G’Day
We are in the process of designing a PCB based on the TI reference design - TIPD126_SCH_revA. The PGA309 will communicate with a 8051 mcu on I2C for access to the eeprom registers. The Vout will be read by an ADS1110, also with I2C.
Parameters are:
Internal Vref: 2.5V
Internal Vexc: 2.075V
Rbridge: 1000ohm
Vout_z: 0.2V
Vout_fs: 1.224V
V/V: 1mV/V
Vsa: 5V
Course Gain: -8.500mV (calculator)
FE PGA Gain: 128 (calculator)
Zero Dac: 1.140V (calculator)
Gain DAC: 642.570m (calculator)
BE PGA Gain: 6 (calculator)
4-20 mA circuit:
R4 + R5 = (Vout_max - Vout_min) / (Iin_max – Iin_min)
= (1.224 – 0.2) / (0.0002 – 0.00004)
= 1.024 / 0.00016
= 6.4Kohm
R4 = 3.16Kohm
R5 = 3.24Kohm
Iin_min = Vout_min / (R4 + R5)
= 0.2 / 6.4Kohm
= 0.00003125mA
Iref = 0.00004 - 0.00003125
= 0.00875mA
R2 = 2.5 / 0.00000875
= 285.714Kohm
~ 287Kohm
C10 = 1 / (2 * pi * R4 * 1.41) * 1000
= 35nF
~ 33nF
fc_act = 1 / (2 * pi * R4 * c10) * 1000
= 1.526KHz
I have some questions regarding the design:
- In the reference design, TIPD126_SCH_revA, there is no resistor in the Vfb. I would have expected to see the same 100ohm as R3? Any reason why?
- Is D1 (SMBJ5V0CA) really necessary since the Vreg from XTR117 supplies a regulated 5V? D1 is quite a big component for small PCB.
- Is the fc_act of 1.526KHz sufficient for a low-pass filter on Vout.
Thanks
Cheers
Dirk