Firing up the pdt11. Dec put 5 volts on an LED?????

[Chris Zach]

Ok, this is weirder: I put the "bad" floppy drive on the bench and 
started to take a look at it. First I checked the LED (yes, it's an 
LED). With a bench voltage of 1.5 volts and a 100ma draw it lit up 
nicely in the IR (detected by phone camera, so nice they can see the 
light) and the photo transistor also seemed to work fine (at the sector 
hole resistance went from infinite down to about 500 ohms). That's good, 
so what is wrong?

I noticed I could crank the LED higher current-wise to 150 ma and the 
voltage was still <2 volts. Interesting. Then I hooked a break-out 
harness to the pdt11 to see what kind of voltage it was putting out to 
the LED.

It's putting out +5v whenever the unit is on. Maybe it's current 
limited? To check I hooked up the drive's plug to the breakout to see 
what the LED was seeing.

+5v. And even weirder, the LED was not lit.

What the heck is going on here?

So I put the LED on the bench for a bit of a destructive test. 
Disconnected the PDT11 from the breakout cable, hooked up the power 
supply, turned up the voltage and the LED came on, then went *off* at 
around 3v. At 5v it was dead off, no IR light as measured by the camera. 
Turn the voltage down, and it comes on again. Up and it goes off.

And unfortunately at 9v it died (CRAP!) as I turned up the current limit 
Yes, I forgot to set the voltage limit on the power supply, my bad, I am 
boo boo the fool...

But this is weird: It looks like DEC put an LED in there with no current 
limiting, and a straight +5 volts. And the LED is always on at this high 
voltage? With no current limiting resistor? This does not make sense, 
but the volt meter don't lie. I'm going to check the working drive to 
see if it is limiting the voltage somehow. I'd say there was a resistor 
in the LED assembly limiting the current, but if that's true my cranking 
the voltage to 9v should not have blown it up, and it should not turn on 
at low voltages then off at 5v.

Maybe the solution is to insert a resistor in series with the second 
drive at around r=e/i or r=5/.1 (100ma) or 50 ohms.

Does this make any sense?

C