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Introductory Digital Electronics CTEC 103 R2R Reduction Supplement Prepared by Mike Crompton (Revised 27 January 2005) D to A Converter R2R reduction At right is the diagram of a binary counter with an R2R network connected to 3 of the Q O/Ps (Q1 pin 9, Q2 pin 8 and Q3 pin 11 of the chip). This changes the circuit into a Digital to Analog (D to A) converter. The resistor values are twice (or ½ ) the size of each other, hence the name R2R. The R2R network will produce a voltage at ‘Analog Vout’ that will be dependant on which Q O/Ps are Hi and which are Lo. This in turn will depend on the binary count. When the count is 0002 all 3 Q O/Ps will be Lo (which is 0V or the same potential as ground), and the analog Vout will be 0V. If the count is 0012, then Q1 the LSB (pin 9) will be Hi or at +5V, and the other two Q O/Ps will remain at 0V or ground. This will give a voltage at Analog Vout that represents the value 1 (0012). That voltage will also be equal to the ‘Step’ voltage which can be calculated as: VQ O/P / Number of steps That is, the typical Q O/P voltage (theoretically 5V) divided by the number of steps, which is calculated as 2N where N is the number of bits (in this case 3 bits, and 23 = 8). So the Step Voltage should be: 0012 = 5/8 = 0.625V If the count is 1002 then Q3 (pin 11) will be Hi and the other two Q O/Ps will be at ground. This means that the 5V is applied to a different point in the R2R network giving a different analog Vout. This time the voltage should be equal to 4 (1002) or 4 times the step voltage. 1002 = 4 * 0.625 = 2.5V. Each time the binary count changes, 5V or 0V will be applied to any 1, 2 or 3 of the Q O/Ps basically changing the R2R network and therefore changing the Analog Vout. On the following pages are a series of circuit reductions showing how the R2R network changes under just 2 of the 8 different combinations of Q O/Ps. The first example, when Q3 (the MSB) is Hi for a binary count of 1002 , is the most straight forward. The second example when Q1 (the LSB) is Hi (binary 0012) is a little more involved. Both examples employ the voltage divider rule, probably the most commonly used rule/formula in electronics and certainly one that will appear time and time again in the Ceng courses. 2 If Q3 O/P is at 5V and Q1 & 2 are at 0V (ground) the circuit on the previous page can be redrawn as shown in diagram ‘A’ at right. Following the normal series/parallel rules for circuit reduction, the first thing to do is to combine the 6k parallel R3 & R6 into one equivalent resistor of 3k as shown in diagram ‘B’. Step 2 is to combine the series 3k R5 and 3k REQUIV into a new equivalent resistance of 6k as depicted in circuit ‘C’. Step 3 is the parallel combination of the two 6k resistances (R6 and the new REQUIV) which give a third new equivalent resistance of 3k as per diagram ‘D’ The last step is to combine the 3k series pair of R4 and the latest REQUIV to give an equivalent resistance for all of resistors R2 to R6 equal to 6k. Diagram ‘E’. It is now a simple matter of applying the voltage divider formula to the circuit shown in the last diagram (‘E’) to calculate the voltage at Analog Vout. In fact, it can immediately be seen that the voltage is 2.5V. However the voltage divider formula can be applied: VRX = (VSUPPLY/ RTOTAL) * RX VREQUIV = 5V/12k * 6k = 2.5V To cross check if this is correct we use the calculated step voltage of 0.625V per step and multiply it by the binary count 1002 or 4: 4 * 0.625V = 2.5V 3 This second example is a little more involved as the voltages at each junction point have to be determined in order to arrive at the Analog Vout. It presumes that Q1 is at 5V and Q2 & Q3 are at ground. The circuit at right, diagram ‘A’, depicts this situation. The voltages shown on diagram ‘A’ cannot be calculated until the circuit has been reduced to the two resistances shown in diagram ‘E’. The reduction process is similar to the first example, step 1 being R1 and R4 in series, giving the REQUIV of 9k shown in diagram ‘B’. The 2nd step is the 9k REQUIV in parallel with the R2 of 6k yielding a new REQUIV of 3.6k, see diagram ‘C’. Diagram ‘D’ shows the result of combining the 2 series resistors of R5 at 3k and REQUIV of 3.6k to give the third REQUIV of 6.6k. The final step, is the parallel pair of R6 at 6k and REQUIV at 6.6k giving the final 2 resistor series circuit of R3 at 6k and REQUIV of R1, 2, 4, 5 &6 at 3.14k, as seen in diagram ‘E’. Applying the voltage divider formula to these final 2 resistances will give us a voltage at their junction, or across REQUIV of: V REQUIV = (5V/9.14k) * 3.14k = 1.7177V Looking at diagram ‘A’ again we see that this is the voltage at the top of R5. We now have to apply the voltage divider formula to the circuit in diagram ‘C’ to get the voltage at the top of R4 or across REQUIV. Remembering that the VSUPPLY for this combination is 1.7177v and not 5V: V REQUIV = (1.7177V/6.6k) * 3.6k = 0.9369V This voltage becomes the VSUPPLY for the R4, R1 combination and will give us the Analog Vout (VR1) by once again using the voltage divider formula: VR1 = 0.9369V/9k * 6k = 0.6246V This is essentially the calculated value for 0012 (or the step voltage) of 0.625 volts. Other binary counts will give different circuit configurations and Analog O/P voltages. 4