power gain

Discussion in 'General Electronics Chat' started by screen1988, May 31, 2013.

1. screen1988 Thread Starter Member

Mar 7, 2013
310
4
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?

2. Externet AAC Fanatic!

Nov 29, 2005
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186
It is not power gain. It is voltage gain.

3. screen1988 Thread Starter Member

Mar 7, 2013
310
4
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.

4. crutschow Expert

Mar 14, 2008
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6,853
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.

5. circuitfella11 Member

May 10, 2013
56
5

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.. Mar 7, 2013
310
4
7. crutschow Expert

Mar 14, 2008
23,116
6,853
You missed the caveat in the Wiki article:
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.

screen1988 likes this.
8. screen1988 Thread Starter Member

Mar 7, 2013
310
4
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?

9. Jony130 AAC Fanatic!

Feb 17, 2009
4,923
1,381
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

screen1988 likes this.
10. screen1988 Thread Starter Member

Mar 7, 2013
310
4
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?

11. Jony130 AAC Fanatic!

Feb 17, 2009
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1,381
Yes, for current we also use 20log. We never use decibels for temperature.

12. crutschow Expert

Mar 14, 2008
23,116
6,853
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.

Last edited: Jun 1, 2013