13
\$\begingroup\$

I am working with a 1970s era computer system with multiple functional boards on its bus (similar to S100 but not S100.) Unlike S100, this system distributes power as already-regulated 5V (and 12V and -5/-12V) to each board in the system.

Every board, near its power connections, has a diode and a 100uF capacitor. I am not a trained engineer and want to understand the purpose of these components in this case:

enter image description here

Sometimes the diode is a 1N4001, sometimes it's a 1N914, but for all boards there's this sort of network around the power connection. (On boards that use multiple of the voltages, it's more complex, with multiple capacitors and diodes across the rails.)

I understand the frequent appearance of the ceramic disc 0.1uF decoupling (or is it bypass?) capacitors near each IC across Vcc and GND as a noise reduction function.

Is the larger (100uF) electrolytic near the power connection serving the same purpose but for the whole board, or is it doing something else?

Is the diode there as a kind of reverse current protection? What circumstances might it be trying to protect against?

\$\endgroup\$
6
  • \$\begingroup\$ It may be useless on it's own as you have drawn it. If you can provide a schematic that shows more, it may start to make sense. For example, many power supply circuits with both positive and negative output voltages typically implement diodes to protect from various conditions that might happen. \$\endgroup\$
    – Justme
    May 29, 2023 at 15:01
  • \$\begingroup\$ Sure. Here's a layout and schematic for one such board. C1 and D1 at the upper right of the layout (and right edge of 1st schematic page) are the ones in question. Thanks sphere.computer/resources/schematics/… \$\endgroup\$
    – BZo
    May 29, 2023 at 15:13
  • \$\begingroup\$ Is the 1N914 mentioned in the documentation somewhere? I would be real nervous using one instead of 1N4001 here, it's a bit weak. \$\endgroup\$
    – pipe
    May 29, 2023 at 15:17
  • \$\begingroup\$ @pipe: On the schematic I linked above, the diode is a 1N914. But on other boards it's a 1N4001 appearing to serve the same function. When you say "weak" can you elaborate, as if I don't know too too much about diodes? :) \$\endgroup\$
    – BZo
    May 29, 2023 at 15:18
  • 1
    \$\begingroup\$ Weak as not being not able to carry much current. Probably opens up thus acting more like a fuse, but across the supply rails, as opposed to being in series. \$\endgroup\$
    – SteveSh
    May 29, 2023 at 15:43

6 Answers 6

19
\$\begingroup\$

The diodes are reverse voltage protection. They conduct if the voltage is reversed shorting out the supply enough to blow a fuse or trip a crowbar. It’s done this way rather than a series diode to avoid the voltage drop a series diode would introduce.

The large capacitors are reservoir caps. They are there to provide a reserve of current when there is a momentary high draw from the circuit that would cause a voltage drop in the resistance between the board and supply.

\$\endgroup\$
4
  • \$\begingroup\$ Thanks. Is the idea on the diode that if there is a problem upstream in the DC supply (or, say, the power connection leads to the board get reversed by accident) it protects the rest of the components on the board and instead shorts with the hope of blowing a fuse upstream? \$\endgroup\$
    – BZo
    May 29, 2023 at 15:16
  • 2
    \$\begingroup\$ I think GodJihyo is right in that it's there for reverse voltage protection. And a '914 makes perfect sense given the year those boards were made. They were the jelly beam of diodes back then. \$\endgroup\$
    – SteveSh
    May 29, 2023 at 15:40
  • 6
    \$\begingroup\$ But, a '914 is a pretty wimpy diode for protecting a power rail against reverse voltage. \$\endgroup\$
    – SteveSh
    May 29, 2023 at 15:41
  • \$\begingroup\$ @SteveSh Thanks, that's all really helpful context. \$\endgroup\$
    – BZo
    May 29, 2023 at 17:03
10
\$\begingroup\$

When you have a system with bipolar power supplies (+/-12, +/-5V), various start-up, shut-down and fault conditions can lead to reversed voltage on one of the supplies, which can destroy chips.

It's cheap insurance to include a reverse-biased diode. See, for example, this schematic from here:

enter image description here

The 100uF caps are just bulk storage.

A smaller diode such as the 200mA-ish glass diodes might be deemed adequate if the loads on the supplies are minimal.

\$\endgroup\$
1
  • \$\begingroup\$ Thank you for this diagram with the added notions of the various scenarios that can introduce momentary bad conditions \$\endgroup\$
    – BZo
    Jun 1, 2023 at 18:31
5
\$\begingroup\$

It's good practice as all cables have inductance. Even if your source have ideal stable 5.00 V at all times, if it's say a meter away, you will have some tens of nH of unwanted inductance in series. If your downstream device have no steep load changes what so ever, you may get away with it from an over/undershoot standpoint and you only need your capacitor for EMI purposes.

Your average power supply will however happily try to respond to load steps as fast as it can and if you can't tolerate much overshoot, you need a capacitor to "terminate" your cable and once again make it appear as stable 5 V despite fast load changes. It will however not take away any resistive drop from the cable resistance.

When you say 1970 era computer, I would expect honking amount of current on the 5 V and any over or undershoot would suffer from ΔU=L*di/dt. L may be small, but di/dt can be large and the allowed ΔV overshoot very small.

As for EMI, capacitors on any constant voltage rail is almost always wanted and the input (or output) of any device is a prime place to put them. 100 uF electrolytic normally have too high ESL to reach high enough up in frequency to cover all your needs, so hopefully there are some MLCCs in parallel somewhere.

Green curve is 10 nF, blue is 100 uF: enter image description here

Diode is as GodJihyo says reverse voltage protection. It requires a fuse or some other kind of current limitation in the system to function.

\$\endgroup\$
1
  • 1
    \$\begingroup\$ So why is the diode necessary in this arrangement? \$\endgroup\$
    – pipe
    May 29, 2023 at 15:15
1
\$\begingroup\$

The diode provides low-energy reverse-polarity ESD protection during handling.

\$\endgroup\$
2
  • \$\begingroup\$ May be a dumb question: in the context where the board is not connected to power (ie during handling) , how does esd current flow such that the diode helps, if there’s no ground connection at the power plug? Thanks! \$\endgroup\$
    – BZo
    May 29, 2023 at 16:10
  • 1
    \$\begingroup\$ If esd or some other potential appears across the power and ground input pins, reversed from normal polarity, the diode conducts and prevents a large negative voltage from appearing across all of the circuit ICs. It's not great, but better than nothing. \$\endgroup\$
    – AnalogKid
    May 30, 2023 at 3:17
1
\$\begingroup\$

Worthy of mention is that tantalum capacitors are very sensitive to reverse voltages. With a multi-rail design (+12, +5, -5 etc) it's quite possible that a momentary reverse voltage could be experienced during power-down. Obviously, a signal diode would quickly die if the supply were connected incorrectly, but it would be able to protect against small residual currents.

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

A very nice blog post that I recently read explains the capacitor:

In the end, even seemingly minor digital switching can cause significant voltage fluctuations and electrical noise across the entire circuit.

[..]

unexplained and hard-to-diagnose issues can creep up with ease.

[..]

the purpose of decoupling capacitors: they are placed across the voltage supply lines and physically close to the offending chip to handle switching transients while preventing high currents and minimizing voltage fluctuations in other parts of the circuitry.

As you describe it, it sounds like the capacitor may be in a suboptimal place for this purpose, but:

[..] the understanding of decoupling caps continues to be hit-and-miss. Some folks skip them altogether and live to tell; others follow ancient lore of uncertain origins, producing monstrosities [..]

\$\endgroup\$
1
  • 1
    \$\begingroup\$ Not only is the cap shown in OP's question not in a good location for typical decoupling of digital chips, it is way too large a value for that function. A 100 uf cap has way too much self inductance to be an effective decoupling cap, even with early 70's (54xx) series logic. \$\endgroup\$
    – SteveSh
    May 30, 2023 at 11:52

Your Answer

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

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