Create a Component in NI Multisim
- Electrogeek Z
- Sep 6, 2019
- 2 min read
Updated: Sep 9, 2020
How To Create a Component in NI Multisim?
In NI Multisim may be some components or IC's as per your requirement are not present, for example if you search for IC LM386 in component search box you can't find it. But you have to simulate a circuit based on that IC; So you have to create that IC or Component in user components box.
Watch below video for step by step component creation in NI Multisim.
Part 1
Part 2
Data Code for IC LM386 -
* lm386 subcircuit model follows: ************************************original* IC pins: 2 3 7 1 8 5 6 4 * IC pins: 1 2 3 4 5 6 7 8 * | | | | | | | | .subckt lm386 g1 inn inp gnd out vs byp g8 ************************************original*.subckt lm386 inn inp byp g1 g8 out vs gnd * input emitter-follower buffers: q1 gnd inn 10011 ddpnp r1 inn gnd 50k q2 gnd inp 10012 ddpnp r2 inp gnd 50k * differential input stage, gain-setting * resistors, and internal feedback resistor: q3 10013 10011 10008 ddpnp q4 10014 10012 g1 ddpnp r3 vs byp 15k r4 byp 10008 15k r5 10008 g8 150 r6 g8 g1 1.35k r7 g1 out 15k * input stage current mirror: q5 10013 10013 gnd ddnpn q6 10014 10013 gnd ddnpn * voltage gain stage & rolloff cap: q7 10017 10014 gnd ddnpn c1 10014 10017 15pf * current mirror source for gain stage: i1 10002 vs dc 5m q8 10004 10002 vs ddpnp q9 10002 10002 vs ddpnp * Sziklai-connected push-pull output stage: q10 10018 10017 out ddpnp q11 10004 10004 10009 ddnpn 100 q12 10009 10009 10017 ddnpn 100 q13 vs 10004 out ddnpn 100 q14 out 10018 gnd ddnpn 100 * generic transistor models generated * with MicroSim's PARTs utility, using * default parameters except Bf: .model ddnpn NPN(Is=10f Xti=3 Eg=1.11 Vaf=100 + Bf=400 Ise=0 Ne=1.5 Ikf=0 Nk=.5 Xtb=1.5 Var=100 + Br=1 Isc=0 Nc=2 Ikr=0 Rc=0 Cjc=2p Mjc=.3333 + Vjc=.75 Fc=.5 Cje=5p Mje=.3333 Vje=.75 Tr=10n + Tf=1n Itf=1 Xtf=0 Vtf=10) .model ddpnp PNP(Is=10f Xti=3 Eg=1.11 Vaf=100 + Bf=200 Ise=0 Ne=1.5 Ikf=0 Nk=.5 Xtb=1.5 Var=100 + Br=1 Isc=0 Nc=2 Ikr=0 Rc=0 Cjc=2p Mjc=.3333 + Vjc=.75 Fc=.5 Cje=5p Mje=.3333 Vje=.75 Tr=10n + Tf=1n Itf=1 Xtf=0 Vtf=10) .ends *----------end of subcircuit model-----------
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