Multi-Stage Amplifier Project

Thread Starter

bigyanengi

Joined Feb 24, 2024
37
I need the following design in LTspice,can someone help?
Please design the circuit in LT spice and show me the waveform. Also clearly indicate each steps.
I need visual LT spice design.

Multi-Stage Amplifier Project

This Assignment aims at verifying and expanding, with design, simulations and measurements, your
creativity and your knowledge and understanding of the differential amplifier circuit.
This is a Project: you must design and build your circuit starting from specifications and constraints.
Please document each step with snapshots, pictures, and your observations. Wherever possible please
include the date and time field and the AD S/N. Please include this page.

1) Using the simulator, design and simulate a multi-stage amplifier using the following specifications and
constraints.
• Use NPN and PNP BJTs (recommended 2N3904, 2N3906)
• Use three supplies: +5V, -5V, 0V
• Use BJTs only in each stage, including loads and current sources (no resistors allowed) except for
current biasing circuits (i.e. use resistors and BJTs only to generate the bias voltages for the
loads and current sources of the amplifier)
• First stage: differential with current < 5mA
• Second stage: common-base to realize, with the output of the first stage, a cascode stage with
current < 1mA
• Third stage: emitter follower with current < 10mA
• Gain > 5,000
• Gain-bandwidth product > 10MHz
• Single dominant pole (compensated if needed for stability)

2) Using the simulator, design and simulate a non-inverting amplifier with gain ~10 by applying a
negative feedback network to the multi-stage amplifier developed in task 1 (i.e. used as operational
amplifier)
a) Simulate the response to a 100mV, 10kHz sinusoidal input, simulate the transfer function and
calculate the -3dB
b) Apply and offset to the sinusoidal input in order to have an average ~0V at the output; explain
why the offset is needed.
 

Thread Starter

bigyanengi

Joined Feb 24, 2024
37
Hi big,
Welcome to AAC,
As this is homework you have to post your best attempt at answering, we can then help.
E
Thank you for the support. I have been really on this and it is my deficiencies. I would really appreciate the help to increase my knowledge base. Once again, thank you.
 

Thread Starter

bigyanengi

Joined Feb 24, 2024
37
Hi big,
Check the 1st current source.
How is the current through the BJT pair is set.??
E
Hi eric,

If Q3's collector is directly connected to Q1's collector, Q3 could be serving a dual role—both as part of the load for Q1 and potentially as a way to influence or control the current through Q1 (and by extension, Q2, due to the differential pair's nature).In this configuration, the current through Q1 and Q2 would still fundamentally depend on the tail current source.
  • Q3 is being used to create a specific voltage or current condition at Q1's collector, which could influence the operation of Q1 and Q2.
  • If another current source or mechanism sets the tail current for Q1 and Q2, Q3's role might be more about shaping the output characteristics or forming part of a feedback or regulation loop.
 
Last edited:

ericgibbs

Joined Jan 29, 2010
18,854
Hi,
The quiescent current through the 1st long-tailed pair is approx 438mA, that cannot be correct.
Check out this video link, it may help, ask if you have a problem.
E
 

Ian0

Joined Aug 7, 2020
9,826
I would suggest simplifying the output stage by replacing Q6 and Q7 and the associated components with a simple common emitter stage.

When you set the constant current through the longtailed pair use the same circuit for Q8.
 

Thread Starter

bigyanengi

Joined Feb 24, 2024
37
I would suggest simplifying the output stage by replacing Q6 and Q7 and the associated components with a simple common emitter stage.

When you set the constant current through the longtailed pair use the same circuit for Q8.
Hi Ian,
Q6 is the common base cascade. So, if I replace that with emitter follower, will that help me achieve my <1mA specification?
 

Thread Starter

bigyanengi

Joined Feb 24, 2024
37
Hi,
The quiescent current through the 1st long-tailed pair is approx 438mA, that cannot be correct.
Check out this video link, it may help, ask if you have a problem.
E
Hello Eric,
Thank you so much. I am going to watch this video now and come back to you if I have more question. Just a question, the second stage is configured as expected based on the assignment constraints?
 

Ian0

Joined Aug 7, 2020
9,826
Hi Ian,
Q6 is the common base cascade. So, if I replace that with emitter follower, will that help me achieve my <1mA specification?
Cascode is a bit complicated. Get it working with a common emitter stage (not emitter follower - that's a common collector stage).
When you get a common emitter stage working, you can add a cascode if you think it will improve things (it will, but only marginally)
Get the constant current sources working first. https://en.wikipedia.org/wiki/Current_source
see figures 4,5,6
 

Thread Starter

bigyanengi

Joined Feb 24, 2024
37
Hi,
I would suggest you set the current by first setting V1 and V2 to 0mV.

Post the value of your new Emitter resistor.
E
Hi Eric,
I watched the video. I did made the modification to my circuit to add the resistor to the Q5 emitter to make a current source. I also set the voltage to 0 as you proposed.


  1. Voltage Divider Equation:
VBQ5=V×RB/(RA+RB)

Where VV is the supply voltage (+5V).

  1. Choosing VBQ5VBQ5:
For simplicity, l want to set VBQ5 just enough to ensure Q5 is in active mode and controls the current adequately, say around 0.7V to 1V above the emitter voltage to account for VBE and provide some margin. Without an explicit emitter voltage target, aiming for a mid-range base voltage might be practical, such as 2V in a +5V system, considering the divider's impact and potential voltage drops.

  1. Selecting Divider Ratio:
Given VBQ5=2V, solving the voltage divider equation for the ratio:

RB/(RA+RB)=2V5V=0.4RA+RBRB=5V2V=0.4

  1. Choosing Current Through the Divider:
To minimize power consumption while ensuring stable biasing, choose a relatively low current for the divider, such as 4mA

  1. Calculating Total Resistance and Individual Resistors:
Total resistance for 1mA at 5V:

RTotal=VI=5V/4mA=1.25kΩ

Given the ratio of RB to RTotal

RB=RTotal×0.4=1.25kΩ×0.4=0.5kΩ

RA=RTotal−RB=1.25kΩ−0.5kΩ=0.75kΩ

Furthermore,

The tail current is set by Rtail in series with Q5, and it's essential to ensure this current is less than 5mA. Assuming Q5 is a transistor acting as a constant current source for the differential pair:

  1. Determine the Voltage Across Rtail:
    • Assuming Q5 is an NPN transistor and the supply voltage (VCC) is +5V, the voltage across Rtail would roughly be VCC−VBE where VBE is typically about 0.7V for silicon BJTs in the active region. Thus, the voltage across Rtail would be approximately 5V−0.7V=4.3V
  2. Calculate Rtail:
    • For a maximum tail current of 5mA, use Ohm's law (V=IRV=IR) to find Rtail:
    • Rtail=V/I=4.3V/5mA=860Ω
    • This calculation provides the resistance needed to ensure a tail current of exactly 5mA. To guarantee it stays under 5mA,
1708807164652.png
 

Thread Starter

bigyanengi

Joined Feb 24, 2024
37
Hi Eric,
I watched the video. I did made the modification to my circuit to add the resistor to the Q5 emitter to make a current source. I also set the voltage to 0 as you proposed.


  1. Voltage Divider Equation:
VBQ5=V×RB/(RA+RB)

Where VV is the supply voltage (+5V).

  1. Choosing VBQ5VBQ5:
For simplicity, l want to set VBQ5 just enough to ensure Q5 is in active mode and controls the current adequately, say around 0.7V to 1V above the emitter voltage to account for VBE and provide some margin. Without an explicit emitter voltage target, aiming for a mid-range base voltage might be practical, such as 2V in a +5V system, considering the divider's impact and potential voltage drops.

  1. Selecting Divider Ratio:
Given VBQ5=2V, solving the voltage divider equation for the ratio:

RB/(RA+RB)=2V5V=0.4RA+RBRB=5V2V=0.4

  1. Choosing Current Through the Divider:
To minimize power consumption while ensuring stable biasing, choose a relatively low current for the divider, such as 4mA

  1. Calculating Total Resistance and Individual Resistors:
Total resistance for 1mA at 5V:

RTotal=VI=5V/4mA=1.25kΩ

Given the ratio of RB to RTotal

RB=RTotal×0.4=1.25kΩ×0.4=0.5kΩ

RA=RTotal−RB=1.25kΩ−0.5kΩ=0.75kΩ

Furthermore,

The tail current is set by Rtail in series with Q5, and it's essential to ensure this current is less than 5mA. Assuming Q5 is a transistor acting as a constant current source for the differential pair:

  1. Determine the Voltage Across Rtail:
    • Assuming Q5 is an NPN transistor and the supply voltage (VCC) is +5V, the voltage across Rtail would roughly be VCC−VBE where VBE is typically about 0.7V for silicon BJTs in the active region. Thus, the voltage across Rtail would be approximately 5V−0.7V=4.3V
  2. Calculate Rtail:
    • For a maximum tail current of 5mA, use Ohm's law (V=IRV=IR) to find Rtail:
    • Rtail=V/I=4.3V/5mA=860Ω
    • This calculation provides the resistance needed to ensure a tail current of exactly 5mA. To guarantee it stays under 5mA,
View attachment 316090
Hi Eric,
I set the input voltages back to desired values and I see the Ic for Q5 is no where to what I want. I do need the current load from Q3 and Q4 since resistor are not allowed in the signal path. What is missing now?1708807538224.png
 

Ian0

Joined Aug 7, 2020
9,826
The lower end of Rb and R1 should go to the negative supply, not ground. The top of Ra should go to ground.
How did you calculate R1? Isn't it supposed to have 5mA through it?
Then do the same for Q8, to set the output current.
 
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