Williams system 6 - mpu power consumption

I think that's good idea to measure current consumption on both boards (I have "calibrated" 0,1 ohm resistor to attach
in series with supply - 100mV ON resistor means 1A consumption )
This method clarified "bad contact / hi current" question

Greetings from Croatia

(btw... clear and precise form of the definition of the problem - GREAT !! )

#5 9 years ago

BMB

gorgarspeaks

Inactive member

>viperrwk
No it was not a typo - I actually put it in IC 14 - will try IC17 this afternoon. Thank you!

>cro_pinman
Thank you but I am not sure i understand completely:
I believe you want me to :
1) put the board on the bench, attach ground
2) Attach +5V through a 0,1 ohm resistor?????
3) Measure voltage on both sides of the resistor?

A) Is this correct?
B) What information will the results give?

#6 9 years ago

MAR

MarAlb

Pinside member

Zeist

9y 53,650 814 19

B)........the voltage drop over the resistor tells you something about the power consumption.....U = I * R, P = I^2 * R, P = U^2 / R

#7 9 years ago

JON

Schwaggs

Gameroom Mods

10y 105K 4,760 10

Measure the voltage across the resistor (one probe on each side of the resistor).

I(current) = E(voltage) / R(resistance)

So to calculate the current consumed by your board, measure the voltage drop across the 0.1 ohm resistor and divide by 0.1. MarAlb is suggesting a 0.1 ohm resistor so you can do the math in your head.

The example he gave is 100mv (0.1 V)

0.1 volts / 0.1ohms = 1 amp

So in this example, .1 volts across a .1 ohm resistor = 1 amp of current.

If you measure a .55 volt drop across the resistor, you have 5.5 amp draw, etc... Using a .1 ohm resistor makes the math as simple as moving the decimal one position to the right.

You will need a high wattage resistor to use this method (5W minimum, 10 watts would be better). If you think there is a short on the board, a 5W or even the 10W resistor may get quite hot so only use the resistor test method for a short time until you know how hot it will get...

#8 9 years ago

Q Q

cro_pinman

Inactive member

Quoted from Schwaggs:
Measure the voltage across the resistor (one probe on each side of the resistor).
I(current) = E(voltage) / R(resistance)
So to calculate the current consumed by your board, measure the voltage drop across the 0.1 ohm resistor and divide by 0.1. MarAlb is suggesting a 0.1 ohm resistor so you can do the math in your head.
The example he gave is 100mv (0.1 V)
0.1 volts / 0.1ohms = 1 amp
So in this example, .1 volts across a .1 ohm resistor = 1 amp of current.
If you measure a .55 volt drop across the resistor, you have 5.5 amp draw, etc... Using a .1 ohm resistor makes the math as simple as moving the decimal one position to the right.

Yes, that's what I wanted to say.

#9 9 years ago

BMB

gorgarspeaks

Inactive member

Havent had time yet - I have both 5w and 10w resistors - will problbly have time in the weekend thank you all!

#10 9 years ago

BMB

gorgarspeaks

Inactive member

Unfortunately changing Leons to IC17 changed nothing: 2 lit LEDs.

PSU supplies 5,04V with nothing attached except my multimeter.
With MPU board attached voltage is 4,12V

Resistor-test:
Using 5W 27ohm resistor:
Voltage across the resistor (one probe on each side of the resistor): 3,53V
Voltage measured on MPU board : 1,24V (dont know if this gives further information.)

3,53V / 27 ohms = 0,130 amp
P = 3,53^2 / 27 ohm = 0,4615

Is that correct and what does it tell me?

#11 9 years ago

Q Q

cro_pinman

Inactive member

We did not understand !! You need resistor value 0,1 OHM (not 27 ohm ) to measure current, or if you haven't
0,1 ohm resistor you can measure current with DMM in series directly (with proper DMM setup )

#12 9 years ago

BMB

gorgarspeaks

Inactive member

I do not have a 0,1 ohm resistor. The sentence "suggesting a 0.1 ohm resistor so you can do the math in your head." made me believe that I could use another resistor value, but then I would not be able to do the math in my head.

When I measure current with DMM in series directly with the MPU board, I get 5.4 in the 200m position.

#13 9 years ago

AKB

Andy_B

Pinside member

Worksop

11y 50,550 516 3

The power consumption of the MPU isn't that relevant. The fact is that without the MPU connected you have good 5 but with it connected only 4.12v which is not high enough for the board to boot.

There is either a problem on the power supply board which means it is failing under load or there is an issue with the MPU board with the 5v shorting to ground somewhere.

1) Do you have some other game you can test the power supply board in?
2) Do any of the chips on the MPU board feel warm when connected to the power supply board?

Andy

#14 9 years ago

JON

Schwaggs

Gameroom Mods

10y 105K 4,760 10

The problem with using a 27 ohm resistor is that the voltage drop is too high. The 3.53 volts you measured across the resistor means the board is only getting 1.47 volts (5v-3.53v=1.47V to the board). None of the chips on that board are running (consuming power) at 1.47V.

The reason you use a .1 ohm resistor is to make the voltage drop as small as possible (and to make the math easy! )

#15 9 years ago

BMB

gorgarspeaks

Inactive member

Andy>> 1) Yes I have acces to a working Gorgar, and everything works in both Gorgars with the MPU board from one of them, and naither of them work with the other MPU board.
Also the bad MPU board is not working on my test bench and my working mpu board boots fine at the workbench.
so I know the problem is on the MPU board.

>There is either a problem on the power supply board which means it is failing under load or there is an issue with >the MPU board with the 5v shorting to ground somewhere.
I feel confident from the test mentioned above that my PSU board is not to blame. Which brings me back to the short-theory.

2) None of the chips (or any other components) on the MPU board feel warm when connected to the power supply board or to workbench.

With DMM I find no dead shorts except the ones mentioned, which I fixed.

I still have the feeling that these questions are important:
3) Am I right in assuming that I still have to find a reason for a large power consumption, before I can do anything else?
4) Is it correct that it is always a good idea to replace original electrolytic capacitors c23? I do not have DSR-meter.
5) Should Tantalum resistors c27 & C31 also be changed?

I have a replacement c23 at hand, and i might also have the tantalums.

Schwaggs>Thanks for the explanation. I will purchase a resistor - meanwhile - I feel that the tests performed shows that the problem is within tke MPU board and that the MPU board consumes too much power. I do not understand what extra information it will give to know how much to much - will this help me find the failing component(s).

#16 9 years ago

AKB

Andy_B

Pinside member

Worksop

11y 50,550 516 3

OK so the power supply board is good

#17 9 years ago

AKB

Andy_B

Pinside member

Worksop

11y 50,550 516 3

Stupid tablet keeps truncating my reply

#18 9 years ago

BMB

gorgarspeaks

Inactive member

Look forward to see what got truncated - hope you have the patience to write it again!

#19 9 years ago

BMB

gorgarspeaks

Inactive member

Andy send the message personally now, but I will reply to all so everyone can help.
I have testet all removable chips/ICs in my working machine. No shorts in them.
No ICs (or other components) get hot even after long power on.

I am inclined to change the electrolytic capasitor as these to my knowledge are the only components sure tu fail over time.

other suggestions?

#20 9 years ago

Q Q

cro_pinman

Inactive member

Did you check J2 connector (power in) on CPU board (solder )

#21 9 years ago

CMX

viperrwk

Inactive member

New Haven, CT

12y 42,650 4,167

You mean the electrolytic at C23 on the MPU board? Probably not the problem unless it is visually damaged. I used to change them as a matter of course but it usually didn't make a difference and now test all caps before changing them. And C27/C31 are tantalum caps that also rarely fail.

You should carefully inspect the MPU board to see if any prior work was done. If there was (likely given how old the machine is - getting harder and harder to find a machine that hasn't been molested somehow), it's possible that some damage was done which could be causing a short.

It's also possible that when you replaced ROM sockets you bridged two leads on the ROMs which would lock the board up as well. Double and triple and quadruple check your work. I've made boneheaded mistakes that I didn't think possible and until the fifth or sixth time I've gone back to check wouldn't of otherwise found.

Unless you have at least a logic probe, further troubleshooting will be very difficult.

viperrwk

#22 9 years ago

Q Q

cro_pinman

Inactive member

Quoted from viperrwk:
inspect the MPU board to see if any prior work was done.

OP
Quoted from gorgarspeaks:
When I measure current with DMM in series directly with the MPU board, I get 5.4 in the 200m position.

That's VERY strange !!!
Especially in "accordance" with voltage drop.... weird (have you another DMM to check this measurement ? )

#23 9 years ago

BMB

gorgarspeaks

Inactive member

I reflew j2 before enything else.
No components look damaged, no components seem to have been changed (except the ones I changed myself - as mentioned above).
I checked my new sockets for shorts (once) there was none. In 1 place there was excatly 1K ohm resistance to neighboring pin - no visible soldering mistakes here - I presume normal - but will check on the working MPU.

I will borrow another DMM to doublecheck

I have ordered a logic probe - but do not have one yet.

I will check my sockets again as suggested - maybe even 2-3 times.

Am I right in assuming that resistance of 1K can not be a short I made - if I made a short - resistance would be lower than that right?

Am I right in assuming that the voltage consumtion could be found by finding somewhere that has lower resistance in the broken board than the working one? If so - how much lower?

#24 9 years ago

ARS

thedefog

Pinside member

10y 93,350 5,673 72

1k resistance would not indicate a short, that is correct. A few ohms resistance in the case of a microscopic bridge is normal (like 3-4 ohms), but it is never as high as 1k. You can measure resistance between your gnd and +5v rail pins on an MPU and it could be lower than that. Then again, you are not directly measuring a resistor when you do that, and there are a number of decoupling capacitors and voltage dividers in the circuit that make that reading sort of pointless, unless it is a very low ohm rating, then you know you've got a short.

#25 9 years ago

BMB

gorgarspeaks

Inactive member