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voltage gain and I/O resistance for ideal amplifier

s55

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Hi, let me try to ask a stupid question.
I'm a beginner for analog circuit design, and I'm reading a following book written by Prof. Sedra.

https://www.amazon.com/Microelectronic-Circuits-Electrical-Computer-Engineering/dp/0195323033

Ideal voltage amplifier model has an infinite value of input resistance (Rin) and a zero output resistance(Rout).
I understood that entire input voltage should be dropped across Rin, and its output voltage needs to be dropped to a load resistance, without dropped voltage in Rout. Please see the figure only, without a question in the following link page.

https://www.chegg.com/homework-help...ble-11-r-5-k2-r-50-k-s2-g-10-ma-v-a-q39955434

How about a voltage gain?
From my derivation,
voltage gain = Vout / Vin = (Iout X Rout) / (Iin X Rin) = (current gain) * (Rout/Rin)
voltage gain will be 0 or very small value, when Rin=infinite (or large value) and Rout=0 (or very small value).

I believe I'm wrong, but I'm not sure what should be corrected. Could you please correct me?
 

KlausST

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Hi,

Generally: R_out is not the same as R_load! Thus don't calculate with "V_out/R_out" at all.
In some cases it's possible to use V_out/V_load = I_out = I_load.

If you want to build a voltage amplifier, then calculate only with voltages.
Don't use I_in (which should be zero at an ideal amplifier) and don't use R_out (gain should be independet of R_out as well as independent from I_out on an ideal amplifier)

Even worse there usually is no R_load but Z_load. --> I_load may depend on frequency.

Klaus
 

FvM

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Your derivation is wrong, Vout ≠Iout X Rout. Current Gain is infinite for an ideal voltage amplifier, still very high for a real amplifier and not useful for calculation of voltage gain.

Voltage is a design parameter of the amplifier that can have different values.
 

s55

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Thanks for the answers.
Actually, I had this question by taking an example from the following book written by Dr. Razavi.
https://www.amazon.com/Fundamentals-Microelectronics-Behzad-Razavi-2008-01-28/dp/B01FKSXI32

In p211, there is a simple example as follws:
----------------
A CE stage must chieve an input impedance of Rin and an output impedance of Rout. What is the voltage gain of the circuit?
answer: since Rin=r_pi = β/gm and Rout = RC, we have
Av = - gm RC
=- β Rout/Rin
----------------

From the answer, β is a current gain, and so voltage gain = (current gain) X Rout / Rin.
Could you explain this answer by relating your previous answers? That would be very helpful for my understanding.
 

KlausST

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Hi,

your first question is about an "ideal amplifier"
your other question is about a simple bjt amplifier .. which surely is far from an "ideal amplifier".

It´s hard to compare both. They differ in a lot of parameters. (e.g. zero output impedance vs high output impedance)

Klaus
 

s55

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They differ in a lot of parameters. (e.g. zero output impedance vs high output impedance)
Thanks for your comments. I think your comment is the key point of my questions. May I ask one more question? Sorry for endless following questions..

Regarding your comments, a beginner like myself may have a stupid thinking: (beyond the circuits...) it is common sense that if there is an ideal model, then a simple example would (try to) have similar values to those of an ideal model, even though a simple example cannot have the exact same values as those of an ideal model.

Basically, I don't understand the reason why we study an ideal amp model, then.
In my stupid opinion, even simple voltage amp may be trying to have similar values, not even close to those of the ideal voltage amp model, because the output impedance in my previous example is a very important parameter.

In the same book written by Dr. Razavi, there are many other amplifier examples, which are not too much simple amp that I mentioned in my previous question. Some of them have large output impedance and small input impedance. So, I'm wondering if those amplifiers have an opposite magnitude of Rout or Rin to that of an ideal amp, simply because those example amplifiers are poorly designed for education purpose for this book.

-s55
 

KlausST

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Hi,

That´s why we have OPAMPs
use a feedbacked OPAMP. It comes pretty close to an ideal amplifier. For sure it has it´s limits.

But first you need to decide your requirements. The requirements of your application.
Be sure: for almost every of your requirements there exists a rather close (ready to buy) solution.


Klaus
 

s55

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Oh, I see. So, each different topology (CE, CB, CC / CG, CS, CD) does NOT need to have ideal parameter values (e.g., high Rin and low Rout), and they are combined together to be an ideal amp or rather close to that. Thank you, KlausSt!
 

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