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

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?

• If that's an idealised opamp, how can you measure the input power? And if you can't measure the input power, how could you measure power gain? Hint-what is the input impedance? Commented Jul 3, 2015 at 17:40
• If the examiner did not specify source and load impedances, there isn't enough information to compute the power gain, therefore he must mean voltage gain. However as Andy points out, that means a gain of 10.
– user16324
Commented Jul 3, 2015 at 17:40

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.

• Shouldn't P1 be 0 as ideal Opamp doesn't draw current? Commented Jul 3, 2015 at 18:33
• @mam: When referencing P1 I am describing dB in general, so has nothing to do with the particular question. Commented Jul 3, 2015 at 18:38
• In general I agree with you,but wrt to question its pretty weird,as Power=V^2/Rin,Iin=0 for ideal condition,as Rin approaches to infinity(~Several Mega Ohm),So is it practical to ask this question? Commented Jul 3, 2015 at 19:00
• @mam: From the context of this question, it is clear that dB is being used to describe a voltage ratio. See my second paragraph. Commented Jul 3, 2015 at 19:19
• dB is neither an SI unit nor a derived unit. Although NIST accept its use it does not specify the 'allowable' application(s). It seems pedantic not to use it, generally, to express ratios, as long as the context is clear.
– Chu
Commented Jul 3, 2015 at 20:01

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.

• I think you have an extra 2 exponent in your formula for dB gain in terms of power. Commented Jul 3, 2015 at 18:24
• @ThePhoton oops Commented Jul 3, 2015 at 18:53

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.

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

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

• why RL is equal to infnity? when you have taken Rin=rin=infinty as internal factor you must also consider Rout=rout=0 for an ideal Opamp. Commented Jul 3, 2015 at 19:51
• @MaMba, because there is no load resistance drawn, i.e, the load is an open circuit. And do not confuse the output resistance with the load resistance. When the output resistance is zero, the open-circuit voltage gain $A_{voc}$ equals the loaded voltage gain $A_v$. In general, $A_v = A_{voc}\frac{R_L}{R_L + R_{out}}$ Commented Jul 3, 2015 at 19:58
• yeah the output is floating,I rather think this question as deprecated model,many things cant still be accounted.You see all the response accounted to the OP accounts the voltage ratio considering Power gain.But if you consider it as Voltage gain all the problem disappear you wont need Rin and Rl,but then Gain=10 Log(Vout/Vin) Commented Jul 3, 2015 at 20:20