I have just read this:
OPEN-LOOP GAIN
Unlike the ideal op amp, a practical op amp has a finite gain. The open-loop dc gain (usually
referred to as AVOL) is the gain of the amplifier without the feedback loop being closed, hence
the name “open-loop.” For a precision op amp this gain can be vary high, on the order of 160 dB
(100 million) or more.

From the section, I can infer that 160 dB here is power gain. But I just wonder why we often use power gain like this not directly amplitude gain.
Is power easy to measure than amplitude?

 

It is not power gain. It is voltage gain.

 

It is not power gain. It is voltage gain.

But from 160 dB, I can use P(dB)= 10 log(Pout/Pin)=20log( Aout/Ain)= 160
=> Aout/Ain = 10^8
Then as in above it is power gain.

 

But from 160 dB, I can use P(dB)= 10 log(Pout/Pin)=20log( Aout/Ain)= 160
=> Aout/Ain = 10^8
Then as in above it is power gain.

I have no idea what this means.

But Externet is correct, the gain is voltage gain, not power gain. Power gain would be near infinite since an op amp's input impedance is very high, thus making the input signal power very near zero.

 

I have no idea what this means.

But Externet is correct, the gain is voltage gain, not power gain. Power gain would be near infinite since an op amp's input impedance is very high, thus making the input signal power very near zero.

yes externet is correct indeed... " open loop DC gain"...its clearly voltage gain..

@TS
maybe got mixed up in formulas on decibel conversion.. now knowing that it is voltage gain, then you can go easily measure it in amplitudes..

 

Can you take a look here?
https://en.wikipedia.org/wiki/Gain

Power gain = 10 log (Pout/Pin) = 20 log(Vout/Vin).

 

Can you take a look here?
https://en.wikipedia.org/wiki/Gain

Power gain = 10 log (Pout/Pin) = 20 log(Vout/Vin).

You missed the caveat in the Wiki article:

This simplified formula is used to calculate a voltage gain in decibels, and is equivalent to a power gain only if the impedances at input and output are equal

and that's clearly not true for an op amp.

Even though the essential units of dB are defined as a power ratio, the voltage ratio verision of dB is often used when the input and output impedances are different, even though that's not technically correct.

Since power gain is essentially meaningless for op amps due to their extremely high input impedance, voltage gain is used.

 

So, there is a definition about voltage gain that:
Gain = 20 log (Vout/Vin)
I always about the coefficient. With power gain it is always 10 but not sure in other measures.
For example, with voltage gain, which one is right:
Gain = 20 log (Vout/Vin)
Gain = 10 log (Vout/Vin)
Is there a way to make sure not mistake?

 

For voltage gain we always use "20 log"
Vout/Vin = 60dB ---> 60/20 = 3 ---> 10^3 = 1000V/V = 1kV/V

160dB = 160/20 ----> 8---> 10^8 = 100MV/V

 

For voltage gain we always use "20 log"

I think, it also applys for current. How about other parameters, such as temperature...?
Now I only know three parameters (power, voltage, current). Do you know any rule for other parameters?

 

Yes, for current we also use 20log. We never use decibels for temperature.

 

So, there is a definition about voltage gain that:
Gain = 20 log (Vout/Vin)
I always about the coefficient. With power gain it is always 10 but not sure in other measures.
For example, with voltage gain, which one is right:
Gain = 20 log (Vout/Vin)
Gain = 10 log (Vout/Vin)
Is there a way to make sure not mistake?

The definition for dB is 10 times the log ratio of the powers. Since power is proportional to the square of the voltage (which is the same as multiplying by 2 with logarithms) then the dB in voltage is 2 x 10 = 20 times the log ratio of the voltages.

Just remember that for power ratios you use a factor of 10 and voltage ratios you use a factor of 20.