BJT circuit with cascaded stages

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    Bjt Cascade Circuit
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SUMMARY

The discussion focuses on a BJT circuit with cascaded stages, comprising an emitter follower and two unbiassed BJTs. Key calculations include base currents (Ib1, Ib2, Ib3), emitter currents (IE1, IE2, IE3), and voltage gains (AV1, AV2, AV3), with a total voltage gain (AV total) exceeding 1000. The output voltage is calculated to be at least 5V based on the total gain. Participants emphasize the necessity of pull-down resistors for biasing and express concerns about the circuit's real-world performance.

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ac7597
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Homework Statement
Use 2N3904 NPN transistors to design an amplifier which will meet the electrical specifications. Several (three) cascaded stages will be necessary.
Amplifier Specifications at 1kHz:
Zin=47k ohms minimum
Zout=4.7k ohms maximum
Overall voltage gain(no load) | Avnl |= 1000 minimum
Output voltage swing capability: at least 5V p-p without visible distortion Power supply voltage available is Vcc=15V DC

Show schematic diagram from Multisim, including reference designators and parts values. The DC voltages from a bias simulation should appear on the schematic. Do a DC bias simulation of all nodes, and then use “Place Text” to enter the values right onto the schematic at each node.

A Multisim plot of the input and output voltages when the input to the amplifier is an AC source at 1kHz with a peak-to-peak amplitude of 5mV. The output voltage should be a clean sine wave with no visible clipping or distortion. The output voltage amplitude should be clearly shown as being at least 5V peak-to-peak, thereby proving that your gain is at least the required value of |1000|.
To start:
For the initial 2N3904 parameters values use beta=150, Vbe=0.7V, ro=50k ohms.
The resistor values should be standard values that can be brought.
Relevant Equations
AV total= AV1 * AV2 * AV3
This cascade is divided into 3 sections thus:
the leftmost section is an emitter follower :
Ib1 = (15V-0.7V)/ ( R1 + 151R2)
IE1= 151*Ib1
re1= 26mV /(IE1)
Zb1= 150( re1+ R2)
AV1= (-150*R1)/ (R1+Zb1)

the center section is an unbiassed BJT :
Ib2 = (15V-0.7V)/ ( R3 + 151(R4+Re2a) )
IE2= 151*Ib2
re2= 26mV /(IE2)
Zb2= 150( re2+ Re2a)
AV2= (-150*R3)/ (R3+Zb2)

the rightmost section is an unbiassed BJT :
Ib3 = (15V-0.7V)/ ( Rb3 + 151(Rc3+Re3a) )
IE3= 151*Ib3
re3= 26mV /(IE3)
Zb3= 150( re3+ Re3a)
AV3= (-150*Rb3)/ (Rb3+Zb3)

AV total= AV1 * AV2 * AV3 >=1000

output voltage= AV total * (0.005V) >=5V

Is this layout correct? How do I proceed further to find the resistors?
 

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What do you mean by "unbiassed BJT"? IMO you need a pullodown resistor on each base to help you set the bias points.

Can you make that correction and show us your simulation results? Thanks. :smile:
 
I must admit that I am too lazy for finding the meaning (and checking the correctness) for symbols like Zb1 and Re2a and...(without a minimum of explanation).
 
berkeman said:
IMO you need a pullodown resistor on each base to help you set the bias points.
I think the operative word there is help.
With no spread of the characteristics, Beta, VBE or resistor tolerance, I see that biasing as valid. I've actually seen a simpler version of it in some consumer products (many years ago anyhow, haven't looked recently).

Real-World, getting the required gain with that circuit could be... interesting; but certainly worth the effort.