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difference in betaac with Spice

When simulating the Spice-file below (a common emittor amplifier), I noticed that betaac determined by means of the voltage sources vmeasc and vmeasb differs from betaac which is shown in the operating point information: betaac=i(vmeasc)/i(vmeasb)=2.405e-2/7877e-05=305.319 whereas betaac from the operating point information = 332.486

can anyone here explain this?

CEC-amplifier VCC 5 0 DC 10 RB 5 3 680K RC 5 6 1K RG 1 2 10K Q1 7 4 0 BC547B C1 2 3 100u VIN 1 0 AC 1 VMEASB 3 4 DC 0 VMEASC 6 7 DC 0 .AC lin 1 1k 1k .PRINT AC i(vmeasb) i(vmeasc) .PLOT AC i(vmeasb) i(vmeasc) .op

*DATABOOK PHILIPS .MODEL BC547B NPN (BF=530 NE=1.3 ISE=9.72F IKF=80M IS=20F VAF=50V
  • BR=10 NC=2 ISC=47P IKR=12M VAR=10
  • RB=280 RE=1 RC=40 TR=.3U
  • CJE=12P VJE=.48 MJE=0.5 CJC=6P VJC=.7 MJC=.33 TF=.5N) .END
Reply to
Candide Voltaire
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"Candide Voltaire" schrieb im Newsbeitrag news: snipped-for-privacy@x10g2000yqj.googlegroups.com...

Hello,

The internal collector current will be divided between the external Rc and and the internal Rce.

Rce = VAF/Ic

If you take this into account, the values are OK. You can check this if you omit VAF in the model or rise it to an extremely high value, e.g. to 5000.

Best regards, Helmut

Reply to
Helmut Sennewald

snipped-for-privacy@x10g2000yqj.googlegroups.com...

VAF=3D50V

TF=3D.5N)

d
y

I checked this out as follows: R_{CE}=3DVAF/IC=3D50V/4.377mA=3D11423.349 ohm i_{ce}=3Dv_c/11423.349=3D24.05/11423.349=3D2.1053371mA betac=3D(2.1053371e-3+2.405e-2)/7.877e-05=3D332.05 which is indeed almost

332.486 as displayed in the operating point information

thank you very much Helmut for sorting this out

candide

Reply to
Candide Voltaire

snipped-for-privacy@x10g2000yqj.googlegroups.com...

VAF=3D50V

TF=3D.5N)

d
y

Dear Helmut, Can you agree with me that betaac as shown in the operating point information is in fact not the real betaac as seen from the transistor- terminals because it does NOT take into account R_{CE}?

best regards, candide

Reply to
Candide Voltaire

snipped-for-privacy@x10g2000yqj.googlegroups.com...

VAF=3D50V

TF=3D.5N)

d
y

Dear Helmut, Even though I now have a correct numerical result, it's not yet clear to me why i(vmeasc) does not "see" the extra current which flows through Rce Can you explain that too?

best regards, Candide

Reply to
Candide Voltaire

s: snipped-for-privacy@x10g2000yqj.googlegroups.com...

F VAF=3D50V

3 TF=3D.5N)

and

ely

I figured it out myself, it's simply because the difference between ic and i_rce as complex quantities results in an addition of their moduli as ib and u_ce have a phase difference of 180=B0

Reply to
Candide Voltaire

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