Voltage Amplifier excercise

Thread Starter

andrew74

Joined Jul 25, 2022
41
An amplifier with +40dB voltage gain, an input resistance of 10k and an output resistance of 1k is used to drive a 1k load.
What is the value of Avo? Also find the power gain in dB.

(solutions: 100V/V ; 44dB)

20220728_100812.jpg

I found Avo = 200 V/V (that's wrong for the solution) and I don't know how to determine the power gain value... I know I should calculate the product of Av and Ai (voltage gain times current gain) but I have no data to find either the output or input current of the circuit.
 

ci139

Joined Jul 11, 2016
1,867
\[{\large \mathbf{40}dB = 10^{\frac{\mathbf{40}}{20}}=?}\]
for power
\[P=\frac{V^2}{R}\quad if \quad \frac{V_{OUT}}{V_{IN}}=? \quad and\quad \frac{R_{OUT}}{R_{IN}}=\frac{1k}{10k}\quad then\quad \frac{P_{LOAD}}{P_{IN}}=\ . . .\ \left({How\ much\ voltage\ is\ there\ on\ the\ output\ load\ resistor?}\right) \]
note!
\[L_P=10\ log\frac{P_{RMS}}{P_{REF.RMS}}=10\ log\frac{P_{OUT.LOAD}}{P_{IN}}\]
https://en.wikipedia.org/wiki/Decibel#Power_quantities
PS! -- assuming you have an Op Amp with the fixed output impedance of 1kΩ - but your output signal amplitude is much less than the ±supply voltage - you get one solution . . . however if your output signal amplitude is near the |±supply rails| you likely get another solution !!!
PS.2 -- i alredy missed it's an \(A_{V.OL}\) - so - it's "another solution"
 
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Thread Starter

andrew74

Joined Jul 25, 2022
41
\[{\large \mathbf{40}dB = 10^{\frac{\mathbf{40}}{20}}=?}\]
for power
\[P=\frac{V^2}{R}\quad if \quad \frac{V_{OUT}}{V_{IN}}=? \quad and\quad \frac{R_{OUT}}{R_{IN}}=\frac{1k}{10k}\quad then\quad \frac{P_{LOAD}}{P_{IN}}=\ . . .\ \left({How\ much\ voltage\ is\ there\ on\ the\ output\ load\ resistor?}\right) \]
note!
\[L_P=10\ log\frac{P_{RMS}}{P_{REF.RMS}}=10\ log\frac{P_{OUT.LOAD}}{P_{IN}}\]
https://en.wikipedia.org/wiki/Decibel#Power_quantities
PS! -- assuming you have an Op Amp with the fixed output impedance of 1kΩ - but your output signal amplitude is much less than the ±supply voltage - you get one solution . . . however if your output signal amplitude is near the |±supply rails| you likely get another solution !!!
PS.2 -- i alredy missed it's an \(A_{V.OL}\) - so - it's "another solution"
Thank you for your reply! I confused Av with Avo ....
 

Thread Starter

andrew74

Joined Jul 25, 2022
41
Thank you for your reply! but I have not yet been able to find the solution :/
I have listed the outline of the problem, so I have no further data beyond what has already been written. Let's assume it's an open loop amplifier (?)

The power gain I was thinking of calculating it (if it's possible to do it in that way) as the product of the voltage gain and the current gain, then transforming it into dB with 10log.

But I still can't find Avo and I don't understand what could be wrong with the calculations I did and already attached. By the way, I have used exactly the formulas and the circuit given in the book in which the exercise is included.
\[{\large \mathbf{40}dB = 10^{\frac{\mathbf{40}}{20}}=?}\]
for power
\[P=\frac{V^2}{R}\quad if \quad \frac{V_{OUT}}{V_{IN}}=? \quad and\quad \frac{R_{OUT}}{R_{IN}}=\frac{1k}{10k}\quad then\quad \frac{P_{LOAD}}{P_{IN}}=\ . . .\ \left({How\ much\ voltage\ is\ there\ on\ the\ output\ load\ resistor?}\right) \]
note!
\[L_P=10\ log\frac{P_{RMS}}{P_{REF.RMS}}=10\ log\frac{P_{OUT.LOAD}}{P_{IN}}\]
https://en.wikipedia.org/wiki/Decibel#Power_quantities
PS! -- assuming you have an Op Amp with the fixed output impedance of 1kΩ - but your output signal amplitude is much less than the ±supply voltage - you get one solution . . . however if your output signal amplitude is near the |±supply rails| you likely get another solution !!!
PS.2 -- i alredy missed it's an \(A_{V.OL}\) - so - it's "another solution"
I am still struggling with calculating the power gain.

I know that: Av = 100 = Vo / Vi. But I don't know Vi (not even the input current) so I cannot find Vo, I cannot find Po = Vo^2 / Ro and calculate the power gain.

If I tried to calculate the power gain as the product of the voltage gain and the current gain, I would still miss the input and output currents, so I don't think this is the right way.
 

ericgibbs

Joined Jan 29, 2010
16,016
Hi Andrew,
Assume that the Vin voltage is, say 1Volt.
You know the Zin=10k, so calculate the Pin Input power.

Do you follow OK.
Post your calculations.

E

Noted on your diagram you show 2k's resistors rather than 1k's stated in the question.??
 
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Thread Starter

andrew74

Joined Jul 25, 2022
41
I am still struggling with calculating the power gain.

I know that: Av = 100 = Vo / Vi. But I don't know Vi (not even the input current) so I cannot find Vo, I cannot find Po = Vo^2 / Ro and calculate the power gain.

If I tried to calculate the power gain as the product of the voltage gain and the current gain, I would still miss the input and output currents, so I don't think this is the right way.
The output resistance is 1k but I have very bad handwriting so it looks like 2k :(

Assuming an input voltage of 1V, I solved the exercise. What got me into trouble was not having the input voltage in the exercise data, and not knowing whether to assume it at a certain value as we did now

SmartSelect_20220731-105658_Samsung Notes.jpg

Thank you all for your advice and answers!
 

ericgibbs

Joined Jan 29, 2010
16,016
Hi Andrew,
Pleased to hear you have now solved it.;)
With regard to choosing the Vin voltage, remember you have an ideal Amplifier, so what would the answer be if you had chosen say a 10V input.??

E

@andrew74

Log of 25000 = 4.39
so how can 20 times that = 43.9.?????

Power = 10log(x/y)
 
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Thread Starter

andrew74

Joined Jul 25, 2022
41
Hi Andrew,
Pleased to hear you have now solved it.;)
With regard to choosing the Vin voltage, remember you have an ideal Amplifier, so what would the answer be if you had chosen say a 10V input.??

E

@andrew74

Log of 25000 = 4.39
so how can 20 times that = 43.9.?????

Power = 10log(x/y)
I realised now that in the picture of the calculations I attached I wrote 20log ... This is obviously wrong and I should have written 10log(25000) = 44dB.
With 10log(25000) the result is correct.

I think I understand: even redoing the calculations with Vi=10V I get the same result. So being an ideal amplifier ... the choice of input voltage is completely arbitrary. I look at my school notes and figure out the reason, because at the moment I don't remember it ... any small suggestions?
:D
 
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dcbingaman

Joined Jun 30, 2021
682
You can calculate the power gain of any voltage amplifier circuit via this derivation:

1659304280741.png

Notice the power gain is proportional to the square of the voltage gain which in log terms becomes 2 times the log of the voltage gain for a multiplier or 40. Also notice the input voltage cancels out and does not effect overall power gain.
 
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Thread Starter

andrew74

Joined Jul 25, 2022
41
Hi Andrew,
Pleased to hear you have now solved it.;)
With regard to choosing the Vin voltage, remember you have an ideal Amplifier, so what would the answer be if you had chosen say a 10V input.??

E

@andrew74

Log of 25000 = 4.39
so how can 20 times that = 43.9.?????

Power = 10log(x/y)
hi,
So what is the answer to this question.?
E
the same as the input = 1V, so the value of the input voltage does not affect the power gain calculation
 

MrAl

Joined Jun 17, 2014
9,349
Hello,

If you have a gain of 10 with output impedance 1k and the load is 1k, the total voltage gain is only 5.
Think about that and why this might be so.

If you need the power gain, you also need to account for the input power, unless you are allowed to assume it is negligible.
 

dcbingaman

Joined Jun 30, 2021
682
Hello,

If you have a gain of 10 with output impedance 1k and the load is 1k, the total voltage gain is only 5.
Think about that and why this might be so.

If you need the power gain, you also need to account for the input power, unless you are allowed to assume it is negligible.
Excellent point.
 

The Electrician

Joined Oct 9, 2007
2,903
Hello,

If you have a gain of 10 with output impedance 1k and the load is 1k, the total voltage gain is only 5.
Think about that and why this might be so.

If you need the power gain, you also need to account for the input power, unless you are allowed to assume it is negligible.
Since power gain is a ratio of powers, if you don't take into account the input power, and you must take into account the output power, what power other than input power will you use to form a ratio with the output power?
 
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MrAl

Joined Jun 17, 2014
9,349
Since power gain is a ratio of powers, if you don't take into the input power, and you must take into account the output power, what power other than input power will you use to form a ratio with the output power?
Hi,

Good Question because of the dogma usually associated with these kinds of circuits and so may not be immediately apparent. I had not thought about that until reading your reply; thanks for the reply.

If the power gain (which can be a loss) is Pout/Pin and we want to know what it is with variable Pin then we might write:
Gp=Pout/pin
where pin is a variable input power.
Taking the limit as pin goes toward zero, we see that Gp=infinity.

A better idea i think maybe is if we consider the internal losses as well as the output power and input power.
Since the output impedance is the same as the output load resistance and with zero input power, the loss in the output Z will match that of the load, so the power gain would be 1/2. This would really be the supply voltage to output efficiency and is important in converters and power supplies and amplifiers of many types. Compare class D to some other class of audio amplifier.

You may be able to come up with some other ideas.
 

Thread Starter

andrew74

Joined Jul 25, 2022
41
Hi,

Good Question because of the dogma usually associated with these kinds of circuits and so may not be immediately apparent. I had not thought about that until reading your reply; thanks for the reply.

If the power gain (which can be a loss) is Pout/Pin and we want to know what it is with variable Pin then we might write:
Gp=Pout/pin
where pin is a variable input power.
Taking the limit as pin goes toward zero, we see that Gp=infinity.

A better idea i think maybe is if we consider the internal losses as well as the output power and input power.
Since the output impedance is the same as the output load resistance and with zero input power, the loss in the output Z will match that of the load, so the power gain would be 1/2. This would really be the supply voltage to output efficiency and is important in converters and power supplies and amplifiers of many types. Compare class D to some other class of audio amplifier.

You may be able to come up with some other ideas.
"This would really be the supply voltage to output efficiency and is important in converters and power supplies and amplifiers of many types". Can you explain again this concept? I did not understand only this sentence.
 

MrAl

Joined Jun 17, 2014
9,349
"This would really be the supply voltage to output efficiency and is important in converters and power supplies and amplifiers of many types". Can you explain again this concept? I did not understand only this sentence.
Hi,

Yes.

If you have a device in which the supply voltage is 10 volts DC and draws 1 amp DC into the circuit and you get 5 volts DC at 1 amp DC out of the circuit, the circuit is consuming power. 10 watts input and only 5 watts output. That's 50 percent efficiency.
This will also happen with a 100 percent efficient device internally with a later added output resistance equal to the load resistance. That happens because the output resistance is the same as the load and so the voltage drops by 1/2. The power, 50 percent of it, is lost in the output resistance.

I mention this because your circuit has significant output resistance (impedance).

I guess i have to wonder what significance of the output power to the actual input power (such as an audio input) would have of the input impedance was very high and so very very little current was drawn into the audio input while the output could be supplying 100 watts to the speaker. If the input power turned out to be 1mw, the power gain would be enormous and would probably have very little applicational use. If however the input consumed 1 watt, then the power gain would be 100 which would probably be practical. I dont want to put limits that are overly strict on this though.
 
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