# Amplifers Maths help

1. Jan 16, 2005

### fcukniles

Hi, im doing some revision and i cant remember how to do these, think i might have missed that lecture :-( shame on me, can anyone give me a hand? any help many thanks (the square is ment to be homes)

1. A microphone is delivering an electrical power of 1mW to the input of an amplifier of
gain 36dB. If the output of the amplifier drives a loudspeaker of 3 resistance calculate the speaker voltage and current.

(3.45 V, 1.15 A)

2. A ‘black box’ voltage amplifier has resistive input and output impedances of 10
kilohms and 100 respectively. It has a voltage gain of 1000 when its output is unloaded. Calculate the output voltage if it is fed from a source of 1mV having a resistive source impedance of (i) 0 and (ii) 1k if the amplifier feeds a resistive load of 1k.

(BB Question)

3. The above amplifier is used with a source resistance of 1k.

Measuring between the load terminals and the amplifier input terminals:

(i) What is the loaded voltage gain as a number?
(ii) What is the loaded gain, expressed in dB?
(iii) What is the loaded current gain?

2. Jan 16, 2005

### fcukniles

anyone?

i have a exam on wednesday so if someone could post up some help before then it would be great.

3. Jan 17, 2005

### chroot

Staff Emeritus
1) The decibel is just another way to express a ratio. The first step to doing calculations with decibels is often to convert them into normal ratios.

The relationship is

$$x \textrm{in dB} = 10 \cdot \log_{10}{x}$$

So, solve the equation

$$36 = \log_{10}{x}$$

for x. You should get a little less than a gain of 4,000.

The output power is thus about 4000 times the input power, or about 4,000 milliwatts. Use

$$P = \frac{V^2}{R}$$

to solve for V, and

$$V = I \cdot R$$

to solve for I.

2) The input and output impedances are simply resistances, internal to the amplifier, that are in series with the source and load, respectively.

An amplifier with an output impedance of 100Ω and a load of 1000&Omega; is effectively driving a total load of 1100&Omega;. The two resistances form a resistor divider, such that 1000/1100ths of the output voltage is actually applied to the load.

The same situation applies to the input; draw a voltage divider and see how much of the input voltage is lost by the amplifier's source impedance.

3) The first two parts are really the same question; the third is just another ratio. I will assume that once you've finished the second problem, the third will be quite simple.

- Warren

Last edited: Jan 17, 2005
4. Jan 17, 2005

### fcukniles

thank you your a star :)

5. Jan 18, 2005

### fcukniles

hi im a little confused.

what do you do to the

36 = log10x to get 4000?

also

v = sqrt [PR] right?
so v = sqrt [0.004 x 3]
v = 0.11V

but it says the answer is 3.45V

thanks for n e help :)

6. Jan 18, 2005

### chroot

Staff Emeritus
To solve 36 = 10 log(x), raise both sides by 10:

$$\begin{equation*} \begin{split} 36 &= 10 \log x\\ \log x &= \frac{36}{10}\\ 10^{\log x} &= 10^\frac{36}{10}\\ x &= 10^\frac{36}{10}\\ &= 3,981.07171 \end{split} \end{equation*}$$

As for the second question, 4,000 milliwatts is 4 watts, not 0.004 watts.

$$V = \sqrt{4 W * 3 \Omega}$$

- Warren

7. Jan 18, 2005

### fcukniles

thnx chroot :)