difference in betaac with Spice

C

Candide Voltaire

Guest
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
 
"Candide Voltaire" <candideguevara@gmail.com> schrieb im Newsbeitrag
news:a3703f4a-7fdd-4a2b-964c-c5bdd521897c@x10g2000yqj.googlegroups.com...
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
Click to expand...
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
 
On May 29, 9:31 pm, "Helmut Sennewald" <helmutsennew...@t-online.de>
wrote:
"Candide Voltaire" <candideguev...@gmail.com> schrieb im Newsbeitragnews:a3703f4a-7fdd-4a2b-964c-c5bdd521897c@x10g2000yqj.googlegroups.com...



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

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
Click to expand...
I checked this out as follows:
R_{CE}=VAF/IC=50V/4.377mA=11423.349 ohm
i_{ce}=v_c/11423.349=24.05/11423.349=2.1053371mA
betac=(2.1053371e-3+2.405e-2)/7.877e-05=332.05 which is indeed almost
332.486 as displayed in the operating point information

thank you very much Helmut for sorting this out

candide
 
On 29 mei, 21:31, "Helmut Sennewald" <helmutsennew...@t-online.de>
wrote:
"Candide Voltaire" <candideguev...@gmail.com> schrieb im Newsbeitragnews:a3703f4a-7fdd-4a2b-964c-c5bdd521897c@x10g2000yqj.googlegroups.com...



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

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
Click to expand...
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
 
On 29 mei, 21:31, "Helmut Sennewald" <helmutsennew...@t-online.de>
wrote:
"Candide Voltaire" <candideguev...@gmail.com> schrieb im Newsbeitragnews:a3703f4a-7fdd-4a2b-964c-c5bdd521897c@x10g2000yqj.googlegroups.com...



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

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
Click to expand...
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
 
On 30 mei, 20:54, Candide Voltaire <candideguev...@gmail.com> wrote:
On 29 mei, 21:31, "Helmut Sennewald" <helmutsennew...@t-online.de
wrote:



"Candide Voltaire" <candideguev...@gmail.com> schrieb im Newsbeitragnews:a3703f4a-7fdd-4a2b-964c-c5bdd521897c@x10g2000yqj.googlegroups.com...

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

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

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
Click to expand...
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°
 
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