What's the Difference Between Amplification and Gain?

AI Thread Summary
Amplification and gain both refer to the increase or decrease in the amplitude of a signal, but they have distinct meanings in electronics. Amplification is a general term for any change in amplitude, while gain specifically measures the ratio of output to input signal, typically expressed in decibels (dB). The discussion clarifies that a gain of 3 from 5V to 15V represents a 200% increase in voltage, not 300%. Additionally, the correct formula for voltage gain is 20 * Log(Vout/Vin), while power gain is calculated as 10 * Log(Pin/Pout). Understanding these differences is crucial for accurate signal processing and analysis.
JJ91
Messages
40
Reaction score
0
Hello,

I can't figure out what's the difference between the two:
Gain - how much the amplitude of the signal was increased/decreased
Amplification - how much the amplitude of the signal was increased/decreased

If we take a basic inverting amplifier circuit:
http://alturl.com/39n78

We can see an amplification of signal by 3 times from 5V to 15V
This is equivalent increase of 300%
300% to dB would mean a gain of 4.77dB according to http://www.lightmachinery.com/percent-to-dB-conversion-calculator.php

Is gain (dB) simply an amplification rate in different units?

Thanks.
 
Engineering news on Phys.org
First of a gain of 3 from 5V to 15V is not a 300% increase, it's a 200% increase. Yes, the calculator you reference does show that a gain of 3 is 300%, but that's not the same as a 300% increase. A 100% increase would be an increase from 5V to 10V. Second the dB calculator is calculating power increase not voltage increase. Power gain varies as the square of voltage increase. A voltage gain of 3 would have a dB value of 9.54 dB.

Gain is often expressed as dB so the gains of multiple stages of other elements can be added together instead of multiplied or divided. It's not different units, it's a different mathematical operation. For voltage gain you should use the formula 20 * Log(Vout/Vin).
 
amplification is what is "occurring ... the gain figure quantifies the amount of amplification occurring


skeptic2 said:
...Gain is often expressed as dB so the gains of multiple stages of other elements can be added together instead of multiplied or divided. It's not different units, it's a different mathematical operation. For voltage gain you should use the formula 20 * Log(Vout/Vin).

and to follow on from skeptic2's formula above ...

power gain = 10 * Log (Pin/Pout)

Dave
 
Last edited:
Super simple, thanks.
 


Hello,

Thank you for your question. To answer your question simply, yes, amplification and gain are essentially the same thing. Both terms refer to the increase or decrease in amplitude of a signal. However, there are some technical differences between the two terms that are worth noting.

Amplification is a more general term that refers to any increase or decrease in the amplitude of a signal. It can be used to describe the increase in voltage in an electronic circuit or the increase in volume in an audio system. Gain, on the other hand, is a more specific term used in the field of electronics and signal processing. It is a measure of the ratio of output signal to input signal and is usually expressed in decibels (dB). In your example, the gain of the amplifier circuit would be 4.77dB, which means the output signal is 4.77 times larger than the input signal.

In summary, amplification and gain are similar concepts that describe the increase or decrease in amplitude of a signal, but gain is a more specific term used in electronics and is measured in decibels. I hope this helps clarify the difference between the two terms. Let me know if you have any further questions.
 
Very basic question. Consider a 3-terminal device with terminals say A,B,C. Kirchhoff Current Law (KCL) and Kirchhoff Voltage Law (KVL) establish two relationships between the 3 currents entering the terminals and the 3 terminal's voltage pairs respectively. So we have 2 equations in 6 unknowns. To proceed further we need two more (independent) equations in order to solve the circuit the 3-terminal device is connected to (basically one treats such a device as an unbalanced two-port...
suppose you have two capacitors with a 0.1 Farad value and 12 VDC rating. label these as A and B. label the terminals of each as 1 and 2. you also have a voltmeter with a 40 volt linear range for DC. you also have a 9 volt DC power supply fed by mains. you charge each capacitor to 9 volts with terminal 1 being - (negative) and terminal 2 being + (positive). you connect the voltmeter to terminal A2 and to terminal B1. does it read any voltage? can - of one capacitor discharge + of the...
Thread 'Weird near-field phenomenon I get in my EM simulation'
I recently made a basic simulation of wire antennas and I am not sure if the near field in my simulation is modeled correctly. One of the things that worry me is the fact that sometimes I see in my simulation "movements" in the near field that seems to be faster than the speed of wave propagation I defined (the speed of light in the simulation). Specifically I see "nodes" of low amplitude in the E field that are quickly "emitted" from the antenna and then slow down as they approach the far...
Back
Top