Question 29.10: Design a 3-bit Flash converter, listing the values of the vo......

The 3-bit converter can be seen in Fig. 29.22. As the values of all the resistors are equal, the voltage of each resistor tap, V_i, will be V_i=V_{REF}\left\lgroup\frac{i}{8}\right\rgroup where i is the number of the resistor in the string for i = 1 to 7. Obviously, V_1 = 0.625 V, V_2 = 1.25 V, V_3 = 1.875 V, V_4 = 2.5 V, V_5 = 3.125 V, V_6 = 3.75 V, V_7 = 4.375 V.

Therefore, when \nu_{IN} first becomes equal or greater than each of these values, a transition will occur in the transfer curve. The transfer curve can be seen in Fig. 29.23 and should look similar to those seen in Ch. 28. The quantization levels and their corresponding thermometer codes are summarized in Fig. 29.24.

The transfer curve of this ADC corresponds to the ADC with quantization error centered about +½ LSB, as discussed in Ch. 28 (Fig. 28.20). To shift the curve by ½ LSB so that the code transitions occur around the LSB values and the quantization error is centered around 0 LSB, the value of the last resistor in the string would have to be adjusted to \frac{R}{2} and the value of the MSB resistor, closest to the reference voltage, would have to be made 1.5R. Then the first code transition would occur at \nu_{IN} = 0.3125 V, and the last code transition would occur at \nu_{IN} = 4.0625, and so the transfer curve would exactly match that of Fig. 28.20.

Based on Fig. 29.24, when \nu_{IN} = 1.5 V, only comparators C_1\ \text{and}\ C_2 will have outputs of 1, since both V_1 and V_2 are less than 1.5 V. The remaining comparator outputs will be 0 since V_3 through V_8 will be greater than 1.5 V, thus generating the thermometer code, 0000011. The encoder must then convert this into a 3-bit digital word, resulting in 010. The same reasoning can be used to construct the data shown in Fig. 29.25. It should be obvious that if the polarity of the comparators were reversed, the thermometer code would be inverted.