1
\$\begingroup\$

I have this circuit running on 5v from a USB charger. When I measure the point on the power rail just before it is connected to the row on the breadboard that power the SN74LS173AN I measure a little more than 5v. When I measure at the other end of the jumper cable that is connected to the same row as pin 16 (Vcc) of the IC I measure less than 5v. If I remove the IC there is almost no difference.

The jumper cable is measured to be 0Ω and has 0v difference if fed with 5v and measured directly at both ends without the rest of the circuit.

I do not have a diagram of the circuit and I suspect that it is not necessary in order the explain the behavior.

Question: why does the power rail and the Vcc rail not show the same voltage?

Screenshot from an ocsilloscope, CH1/yellow is plugged into the power rail, CH2/blue is plugged into Vcc rail of the IC Screenshot from an ocsilloscope, CH1/yellow is plugged into the power rail, CH2/blue is plugged into Vcc rail of the IC

Photo of both probes plugged into the breadboard Photo of both probes plugged into the breadboard

Photo of the complete circuit Photo of the complete circuit

Photo of the circuit from another angle Photo of the circuit from another angle

Photo of the circuit from yet another angle Photo of the circuit from yet another angle

\$\endgroup\$
7
  • 1
    \$\begingroup\$ can you plug the jumper cable in 2 different slot of your breadboard? sometime the contacts of slot can be damaged and become resistive. \$\endgroup\$ – vrleboss Apr 9 '17 at 12:41
  • 1
    \$\begingroup\$ I have had connection problems with breadboards and bare copper wires. I now use tinned wires only. \$\endgroup\$ – Bruce Abbott Apr 9 '17 at 14:30
  • 1
    \$\begingroup\$ @vrleboss, I just tried connecting the oscilloscope directly to the jumper terminals while the jumper was connected to the breadboard and that yielded the voltage I was expecting so there must be something wrong with the breadboard. Thank you. \$\endgroup\$ – Robin Theilade Apr 9 '17 at 16:37
  • 1
    \$\begingroup\$ @BruceAbbott, I just tested using what I believe to be a tinned wire and that remedied the issue. The original wire was copper. Thank you. \$\endgroup\$ – Robin Theilade Apr 9 '17 at 16:40
  • 1
    \$\begingroup\$ As an aside: Don't stick scope probes into your breadboard like that. If the probe falls off to the side, it'll destroy the contacts on the board! If you need to measure a voltage on the breadboard, connect a short wire and use the clip attachment on the probe. \$\endgroup\$ – user39382 Apr 9 '17 at 18:07
1
\$\begingroup\$

Question: why does the power rail and the Vcc rail not show the same voltage?

You are seeing the effect of the current consumption of your 74LS173. Using Ohms Law, the voltage drop you are seeing, which appears to be up to (5.08 V - 4.92 V =) 0.16 V, means two things:

  • There must be a resistance between the measuring points.
  • There must be a current flowing between the measuring points.

One point to note is that you have no appropriate decoupling capacitors close to the TTL ICs. While it's not the cause of the behaviour you observe, you may get incorrect IC behaviour as a result of their omission.

Now back to your measured voltage difference. You said:

The jumper cable is measured to be 0Ω

Although that isn't true in an absolute sense, if we assume the jumper wire itself has a truly negligible resistance, then there is non-negligible resistance elsewhere, likely in the breadboard contacts (as mentioned in the comment by vrleboss).

Since you have only measured the voltage difference, there are two "unknowns":

  1. the current consumption of the IC, and
  2. the resistance between the measuring points (not just the resistance of the jumper wire).

Having two unknowns means we cannot know which of them is unexpectedly high (or perhaps they both are) until you take more measurements.

Looking at the breadboard, there seems to be the potential (no pun intended) for a higher-than-expected current consumption due to floating inputs. The jumper wires connected to 74LS173 pin 15 and pins 9&10 (which are all inputs) seem to be unconnected and therefore floating. This situation can lead to an LS TTL IC drawing excessive current. If the IC is drawing excessive current, then a relatively small resistance could produce the voltage drop you are reporting, and the IC itself would also be noticeably warm to the touch.

Depending on your available test equipment, I suggest:

  • careful resistance measurement between the points where you measure the voltage drop (with power off, of course).

    Try using different breadboard contacts in that IC's power path, to see if the measurements change.

  • careful current measurement between the points where you measure the voltage drop.

    When measuring the current, you need to minimise the effects of the "burden voltage" to avoid the current measurement causing other side effects.

The TI 74173 and 74LS173 datasheet mentions a typical 74LS173 supply current of 19 mA (page 7) under specific conditions. However, if my concerns about floating inputs are correct, your circuit does not meet the conditions required to expect that current consumption, and I wouldn't be surprised if it is higher than that value on your board.

\$\endgroup\$
5
  • \$\begingroup\$ You're right about the capacitor, it exists on another module which is not visible in my question. \$\endgroup\$ – Robin Theilade Apr 9 '17 at 17:18
  • \$\begingroup\$ @RobinTheilade - FYI if the capacitor you mention exists on another module, then it won't be close enough to those TTL ICs in your photo to be an effective decoupling capacitor for them. As far as those TTL ICs on your breadboard are concerned, a decoupling capacitor on another module doesn't exist (due to the inductance of the wiring between them and the other module). \$\endgroup\$ – SamGibson Apr 9 '17 at 17:27
  • \$\begingroup\$ vrlsboss was on the right track, but Bruce Abbott had the answer, copper wires together with the breadboard yields 2Ω while tinned wires yield 0.1Ω both measured over a 5 point rail. About the unconnected pins, that was me removing components to simplify the circuit for debugging. Thank you for your detailed answer. \$\endgroup\$ – Robin Theilade Apr 9 '17 at 17:30
  • \$\begingroup\$ I haven't thought about the inductance. I'm following Ben Eater's guide on how to build an 8-bit computer (link) and while he does not always do it by the book it seems to still work for him. But I guess it doesn't hurt to add a few more capacitors. \$\endgroup\$ – Robin Theilade Apr 9 '17 at 17:33
  • \$\begingroup\$ @RobinTheilade - I'm glad my answer was useful. As I explained, resistance was involved somewhere, as you have confirmed by performing the resistance measurement I suggested. In that video you linked, although some power-related wiring is still hidden, there appears to be a decoupling capacitor on the breadboard, which is hidden in the initial view. Look at around 5:40 in the video, top-left corner. He certainly isn't following best-practices for decoupling, but slow clock speed & good PSUs might allow the bare minimum decoupling. You may or may not be so lucky. Your choice whether to risk it! \$\endgroup\$ – SamGibson Apr 9 '17 at 17:49

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.