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# Amplifier Gains – Solved Problems on Power & Voltage Ratios

Let us look at some numerical on amplifiers to get a better grip of the theoretical understanding.

**The ratio of output to input voltage of an amplifier is 20. Compute the voltage gain.**

**Solution **− The voltage ratio is given as 20. We need to convert this linear ratio into logarithmic ratio to arrive at the solution.

$$Voltage\:gain(dB)=20log_{10}(20)\sim\:26dB$$

The positive voltage gain indicates that the output power is greater than the input power.

**What does amplification in power by 3 dB indicate?**

**Solution **− Let the output power be ‘y’ and input power be ‘x’. Representing this as an equation, we have

$$3dB=10log_{10}(\frac{y}{x})$$

$$log_{10}(\frac{y}{x})=0.3dB\Rightarrow\:\frac{y}{x}=10^{0.3}\sim\:2$$

Thus, the output power is twice the input. Amplification in power by 3 dB therefore indicates that the output power is twice the input power.

**A cascaded amplifier system produces an output of 16 kW for an input of 1 kW. How many ‘3-dB’ amplifiers were used for this amplification?**

Solution − Four amplifiers each with gain of 3 dB were used. We know that after each stage of amplification, the power available is twice the power available at the preceding stage of amplification. Let us look at the table below to understand the power available after each stage of amplification.

Stage of amplification | Output power |
---|---|

First | 2 kW |

Second | 4 kW |

Third | 8 kW |

Fourth | 16 kW |

$$1kW+3dB\longrightarrow\:2kW$$

$$2kW+3dB\longrightarrow\:4kW$$

$$4kW+3dB\longrightarrow\:8kW$$

$$8kW+3dB\longrightarrow\:16kW$$

**What are the power and voltage ratios for respective gains of 1 dB?**

**Solution **− Finding power ratio − We are not provided with the reference power. Let us take 1 W as the reference. Using the equation, we have

$$10log_{10}(\frac{P}{1W})=1dB$$

$$(\frac{P}{1W})=10^{0.1}\sim\:1.26\Rightarrow\:P=1.26W$$

For finding voltage ratio, we take reference as 1 V and proceed with the equation.

$$20log_{10}(\frac{ab}{cd})=1dB$$

$$(\frac{V}{1V})=10^{0.05}\sim\:1.122\Rightarrow\:V=1.122V$$

Thus, voltage ratio is 1.122 and power ratio is 1.26 for this numerical.

What else do we infer from this? It can be observed that the voltage ratio is square root of the power ratio. This is because we have the same reference levels for both voltage and power.

**It is important to note that 1 V, 1 W, 1 mW and so on may not always be the fixed reference levels. Some of the reference power levels used intelephone and broadcasting industries include 6 mW, 10 mW, 12.5 mW and so on. It basically depends on the requirements of the application.**

**The input voltage of an amplifier is 2.5 mV. If the output were to be 12 dB above 1 mV reference, the voltage gain of the amplifier is ______**

**Solution **− 1 mV reference corresponds to 0 dBmV.

$$For\:input\:power=reference\:power\:of\:1mW,power\:in\:dBm\:is\:0$$

$$P_{dBm}=10log_{10}(\frac{1mW}{1mW})=0dBm$$

It is given that the output is 12 dB above the reference. Hence the output voltage is 0 dBmV + 12 dB which is 12 dBmV. What does this 12 dBmV correspond to in the linear scale?

$$12dBmV=20log_{10}(\frac{V}{1mV})$$

$$(\frac{V}{1mV})=10^{12/20}\approx\:4\Rightarrow\:V=4mV$$

Now, it is easy to determine the voltage gain. The output to input voltage ratio is 1.6 and with this, we can find out the voltage gain.

$$20log_{10}(1.6)=4.1dB$$

**The output power of an amplifier with reference level of 1 W is 7 dBW. What is the output power in linear scale for a reference power of 0.5W considering the same output power?**

**Solution **− We are provided with the reference power as 0.5 W. By equation,

$$7dBW=10log_{10}(\frac{P}{0.5W})$$

$$log_{10}(\frac{P}{0.5W})=0.7$$

$$(\frac{P}{0.5W})=10^{0.7}\sim\:5W\Rightarrow\:P=2.5W$$

**What is the reference voltage level if the voltage gain is 15 dB and the output voltage is 16.87 mV? Repeat the same if the voltage gain is 20 dB.**

**Solution **− Going by the equation, we have,

$$20log_{10}(\frac{16.87mV}{P_{ref}mV})=15dB$$

Upon simplification, we get the reference power level as 3 mV.

If the voltage gain is 20 dB, we have,

$$20log_{10}(\frac{16.87mV}{P_{ref}mV})=20dB$$

The reference level voltage level is 1.687 mV

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