3
\$\begingroup\$

I would like to preface this post by saying I have no formal education or training with electronics and am completely self-taught. I really enjoy designing home brew computers from the late 70's and early 80's as my hobby. I have made a couple in the past (6800, 6502, and now 8080) and they have all be successful, but all on two-layer PCBs. For this project, I have created an 8080 based computer (don't ask why lol) on a four-layer PCB but am having a problem with the power input.

It is my first four-layer PCB and everything works perfectly except on the power input. For some reason, when I apply a constant 5 volts with my bench power supply, the input voltage is always lower, depending on how many components are installed on the board. With no components, there is no voltage drop. When I install all the components into their sockets, it drops by about 1.2 volts. To make the circuit run properly, I have to increase the power supply to 6.2 volts so that the board power is exactly 5 volts.

I know that voltage drop is usually caused by resistance in the circuit, so that was my first point of testing. I measured several points across the entire board, and at any point between any two VCC pins or two ground pins, the resistance was always zero ohms. Between the power input pins and any pins are zero ohms. And in the wires from the power supply to the board is also zero ohms. (I know there is SOME resistance, but my multimeter read 0.01 at the highest point.)

I am not sure what else could be causing the board. The board is quite large at 8in x 10in. Luckily PCB prices have come down a bunch from what I have heard because I only had to pay $65 for it! Anyways, I did a lot of research when designing it so I could make the best board possible. The layers are as follows:

  1. Signals
  2. GND
  3. VCC (5 volts)
  4. Signals

I have several larger 100 uf capacitors at the power input and 0.1 uf bypass capacitors on all ICs on the board. There is also a -5v generator and a 12v boost converter, if that makes any difference. I am not sure what else to test here. Maybe the power supply is wonky, but I doubt it because the voltage is correct when read with the multimeter. Again, 6.2 volts at the power supply, 5.0 volts on the board.

Thank you so much for reading my post and any help would be greatly appreciated!

\$\endgroup\$
5
  • 3
    \$\begingroup\$ If the power supply is 6.2 volts when the voltage measured on the board is 5.0 volts, the voltage drop must be in the wiring between the power supply and the board. \$\endgroup\$ Commented Dec 14, 2021 at 23:52
  • 1
    \$\begingroup\$ A single dodgy component may be pulling more current than expected and causing the voltage drop. \$\endgroup\$
    – JYelton
    Commented Dec 14, 2021 at 23:55
  • 3
    \$\begingroup\$ So, what's the measured voltage at the connector, and at the power supply? And have you measured the current? \$\endgroup\$
    – TimWescott
    Commented Dec 15, 2021 at 0:01
  • 1
    \$\begingroup\$ Measure current, sounds like you have a short. Is the board also getting hot? \$\endgroup\$ Commented Dec 15, 2021 at 0:39
  • 1
    \$\begingroup\$ Can you post images of your board, with all 4 layers individually? \$\endgroup\$
    – awjlogan
    Commented Dec 15, 2021 at 20:51

4 Answers 4

0
\$\begingroup\$

To get a voltage drop of 1.2V across a cable with less than 10mΩ of resistance means you'd have 1.2V/0.01Ω=120A (or more!) of current, so something is definitely off. At this point you need to question everything.

  1. Make sure the test equipment is working (get known values for resistance and voltage and make sure it checks out). Make sure the multimeter is set to the right range.

  2. Make a better resistance measurement of the cable with a better DMM

  3. Make an AC measurement -- sometimes DMM's read AC with DC, the AC content should be low, in the 100mA to 10mA range. (put your meter in AC mode and measure the voltage across the cable.

And in the wires from the power supply to the board is also zero ohms

No cables are zero ohms 24AWG is 25mΩ per 1 foot of cable, and connectors also add several miliohms minimum so I would expect that you would see more than 10mΩ

\$\endgroup\$
1
\$\begingroup\$

Thanks for ally your replies! It looks like several of you were correct!

  1. It looks like the problem was a couple of things: firstly, the wires were getting really hot even though I could not measure a huge amount of current. I guess they could not handle the 1 amp required for these vintage electronics :) They were getting really hot while in use, but cooled down quickly once the power was disconnected. This is why I never found them to be a problem because I never touched them while it was on. They also weren't hot enough to be burning or create a smell, so this was an accidental find while trying to measure the current as you suggessted.

  2. Another problem was a rouge chip on the board drawing way too much current. It was right next to the power input and replacing it relieved a 0.4 voltage drop. The remaining voltage drop was relieved by replacing the cables with thicker ones.

Now that these problems are alleviated, there is only a 0.1 volt drop on the power input, which is totally acceptable. Thanks again for all your help!

\$\endgroup\$
1
  • \$\begingroup\$ That's a short circuit in that chip. I never turn on a new board design without having a current meter on it and an expectation of the correct current value. That way if there is a short I can turn it off before things are damaged. \$\endgroup\$ Commented Dec 16, 2021 at 20:26
0
\$\begingroup\$

If you are using a current-limiting power supply, and the current setting is lower than needed, it will begin to drop the output voltage. It may not be a problem, just need to adjust the power supply for your board.

\$\endgroup\$
0
\$\begingroup\$

The number of layers will probably not be related to your problem. If you have a proper PCB design.

The solution is a power supply that has sense lines and you connect the sense lines where you want to apply the specified voltage in order to compensate the voltage drop. But in any case don't do so if you suspect a fault in your circuit.

If you suspect an error, it is better to limit the current to the expected value. Look for hotspots with a thermal imaging camera.

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.