Expressing Gain in Decibels?
- It's just a convention. Decibels always refer to power, and power is proportional to voltage squared and current squared. The math works like this:
$$\begin{align} A_{v,dB} &= 10\cdot\log \frac{V_o^2}{V_i^2} \\ &= 10\cdot\log \left(\frac{V_o}{V_i}\right)^2 \\ &= 10\cdot 2 \cdot\log \frac{V_o}{V_i} \end{align} $$
EDIT: Power is proportional to voltage/current squared for all linear circuits. In AC circuit analysis, voltage, current, and power all become phasors. In nonlinear circuits, power may not be proportional to voltage/current squared, but the convention is still used for decibels. It even holds in an abstract field like signal processing, where the "power" of a signal is defined to be the average of the amplitude squared.
- Decibels are used to give the magnitude of the gain. If you want to talk about power loss, you use negative decibels, which represents a fractional gain. For example:
$$A_p = 0.01$$ $$A_{p,dB} = 10\cdot\log 0.01 = -20\:\mathrm{dB}$$
You can also have an actual negative gain, like what you get from an inverting amplifier. The negative there is described as a phase shift. To fully describe an amplifier, you usually need both the magnitude and the phase of the gain. (This might be a bit advanced for you, but you could try looking up Bode Plots to see how this is used in real life.) Anyway, here's how to describe an inverting amplifier with a voltage gain of -2.4:
$$A_v = -2.4$$ $$|A_v| = 20\cdot\log 2.4 \approx 7.6\:\mathrm{dB}$$ $$\angle A_v = 180^\circ$$
Decibels is always about power. The expressions for voltage and current use a factor of 20 instead of 10 because power ratios are proportional to the square of the voltage or current ratio.
Power gain can be negative, but then it's generally called "loss". The logarithm of any number less than unity is negative.
Decibel is a measure of power ratio.
It is 10x (for power ratios) because the unit is bel (deci is one of the 'approved' prefix for 1/10).
It is 20x (ie 2*10) when dealing with voltages and current as the dissipated power is typically proportional to the square of these quantities. Due to log rules a \$10*log(\frac{V_1}{V_2})^2\$ would see the power2 come down as a scalar gain \$2*10*log(\frac{V_1}{V_2})\$
dB is an indication of the ratio of power between two points. If you cannot take credit for a load impedance being held constant then actual power needs to be used