Interesting that in that table they don't show which Turbo pin is Ground. Maybe they switch round, in which case most of the work is already done (or would be if the case LED was a two leg bi-colour LED). Can you measure the voltage on both those pins, compared to Ground, when the turbo is on and off? Have you got any pictures showing the Turbo LED and connector?

I can provide some pictures and take some measurements after new years eve 😀

snufkin wrote on 2021-12-31, 16:09:

Interesting that in that table they don't show which Turbo pin is Ground. Maybe they switch round, in which case most of the work is already done (or would be if the case LED was a two leg bi-colour LED). Can you measure the voltage on both those pins, compared to Ground, when the turbo is on and off? Have you got any pictures showing the Turbo LED and connector?

PIN 1: +5v. Does not change even if turbo is on or off.
PIN 2: 0.2v when turbo is on, 2.9v when turbo is off (measured against PIN 4).
PIN 3: +5v POWER
PIN 4: Ground

I did not expect that behavior from the turbo pins. I was sure PIN 2 was ground. No matter how it works, it still does the intended effect of turning off the LED when turbo is OFF.

Here is a picture of the current setup:

If I flip the 3-pin connector for the power LED around, I get yellow color.

Ok, any chance of a picture of the back of the Green/Yellow LED? I'd like to check how many pins is has.

Just to check I understand that photo correctly of the connector correctly: the red wire goes to pin 1, with its black pair going to pin 2. Then either of the black wires on the other connector go to pin 3, with the Green wire always on pin 4.
Is that right?

I think the turbo indicator is open-collector wired, which means pin 2 is either floating (so no current flows, so LED is off) or connected to Ground (current can flow through the LED from +5V to Ground. I'm more confused by the power LED since I was assuming that the black wires would be Ground, in which case connecting them to the +5 pin 3 would mean the LED wouldn't turn on as the LED would be reverse biased.

I'm wondering if it'll be possible to somehow use the Green/Yellow connector on pins 2,3,4 which would then free up the Red connector the HD.

you need to figure out if your 2 color led is a common cathode or common anode. type. 😀

2 color LED with three leads will have a common pin either cathode or anode. The ones with 2 leads will have bi directional LED in parallel can be same color (flicker-free if on AC) or different color.

Cheers,

weedeewee wrote on 2022-01-01, 21:41:

you need to figure out if your 2 color led is a common cathode or common anode. type. 😀

Well, we know (I think) at the moment that with the green wire on pin 4 (Ground) then one of the black wires on pin 3 (+5) produces Green and the other produces Yellow. So that sounds like the Green wire is a common cathode (-ve). But the colours seem weird for that. Why not have Green, Yellow and Black wires? Or two green and one black if they wanted to save on using different colour wires. Using Black for a +ve supply is odd. So I wanted to check a bit further before starting to connect anything to power.

snufkin, considering at the moment the double color led is only used for one color and that staring at the wire color is a bad idea, I can only add that we'll just have to wait for further info 😀

I hope it's a common anode.

weedeewee wrote on 2022-01-01, 22:42:

snufkin, considering at the moment the double color led is only used for one color and that staring at the wire color is a bad idea, I can only add that we'll just have to wait for further info 😀

Ah, I took when Vetz said "If I flip the 3-pin connector for the power LED around, I get yellow color." that would mean flipping the black-green-black connector, and the only way that fits on pin 3&4 (given the turbo LED on pins 1&2) is with the middle pin of the 3-pin connector (green) on pin 4 (Ground) of the header. I could have misinterpreted that though.

I hope it's a common anode.

Oh, so do I. That'd mean just bung the connector as is on pin 2,3,4 and call it done. I think. But at the moment I've got a nasty feeling it's not going to be that easy.

snufkin wrote on 2022-01-01, 21:28:

Ok, any chance of a picture of the back of the Green/Yellow LED? I'd like to check how many pins is has.

Just to check I understand that photo correctly of the connector correctly: the red wire goes to pin 1, with its black pair going to pin 2. Then either of the black wires on the other connector go to pin 3, with the Green wire always on pin 4.
Is that right?

That is correct. I've attached some pictures of the LED's:

snufkin wrote on 2022-01-01, 21:28:

I think the turbo indicator is open-collector wired, which means pin 2 is either floating (so no current flows, so LED is off) or connected to Ground (current can flow through the LED from +5V to Ground. I'm more confused by the power LED since I was assuming that the black wires would be Ground, in which case connecting them to the +5 pin 3 would mean the LED wouldn't turn on as the LED would be reverse biased.

I'm wondering if it'll be possible to somehow use the Green/Yellow connector on pins 2,3,4 which would then free up the Red connector the HD.

This case and the motherboard does not match. The case is from a Commodore PC 10, but a previous owner swapped out the mainboard with a generic XT turbo board. The POWER led was originally the RED one and the 3 pin black-green-black cable with the duo colored yellow/green was indicating if the floppy drives were in use or not (from what I read from the manual).

You can see the cables clearly in this photo of the original board (not my photo/computer):

snufkin wrote on 2022-01-01, 23:03:

Ah, I took when Vetz said "If I flip the 3-pin connector for the power LED around, I get yellow color." that would mean flipping the black-green-black connector, and the only way that fits on pin 3&4 (given the turbo LED on pins 1&2) is with the middle pin of the 3-pin connector (green) on pin 4 (Ground) of the header. I could have misinterpreted that though.

That is correct interpretation.

snufkin wrote on 2022-01-01, 23:03:

That'd mean just bung the connector as is on pin 2,3,4 and call it done. I think. But at the moment I've got a nasty feeling it's not going to be that easy.

I've already tested using the current cable on pin 2,3,4, but that didn't work. I get no light on the LED when turbo is on or off, and flipping the connector does not make a difference. I'd guess for to even work the connector would need to be rearranged so that green is on pin 4 (ground). From what I understand the LED is a common cathode since ground is the middle pin (green cable)?

I don't think a common cathode led will work for your purpose.
Rearranging the pins in the connector should be straightforward.

Ok, so we can be fairly certain that it's common cathode, which is a shame. I don't think rearranging the pins on its own helps, since we need the power LED on all the time (so one ground) and the other switched via the turbo to ground (so a second ground pin), and we've only got the one ground pin on the LED.

I figure there are a few options here:
1) Buy a 3 pin common anode LED, then connect that to pins 2,3,4. Probably easiest option. Not actually all that great as the two LEDs would share the same current limiting resistor. You can see the two 330 ohm resistors next to the connector, probably one goes to pin 1 and the other to pin 3. If both LEDs are driven from the +ve supply on pin 3 then they will be a bit dimmer. On the other hand, maybe that would appear correct, as in a colour change without a brightness change since maybe 2 dim LEDs are as bright as 1.

2) Go with your original plan and make a changeover device. Could probably cobble something together with a pFET and a resistor. Connect Green to Pin 4 Ground, one of the Black to pin 3 (+5 via a 330R), the other Black to Drain on a pFET. Connect pin 2 (Turbo Ground) to Gate on the pFET, then connect a 100k resistor from Gate to Source, and connect the Source to pin 1 (separate +5 via a 330R). Might work (simple simulation says it doesn't blow up), bit fiddly. If you have to buy a pFET then may as well just buy an LED.

3) Rejig your plans slightly. Use the Red LED as the turbo indicator (pins 1,2) as it needs its own ground. Connect the Green to pin 4 and one of the black wires pin 3. Connect the other black wire to a 330 ohm resistor, and connect the other end of the resistor to whichever is the activity switched output on the XT-CF-Lite. Depends on the exact version, but looking at one schematic shows that it will actively switch +ve and -ve, so it can drive either way unlike the Turbo pin which will only drive -ve/Ground. Still a bit fiddly and the LED might be inverted, so it's normally lit and goes out for activity, but probably the easiest option if you can find a resistor to hand.

From the look of the LED, there may be sleeved resistors on the outer 2 pins, are the sleeves lumpy enough to be hiding them.
I'm having half an idea for a circuit with 2 small PNP transistors, TR1 is driven by a resistor to base from turbo, emitter to +v, collector to LED green (via a resistor if there isn't one in the LED wire). TR2 driven by a resistor from C of TR1, and driving the yellow LED - common to ground, have to work out suitable R values to achieve a good saturation drive level - as TR1 must hit saturation to ensure TR2 turns off.
Kind of thinking the diagram in my head at the moment

Thanks for the replies, really appreciate it.

snufkin, I'm a bit worried about going for option 1 as the current LEDs fit perfectly to the sockets in the case. Please let me know if they are totally standard, if that is the case, then this would be a viable option.

Regarding option number two. From how I read it (please correct me if I misunderstand something), one of the black cables would always be connected to pin 3 (power). That would mean in a turbo OFF scenario the LED would get 5v from both black cables at the same time and that would need to trigger a color switch (or opposite, 5v in both is the standard at turbo ON and gives green color, turning it off would swap to yellow). Do you know how bi colored LEDs would normally behave in such a scenario?

Matth79 wrote on 2022-01-02, 19:13:

From the look of the LED, there may be sleeved resistors on the outer 2 pins, are the sleeves lumpy enough to be hiding them.
I'm having half an idea for a circuit with 2 small PNP transistors, TR1 is driven by a resistor to base from turbo, emitter to +v, collector to LED green (via a resistor if there isn't one in the LED wire). TR2 driven by a resistor from C of TR1, and driving the yellow LED - common to ground, have to work out suitable R values to achieve a good saturation drive level - as TR1 must hit saturation to ensure TR2 turns off.
Kind of thinking the diagram in my head at the moment

Thanks for thinking of an idea/diagram! No resistors in the sleeves.

vetz,

What are the measurements for the led that fits ?

edit: just checked on mouser and the only rectangular bicolor leds they have are common cathode. 🙁
- digikey only has common anode rectangular green/red @2mmx5mm in stock

vetz wrote on 2022-01-02, 20:34:

snufkin, I'm a bit worried about going for option 1 as the current LEDs fit perfectly to the sockets in the case. Please let me know if they are totally standard, if that is the case, then this would be a viable option.

Oh, it'll almost certainly be some sort of standard.

Ha, just did the same as Weedeewee. Went to mouser, over 4,500 through-hole LEDs. 'Great' I thought. Select common anode. Results remaining? 13. With rectangular case? 0.

Went to Kingsbright (big LED maker). 0.

Ok, that's probably not an option.

Regarding option number two. From how I read it (please correct me if I misunderstand something), one of the black cables would always be connected to pin 3 (power). That would mean in a turbo OFF scenario the LED would get 5v from both black cables at the same time and that would need to trigger a color switch (or opposite, 5v in both is the standard at turbo ON and gives green color, turning it off would swap to yellow). Do you know how bi colored LEDs would normally behave in such a scenario?

As I understand it (from looking at datasheet, I haven't tried this), there are two LEDs in the package and they can be turned on separately via the +ve side (the two black wires). So there's a Yellow and Green LED, of which you'd have one on permanently as the power LED, then the second added when the turbo was on, which would change the colour. You might need to experiment to find out which combination you preferred (Yellow, then Yellow+Green, or Green, then Green+Yellow). It's a shame the colours are yellow and green, since yellow already has green in it, so the colour shift won't be as big as if it was a red/green LED.

It'd be a bit more work (although not too much, I think) to make it in to a complete change over.

Sure you don't want to go with option 3?

[edit: I think you need one of these: https://uk.farnell.com/diodes-inc/dmc2710udwq … -sot/dp/3828400 and a 100k(ish) resistor. And a bit of fiddly wiring.
No idea if the link will work, but anyway, I had a go at a simple simulation, there's a clickable switch by pin 2:
https://www.falstad.com/circuit/circuitjs.htm … uh5TuS6RlNE1hAA

]

[another edit: It's possible the nCh FET won't turn off enough and whichever colour LED is connected to it might get dimmer but not turn off. Might be able to dump the nCh and just bung a diode from pin 3 to pin 2, so pin 3 gets pulled fairly close to Ground when the turbo switch causes pin 2 to go to ground. Vf on an LED will be lower than Vf on a normal diode. Could even get a schottky to get the Vf lower. Still not perfect, uses more components, not efficient (added diode just burns power when the LED is off) but probably good enough if the first doesn't work:
https://www.falstad.com/circuit/circuitjs.htm … AAuiRGZYGGYB+QA
]

With no resistors in the LED wires, adding a diode voltage drop can make a much simpler switch
Also, I don't know if "Elektor" or their TUP TUN DUG DUS definitions are still around, but the transistor is TUP (universal PNP) and the diode is DUS (universal silicon), your favourite 2N series, BC series, 2SC series etc.

Matth79 wrote on 2022-01-04, 00:12:

With no resistors in the LED wires, adding a diode voltage drop can make a much simpler switch
Also, I don't know if "Elektor" or their TUP TUN DUG DUS definitions are still around, but the transistor is TUP (universal PNP) and the diode is DUS (universal silicon), your favourite 2N series, BC series, 2SC series etc.

That looks similar to the pFET option, although parts may be easier to get hold of or borrow from other boards. And it's definitely the easier option to have one LED lit all the time rather than trying to do a change-over. A slight thing to watch out for is that the two +5 pins on the header are both connected to +5 through separate 330 ohm resistors. Not really a problem as 5V can be taken from a molex power connector, or anywhere else. I was trying to keep connections to just the motherboard header. Can nearly get away with just 3 pins of the header when doing the change-over version, except that causes a problem getting the nCh to turn on fully because of the voltage drop over the 330 ohm resistor.

I've also just realised I was being dumb with the original changeover circuit. I should have just connected the gates of the pFET and nFET directly together. That way there's no issue with the nFET turning off.

So, a possible (looks to work) way to do a change-over circuit, with one resistor and one dual n+p channel FET inserted between the 4 pin motherboard header and the 3 pin LED plug:

https://www.falstad.com/circuit/circuitjs.htm … uh5TuS6RlNE1hAA

[edit: I got the link for the simulation wrong:
https://www.falstad.com/circuit/circuitjs.htm … IsQvQSOJvVcGogA

Thanks for the replies guys! You're awesome

I'll look more into your diagrams and suggestions once I get some free time later today.