5V PRECISION DATA ACQUISITION
SUBSYSTEMS
1
2
3
4
5
6
7
8
TC530
TC534
Timing
Status
Power-up RESET
AZ
Undefined
Write LOAD VALUE to Serial Port
Converter in Normal Service
Conversion
Phase
AZ
INT
DINT
IZ
AZ…
Continuous Conversions
R/W brought LOW during AZ
for serial port write cycle
R/W
Converter held in AZ
state due to RESET = 1
R/W = HIGH strobes
LOAD VALUE into
timebase and starts
conversion
RESET
DCLK
DIN
1
1
0
0
1
1
1
1
MSB
LOAD VALUE
LSB
EOC
Figure 4. TC530/534 Initialization and Load Value Write Cycle
verter input. A0 is the least significant address bit (i.e.,
channel 1 is selected when A0 = 1 and A1 = 0). The
multiplexerisdesignedtobeoperatedinadifferentialmode.
For single-ended inputs, the CHx– input for the channel
under selection must be connected to the ground reference
associated with the input signal.
The charge pump clock operates at a typical frequency
of 100kHz. If lower quiescent current is desired, the charge
pump clock can be slowed by connecting an external ca-
pacitor from the OSC pin to VDD. Reference typical charac-
teristics curves.
APPLICATIONS
Design Example
R/W
Figure6showsatypicalTC534interrupt-drivenapplica-
tion. Timing and component values are calculated from
equations and recommendations made in the Dual Slope
Integrating Converter and Programming the TC530/534
sections of this document. The EOC connection to the
processor INT input is for interrupt-driven applications only.
(In polled systems, the EOC output is available on DOUT).
DCLK
DOUT
EOC OVR POL MSB
LSB
GIVEN
Figure 5. Serial Port Data Read Cycle
REQUIRED RESOLUTION: 16Bits(65,536counts.)
MAXIMUM VIN:
±2V
POWER SUPPLY VOLTAGE: +5V
60Hz SYSTEM
DC/DC Converter
An on-board, TC7660H-type charge pump supplies
negative bias to the converter circuitry, as well as to external
devices. The charge pump develops a negative output
voltage by moving charge from the power supply to the
reservoir capacitor at VSS by way of the commutating
capacitor connected to the CAP+ and CAP– inputs.
1. Pick Integration time (tINT
)
66msec
2. Estimate crystal frequency
FIN = 2R/tINT = 2 x 65536/66 x 10 –3 = 1.98MHz
(use 2MHz)
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