# [Design] Two Stage Amplifier

Discussion in 'Homework Help' started by jegues, Apr 19, 2011.

1. ### jegues Thread Starter Well-Known Member

Sep 13, 2010
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Hello everyone.

I'm trying to design the following two stage amplifier to meet a given set of criterion for numerous different values throughout the circuit.

I'm little overwhelmed with the task and I'm not entirely sure where to start.

Can someone give me a better idea of what exactly I'm looking at, and the right approach to attack the problem?

Is there a name for this specific type of amplifier?

Where should I start?

According to our professor, it is pretty hard to meet all the constraints within the given ranges so we were told that we can "loosen up" the criterion a bit if need be.

Thanks again!

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2. ### Jony130 AAC Fanatic!

Feb 17, 2009
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I don't think that there is only one "right way" to solve this type of problem.
And maybe you start form CC amplifier.
To get Rout < 50Ω
50Ω < (re + Rc/(β+1))
re = 26mV/Ic
And do you know the β (Hfe) for T1 and T2 ?

3. ### jegues Thread Starter Well-Known Member

Sep 13, 2010
735
45

What is a "CC amplifier"? As I've said, we haven't been given a formal explanation of what we're dealing with here, so I need some details/info before we dive into the analysis.

Dec 26, 2010
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CC = Common Collector = Emitter Follower, as used here in the second stage.

The input and output impedances constrain the design pretty severely, (not altogether a bad thing, as it helps to define circuit values). For instance, the base bias potential divider must have more than 25kΩ equivalent AC resistance, so that the total impedance will still be more than 25kΩ allowing for the transistor input impedance.

This is a pretty severe requirement: you can get a good idea of how difficult it is by working back from the output impedance (assume a reasonable value for T2 Ie: you have been told the dissipation).

You can get an upper bound on the first stage collector load RC in terms of T2 β, and from this use the voltage gain to get (re + RE1). The required input impedance will give a minimum requirement for T1 β. I think you will find that both transistors will need to have quite large β for this circuit to be feasible. You probably won't be able to be very generous with the base bias chain current either, even allowing for a pretty big β.

5. ### t_n_k AAC Fanatic!

Mar 6, 2009
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I've run several simulations using BC549 and BC559 transistors. These allowed me to achieve the design goals using a 9V DC supply.

It's quite a challenge - particularly for jegues who appears not to have had much experience with this sort of design.

6. ### jegues Thread Starter Well-Known Member

Sep 13, 2010
735
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When you talk about a base bias potential divider, where is this on the circuit? Are you talking about Rc and RB1 at the upper portion of the first transistor? You are talking about the bounds on the input resistance here, correct?

What are you reffering to when you talk about "the base bias chain current"? I'm still quite new to most of this stuff so I'm still pretty confused.

So should my Vcc be around 9V as well?

I'm still pretty confused as to how I go about preforming the design and manage all my constraints at the same time.

How should I start my design process? (i.e. what equations should I be calculating first? Input resistance?)

I think if I can get things moving in the right direction, I will have a better idea about how to proceed.

Is there any valid assumptions I can make about any of the voltages/currents in this circuit?

Thanks again!

7. ### Jony130 AAC Fanatic!

Feb 17, 2009
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RB1 and RB2 is a voltage divider (base bias potential divider)

Typically when we design Common Emitter amplifier we choose current that is flow through RB1 and RB2 ten times the base current Id = 10*Ib.
But your circuit must provide relatively high input inpedance.
So to maximize Rin you need use high values of RB1 and RB2 (low current divider current).
Because the base bias voltage divider must have more than 25kΩ equivalent resistance RB1||RB2

For 9V
Ic1+Ic2 = 25mW/9V = 2.7mA
For Vcc = 12V
Ic1+ Ic2 = 15mW/12V = 2mA
So it is up to designer choose

Firs of all do you understand how this two amplifiers work separately ??

8. ### jegues Thread Starter Well-Known Member

Sep 13, 2010
735
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Thank you for the reply Jony130, that cleared up alot of my confusion!

I can honestly admit that I don't have a comprehensive understanding of how these two amplifiers work. However, I am in the process of trying to learn them both and I am following the theory in my textbook very closely.

Can you recommend where I should start? If I know what the two stages of the amplifier are called, I can look them up in textbook and get an idea of their configuration and how it works.

I'm still lost as to where to start my analysis though.

Can you nudge in the right direction so I can attempt to generate some equations/analysis and post my results for you?

Thanks again!

Feb 17, 2009
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10. ### jegues Thread Starter Well-Known Member

Sep 13, 2010
735
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Here's my first crack at the common collector stage. Am I okay in reducing the leftmost portion of the circuit into its thevenin equivalent?(I wasn't sure how to deal with Vin connected at the center node)

Also, can I ignore the presence of RE2? (The capacitor functions as a short circuit, correct?)

I think there's some mistakes in my work but I'd love to know what I did wrong so I can fix it and give this thing a better shot!

Thanks again!

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11. ### Audioguru Expert

Dec 20, 2007
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Since your professor is not teaching you anything then why is he being paid?

12. ### jegues Thread Starter Well-Known Member

Sep 13, 2010
735
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Good question.

His response was that there is something called "active learning" and that's what we should be doing. I'm doing my best with what I got at the moment.

Can you help me understand what I'm doing wrong or what I'm misunderstanding?

13. ### Audioguru Expert

Dec 20, 2007
10,241
1,118
It is difficult (and boring) for me to teach the fundamental basics to you since you didn't know the function of the first two resistors in the circuit that are the "base bias potential divider" and you didn't know what is called "a CC amplifier".

I think first you should be taught the basics about how a transistor amplifier stage works before you memorize some equations. The basics are probably written in an E-Book on this site.

It is too bad that your professor doesn't teach you the basics. Maybe he should get a new career selling cars or something.

14. ### jegues Thread Starter Well-Known Member

Sep 13, 2010
735
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Oh one piece of information that I forget to mention is that we are also given the following,

$V_{cc} = 9V, \quad C_{c1} = C_{c2} = 100nF, \quad C_{b} = 10uF$

15. ### jegues Thread Starter Well-Known Member

Sep 13, 2010
735
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Giving it another crack, since we now know Vcc, we can determine total current from Vcc.

$I_{C} + I_{B1} + I_{E3} = \frac{25mW}{9V} = 2.78mA$

Now if we were to apply the rule of thumb for H biasing we should obtain,

$V_{o} = 3V$

Now I choose a value for my IE3 current and RE3 resistance such that this criteria is met.

However, how do I go about determining the IE3 current and/or resistance should be? I know they both better be giving me 3V at the output, but how do I deicde how much of 2.78mA to allocate to this portion of the circuit?

If I can get this figured out, I know T2 should be in active mode and thus the voltage drop from emitter to base should be 0.7V, this gives me VB2=VC1.

Since I now know VC1, I can obtain IC and thus have obtained IB1 by KCL.

With IB1 in mind, we being to attack the idea of appropriately selecting RB1 and RB2 such that they meet input resistance requirements.

Thanks again!

16. ### Jony130 AAC Fanatic!

Feb 17, 2009
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You can get IE3 from Rout

$R{out} = re + \frac{Rc}{\beta +1} || R{e3}\approx re$

$re = \frac{26mV}{Ie2}$

So if you want Rout<50Ω Ie2 must be larger then

$Ie2 \ge\frac{26mV}{R{out}} = \frac{26mV}{50\Omega}= 520\mu A$

17. ### jegues Thread Starter Well-Known Member

Sep 13, 2010
735
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Is that a typo? Do you mean Ie3 here?

18. ### Jony130 AAC Fanatic!

Feb 17, 2009
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In your diagram we have two BJT first NPN label T1 and the PNP label T2.
So emitter current of a T2 transistor is Ie2.

19. ### jegues Thread Starter Well-Known Member

Sep 13, 2010
735
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Oh okay thats much more clear, I'll do the same.

20. ### jegues Thread Starter Well-Known Member

Sep 13, 2010
735
45
Now that we have IE2, we can go backwards and obtain an equation for VC1.

$V_{out} - V_{C1} = 0.7, \quad V_{C1} = 2.3V$

$I_{C1} = \frac{V_{CC} - V_{C1}}{Rc} = \frac{6.7V}{Rc}$

Is there another equation we can incoporate Ic and/or Rc so we can establish these two values?