Is it voltage or power of the signal amplified when the opamp amplification is given in dB unit?

Question

Today, one of my colleague said that he entered a preliminary examination in a company after his job application. One of the questions is exactly "Design a non-inverting amplifier with 20dB gain.". The goal of the question is to find the resistor values \$R_1\$ and \$R_2\$.

^{simulate this circuit – Schematic created using CircuitLab}

If we take this 20dB as "power amplification", then

$$ 20dB = 10 log_{10}\dfrac{P_o}{P_i} \implies \dfrac{P_o}{P_i} = 100 = \dfrac{V_o^2}{V_i^2} \implies \dfrac{V_o}{V_i} = 10 \implies \dfrac{R_1}{R_2} = 9. $$

If we understand it as "voltage amplification", then

$$ 20dB = 10 log_{10}\dfrac{V_o}{V_i} \implies \dfrac{V_o}{V_i} = 100 \implies \dfrac{R_1}{R_2} = 99. $$

Which one is the correct approach?

Answer 1

Deci-Bels (dB) always represent a power ratio. Specifically, the power ratio of P2 with respect to P1 expressed in dB is:

  dB = 10Log₁₀(P2/P1).

Sometimes, as in the question you show, we use dB as a short form for voltage ratios. However, since power is proportional to the square of the voltage, a voltage ratio expressed in dB is:

  dB = 10Log₁₀((V2/V1)²) = 20Log₁₀(V2/V1)

Therefore the correct answer to the question was to realize that a voltage gain of 10 was being asked for, and R1/R2 needs to be 9.

Answer 2

No, voltage gain is 20 log the ratio and not 10 log the ratio.

20dB = voltage gain of 10 and a power gain of 100. For the op-amp circuit it only makes sense to set it to have a voltage gain ratio of ten

Voltage gain = 20\$log_{10}(\dfrac{V_O}{V_I})\$ or 10\$log_{10}(\dfrac{V_O}{V_I})^2\$ = 10\$log_{10}(\dfrac{P_O}{P_I})\$

Just think about the power in a resistor if you made the voltage ten times bigger, the power goes up by the square of the voltage increase.

Answer 3

Decibels technically always measure power gain according to the formula

$$\mathrm{Gain(dB)}=10\log_{10}{\frac{P_o}{P_i}}$$

When the input and load impedances are the same we have

$$ \frac{P_o}{P_i}=\frac{V_o^2}{V_i^2}$$

so

$$\mathrm{Gain(dB)}=10\log_{10}{\frac{V_o^2}{V_i^2}}=20\log_{10}{\frac{V_o}{V_i}}$$

Sometimes we use decibels imprecisely when the input and load impedances aren't equal or arent' known to give the voltage gain, but always with the pre-factor 20, not 10.

So even if your questioner meant to use dB in this imprecise way, 20 db still gives a voltage gain of 10, not 100.

Answer 4

If we take this 20dB as "power amplification", then

The power gain of a voltage amplifier depends on the input resistance and the load resistance.

For a non-ideal voltage amplifier, the power gain is related to the voltage gain as follows:

$$G_P = \frac{P_L}{P_{in}} = \frac{V_LI_L}{V_{in}I_{in}}=\frac{V^2_LR_{in}}{V^2_{in}R_L}{} = A^2_v\frac{R_{in}}{R_L}$$

where \$A_v\$ is the (loaded) voltage gain. In dB, the power gain is thus

$$G_P (\mathrm {dB}) = 10\log \left( A^2_v\frac{R_{in}}{R_L}\right ) = 20\log \left( A_v\right ) + 10 \log \left( \frac{R_{in}}{R_L} \right)$$

In the given circuit, the load is an open circuit (\$R_L = \infty\$). Thus, for finite input resistance

$$R_L = \infty \Rightarrow G_P = 0 = -\infty\; \mathrm{dB}$$

(When both the load and input resistance are infinite, the power gain is indeterminate.)

In other words, it must be voltage gain that is being specified in this case. As other answers have pointed out, voltage gain in dB is given by

$$A_v (\mathrm {dB}) = 20\log \left( A_v\right )$$