[04-MAY-22] The Direct Fiber Positioning System (DFPS) Controller Board (A3045) is a rectangular circuit that generates the control voltages for a single two-axis fiber positioner. The board takes as input ±250 V, 0 V, and 3.3 V power supplies and two single-ended 3.3-V logic levels Serial Clock (SCK) and Serial Data In (SDI). For outputs it has ±250-V control voltages North (VN), South (VS), East (VE), and West (VW) as well as a single-ended 3.3-V logic level Serial Data Out (SDO). The A3045 is designed to plug into a DFPS service board such as the DFPS Base and Service Board (A3043).
The A3043A controller could, in principle, be deployed in a fiber positioning system of any size. Our prototype design deploys sixteen A3045As in a 4×4 array, as shown below.
Each controller board holds four high-voltage amplifiers with gain ×200 and output range ±250 V. These are driven by four slow digital to analog converters (DACs) with 0.1% precision. The DACs are driven by the controller logic, which communicates with the backplane through the service board over a three-wire, single-ended communication bus.
The following versions of the DFPS Controller Board (A3045) are defined.
|A||A3043||9 mm wide, 60 mm long,||Single-fiber controller with 12-way connector.|
[04-MAY-22] We arrange components on the A304301A printed circuit board: four high-voltage amplifiers, logic, and oscillator. We have plenty of space with a 60-mm long, 9-mm wide board, provided we can place P0402 resistors on the bottom side. These resistors are 0.35 mm high. We drop the board thickness to 0.53 mm (0.021"). We plan to stiffen the boards with an epoxy coating, and we will mount capacitors perpendicular to the board length so they do not crack when the board flexes during assembly. Our SOT-23 transistors are 1.1 mm high. In our arrangement on the service board, we have 2.6 mm between the top sides of A3043A boards, and we need 2.2 mm for two sets of SOT-23 transistors, so we have 0.4 mm clearance. The bottom sides of neighboring controller boards are separated by 1.1 mm. We need 2 × 0.35 mm = 0.7 mm for two sets of P0402 resistors. We have 0.4 mm clearance.