for 0 ≤ θ < β and zero afterwards. β is the first nontrivial zero of g after π, found by a bracketed root solver. D1 conducts from 0 to β; vo = vs there and zero otherwise.

Vo,dc = Vm(1 − cos β)/(2π),   β = π + σ,   Io,dc = Vo,dc/R

With DF: set a = e−π/q, Iπ = B/(1 − a), and I0 = aIπ. Periodicity is imposed exactly, io(2π) = io(0), rather than approximated by simulating a finite number of start-up cycles.

io(θ) = g(θ) + I0e−θ/q   for 0 ≤ θ < π;
io(θ) = Iπe−(θ−π)/q   for π ≤ θ < 2π.

D1 conducts in the positive half-cycle. DF provides the load-current path in the negative half-cycle. Then vo = max(vs, 0), Vo,dc = Vm/π and Vo,rms = Vm/2. For finite positive R and L, the ideal freewheeling current decays exponentially without a finite-time zero; extremely small tails may round or underflow to zero. Display rounding does not define a physical conduction threshold.

For R alone: vo = max(vs, 0) and io = vo/R. Adding DF carries no load current and has no effect on these waveforms.

General RE/RLE solution: opposing E and diode drop VF

Define U = E + VF and assume U > 0 in the formulas containing 1/c below. A conducting diode has iD ≥ 0 and vD = VF; an off diode has iD = 0 and vD ≤ VF. With no current, vo = E. If U ≥ Vm, the unique periodic steady state has no current. For U = 0, use the R/RL reference above.

D1 on: vo = Vm sin θ − VF
DF on: vo = −VF
io = 0: vo = E

With L = 0 and Vm > U: io = max(Vm sin θ − U, 0)/R and vo = Rio + E. Turn-on from zero is α = asin(U/Vm); turn-off is β = π − α. Thus Io,dc = [2Vm cos α − U(π − 2α)]/(2πR). Adding DF has no effect without inductance and with this polarity of E.

For L > 0, let q = ωL/R, c = (E + VF)/R, A = (Vm/R)/(1 + q²) and B = Aq. On a source-conduction interval, L dio/dt + Rio = vs − E − VF. Starting at angle a with current Ia, the exact solution is:

F(θ) = A sin θ − B cos θ − c
io(θ) = F(θ) + [Ia − F(a)] e−(θ−a)/q

Without DF, start from zero at α = asin(U/Vm) and locate the first subsequent current zero β. With DF, any current remaining at π commutates to it because both diode drops are equal; E does not shift that commutation. During freewheeling, L dio/dt + Rio = −U. From current Ib at angle b:

io(θ) = (Ib + c)e−(θ−b)/q − c
θzero = b + q ln(1 + Ib/c)

The freewheeling interval ends at this zero or at the next source commutation, whichever comes first. Negative currents are not allowed. The solver imposes io(2π) = io(0), rather than simulating a finite number of start-up cycles. If current survives through phase zero, D1 takes over at zero, not at α. That early current can still extinguish before α, leading to a zero-current gap and restart at α; the solver retains both source intervals.

For uninterrupted current with DF, set ae = e−π/q. Then Iπ = B/(1 − ae) − c; I0 = Bae/(1 − ae) − c. This candidate is valid only if current is nonnegative throughout the period, not just at its endpoints. In this regime Vo,dc = Vm/π − VF and Io,dc = (Vm/π − VF − E)/R. A positive mean is necessary but not sufficient for uninterrupted current. Otherwise include every zero-current gap explicitly.

Capacitor filter (R‖C): exact periodic solution

With an ideal source, D1 conducts while vs − VF can supply both resistor R and capacitor C. During conduction, vo = Vm sin θ − VF, and the diode current is the sum of the resistor and capacitor currents:

iD1(θ) = (Vm sin θ − VF)/R + ωCVm cos θ

D1 turns off at β, the first angle after the voltage peak where iD1(β) = 0. From then on, only the capacitor supplies R:

vo(θ) = (Vm sin β − VF) e−(θ − β)/(ωRC)

Conduction restarts at α in the next period, when the rising source catches up with the capacitor voltage. The solver finds α numerically from:

Vm sin α − VF = (Vm sin β − VF) e−(2π + α − β)/(ωRC)

Then ΔVo = Vo,max − Vo,min = Vm(1 − sin α). For ωRC ≫ 1 the discharge is almost linear and lasts almost a whole period, which gives the usual estimate ΔVo ≈ (Vm − VF)/(fRC); it overestimates the ripple, because the real discharge is shorter and curves upwards. The capacitor current has zero mean, so ID1,avg = Io,dc. The peak diode current occurs at α, or at θ = atan(1/(ωRC)) if that angle lies inside the conduction interval. While D1 blocks, vD1 = vs − vo, which approaches −(2Vm − VF) when the ripple is small.

Power: distinguish load heating, DC-load absorption and converter loss

For a purely resistive load (L = 0 and E = 0), io = vo/R, so Io,rms = Vo,rms/R and Pout = Vo,rmsIo,rms. These shortcuts require instantaneous voltage/current proportionality. Do not use them as general rules with inductance or a nonzero DC voltage; use the average instantaneous power and show the RMS product separately:

Pout = ⟨voio⟩ = RIo,rms² + EIo,dc,   Sout = Vo,rmsIo,rms,   Pdc = Vo,dcIo,dc,   η = Pdc/Pout.

In periodic steady state, ⟨vL⟩ = ⟨pL⟩ = 0, so Vo,dc = RIo,dc + E and Pout = RIo,rms² + EIo,dc. The DC-source term is absorbed power, not loss. Instantaneously ps = pR + pL + pE + pD1 + pDF; its average gives Pin = Pout + PD1 + PDF. Setting VF = 0 removes diode losses, not the load-power term Eio, which is positive when E > 0 and current flows. With all currents zero and E > 0, vo is constant, FFv = 1 and RFv = 0; current ratios, PF, THD and power-efficiency ratios are undefined. A flat battery-voltage trace therefore does not establish useful power transfer.

Ripple and shape metrics
Vo,ac = Vo,rms2−Vo,dc2,   FFv = Vo,rms/Vo,dc,   RFv = Vo,ac/Vo,dc.
Io,ac = ⟨(io−Io,dc)2⟩,   FFi = Io,rms/Io,dc,   RFi = Io,ac/Io,dc.
Harmonics: distinguishing two current-distortion ratios

Here the analyzed current is the single-phase source current. In the equations, Irms, Idc, and I1,rms refer to its full-period RMS, signed mean, and fundamental RMS. Both ratios use the fundamental-current RMS as denominator, never total RMS. DC excluded removes the mean-current contribution only from the numerator; DC included retains it. The readouts are percentages (100 × the ratio); exports retain dimensionless ratios unless explicitly labeled %. Both are undefined when the fundamental is zero. The rectifier values use the complete source-current waveform, including capacitor charging current, independently of the number of spectrum bars displayed.

DPF = cos φ1,   kd = Is1,rms/Is,rms,   PF = Pin/(Vs,rmsIs,rms) = DPF · kd.
Source-current THD — DC included = Is,rms2−Is1,rms2/Is1,rms.
Source-current THD — DC excluded = Is,rms2−Is,dc2−Is1,rms2/Is1,rms.

φ1 is the lag of the source-current fundamental relative to the sinusoidal source voltage; it is not generally atan(ωL/R). CF = Is,pk/Is,rms. TUF is defined here as Pdc/(Vs,rmsIs,rms). No primary winding or transformer magnetization is modeled.

Device values, numerical accuracy and reference check

Diode means and RMS values use a full period. Reverse-voltage peaks include both sides of switching jumps. With no current, vD1 = vs − E; when DF is present, vDF = −E. If the circuit is off at the negative source peak, D1 blocks Vm + E. It can block less if current still flows then; use the actual waveform maxima. β is the last D1 turn-off, not necessarily the end of load current. The interval list retains any brief initial conduction, gap and restart. These peaks are operating stresses, not component ratings.

The solution is analytic on each conduction interval. Device switching instants are located explicitly. Event-split 16-point Gauss–Legendre quadrature, with additional resolution near exponential tails, evaluates current and power metrics and Fourier coefficients. Charts use a separate drawing grid with the switching events inserted. Values at an ideal switching discontinuity follow the right-hand interval. The live consistency checks below compare independently integrated and analytic quantities; displayed values are rounded.

Ideal R-load benchmark (E = VF = 0)
Ideal R-load reference — E = VF = 0; static benchmark, not the current circuit result: Vo,dc = Vm/π,   Vo,rms = Vm/2,
η = 4/π² ≈ 40.53%,   FF = π/2 ≈ 1.5708,   RF ≈ 121.14%,
TUF = 22/π² ≈ 0.2866,   PF = 1/2 ≈ 0.7071,   Source-current THD — DC excluded ≈ 43.52%,   Source-current THD — DC included = 100%.

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Continue to Intermediate to add the diode forward drop VF, the opposing voltage E and a capacitive filter with capacitance C, together with power, ripple and more experiments; then Extended for energy flow and diode electrical stresses, and Advanced for the full analysis. Your circuit settings stay the same.

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