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5_6_4_Einweggleichrichter

FE

feslehrer

·

Created Nov 9, 2026

·

Updated Nov 9, 2026

Schematic preview for 5_6_4_Einweggleichrichter
Details

Components

4


Schematic

Yes


Simulation

Transient analysis


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Circuit notes

Diode im Gleichstromkreis


Die Halbleiterdiode besitzt 2 Anschlüsse, Anode (A) und Katode (K). Zunächst ist die Anode an den Pluspol der Spannungsquelle angeschlossen.

Arbeitsauftrag

  1. Starten Sie die interaktive Simulation.
    Beobachten Sie E1.
  2. Drehen Sie die Diode R1 im Stromkreis, so dass die Kathode Richtung Puspol und die Kathode zur Lampe E1 zeigt.
    Führen Sie die Simulation erneut aus.
  3. Beschreiben Sie das Verhalten der Diode in der Schaltung.

Components used

Components used
4 components
NameManufacturerMPNCountType

Diode

Generic

—

1

Official

Resistor

Generic

—

1

Official

RMS AC Voltage Source

Generic

—

1

Official

SPICE netlist

* **************************************************
* Generated from CircuitSim Schematic
**************************************************
* net label L1 joins NET_005::component:node-7216812e-2923-46f4-a103-897d7f879240:T0 NET_005::node:junction-e5a710e4-2bfe-4eef-be00-f4417aa21645
* net label N joins NET_006::node:junction-0376c36a-d530-4e5d-b886-284d946fefeb NET_006::node:junction-e6f9e1ee-0694-4c09-9a7f-7360576e1783
X_R1 N4 N3 DIODE_25d079d4 IS=1e-14 RS=0 N=1 BV=1e30 IBV=100p CJO=0 VJ=1 M=0.5 TT=0 EG=1.11 XTI=3 KF=0 AF=1 FC=0.5 ISR=0 NR=2 IKF=1e30 TNOM=27 IC=0 TEMP=27
R_RL N3 0 1K Resistor_1d9c8cf9a__R_RL
X_T1 N5 0 N4 0 Param1P1S_e1d792b5 PTurns=2.147K PNegativePolarity=0 STurns=112 SNegativePolarity=0
V_Unetz N5 0 dc 0 ac 1 0 distof1 0 0 distof2 0 0 sin ( 0 {230*1.414213562} 50 0 0 0 )
E_VP1 VP1 0 N5 0 1
E_VP2 VP2 0 N4 0 1
E_VP3 VP3 0 N4 N3 1
E_VP4 VP4 0 N3 0 1

**************************************************
* Models and supporting definitions follow
**************************************************
* Source-compatible two-terminal junction diode
.subckt DIODE_25d079d4 A K params: IS=1e-14 RS=0 N=1 BV=1e30 IBV=100p CJO=0 VJ=1 M=0.5 TT=0 EG=1.11 XTI=3 KF=0 AF=1 FC=0.5 ISR=0 NR=2 IKF=1e30 TNOM=27 IC=0 TEMP=27
    .model DIODE_25d079d4_CORE D(IS={IS} RS={RS} N={N} BV={BV} IBV={IBV} CJO={CJO} VJ={VJ} M={M} TT={TT} EG={EG} XTI={XTI} KF={KF} AF={AF} FC={FC} ISR={ISR} NR={NR} IKF={IKF} TNOM={TNOM})
    DCORE A K DIODE_25d079d4_CORE IC={IC} TEMP={TEMP}
.ends DIODE_25d079d4
.model Resistor_1d9c8cf9a__R_RL r(tc1=0 tc2=0 tnom=27)
.subckt Param1P1S_e1d792b5 p1pos p1neg s1pos s1neg params: PTurns=10 PNegativePolarity=0 STurns=1 SNegativePolarity=0
    .param PTurnsEffective = {PTurns * ((PNegativePolarity < 0.5) ? 1 : -1)}

    .param STurnsEffective = {STurns * ((SNegativePolarity < 0.5) ? 1 : -1)}

    G1 p1pos p1neg value={-1/PTurnsEffective*(STurnsEffective*I(Es1))}
    Es1 s1pos s1neg value={V(p1pos,p1neg)*STurnsEffective/PTurnsEffective}
.ENDS

**************************************************
* Simulation settings
**************************************************
.tran 400u 0.03 0 400u
.save V(VP2) V(VP4) V(VP3) V(VP1)
.end

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