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It is pretty standard practice to clamp a signal to Vcc by using a diode and resistor. However, when this happens, some small amount if current (maybe ~50mA) is flowing back into the Vcc line. When using a switching regulator to supply the Vcc line, what implications does introducing this current source have on the system? Is this safe?

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  • \$\begingroup\$ If you're going to do this, at least clamp Vcc with a Zener. \$\endgroup\$ – Matt Young Jun 6 '19 at 15:12
  • \$\begingroup\$ @MattYoung The specific application is using a MAX9926 which has the diodes internally. It is the only chip on the 5V line. \$\endgroup\$ – user8908459 Jun 6 '19 at 18:01
  • \$\begingroup\$ That doesn't matter. If you're going to put more current into Vcc than your load current, the rail voltage is going to rise. You need to give that excess current a path GND. \$\endgroup\$ – Matt Young Jun 6 '19 at 18:03
  • \$\begingroup\$ Oh sorry, I misread your last statement. You are talking about putting a zener between Vcc and ground. That makes plenty of sense. \$\endgroup\$ – user8908459 Jun 6 '19 at 18:21
  • \$\begingroup\$ A fuller picture of your configuration would help prevent answers looking like shots in a shooting gallery. Knowing the energy source and magnitude and clamping rail characteristics etc are ESSENTIAL parts in the design - so not telling us them makes a good solution approximately impossible. \$\endgroup\$ – Russell McMahon Jun 7 '19 at 1:58
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It is pretty standard practice to clamp a signal to Vcc by using a diode and resistor. However, when this happens, some small amount if current (maybe ~50mA) is flowing back into the Vcc line. When using a switching regulator to supply the Vcc line, what implications does introducing this current source have on the system? Is this safe?

Brief: Energy sourced from another supply MAY cause clamping supply rail "pump up" if the clamped energy exceeds the mean energy drawn from the supply. In any case, noise spikes/surges from the clamped energy may have effects on system operation if not well filtered.

Detail:

Your question is too general to be answered with a single hard answer. You would need to know the energy source for the switching regulator and the characteristics of the Vcc supply, and in specific circumstances other factors may also be relevant, but some general principles can be established.

When energy from a inductive source (be it a switching regulator or an inductor carrying a current which is interupted) is transferred to another location, such as a power supply rail, then it will add to the energy able to be supplied by that rail.
"Excessive" energy from a switching regulator is usually supplied in "bursts" - usually one per switching cycle. So both the energy peaks and the average energy transfer rate need to be considered.

If the regulator is supplied from an energy source different from the rail that it is clamped to (eg 5V input supply with output clamped to a 12V supply) then the potential exists for the excess energy to "pump up" the supply. For example, if the clamped energy on average provides 10 mA at 12V, and the 12V supply has a load of only 1 mA, then the additional 9mA will cause the 12V supply voltage to rise. How fast and how far it rises will depend on the supply capacitor sizes and on the nature of the load. If left long enough the supply voltage MIGHT be pumped up to a substantially higher voltage - maybe 15V or 20V or more. If the 12V supply was supplied by a linear regulator it would be usual for the regulator NOT be designed to deal with excess energy on its output. This may lead to damage to or destruction of the regulator and/or back feeding of energy through the regulator to its input circuitry. While the 9 mA excess in this example would generally not cause damage due to current flow, a voltage of 20C or more on a 12V rail may damage active components or convince eg Tantalum capacitors that their life had come to an end and that they should emit magic smoke and flames. As happens.

If the switching regulator was powered from the same rail that the excess energy was being returned to "pump up" cannot occur as it is simply returning energy to the place from whence it has come.

However, in both cases the surges/spikes of energy my cause noise on the supply and maloperation of functionality. Worst case this may cause a "crash" due to the energy spikes.

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Related:

I have seen manufacturer's application notes where mains input circuitry clamps input current to supply rails either with clamp diodes or via IC body diodes. This practice risks "pumping" the local supply to voltages far above the intended rail voltage under some conditions.

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  • \$\begingroup\$ I am definitely sourcing more energy back than I am drawing from the supply. Does you answer also hold true for a linear regulator? \$\endgroup\$ – user8908459 Jun 6 '19 at 14:39
  • \$\begingroup\$ @user8908459 Linear regulators are also not designed t accept energy on their outputs. If the vclamped energy exceeds the energy take from the bus it will "pump up. Some linear regulators are damaged by backfeeding current and you could generally expect them to "behave badly". When backfeed is likely it is common to place a reversed diode from output to input so energy is transferred to the input bus. If this powers the source of the pumpup energy all is prov=bably OK enough - but if the clamped energy comes from elsewhere it may still cause issues. \$\endgroup\$ – Russell McMahon Jun 8 '19 at 13:29
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You are not back powering the SMPS regulator you are back powering the regulator load. Providing the load current on the switching regulator (Buck or Boost) is much higher than any current from a clamp there should be no problems at all.

However if the current drawn from the regulator is less than the current from the clamp (or many clamps) then the VCC will rise. What effect that may have will vary depending on the design.

Have a look at this answer on (relatively) simple input clamping where the current in the clamp IS NOT returned to the VCC supply. Be aware that if you have a lot of inputs being protected by just a diode clamp to VCC, the currents are additive under fault conditions. This becomes very important when your supply current from your regulator drops ...for example if you put your MCU to sleep it may be that you can only tolerate a few mA of clamp current before the VCC/VDD starts to rise. It's very easy to kill an MCU in this manner.

You might also gain by Reading this answer for a more effective way to handle high input voltages.

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