VOGONS


Turbo LED with 4 pins?

Topic actions

First post, by pewpewpew

User metadata
Rank Oldbie
Rank
Oldbie

J3 is marked 1 2 3 4. There is nothing online or in my parts bin has only 4 pins.

1 and 4 are common. 2 goes to two resistors on the LCD. 3 goes to the emitter legs of the pair of C945 NPN.

Maybe?: 1 2 are power, - +. Then maybe 3 4 are spliced into the positive wire from the board's TURBO LED jumpers to the LED itself?

I'm thinking that perhaps the transistors are rigged as a switch set to flip when it senses 5v.

But I don't know, and I'd rather not risk my only unit by simply plugging it in to find out. Also something is nagging me about that 4 being common with 1. I don't see you why you would do that in the arrangement I suggested, and suspect it would cause a short. Possibly the diode aspect of the LED is used in some way to avoid that?

file.php?id=15723

Reply 1 of 4, by pewpewpew

User metadata
Rank Oldbie
Rank
Oldbie

Powered breadboards rock.

file.php?id=15726

So, I'm not sure at all now.

Power applied to 3 4 as G + shows '33'. A two-socket power plug applied any other way does nothing.

By fiddling, I've figured out a system that uses three of the four pins. Shorting #2 to ground switches the state to '16'. Since the Turbo switch is the 3-pin type, I can attach the remaining pin to the LED lamp, which in turn is attached to 5v.

PROBLEMS

- I can't think of a way that uses all four pins.

- In each LED character state, I have faint bleed-in of the 'off' state. It doesn't really show in the photo. The left vertical of that second character is faintly lit -- just enough to be visually undeniable and a little annoying. You get the equivalent ghost of 33 behind a 16.

I presume that's voltage bleed-through in the circuit. Is this normal? Should I be concerned?

NOTES

The original LED states were 16 and 8.

MINOR QUESTION

Should I have set the half-state for 16 or for 17?

EDIT -- screwed up already. Demonstrating my utter ignorance, I burned out the LED lamp in my test setup. Vague memory tells me i should have had a resistor inline. But there won't be anything like that when things are installed, right? Why would the 5v of the computer be different from the 5v off this breadboard? (Cheap multimeter reads 5.04v for the breadboard.)

Reply 2 of 4, by SquallStrife

User metadata
Rank l33t
Rank
l33t

The "bleed through" is probably just because the circuit design is minimal (read: lazy), and some current flows base-emitter or base-collector through the the transistor on the "off" side.

pewpewpew wrote:

EDIT -- screwed up already. Demonstrating my utter ignorance, I burned out the LED lamp in my test setup. Vague memory tells me i should have had a resistor inline. But there won't be anything like that when things are installed, right? Why would the 5v of the computer be different from the 5v off this breadboard? (Cheap multimeter reads 5.04v for the breadboard.)

I doubt it's because you powered it from 5v.

More likely, it's because the pins aren't meant to be shorted, rather, they're meant to be driven high or low by the motherboard's "Turbo LED" output, which is already current limited.

VogonsDrivers.com | Link | News Thread

Reply 3 of 4, by pewpewpew

User metadata
Rank Oldbie
Rank
Oldbie

Thank you.

I expected to follow up with some sort of solution too, but that's still eluding me. I can figure out a method that uses three of the pins, and the other half of the Turbo switch, but it is inelegant. I still haven't figured how they meant this to work using all four pins.

This LED panel and the switch are original to the case, which at least tells me they did not use both sides of the switch.

pins 234 are --+ just like the typical three-socket connector that runs off a Molex. In standard practice the -+ pair are powered by the Molex, and the final - runs off to a single socket.

A lot of turbo panel jumper-info label that one as 'S' (for single?) and say it goes to the neg Turbo LED header on the motherboard. But these panels all have two more pins to power the Turbo LED themselves.

I could kinda see this panel working if that single socket goes to the neg Turbo LED header AND that negative pin was only connected to ground when the Turbo switch is closed. My motherboard doesn't do that, but I've no idea about the 386 that sat in there originally.

And hitching up the LED would remain a mystery. The final pin '1' on the panel is a +5v, but you would still need to switch this somehow, and the only leftover pin is the +5v of the motherboard's Turbo LED.

So I remain chasing my tail with this one. Things don't quite fit.

I /am/ fixing up the other bits while I think about it, but it's just about at the point where I'll go with the inelegant solution so I can button up the case.

Reply 4 of 4, by pewpewpew

User metadata
Rank Oldbie
Rank
Oldbie

FAIL. I didn't figure it out. Instead I've fudged it as follows.

  1 | 2 | 3 | 4 
5v | G | G | 5v
n/c| S | - | +

The S-+ is the usual 3-socket connector that is a -+ spliced from a Molex, and the S is the solo black wire that usually goes to the motherboard Turbo LED neg header.

With that setup connected to my board, selecting Turbo will show the LOW value on the LED panel, so I've rearranged the jumpers to reverse things.

This would leave two +5v sources to power a Turbo LED -- pin #1 at the panel, and the remaining half of the motherboard Turbo LED header. To use either I need a now-unavailable neg pin.

So fine then. Made a little 2-to-3 adapter to give me a pair of neg pins at the motherboard Turbo LED header. Then it's just jack-in the regular two-socket Turbo LED plus that solo black socket from pin #2.

Works fine but for the lingering worry that follows faking it because you couldn't figure out the Right way.

Should remark that the +5v on the panel's #1 pin is possibly as good but the traces show its voltage is coming direct from the PSU - no resistor. There may be some sort of resistance on the neg leg somewhere, but also perhaps not, because neg sides are typically wired as 'ground'.