Problem with schema on NI Multisim

Thread Starter

rio7

Joined Jan 1, 2015
40
Good morning, I have schema ( picture ), but I don't know how to created it on NI Multisim ( I do it, but this schema looks like... You can see in my attached files ).
I must have and component who can show me the oscilograms. How should I do it ?
And one more problem: I don't find component - VT2, VT3 and VT4 ( in picture ) in NI Mulstisim ( 13 ) program.



Thank You!
 

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JoeJester

Joined Apr 26, 2005
4,390
What is the source of your schematic picture? It looks like you drew it in mspaint. Are you not showing all the information available to you?

VTx where x is a number is the component designation. Multisim identifies transistors differently. Cx would be capacitor, Rx would be resistor, Qx or Tx or in the case of your picture, VTx is transistors. Lx is inductors. Of course the V could mean Virtual ... or perfect. VD1 is a diode.

You can insert some NPN and PNP transistors as appropriate to make the schematic in multisim.
 
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Thread Starter

rio7

Joined Jan 1, 2015
40
What is the source of your schematic picture? It looks like you drew it in mspaint. Are you not showing all the information available to you?

VTx where x is a number is the component designation. Multisim identifies transistors differently. Cx would be capacitor, Rx would be resistor, Qx or Tx or in the case of your picture, VTx is transistors. Lx is inductors. Of course the V could mean Virtual ... or perfect. VD1 is a diode.

You can insert some NPN and PNP transistors as appropriate to make the schematic in multisim.

Yea, I drew it on windows paint. I have just these numbers of it:

P(out)= 2W
R(load/strain/workload) = 12Ω
Uin = 20mV
F = 300 Hz to 7 kHz
Rin = 2 kΩ
M <= 3dB
And I have created audio amplifier.

I attachement better schema of it.

But I have problem:

I add 20mV ( I try and 20, 100 V ) in AC voltage, but osicilograms showing something wrong.. I attachement some picture of it.
What can be problem at ?
 

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ScottWang

Joined Aug 23, 2012
7,400
The value of resistors quite strange, you could change the position first,.
1. Cut off the connection from R5 and C4.
2. Connecting R5 to GND.
3. Connecting C4 to the e of Q1.
And try again, when it can do the amplifying then you can adjust the value of resistors.
 

Thread Starter

rio7

Joined Jan 1, 2015
40
The value of resistors quite strange, you could change the position first,.
1. Cut off the connection from R5 and C4.
2. Connecting R5 to GND.
3. Connecting C4 to the e of Q1.
And try again, when it can do the amplifying then you can adjust the value of resistors.
I do it. I attachment picture. But I see what C4 ir going/flashing to red color.
And oscilograms showing still not really sinusoid.
P.S. What wrong with resistors position ?
 

Attachments

ScottWang

Joined Aug 23, 2012
7,400
The negative of C4 and the c of Q2, R4 and 0V of input signal, they must be connecting to GND, I 'm sure why you disconnected them?

And set the Vin max to 2V.
 
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Jony130

Joined Feb 17, 2009
5,488
The problem of R5 and C5 as I mentioned on #9, do you think is it ok?
R5 and C5 should be connected in series.
Because this circuit should work as a poor's man op amp. And Q1 is a differential amplifier (base a "+" input and emitter as a"-" input) and the closed loop gain is equally to
Av =(1+ R6/R5) ≈ R6/R5

Here you have a very similar circuit

 

ScottWang

Joined Aug 23, 2012
7,400
R5 and C5 should be connected in series.
Because this circuit should work as a poor's man op amp. And Q1 is a differential amplifier (base a "+" input and emitter as a"-" input) and the closed loop gain is equally to
Av =(1+ R6/R5) ≈ R6/R5

Here you have a very similar circuit

No, they are different, in your new circuit that the R4 and C3 is in parallel with the Vce, and the c of 3906 has 8.2k to provide the dc quiescent current to ground, the R3(2.7K) provide the feedback current.

The circuit of bertus on #15, R6 is the feedback current, R5 provide the dc quiescent current to ground, you can't using the feedback resistor to replace the e resistor, if you do that then the e will loss the dc quiescent current.
 
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