- 2076 Chaitra · 8 marks
Interfacing 1408 DAC for 0 to 10 V
Short answer
DAC1408 sinks current; Iref = 2 mA from +10V via 5 kΩ, feedback resistor 5 kΩ yields Vout = 0–9.96 V. Address AFH is decoded, and data is latched by a 74LS373. A 0.1 µF compensation capacitor is placed from pin 16 to VEE.
Design the interfacing of 1408 DAC with an output port of address AF H for 0v to 10v range. Note that take appropriate values for resistors and capacitors.
DAC1408 (MC1408) basics
The MC1408 is an 8-bit multiplying DAC with a current output. Its full-scale output current is Io = Iref x 255/256, where Iref = Vref / R14. Io is a current sink into pin 4, so the op-amp (inverting input at pin 4) supplies this current through Rf. The op-amp output is therefore positive: Vout = +Io x Rf.
Design values for 0-10 V
- Vref(+) = +10 V, R14 (pin 14) = 5 kohm, so Iref = 10 V / 5 kohm = 2 mA. Pin 15 (Vref-) is connected to ground through R15 = 5 kohm (equal to R14).
- Vout = Io x Rf, with Rf = 5 kohm: Vout = 2 mA x 5 kohm x N/256 = 10 V x N/256. For N = FFH, Vout = 9.96 V; for N = 00H, Vout = 0 V.
- Supply: VCC = +5 V (pin 13), VEE = -15 V (pin 3). A compensation capacitor of 0.1 uF goes from pin 16 to VEE. The op-amp (LM741/LF356) runs on +/-15 V, with feedback resistor Rf = 5 kohm (a 10 pF capacitor across Rf if needed).
- Latch: a 74LS373/74LS374 holds the data at the output port.
Port address decoding AFH (8085, I/O write)
AFH = 1010 1111. For I/O instructions, the address lines A7-A0 appear on both A15-A8 and AD7-AD0. IOW' is low only during an I/O write, so it already includes IO/M.
A7,A5,A3,A2,A1,A0 -----+
A6,A4 --[NOT]----------+--[8-input NAND]-- output L (low only when address = AFH)
L and IOW' ---[NOR]---> G (clock) of 74LS373 (latch); G is high only when both are low
8085 AD0-AD7 -> 74LS373 (D0-D7 -> Q0-Q7) -> DAC1408 A8..A1 (MSB = A1)
+10 V --[R14=5k]--> pin 14 ; pin 15 --[5k]--> GND
pin 4 (Io) --> op-amp (-) ; Rf = 5k between output and (-)
op-amp (+) -> GND ; Vout = 0 to +9.96 V
pin 16 --[0.1 uF]--> VEE (-15 V)
Program
MVI A,80H ; 80H -> about 5 V
OUT AFH ; send to output port AFH
HLT
To generate 0-10 V in steps, send codes 00H to FFH with OUT AFH. The output voltage for code N is 10 x N / 256 V (resolution 39 mV).
✓ Answer checked by Claude Sonnet 5.5
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