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Need help with a Compaq SLT 286 power board

Coder

Coder

Experienced Member
Joined
Apr 17, 2023
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I got this the other day and it won't turn on. I think it may be the caps on the DC/DC power board as I've read they are a common problem, as is usually the case in our hobby. I have replacements in my shopping cart on Mouser, but I wanted to verify the problem before ordering. I started poking around the board and to my surprise, I see the power rails I would expect to see. I have +12v, +5v, and -12v (roughly). But I'm not sure if they are all getting to the proper pins on J1. I was hoping that someone could verify some things for me on a known-good board, or could at least double check my logic. I also saw something I didn't expect and was hoping someone with more knowledge could tell me if it looks ok or if it might be a problem.

20231027_105913.jpg


I used the ( + ) side of the capacitors for my readings. I know this doesn't test the capacitors, but I also figured they were likely to be on the main power rails. At least the big ones. The values seemed ok to me, but my knowledge is limited. Here's the readings:

C500+4.90v
C510+18.37v
C511+18.89v

These values weren't under load, though. So I'm not sure if that voltage is stable.

20231027_110028.jpg

The two smaller caps above J1 had the following values on their ( + ) leads.

C504-12.90v
C503nothing (later confirmed it's shorted to ground)

Which brings us to Question #1: I noticed that the ( + ) and the ( - ) of C503 are connected to the same trace with a diode (CR503 between the two capacitors). Checking the ( - ) lead of C503, I get -12.90v as it comes directly from the ( + ) lead of C504. My question is this: Does it make sense for C503 to have it's positive lead connected to ground or does this indicate a possible fault?

I also made the following observations about the pins on J1. Pin 1 is lower left corner and pin 20 is upper right.

PinValue
-------------------
1GND (possibly connected to ( + ) of C503)
2GND
3GND
4GND
5GND
6+12v
7Battery Sense? I forgot I had the power brick plugged in (power switch was off) when I was checking for grounds. When I touched this pin in diode mode with the other lead on ground, the power brick's red light lit up. I assume this light is to indicate the battery is charging.
8+5v? I swear I got +5v on this pin during my first pass. But all subsequent passes read 0v. Not ground. Zero volts. Following the trace for this pin, it appears to terminate somewhere beneath a CD4049UBCM "Hex inverting buffer". The spec sheet lays out several usages beyond just as a hex buffer though.
9?
10? This pin also appears to terminate under the same CD4049UBCM hex inverting buffer as pin 8. Both are on output pins.
11GND
12-12v
13GND
14-1v? Not sure what this would be.
15+18v
16+18v
17+5v
18+18v
19+5v
20+5v

If you have a good power board for one of these and some free time, could you check what you get on yours? Especially pins 8 and 10. If I'm reading the schematic for the inverting buffer correctly, I may have a bad chip. The input on the inverting buffer for pin 10 is at just under 15v. I was expecting to see -15v, but it's not reading anything. Is my understanding of how these inverting buffers work correct?

Thanks for any help you guys can give. I appreciate it.
 
I tried to find the schematics for this beast, but I have not succeeded so far.

I would expect +/- 5V and (possibly) +/- 12V to be generated by the power supply.

It would not be unusual for the positive (+) side of an electrolytic capacitor to be connected to 0V/GND if it was a reservoir/smoothing capacitor associated with a negative supply rail.

Have you tried to find the schematics or a maintenance manual online?

If you can find one, I can help out.

Dave
 
A CD4049 is (indeed) an inverter package.

However, the power supply on the output of a gate will only swing as far as the power rails it is fed from.

If the 4049 is fed from 0V and +15V (say) and an input is +15V, then the output will be 0V.

Dave
 
I'm not sure I understand. When you say "the power rail it is fed from" are you talking about "VDD" (pin 1)? Because that is getting +5v. But the pin 7. labeled "C" on the datasheet diagram, is what's getting ~15v. I thought that pin 6 would be "inverse" of +15v, but I'll be honest, I don't really understand electronics enough to know what the "inverse" of a particular voltage would be.

Here's the pin diagram:
diagram.JPG
and here's a link to the datasheet: https://www.mouser.com/datasheet/2/308/cd4049ubc-1190588.pdf

As far as the schematics go, I haven't found anything either. Best I could find was the technical service manual from the excellent minuszerodegrees website: https://www.minuszerodegrees.net/manuals/Compaq/Compaq SLT286 and SLT386s20 - Maintenance and Service Guide.pdf

The only really helpful information I could glean from that is it lists the output voltages of the internal PSU as follows:

dc-dc output.JPG

Unfortunately, no information on which pins those voltages should be on. But it does show that I'm missing a -26v output somewhere. So that's something at least.
 
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In your particular case, the output can never go lower than Vss (0V/GND) or higher than Vdd (+5V).

+15V in, 0V out.

0V in, +5V out.

Checkout the transistor equivalent circuit for details of each gate in the datasheet.

Dave
 
I've got another question for you. I recently got a thermal imaging camera attachment that plugs into my phone. When I turn on the supply, I see a SMD resister climbing to over 100°C. Am I correct that under normal operation, a resistor, especially one that small, shouldn't get nearly that hot? It's marked "150" and when I check it with my meter in 200Ω scale, it reads .150. Would that indicate that a component further down the pike is drawing more current than the resistor was designed for? Or would you assume the resistor itself was failing? Or am I just overthinking things and hot-enough-to-scald-you is normal in this instance?
 
100 degrees C is too hot for my liking.

The measured resistance appears to be correct - so it is unlikely to be failing (yet)!

It certainly may have a much too high current flowing through it. Measure the voltage across the resistor and then use Ohm's Law to calculate the current and power dissipation and compare those to the specifications for the resistor in the associated datasheet.

Dave
 
Yeah, that was in circuit. I've ordered the caps from mouser. They should be here tomorrow. I'll take the resistor out of circuit and see what it reads.
 
I changed the capacitors out and that hasn't changed the behavior of the computer. The resistor now stays around 35C. I haven't checked the voltages yet, but kinda wonder if the PSU has been a red herring. When the computer is turned on, the keyboard lights flash like you'd expect. The power light comes on and stays on, but nothing displays on the screen. When I turn off the power switch, the screen momentarily flickers. I tried an external display and got nothing there. Is there a key combination necessary to switch to an external display?

I think it's not getting all the way through the boot process because hitting the caps lock or num lock key doesn't seem to do anything. But I know some computers don't respond to those keys until you get past the boot screen, so I'm not sure.

Does anybody have one of these? If you do can you tell me what should be happening at boot up with the keyboard?
 
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