# Volume III - Semiconductors, Chapter 1: Decibels

#### David Bridgen

Joined Feb 10, 2005
278
The treatise, while scholarly, is, I think, too verbose. The timber is obscured by the arborescence. (Look it up.)

Be that as it may, there is a serious omission.

It concerns expressing amplifier gain in dB.

This should be done when, and only when, the input and output impedances are equal.

Consider an amplifier about which we are told only that its voltage gain is 10.

Not only neophytes, but many experienced practitioners of the electronic art, will jump in with both feet and declare that the gain is also 10dB.

Now let's reveal the amplifier's input and output impedances to be 1000 and 600 ohms respectively.

Power in the input impedance is 1mW. That in its output impedance is 166.7mW. That's a gain of 22dB.

Change the output impedance to 50 ohms. Output power 2W. Gain 33dB.

Now you see the fallacy of ignoring or overlooking the impedances.

An example in the text seems to indicate confusion on its author's part.

Example: The voltage into a 600 Ω audio line amplifier is 10 mV, the voltage across a 600 Ω load is 1 V. Find the power gain in dB.

The word "power" is irrelevant. Given that both input and output impedances are equal, as they must be in order not to invalidate the use of dB notation, it matters not whether voltages or powers are being considered. The result is the same.

#### KL7AJ

Joined Nov 4, 2008
2,229
The treatise, while scholarly, is, I think, too verbose. The timber is obscured by the arborescence. (Look it up.)

Be that as it may, there is a serious omission.

It concerns expressing amplifier gain in dB.

This should be done when, and only when, the input and output impedances are equal.

Consider an amplifier about which we are told only that its voltage gain is 10.

Not only neophytes, but many experienced practitioners of the electronic art, will jump in with both feet and declare that the gain is also 10dB.

Now let's reveal the amplifier's input and output impedances to be 1000 and 600 ohms respectively.

Power in the input impedance is 1mW. That in its output impedance is 166.7mW. That's a gain of 22dB.

Change the output impedance to 50 ohms. Output power 2W. Gain 33dB.

Now you see the fallacy of ignoring or overlooking the impedances.

An example in the text seems to indicate confusion on its author's part.

Example: The voltage into a 600 Ω audio line amplifier is 10 mV, the voltage across a 600 Ω load is 1 V. Find the power gain in dB.

The word "power" is irrelevant. Given that both input and output impedances are equal, as they must be in order not to invalidate the use of dB notation, it matters not whether voltages or powers are being considered. The result is the same.

Hi David:

Very good poiint. I always demonstrate to my students that, if you violate the constant impedance rule, you can get power gain out of a tranformer....which is, of course, impossible. An alternative is to use the dbV notation, using 20 log(v2/v1), IF the impedances are known to be equal.

As repugnant as "voltage dB" is to me (and not a small number of other instructors) the fact of the matter is that most direct measurements are of voltage, especially in R.F. circuits. Fortunately, for most R.F. circuits, the impedances ARE normalized, generally to 50 ohms.

#### Dcrunkilton

Joined Jul 31, 2004
422
Let me know if you have any proposed changes to the Amplifiers chapter, for example the addition of a dB gain problem for an amplifier with un-equal input and output impedances.