digital logic - Signal to Noise ratio problem

A 200mv peak to peak sinusoidal signal is applied to an ideal 12 bit A/D converter, for which Vref(v p-p full scale) is 5v. Find signal to noise ratio.

This is a problem worked out by one of my lecturer and the answer seems to be 46dB. But I am confused on his method. Is there anyone who could explain me how the result was worked out?

Answer

If \$N\$ is the number of bits then your input signal range is divided into quantization intervals of size

$$q = \frac{V_{ref}}{2^N}$$

The maximum quantization error is \$q/2\$ and it is usually assumed that the quantization error is uniformly distributed between \$-q/2\$ and \$q/2\$. So the PDF of the quantization error is constant between \$-q/2\$ and \$q/2\$ with height \$1/q\$. With this assumption the quantization noise power is

$$P_q = \frac{1}{q}\int_{-q/2}^{q/2}x^2dx = \frac{q^2}{12} = \frac{V_{ref}^2}{12\cdot 2^{2N}}$$

The signal power is given by

$$P_x = \frac{V_{max}^2}{2} \text{ with } V_{max} = 100mV$$

The factor \$1/2\$ in the signal power comes from the fact that the input is sinusoidal, i.e. its power is given by half its maximum value. Note that the maximum value is the amplitude which is half the peak-to-peak value. Putting everything together we get

$$SNR = 10\log\frac{P_x}{P_q} = 10\log\frac{6V_{max}^22^{2N}}{V_{ref}^2} = \\ = 10\log\frac{6V_{max}^2}{V_{ref}^2} + 10\log 2^{2N} = 10\log\frac{6(0.1)^2}{5^2} + N\cdot 20\log 2 =\\ = -26.2 + 12\cdot 6.02 = 46\text{dB}$$

EDIT: To see how the \$1.76\$dB pop up we now use \$V_{pp}\$ instead of \$V_{max}\$, where \$V_{pp}=2V_{max}\$ is the peak-to-peak input voltage. The SNR can then be written as

$$SNR = 10\log\frac{3V_{pp}^2}{2V_{ref}^2} + 10\log 2^{2N} =\\ = 10\log\frac{V_{pp}^2}{V_{ref}^2} + 10\log\frac{3}{2} + N\cdot 20\log 2 =\\ = 10\log\frac{V_{pp}^2}{V_{ref}^2} + 1.76 + 6.02\cdot N$$

So if we use the maximum input range, i.e. \$V_{pp}=V_{ref}\$ we get the formula that was mentioned in the question.