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Cougar stick's pcb, shift register's pins.


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Posted

Does anyone know what are pins in those white connectors.  I'm "recycling" old Cougar system and building a collective with electronics from a donator stick.

IMG_20221123_115857543_2.jpg

Win10, HP Reverb G2,

Posted (edited)

Here's how to decode them.

1. Look at the chip spec itself, see picture #1. There are the Parallel in pins, numbered 1 to 8. Bit #8 will go out first. That is, it's an MSB first scheme. So, 3 chips forms a chain of 24 bits parallel in, serials out. Normally, the values in the buffer changes according to the values on each lead... Whatever they are, the bits in the buffer changes with it.

2. When the Select pin goes active, 0V, the buffer locks up whatever values are in the buffers. Then, the caller sends in 24 pulses (clock), and each pulse sends out 1 bit of value. So, P8 on U1 will be send out as bit 0, etc.

3. In Hempstick User's Guide,http://www.hempstick.org/download/manual/Hempstick-UserGuide.pdf, I have listed all the values. For your convenience, I have included a screenshot in the 2nd picture. In that table, use the column Position in Hempstick Internal. I don't remember which end is which end. But if I have to guess, #0 is the first bit out. I usually write cod that way. way.... But, to confirm, all you have to do is to take a multimeter, and test the conductivity of them.... I happened to have one Cougar PCB disassembled... I tested it.

 

It works like this.

U3 <- U2 <- U1

This forms a cascaded of 3x 8bits. This is an MSB arrangement. So, parallel in bit 8 of U3 is bit 23, bit 7 of U3 is bit 22, etc. However, it is sent out MSB first... so first bit that comes out is store in Hempstick as bit 0 (b/c I am writing software, not using a hardware shift register, so it's as convenient for me to store either way.

Each J connector from J1 to J6, pin 1 is power, Vdd. Pin 5 is MSB bit 

So, here's the list.

The format is Internal Cougar J-Connect-Pin# (Cougar internal Bit #, Position in Hempstick Internal).

J6P5 (23, 0)

J6P4 (22, 1)

J6P3 (21, 2)

J6P2 (20, 3)

J5P5 (19, 4)

J5P4 (18, 5)

J5P3 (17, 6)

J5P2 (16, 7)

J4P5 (15, 6)

J4P4 (14, 9)

J4P3 (13, 10)

J4P2 (12, 11)

J3P5 (11, 12)

J3P4 (10, 13)

J3P3 (9, 14)

J3P2 (8, 15)

J2P5 (7, 16)

J2P4 (6, 17)

J2P3 (5, 18)

J2P2 (4, 19)

J1P5 (3, 20)

J1P4 (2, 21)

J1P3 (1, 22)

J1P2 (0, 23)

 

 

 

 

image.png

image.png

Edited by Hempstead
Posted
23 hours ago, Habu_69 said:

These might help.

Cougar Joystick Handle wiring 1 of 2 Final_s.jpg

Cougar Joystick Handle wiring 2 of 2 Final_s.jpg

 

17 hours ago, Hempstead said:

Here's how to decode them.

1. Look at the chip spec itself, see picture #1. There are the Parallel in pins, numbered 1 to 8. Bit #8 will go out first. That is, it's an MSB first scheme. So, 3 chips forms a chain of 24 bits parallel in, serials out. Normally, the values in the buffer changes according to the values on each lead... Whatever they are, the bits in the buffer changes with it.

2. When the Select pin goes active, 0V, the buffer locks up whatever values are in the buffers. Then, the caller sends in 24 pulses (clock), and each pulse sends out 1 bit of value. So, P8 on U1 will be send out as bit 0, etc.

3. In Hempstick User's Guide,http://www.hempstick.org/download/manual/Hempstick-UserGuide.pdf, I have listed all the values. For your convenience, I have included a screenshot in the 2nd picture. In that table, use the column Position in Hempstick Internal. I don't remember which end is which end. But if I have to guess, #0 is the first bit out. I usually write cod that way. way.... But, to confirm, all you have to do is to take a multimeter, and test the conductivity of them.... I happened to have one Cougar PCB disassembled... I tested it.

 

It works like this.

U3 <- U2 <- U1

This forms a cascaded of 3x 8bits. This is an MSB arrangement. So, parallel in bit 8 of U3 is bit 23, bit 7 of U3 is bit 22, etc. However, it is sent out MSB first... so first bit that comes out is store in Hempstick as bit 0 (b/c I am writing software, not using a hardware shift register, so it's as convenient for me to store either way.

Each J connector from J1 to J6, pin 1 is power, Vdd. Pin 5 is MSB bit 

So, here's the list.

The format is Internal Cougar J-Connect-Pin# (Cougar internal Bit #, Position in Hempstick Internal).

J6P5 (23, 0)

J6P4 (22, 1)

J6P3 (21, 2)

J6P2 (20, 3)

J5P5 (19, 4)

J5P4 (18, 5)

J5P3 (17, 6)

J5P2 (16, 7)

J4P5 (15, 6)

J4P4 (14, 9)

J4P3 (13, 10)

J4P2 (12, 11)

J3P5 (11, 12)

J3P4 (10, 13)

J3P3 (9, 14)

J3P2 (8, 15)

J2P5 (7, 16)

J2P4 (6, 17)

J2P3 (5, 18)

J2P2 (4, 19)

J1P5 (3, 20)

J1P4 (2, 21)

J1P3 (1, 22)

J1P2 (0, 23)

 

 

 

 

image.png

image.png

 

Thank you both for these answers.

Habu_69: I have those schematics from Cougar World, but those don't excatly match to my Cougars wiring...

Hempstead: Your answer is way out of my league... 🙂 But I'm gonna still try to understand it.

So,I have a problem with TG1 and TG2, how to separate those and connect individually. Now as it is,  TG1 has to be pressed simultanous with TG2 to get the signal from TG2.

Win10, HP Reverb G2,

Posted
2 hours ago, F4ttyRider said:

 

Thank you both for these answers.

Habu_69: I have those schematics from Cougar World, but those don't excatly match to my Cougars wiring...

Hempstead: Your answer is way out of my league... 🙂 But I'm gonna still try to understand it.

So,I have a problem with TG1 and TG2, how to separate those and connect individually. Now as it is,  TG1 has to be pressed simultanous with TG2 to get the signal from TG2.

Simple...

Take this entry as an example.

J6P5 (23, 0)

That means it's J6 socket on the board, pin 5, in your picture. Take '0' and look up into my table in the Position in Hempstick Internally. That gives you the Trigger 1st Stage from the Column Cougar Stick Function.

Another example.

J2P5 (7, 16)

That means, J2 socket on the PCB, pin 5. Row 16 in my table gives you DMS Up for Cougar, meaning the DMS HAT switch up.

If you have a Warthog, use the Warthog Stick Function column.

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