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I feel stupid for asking this, since I know it's a relatively simple procedure, but who can tell me the *proper* way to add LEDs to my FB2 for power, Difficulty A, and Difficulty B switches? If I am thinking about this right, I'll want to cut one of the switch connections and insert my LED there (facing proper direction of course). But how do I choose a proper value for LED voltage, and do I need to worry about possibility of drawing too much current?

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I feel stupid for asking this, since I know it's a relatively simple procedure, but who can tell me the *proper* way to add LEDs to my FB2 for power, Difficulty A, and Difficulty B switches? If I am thinking about this right, I'll want to cut one of the switch connections and insert my LED there (facing proper direction of course). But how do I choose a proper value for LED voltage, and do I need to worry about possibility of drawing too much current?

 

I don't believe you'd want to cut one of the switch connections. I can't see any reason to put a power LED in series with the switch. You would need to tap off the side of the switch that feeds power out to the system when the switch is on and is dead when the switch is off. You'd pick up your switched power there to feed the LED in series with the REQUIRED current limiting resistor and then to ground. It doesn't actually matter whether the resistor is between power and the LED or between the LED and ground.

 

You'll need to do a little research on how to hook up LED's in general to make sure you understand about the proper polarity and how to size the current limiting resistor according to the voltage that you'll be using.

 

I'd have to double check how the difficulty switches are wired. Those LED's may be considerably trickier to implement. Somebody may know that implementation more readily than I do, so I'll wait to see if someone chimes in on that. What is your plan there? One LED that lights up to indicate, say, A difficulty? I would expect, ideally, to see two LED's: one lit for A, the other for B. Or, maybe even a bi-color.

Edited by BigO
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Right, I'm thinking one for A and one for B.

 

 

EDIT: After thinking about this I came to exactly the same idea as you did for the schematic. Resistor value is 150 ohms.

      switch         rest of circuit
+5---o/ o--------------------o
       |
       |
       |---------------->|-----WMW------GND
                       LED    resistor   ground

 

Thanks, I think I can get it from here!

 

EDIT2: And here's a handy calculator. http://led.linear1.org/1led.wiz

Edited by Hornpipe2
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  • 1 month later...

Plugging away at this slowly. My diagram above works perfectly for the power light. However, I stuck a multimeter on the difficulty switches and now I'm a little worried about how I'd add lights to those. The way it works is, a pin from the CPU provides 3.3V to the switch, which is then either left hanging or grounded. Somewhere inside the blob, the CPU is checking to see if that pin carries 3.3V or 0.

 

Can someone tell me if this is the 'proper' way to fix this:

* Cut the trace going from switch contact to ground and insert a 68ohm resistor-LED series. This essentially turns that GND connection into a 68ohm (plus LED resistance) pulldown.

e.g. convert the current setup:

contact
o------------GND

to

contact
o------------/\/\/\---->|-----GND

 

I would be worried that doing this would cause the CPU to somehow read the voltage wrong and my difficulty switch would be permanently broken. This looks right to me, but I'm no expert...

Edited by Hornpipe2
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  • 4 months later...

After chopping traces and trying the method above for difficulty switches, I have found that it DOES NOT WORK. I guess fiddling around with CMOS logic is tricky business : )

 

However, here is a method that SHOULD work. I have not tried it yet but will probably do so in the next couple days. It will require re-wiring the difficulty switches a bit, probably cutting a couple traces, and bringing +5V on a wire to the switch board.

 

The difficulty switches on the FB2 are double-throw switches, so it would be possible to use the switch to connect GND to either difficulty logic, or LED circuit. When set in one position, the LED is disconnected from GND (won't light) while the difficulty logic line is grounded. Set to the other position, and the difficulty line is left hanging (3.3V) whle the LED circuit is grounded and complete, allowing the light.

 

Use a 150ohm resistor here (same as the power light above). Diagram:

+5V      LED      150ohm              switch
o-------->|------/\/\/\--------.
                              |___o
                                      \
                                        o------- GND

(diff. line from CPU)          .___o
o------------------------------'

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  • 4 weeks later...

Just out of curiousity -- did you ever get the difficulty LEDs to be on when advanced (difficult w/switch open) was selected?

 

I tried all sorts of ways of connecting the LED up in series with the switch and could only get it to light when it was closed which is beginner (not difficult) selected. Its better than nothing, but I suspect the switches aren't really double throw. I'm thinking the LEDs need to be hooked up in parallel with the switch that way they'd turn off when the switch was closed (beginner) and be on with it open (advanced.)

 

post-7533-1210508010_thumb.pngpost-7533-1210507878_thumb.jpg

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Just out of curiousity -- did you ever get the difficulty LEDs to be on when advanced (difficult w/switch open) was selected?

 

I tried all sorts of ways of connecting the LED up in series with the switch and could only get it to light when it was closed which is beginner (not difficult) selected. Its better than nothing, but I suspect the switches aren't really double throw. I'm thinking the LEDs need to be hooked up in parallel with the switch that way they'd turn off when the switch was closed (beginner) and be on with it open (advanced.)

 

post-7533-1210508010_thumb.pngpost-7533-1210507878_thumb.jpg

 

The switches are not double throw in the 2600 schematic and I'm pretty sure they're not double throw switches in the FB2.

 

Assuming that the functionality matches the original Atari 2600 functionality (a big assumption), let's talk as if we're modifying a real 2600. The difficulty switches connect pins from the I/O-timer-RAM (RIOT, 6532) chip to ground. The logical conclusion from that arrangement would be that the RIOT input pins are pulled up to a logic 1 internally or are treated as logic 1 when floating.

 

On the basis of those assumptions, I'd try hooking the input of a buffer chip to the non-grounded (input to the RIOT) side of the switch and use that buffer's output to drive the status indicator LED. It might require a pullup resistor if those RIOT inputs float. You'll, of course, need to power the buffer chip by pulling power from somewhere in the FB2 circuitry.

 

If the logic is reversed (the LED comes on in the opposite state of what you want), use an inverting buffer.

 

(One could also implement a similar LED driver with a transistor and resistors.)

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The switches are not double throw in the 2600 schematic and I'm pretty sure they're not double throw switches in the FB2.

 

Assuming that the functionality matches the original Atari 2600 functionality (a big assumption), let's talk as if we're modifying a real 2600. The difficulty switches connect pins from the I/O-timer-RAM (RIOT, 6532) chip to ground. The logical conclusion from that arrangement would be that the RIOT input pins are pulled up to a logic 1 internally or are treated as logic 1 when floating.

 

On the basis of those assumptions, I'd try hooking the input of a buffer chip to the non-grounded (input to the RIOT) side of the switch and use that buffer's output to drive the status indicator LED. It might require a pullup resistor if those RIOT inputs float. You'll, of course, need to power the buffer chip by pulling power from somewhere in the FB2 circuitry.

 

If the logic is reversed (the LED comes on in the opposite state of what you want), use an inverting buffer.

 

(One could also implement a similar LED driver with a transistor and resistors.)

Thanks! -- I took your advice and used a couple of transistors to 'switch' the leds off when the pin is low (beginner) and on when the pin is high (advanced) which is what I wanted. Works perfectly and I didn't need a pull-up so it must go high when not grounded -- I'm guessing the real hardware behaves the same. The FB2 even has little solder pads installed and labeled for the difficulty switches. I assumed it would be a bit simpler though.

post-7533-1210584827_thumb.jpg

 

edit: I ended up having to add 10k pull-ups to +5v. The LEDs lit w/o them but the transistors dropped the voltage a bit and the FB2 didn't recognize it as high for the advanced' setting.

post-7533-1210656397_thumb.jpg post-7533-1210757251_thumb.jpg

Edited by Prodos8
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  • 8 years later...

Sorry for resurrecting an old thread, but it's a good read as well as a fun mod. I added the power LED only, with a red LED and a 150 Ohm resistor. According to the store the LED had a 1.7V forwarding voltage with a forwarding current of 25mA, so I had used the website http://led.linear1.org/1led.wiz to figure out the appropriate resistance.

 

My ground connection went to J19 (either side will do) and I drew the 5V from S1 (on the power switch PCB). I have a Rev B Flashback 2.

 

My only regret was that I used a 5mm LED with a black LED holder to mount it. I should have used a smaller LED (3mm) because the light is too bright. Nonetheless it was worth the effort.

 

I also did the soft reset mod that was posted here. I think people who are novices like me will enjoy doing these little projects, as you'll get the most out of your device and feel good about it too! I don't plan on doing the cartridge mod but I already have another FB2 with this mod done by someone else.

post-47549-0-65775100-1475289308_thumb.jpg

Edited by rednakes1
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