What is the most indestructible voltage regulator that won't fail short?

[seqcst]

I am extremely worried about my voltage regulator failing short and frying my electronics see: AMS1117 short circuit protection not working (https://electronics.stackexchange.com/questions/530181/ams1117-short-circuit-protection-not-working)

Some context: I'm a beginner at electronics I want to power a handful of 5V electronics - some passive infra red sensors which draw only 100-200 uA each, as well as some active radars which may draw around 100mA each.

Maximum current draw will be less than 300mA.

The datasheet on my PIR sensors say that a stabilized power supply is needed so I think I do need a voltage regulator to regulate the already-regulated output of my wall wart. Now I'm wondering which voltage regulator I should get and whether I should get a 6V or 7.5V wall wart.

The advantage of the 6V wall wart is that even if the voltage regulator fails short, it won't fry my components (which can tolerate up to 6V), whereas if voltage regulator fails short with the 7.5V power supply then the 7.5V input may fry my sensors.

But the advantage of the 7.5V power supply is that it allows me to chain two voltage regulators - maybe one LDO to step it down from 7.5V to 6V, and then another LDO to step it down from 6V to 5V? That sounds more complicated but it should result in a less noisy power supply? And also, having two voltage regulators provides some redundancy in case one of them fails short?

Originally I was thinking of using an Arduino Uno for powering my electronics but I've read many reports of the AMS1117 failing short so now I'm worried about buying an Arduino Uno since the Arduino Uno internally uses one of those AMS1117 voltage regulators and if that fails then my precious electronic components will be fried.

My next thought was to buy a breadboard power supply like the DFRobot DFR0140 but I saw that one also has a reputation for failing short and internally it also uses the AMS1117 voltage regulator.

I've seen many people online saying that the 7805 voltage regulator will not fail short but some other people have said that it is possible that even the 7805 can also fail short.

Other people have said that the OFFICIAL AMS1117 should have built-in over-current protection and thermal shutdown and thus all of those online horror stories are due to people using Chinese AMS1117s. But aren't most electronic components made in China now? And DFRobot is a reputable company, it presumably uses properly sourced components including the AMS1117 - surely DFRobot is not using the fake AMS1117s. Plus, there have been reports of the Arduino's voltage regulator failing too - surely not all of the failed Arduinos were also using the fake AMS1117s??

I see many people online saying that the 7805 is "bulletproof" and "indestructible" but it has quite a large voltage dropout - 2 volts - compared to modern LDO voltage regulators which means it has to dissipate a lot of heat, so I'm worried that it will overheat with 300mA from 7.5V voltage (2.5V * 0.3A = 0.75W which is quite a lot).

If I instead buy a 6V wall wart and then use a low dropout voltage regulator then the voltage regulator won't have to dissipate as much heat (1V * 0.3A = 0.3W which is much less) so maybe I won't need a heat sink in that case?

But I have only seen online comments about the 7805 and how bulletproof and indestructible it is. I can't find online comments about the modern low dropout voltage regulators so I don't know how reliable these are. Are they as bulletproof and indestructible as the 7805/LM340A?

I also thought of buying some automotive-grade voltage regulators. Reading the datasheets, it sounds like those should be even more indestructible than the 7805, but those have so many pins and look really complicated to wire up. Also I couldn't find online comments about the reliability of those things either.

What is the most indestructible voltage regulator that I can get so that I won't have to worry about it failing short and frying my electronics? And should I get a 7.5V or 6V wall wart?

Thanks a lot!

[CaptDon]

I doubt you can find any that have a 100% guarantee against failing shorted. The only protection in military and hi-rel industrial settings was an SCR crowbar set to trigger at about 5.4 to 5.5vdc. This means your input power may also need to be fused since the output will appear to be dead shorted. This will protect your electronics but you may also get nuisance trips and you never want to slow down or delay the trigger event.

[BrokenYugo]

Yeah, adding a crowbar circuit is about as good as it gets. No regulator is fail proof, I've seen a couple 7805s in particular fail shorted in-out. All linear regulator ICs have a pass transistor in them and that transistor can and will fail if overloaded/overheated like any other.

I've been meaning to make a bunch of crowbar circuits and add them to everything I own with a regulated 5 volt rail, just as insurance. Be careful selecting the SCR, they aren't all rated to clear fuses.

[Doctorandus_P]

Both LM317 and  the 7805 types are quite robust. But the "indestructibility" is a property of the complete circuit design, and not of a single component. Apart from part failure, solder joints can also fail, and this can also result in a too high output voltage.

Extra ESD protection is also a way to make a circuit more robust. And it is common to add more circuitry to make it more robust. The crowbar example already mentioned is a good example of this.

[CountChocula]

I don't think that “bulletproof” exists (if for no other reason that you could just get a bad part that goes up in smoke the first time you power it up)… it's all about layers of protection.

In a production environment, your first line of defence is picking the right parts for the job. An LDO's reliability isn't just dictated by the amount of current it supplies, but also by the voltage drop it needs to effect, since the difference between input and output is going to be dissipated as heat. If you have to deal with a large input/output voltage difference and significant current draw, without some form of heat sinking, the life of the LDO will be shortened at best, and at worst it will simply heat up and self-destruct.

If, on the other hand, you leave yourself ample margin, most jellybean linear regulators will work well for a really long time. Failures do occur, I suppose (especially with older parts that have been in production for decades), but the reason why these ICs are so cheap isn't so much that manufacturers are cutting corners—it's just that the designs have been around for so long that they have been optimized to the point where they can be manufactured in huge quantities at the lowest possible price.

Now, it's a different story when you're developing something; here, the most likely cause of failure isn't a bad part… it's your hands

If you read through the reports of 1117s and 7805s failing, I bet you'll find that, in most cases, the problem was caused by someone miswiring them and accidentally shorting their outputs to ground. In these circumstances, whatever protection is built into the ICs is going to be of limited help, but the problem isn't the reliability of the part… it's the engineer using it!

Given that accidents happen, particularly when you're a beginner, you simply need an additional layer of protection, like a good bench power supply that can provide both constant voltage and constant current. When you first power up your circuit, you choose the minimum voltage at which you expect it to work, and only allow a tiny amount of current before CC mode kicks in. If you happen to have made a mistake, this inserts an upstream safety mechanism that helps you fail safely and protects your circuit from releasing the magic smoke.


—CC

[seqcst]

Yeah, adding a crowbar circuit is about as good as it gets. No regulator is fail proof, I've seen a couple 7805s in particular fail shorted in-out. All linear regulator ICs have a pass transistor in them and that transistor can and will fail if overloaded/overheated like any other.

I've been meaning to make a bunch of crowbar circuits and add them to everything I own with a regulated 5 volt rail, just as insurance. Be careful selecting the SCR, they aren't all rated to clear fuses.

Wow I did not know that even the 7805 can fail short! Thanks for sharing that.

Just to be clear: You are talking about the current rating on the SCR? The current rating for the SCR just needs to be higher than the current rating on the fuse, right?

[Zero999]

Just about all semiconductors can fail short circuit.

The SCR needs to be rated to a much higher current than the fuse. Semiconductors often protect fuses, rather than the other way round. Put a 2A fast blow fuse on the regulator's input and use an SCR rated to 6A or more.

[mawyatt]

I doubt you can find any that have a 100% guarantee against failing shorted. The only protection in military and hi-rel industrial settings was an SCR crowbar set to trigger at about 5.4 to 5.5vdc. This means your input power may also need to be fused since the output will appear to be dead shorted. This will protect your electronics but you may also get nuisance trips and you never want to slow down or delay the trigger event.

Supply Crowbars were often used in military electronics, we even used them for equipment nuclear survivability as well. Here all the system PN junctions would "turn on" due the ionizing radiation from a nuclear blast, and as little as 10mJ of energy on a single pin could damage/destroy a sensitive IC. This limited the amount of local decoupling capacitance and the main supplies were quickly Crowbarred during an "event".

The Crowbar was an high current SCR with 3 large PN Junctions (recall we used 2N3055s) in X,Y and Z axis to "detect" the ionizing radiation. These PN junctions were wired OR to the SCR Gate in reverse bias state and would conduct when exposed to the radiation and turn on the SCR.

We had an in-house massive Flash Xray system that was utilized for ionizing radiation simulation testing and compatibility verification.

Lead shielding and all, we never wanted to "be around" when they ignited that Xray system 

Best,

[Ian.M]

Just to be clear: You are talking about the current rating on the SCR? The current rating for the SCR just needs to be higher than the current rating on the fuse, right?

No,  Fuse characteristics are *far* more complicated than that.  Lets take a look at an example one,  Bussmann GMA series, fast 20mm glass fuse: https://www.sparkfun.com/datasheets/Components/General/glassfusedatasheet.pdf
and specifically the 1A rated one. Key parameters for our application are its cold resistance of 0.163 ohms and its I²t rating of  0.48, which is a measure of the energy it takes to blow it (though not dimensionally correct as energy, as it would need to be multiplied by its effective resistance).  The graph on page 2 shows that this 1A fuse needs a little over 7A to blow in 10 ms, and extrapolating from the I²t rating you'd want over 22A to blow it in 1 ms.  The TRIAC or SCR must have a pulse current rating high enough for long enough  to clear the fuse, and if you don't want it to be a sacrificial one-shot, that current rating must be sustained  for long enough to clear the arc.  You need some margin as well as the TRIAC or SCR may well be called on to crowbar the fuse, starting off considerably warmer than the nominal 25 °C its rated at.  Also you need to consider the inductance and total DC resistance of the crowbar loop and on state voltage drop across the TRIAC  or SCR to ensure there is enough current to clear the fuse in the desired time, but not so much that it exceeds the TRIAC  or SCR's pulse rating.

There's an additional fly in the ointment - The failed regulator may has enough DC resistance to limit the current and prevent the fuse clearing in the desired time, which may result in the TRIAC or SCR cooking itself, and if the current doesn't reach the min. fusing current may heat the failed regulator enough to be a fire hazard or to cause heat damage to the board or nearby parts.  To avoid this you may need to crowbar upstream of the regulator, but sensing the downstream voltage, with an adequately pulse rated diode from regulator output to input, cathode positive, so the crowbar can still remove the overvoltage from the load quickly.

Zero999 has suggested a factor of six greater SCR continuous rating vs the fuse rating, but you'd need to check datasheets and do the math to confirm that.

[bdunham7]

I am extremely worried about my voltage regulator failing short and frying my electronics see: AMS1117 short circuit protection not working (https://electronics.stackexchange.com/questions/530181/ams1117-short-circuit-protection-not-working)

Some context: I'm a beginner at electronics I want to power a handful of 5V electronics - some passive infra red sensors which draw only 100-200 uA each, as well as some active radars which may draw around 100mA each.

Maximum current draw will be less than 300mA.

A series regulator shorting or going rogue will cause an overvoltage, but a shunt regulator going short will simply blow the input fuse if there is one and overload the supply if there isn't.  Since your current requirements are reasonable, you could use a simple zener shunt regulator after the series regulator as a protective device.  The common saying that fuses don't protect semiconductors is only half true.  Fuses often can't act quickly enough to save semiconductors from a fault that causes overvoltage, but if you can turn the overvoltage fault into an overcurrent then fusing will work very well.  Simple devices like rectifiers can usually take very high peak currents without failing as long as the duration is short--the failure mechanism is entirely thermal (to a point). 

For your application, a 5-watt 1N5339B zener after the regulator and an MDL-1/2-R delay fuse before the regulator would do a good job of protecting against both startup transients and regulator failure without spurious tripping issues.  Since this is a one-off situation, you can take the time to hand-select the regulator and the zener to make sure that the voltages are what you want.  There should be some leakage current through the zener in normal operation, this just indicates that the zener is in a lower impedance zone and ready to step in should the voltage rise.

https://www.mouser.com/datasheet/2/308/1N5333B_D-2309169.pdf

[bdunham7]

Zero999 has suggested a factor of six greater SCR continuous rating vs the fuse rating, but you'd need to check datasheets and do the math to confirm that.

As much as I admire overkill, I can't agree with that line of thinking.  First, you have to consider the input source impedance or current capability--is the OP running this off of a car battery?  If so, some resistive ballast to limit the fault current to a non-ludicrous value would be the first step towards design sanity.  The 1A GMA fuse you chose for an illustration has an I^2T of 0.48, you can get TO-92 package SCRs with an I²T of double that @ 8.3ms.  And those devices--both fuse and SCR--are intended for mains power, whereas low voltage DC and low fault currents will be even more gentle.  There's no need for the fuse to blow quickly here as the SCR will have clamped the load side to some low voltage and can stay that way for a while.

[Zero999]

Zero999 has suggested a factor of six greater SCR continuous rating vs the fuse rating, but you'd need to check datasheets and do the math to confirm that.

As much as I admire overkill, I can't agree with that line of thinking.  First, you have to consider the input source impedance or current capability--is the OP running this off of a car battery?  If so, some resistive ballast to limit the fault current to a non-ludicrous value would be the first step towards design sanity.  The 1A GMA fuse you chose for an illustration has an I^2T of 0.48, you can get TO-92 package SCRs with an I²T of double that @ 8.3ms.  And those devices--both fuse and SCR--are intended for mains power, whereas low voltage DC and low fault currents will be even more gentle.  There's no need for the fuse to blow quickly here as the SCR will have clamped the load side to some low voltage and can stay that way for a while.

You're talking about the maximum prospective-short circuit current.

One needs to consider the purpose of the fuse, which is to protect the SCR. In this case, it's not the battery, or even the wiring, which will determine the short circuit current, but the fuse, the failing regulator itself and the SCR. I very much doubt the battery impedance even comes into it, unless it's a small battery.

There's a lot more I could add, but I don't have time at the moment.

[ejeffrey]

I think you are over thinking this a bit.  That's understandable as you are new at this and trying to make sense of confusing and somewhat contradictory online advice.  So here is my own confusing advice.

Any regulator can fail short circuit.  That's just a fact of life with semiconductors.  However if used correctly it should be so rare that you can ignore it.  Not everyone can ignore this, but you are working on low voltage low current non safety critical electronics.  And as a beginner, if you blow stuff up the overwhelming probability is that it will be because you did something wrong (aka a learning experience) rather than a random failure of a regulator.  So my advice is to worry less about extreme robustness and more about not using parts you can't afford to destroy.  The  when things happen post here asking for help if needed.

That said, the AMS1117 is not the easiest part to use.  It has a minimum load current and specifies tanalum output capacitors for stability.  Given the choice  between only these two I would pick the lm7805 unless I actually needed the low dropout voltage, but there are better LDOs available.

A common trap for hobbyists is input voltage range.  People get normal operation  right but neglect the turn on transient.  Especially in the case of using an external wall wart with a long cable there can be a pretty serious turn on transient.  The ams1117 is not to bad here with a 15 V input limit and you considering a 6 or 7 V wall wart but if you were using a 12V input you would want to have some input protection.

The LM7805 does have one major trap which is that it is not protected against input short circuit, especially with large output capacitors.  You need a diode from output to input with the LM7805.  Luckily this is almost always shown in example circuits.

[bdunham7]

You're talking about the maximum prospective-short circuit current.

Yes, and I understand all of the factors that might limit fault current.  Keep in mind that I'm replying to IanM, including his slightly erroneous paraphrasation of what you said.  The only real quibble I have with what you said is that the OP needs a very reasonable 300mA with a small wall-wart as a source and you've suggested a 2A fast fuse.  IanM suggested a 1A fuse and a very high fault current to clear it quickly.  I'm suggesting deliberately limiting the fault current and letting the fuse take its time.  It shouldn't be that difficult to limit the fault current to 5X the design current when it is a one-off without any efficiency objectives or appreciable cost constraints. 

[seqcst]

I am extremely worried about my voltage regulator failing short and frying my electronics see: AMS1117 short circuit protection not working (https://electronics.stackexchange.com/questions/530181/ams1117-short-circuit-protection-not-working)

Some context: I'm a beginner at electronics I want to power a handful of 5V electronics - some passive infra red sensors which draw only 100-200 uA each, as well as some active radars which may draw around 100mA each.

Maximum current draw will be less than 300mA.

A series regulator shorting or going rogue will cause an overvoltage, but a shunt regulator going short will simply blow the input fuse if there is one and overload the supply if there isn't.  Since your current requirements are reasonable, you could use a simple zener shunt regulator after the series regulator as a protective device.  The common saying that fuses don't protect semiconductors is only half true.  Fuses often can't act quickly enough to save semiconductors from a fault that causes overvoltage, but if you can turn the overvoltage fault into an overcurrent then fusing will work very well.  Simple devices like rectifiers can usually take very high peak currents without failing as long as the duration is short--the failure mechanism is entirely thermal (to a point). 

For your application, a 5-watt 1N5339B zener after the regulator and an MDL-1/2-R delay fuse before the regulator would do a good job of protecting against both startup transients and regulator failure without spurious tripping issues.  Since this is a one-off situation, you can take the time to hand-select the regulator and the zener to make sure that the voltages are what you want.  There should be some leakage current through the zener in normal operation, this just indicates that the zener is in a lower impedance zone and ready to step in should the voltage rise.

https://www.mouser.com/datasheet/2/308/1N5333B_D-2309169.pdf

Wow, this looks complicated.

So if I understand correctly, the shunt regulator works by getting rid of the excess voltage using reverse current flow? So the current across the zener diode will be higher than the rest of the circuit? The datasheet says it can dissipate 5W but how much heat is produced?

Also, I searched through that datasheet but I couldn't find info about the noise, line regulation, or load regulation anywhere.

Does noise, line regulation, and load regulation not apply to shunt zener regulators?

Zero999 has suggested a factor of six greater SCR continuous rating vs the fuse rating, but you'd need to check datasheets and do the math to confirm that.

As much as I admire overkill, I can't agree with that line of thinking.  First, you have to consider the input source impedance or current capability--is the OP running this off of a car battery?  If so, some resistive ballast to limit the fault current to a non-ludicrous value would be the first step towards design sanity.  The 1A GMA fuse you chose for an illustration has an I^2T of 0.48, you can get TO-92 package SCRs with an I²T of double that @ 8.3ms.  And those devices--both fuse and SCR--are intended for mains power, whereas low voltage DC and low fault currents will be even more gentle.  There's no need for the fuse to blow quickly here as the SCR will have clamped the load side to some low voltage and can stay that way for a while.

I'm going to be running this from mains power. I'm literally going to just plug the wall wart directly into my mains power supply.

I think you are over thinking this a bit.  That's understandable as you are new at this and trying to make sense of confusing and somewhat contradictory online advice.  So here is my own confusing advice.

Any regulator can fail short circuit.  That's just a fact of life with semiconductors.  However if used correctly it should be so rare that you can ignore it.  Not everyone can ignore this, but you are working on low voltage low current non safety critical electronics.  And as a beginner, if you blow stuff up the overwhelming probability is that it will be because you did something wrong (aka a learning experience) rather than a random failure of a regulator.  So my advice is to worry less about extreme robustness and more about not using parts you can't afford to destroy.  The  when things happen post here asking for help if needed.

That said, the AMS1117 is not the easiest part to use.  It has a minimum load current and specifies tanalum output capacitors for stability.  Given the choice  between only these two I would pick the lm7805 unless I actually needed the low dropout voltage, but there are better LDOs available.

A common trap for hobbyists is input voltage range.  People get normal operation  right but neglect the turn on transient.  Especially in the case of using an external wall wart with a long cable there can be a pretty serious turn on transient.  The ams1117 is not to bad here with a 15 V input limit and you considering a 6 or 7 V wall wart but if you were using a 12V input you would want to have some input protection.

The LM7805 does have one major trap which is that it is not protected against input short circuit, especially with large output capacitors.  You need a diode from output to input with the LM7805.  Luckily this is almost always shown in example circuits.

Thanks! I'm thinking of using the LM340A instead of the 7805 since it has better accuracy. Is the LM340A basically the same as 7805 except for having better accuracy?

Thanks for warning me about the transient. THe datasheet for the LM340/7805 says max input voltage is 35V, is that good enough?

I see from the datasheet that it says:

If transients exceed the maximum rated input voltage of the device, or reach more than 0.8 V below ground and
have sufficient energy, they will damage the regulator. The solution is to use a large input capacitor, a series
input breakdown diode, a choke, a transient suppressor or a combination of these.

So as I understand it, a large input capacitor will be sufficient to protect the voltage regulator from transient voltage spikes?

How big does the input capacitor need to be?

I see in the 7805 datasheet there is a circuit, I'm not exactly sure what the circuit is meant to be. Is it an example circuit to show how to use the 7805?

In that circuit, there are 2 capacitors C1 and C2. The C1 is connected to the input is 0.22uF while C2 is optional but suggested 0.1 uF ceramic. Would bigger capacitors be better in this case?

It doesn't say what the type of the input 0.22uF capacitor should it be also ceramic?

Thanks a lot!

EDIT: Actually, I think TVS diodes are used for suppressing transients, so maybe I should be using them instead of a large input cap?

[tszaboo]

Just put a resistor in front of the regulator. Size it based on the max current. For example, 0.1A with 1V voltage drop would be 10 Ohm. If you have a 7.4V power supply, your max fault current from the wall adapter is 740mA. Or your max power in your circuit is 3.2V*0.32A=1W. You will need a 1W resistor.
The capacitor type depends on the regulator type. Read the datasheet. Some doesn't like ceramic capacitors.

[bdunham7]

I see many people online saying that the 7805 is "bulletproof" and "indestructible" but it has quite a large voltage dropout - 2 volts - compared to modern LDO voltage regulators which means it has to dissipate a lot of heat, so I'm worried that it will overheat with 300mA from 7.5V voltage (2.5V * 0.3A = 0.75W which is quite a lot).

What is the most indestructible voltage regulator that I can get so that I won't have to worry about it failing short and frying my electronics? And should I get a 7.5V or 6V wall wart?

I missed this important part of your question.  A genuine LM7805 or LM340 (don't buy from Aliexpress, eBay or the like) is very reliable and will dissipate up to about 4W if it is in a room temperature environment.  However, it will get quite hot and 4W is really pushing it.  2W and below is probably OK without any heatsink, but heatsinks are easy enough to add.  You want the input voltage to be about 7.5V at full load, so you have to know more about your wall wart than its nominal rating.  The actual output of a "7.5V" wall wart is not likely to be precisely 7.5V and it will vary under load.  Also, 7.5V is probably too low.  It's OK if the voltage is higher at low loads since at lower currents there is less heat produced.  In your case with a 300mA maximum load, your regulator would easily tolerate 10V or so input voltage so there's no need for great precision.  What I typically do is once I've gotten my transformer and stuff for the input of a linear supply, I'll select a power resistor to use in series that drops the voltage down to 7.5 to 8.0 volts at the rated load.  So, for example, lets say your wall wart puts out 9.5V when loaded to 300mA.  You want to drop about 2V, so you put a 6.3R resistor in series for 1.9V drop and now you have 7.6V at the regulator.  The resistor is dissipating 0.6W and your regulator is dissipating 0.8W and everything is cool.  The resistor also limits the short circuit current to less than 1.5A, which makes the protection issues discussed earlier a lot easier to deal with.

Don't obsess over efficency, heat, low dropout and things like that when making a linear supply like this, especially if your main concerns are fault protection and reliability.  Regulation by dissipation is how these things work and 50% efficiency is probably the best you're going to get.  Trying to improve on that will get you into trouble.

[bdunham7]

Wow, this looks complicated.

Does noise, line regulation, and load regulation not apply to shunt zener regulators?

Sorry if I wasn't clear--I'm only proposing the shunt as a protection system.  Using only a shunt regulator in this case would be more complicated.

I'm going to be running this from mains power. I'm literally going to just plug the wall wart directly into my mains power supply.

Mains power is what goes into your wall wart, what comes out is not unless there is a terrible failure inside of it.  Mains power has fault currents up to thousands of amperes, your wall wart probably has a fault current of 1 or 2 amperes.  The fuse and SCR being discussed are capable of working with mains power directly which is much more difficult.

If transients exceed the maximum rated input voltage of the device, or reach more than 0.8 V below ground and
have sufficient energy, they will damage the regulator. The solution is to use a large input capacitor, a series
input breakdown diode, a choke, a transient suppressor or a combination of these.

The zener and the other diode in the circuit I've attached will provide some protection against those startup transients, but if you are concerned about them you can add another regular 1N4001 in front of the regulator for complete reverse polarity protection.  The values of the capacitor and resistor will depend on your power source.

[Ian.M]

Note that I chose to consider the 1A fuse not because it is the right one for our O.P's application, but solely because it was the first datasheet I found for a 20mm glass fuse from a reputable manufacturer, and made the point that to blow it quickly needs a lot more current than its rating.   

[bdunham7]

Note that I chose to consider the 1A fuse not because it is the right one for our O.P's application, but solely because it was the first datasheet I found for a 20mm glass fuse from a reputable manufacturer, and made the point that to blow it quickly needs a lot more current than its rating.

OK, fair enough.  And it's true of course about fusing time vs current, but that's why it is so difficult to protect semis with fuses, right?  They're never quite fast enough. 

[seqcst]

Wow, this looks complicated.

Does noise, line regulation, and load regulation not apply to shunt zener regulators?

Sorry if I wasn't clear--I'm only proposing the shunt as a protection system.  Using only a shunt regulator in this case would be more complicated.

I'm going to be running this from mains power. I'm literally going to just plug the wall wart directly into my mains power supply.

Mains power is what goes into your wall wart, what comes out is not unless there is a terrible failure inside of it.  Mains power has fault currents up to thousands of amperes, your wall wart probably has a fault current of 1 or 2 amperes.  The fuse and SCR being discussed are capable of working with mains power directly which is much more difficult.

If transients exceed the maximum rated input voltage of the device, or reach more than 0.8 V below ground and
have sufficient energy, they will damage the regulator. The solution is to use a large input capacitor, a series
input breakdown diode, a choke, a transient suppressor or a combination of these.

The zener and the other diode in the circuit I've attached will provide some protection against those startup transients, but if you are concerned about them you can add another regular 1N4001 in front of the regulator for complete reverse polarity protection.  The values of the capacitor and resistor will depend on your power source.

Wow, thank you so much for the schematic! I spent hours staring at your schematic and now I finally think I have begun to understand a bit of it. But I'd like to ask some questions just to be sure I really understand it.

1). The resistor in front of the fuse. What's the purpose of this? Is it just for dropping the voltage of the power supply? My wall wart says it's a regulated 7.5V so I'm worried any drop in voltage before the voltage regulator might drop the voltage too low.

2). The fuse. Am I correct in assuming that the fuse is there for only 2 reasons: 1). Reverse polarity protection. 2). Over-voltage protection.

3). The optional (dotted line) diode I assume is for reverse polarity protection so that if the input voltages are reversed then it will short, preventing current from going to the rest of the circuit while also blowing the fuse? But while it's shorting, a little bit of current will still reach the rest of the circuit right? Maybe not mA but uA? Will that be enough to damage the rest of the components? Some of my PIR sensors are extremely low power - they draw only single digit uA, I'm worried there might be enough current even in a short to damage those?

4). The diode connecting the output of the LM340 to the input. I do not understand the purpose of this diode. As I understand it, current always flows from high voltage to lower voltage. The output of the LM340 should always be a lower voltage than the input, so no current should ever flow through that diode. Even if the LM340 fails short, the voltage should be the same...so how could any current even pass through this diode? Even in a reverse polarity situation, the optional diode (dotted line) should form a short circuit, so again I cannot see how any current could ever pass through the diode.

====== Tangent about reverse polarity protection =======

For reverse polarity protection, why not put the diode in series next to Vin instead of in parallel? It would still work right? I guess because of the voltage drop? But Schottky diodes like the 1N5817 only drops like 0.25V right? Would that still be too much?

Actually, I guess I can't use Schottky diodes because the reverse leakage current of multiple microamps might be too much. Some of my PIR sensors draw only single digit microamps, so I guess microamps in the reverse direction will probably damage it.

I found this normal diode 1N4148 that has only 25nA reverse leakage current. But the voltage drop varies quite significantly with load (from 1 to 1.6V for the 1N4148 from 0-300mA), which makes it somewhat problematic to connect it in series.

Also, the 1N4148 will probably overheat at 300mA. The datasheet says its power dissipation is 440-500mW. At 300mA, the voltage drop will be 1.6V, and 1.6V * 300mA = 480mW which is very close to the limit.

For the 1N4001 the power dissipation is 3W so it's much better, but its leakage current is like 5-30uA which worries me.

So I actually cannot find any diodes that I can connect in series that fit my requirements:

1). Low reverse leakage current
2). Doesn't overheat at 300mA

Also if I connected a diode in series, I would have to use a higher voltage power supply, since 7.5V is just enough for the voltage regulator, if I put a diode in front of the voltage regulator then it will drop the voltage too much for the voltage regulator to work.

From the LM340A datasheet:

The input voltage must remain typically 1.7 V above the output voltage even during the low point on the input ripple voltage.

The datasheet for the 7.5V wall wart says noise is 50mV and line regulation is 0.5%, load regulation is 5%. Overall tolerance is 5%.

So maybe I should get a 9V power supply instead just to be on the safe side? I guess I would be wasting even more power then...should I just add more diodes to drop the voltage more? The voltage drop would vary quite significantly with load, and makes me worried about the diode possibly overheating due to how much power it has to dissipate.  voltage regulator (since each diode drops 0.25V)...seems a bit excessive...maybe I should use some higher-voltage-drop diodes so that I don't need so many diodes.

All this trouble just to connect a diode in series for reverse polarity protection...I guess the voltage fluctuation due to load will add noise too to the voltage regulator...is that why people don't connect the diode in series for polarity protection?

======== End of tangent about reverse polarity protection ==========

Your circuit looks just like the crowbar circuit without the thyristor and capacitors.

Is my understanding correct that the difference between your circuit and a crowbar circuit is that, in the event of a short voltage spike, the crowbar will "lock in" and short until the fuse is blown, while your circuit will simply "clamp" the voltage to 5V?

In over-voltage condition the zener diode will begin to allow current through, right? But only so much current until the voltage drops back down to 5V? Maybe the current is all used up for heating the components and won't be enough to blow the fuse??

Actually, the major difference I see between your design and crowbar is that in a crowbar circuit, over-voltage will create an actual short, but in your circuit, over-voltage will allow current to flow through the voltage regulator, the zener diode, and the fuse.

Since the current still has to go through the voltage regulator, it's not a short and won't actually blow the fuse?

If the LM340 fails and becomes a dead short, the fuse will indeed blow. But if the LM340 fails and becomes a resistor, then the fuse won't blow...what will happen in that situation? I guess the components in the "protection circuit" (zener, voltage regulator, and fuse) will just slowly get warmer until something fails?

[bdunham7]

1). The resistor in front of the fuse. What's the purpose of this? Is it just for dropping the voltage of the power supply? My wall wart says it's a regulated 7.5V so I'm worried any drop in voltage before the voltage regulator might drop the voltage too low.

The resistor is to progressively drop the regulator input voltage as the load increases with the target of having it reduce to near the minimum allowable at the full anticipated load.  If you have a regulated 7.5V input, then no resistor is needed. 

2). The fuse. Am I correct in assuming that the fuse is there for only 2 reasons: 1). Reverse polarity protection. 2). Over-voltage protection.

The fuse is for overcurrent protection in case an improper input or regulator failure causes the other protection devices to draw too much current.  The fuse doesn't need to blow for the protection to work.  For example, if your 7.5V supply can only put out 500mA and the regulator shorts for some reason, the zener will simply take the current indefinitely without blowing the fuse.   Same for reverse polarity.

3). The optional (dotted line) diode I assume is for reverse polarity protection so that if the input voltages are reversed then it will short, preventing current from going to the rest of the circuit while also blowing the fuse? But while it's shorting, a little bit of current will still reach the rest of the circuit right? Maybe not mA but uA? Will that be enough to damage the rest of the components? Some of my PIR sensors are extremely low power - they draw only single digit uA, I'm worried there might be enough current even in a short to damage those?

The optional diode would be to protect the regulator (mostly) against random transients and such that may occur when using a wall wart with plugs and connections.  The normal diode mode of the zener already protects the output.  It probably isn't necessary, especially if you use a decent size capacitor (say 22µF) on the input.  If you want more robust reverse polarity protection without any risk of fuse-blowing, you could also put in a series diode, but that would really be serving a different function.  A series diode alone may not help with transients in certain situations. If your devices are extraordinarily sensitive to reverse current and limiting the reverse voltage to one diode drop (0.7V or so) isn't good enough, you could use Schottky diodes instead.  The leakage of the Schottky diode would not be an issue because the resulting voltage would be very low due to the built-in voltage divider resistor in the regulator.

4). The diode connecting the output of the LM340 to the input. I do not understand the purpose of this diode. As I understand it, current always flows from high voltage to lower voltage. The output of the LM340 should always be a lower voltage than the input, so no current should ever flow through that diode. Even if the LM340 fails short, the voltage should be the same...so how could any current even pass through this diode? Even in a reverse polarity situation, the optional diode (dotted line) should form a short circuit, so again I cannot see how any current could ever pass through the diode.

Read the datasheet for the LM340 carefully--they explain it.  If your output has any significant capacitance (in the devices typically) and the power goes out, those capacitors may try to 'backfeed' through the regulator and it can't tolerate that.  The diode bypasses the regulator and allows that current to flow through it instead.  This is a very common setup with LM-style regulators.

Also if I connected a diode in series, I would have to use a higher voltage power supply, since 7.5V is just enough for the voltage regulator, if I put a diode in front of the voltage regulator then it will drop the voltage too much for the voltage regulator to work.

The actual dropout voltage of the LM340 at 300mA is probably more like 1.3-1.4V, so you'd probably get away with a diode.  Using a Schottky would help, as mentioned.  However, the regulator will regulate a bit better with a higher input voltage, so if you have very dynamic loads perhaps a 9V suppy plus an appropriate resistor would be better.  Don't use multiple diodes for voltage dropping, they don't have the desired current tapering effect in case of overload or failure of the regulator.

Your circuit looks just like the crowbar circuit without the thyristor and capacitors.

Is my understanding correct that the difference between your circuit and a crowbar circuit is that, in the event of a short voltage spike, the crowbar will "lock in" and short until the fuse is blown, while your circuit will simply "clamp" the voltage to 5V?

In over-voltage condition the zener diode will begin to allow current through, right? But only so much current until the voltage drops back down to 5V? Maybe the current is all used up for heating the components and won't be enough to blow the fuse??

Actually, the major difference I see between your design and crowbar is that in a crowbar circuit, over-voltage will create an actual short, but in your circuit, over-voltage will allow current to flow through the voltage regulator, the zener diode, and the fuse.

Since the current still has to go through the voltage regulator, it's not a short and won't actually blow the fuse?

If the LM340 fails and becomes a dead short, the fuse will indeed blow. But if the LM340 fails and becomes a resistor, then the fuse won't blow...what will happen in that situation? I guess the components in the "protection circuit" (zener, voltage regulator, and fuse) will just slowly get warmer until something fails?

Yes, the zener clamps as opposed to shorting.  That's a lot handier in case of transients and if there is a failure but insufficient current to blow the fuse, it will simply continue to function, albeit with a somewhat higher output voltage.  Which solution you choose depends on the desired behavior.  There is an MDL-3/10-R fuse that is guaranteed to blow at 500mA (but never at 300mA) but you still could have a fault condition in between there.  You might add a voltage indicator circuit that tells  you your output is low, good or high.  However much effort you put into making something bulletproof or foolproof, there's always something that you didn't anticipate that will trip you up.  Personally, I think the original circut I drew, without the optional diode and with a 22µF input capacitor, all powered by your 7.5V supply, will be robust enough that you really ought to worry about other things.

[David Hess]

Input voltage surges which destroy a regulator can happen.  If the input capacitance is low and the load current suddenly decreases, then the transformer's leakage inductance can cause a voltage spike.  This usually shows up with cheap consumer stuff where the AC adapter has high leakage inductance to limit current and the regulator has low input capacitance because it is used to only limit the voltage.

If a higher voltage may be applied to the regulator's output, or the input can be shorted, then a reverse protection diode is required unless the regulator is specifically designed to handle this like some PNP pass transistor regulators.  If present, the adjustment pin may also need to be protected with a diode.

Regulators are not really designed to routinely operate with their short circuit current limit or thermal protection active.  Some regulators support adjustable current limiting or it can be implemented externally.

[Gyro]

If you want something 'less destructible' when it comes to input surges etc. why not go for an LM2940-5.0? It has an automotive spec input - 60V input surge and reverse polarity protection. You can even put it in backwards without damage. It's an LDO too.

[seqcst]

If you want something 'less destructible' when it comes to input surges etc. why not go for an LM2940-5.0? It has an automotive spec input - 60V input surge and reverse polarity protection. You can even put it in backwards without damage. It's an LDO too.

Thanks! That does indeed look more indestructible than the LM340, however the noise is also significantly more than the LM340A...is there a lower noise version of this? I've looked for a 3-pin automotive-grade low-noise voltage regulator that allows over 300mA current and couldn't find any that had as low noise as LM340A.

Am I overthinking the noise? I don't actually know how much noise is too much for my PIR sensors. If they're interchangeable should I just get both and try both of them out?

[seqcst]

1). The resistor in front of the fuse. What's the purpose of this? Is it just for dropping the voltage of the power supply? My wall wart says it's a regulated 7.5V so I'm worried any drop in voltage before the voltage regulator might drop the voltage too low.

The resistor is to progressively drop the regulator input voltage as the load increases with the target of having it reduce to near the minimum allowable at the full anticipated load.  If you have a regulated 7.5V input, then no resistor is needed. 

2). The fuse. Am I correct in assuming that the fuse is there for only 2 reasons: 1). Reverse polarity protection. 2). Over-voltage protection.

The fuse is for overcurrent protection in case an improper input or regulator failure causes the other protection devices to draw too much current.  The fuse doesn't need to blow for the protection to work.  For example, if your 7.5V supply can only put out 500mA and the regulator shorts for some reason, the zener will simply take the current indefinitely without blowing the fuse.   Same for reverse polarity.

3). The optional (dotted line) diode I assume is for reverse polarity protection so that if the input voltages are reversed then it will short, preventing current from going to the rest of the circuit while also blowing the fuse? But while it's shorting, a little bit of current will still reach the rest of the circuit right? Maybe not mA but uA? Will that be enough to damage the rest of the components? Some of my PIR sensors are extremely low power - they draw only single digit uA, I'm worried there might be enough current even in a short to damage those?

The optional diode would be to protect the regulator (mostly) against random transients and such that may occur when using a wall wart with plugs and connections.  The normal diode mode of the zener already protects the output.  It probably isn't necessary, especially if you use a decent size capacitor (say 22µF) on the input.  If you want more robust reverse polarity protection without any risk of fuse-blowing, you could also put in a series diode, but that would really be serving a different function.  A series diode alone may not help with transients in certain situations. If your devices are extraordinarily sensitive to reverse current and limiting the reverse voltage to one diode drop (0.7V or so) isn't good enough, you could use Schottky diodes instead.  The leakage of the Schottky diode would not be an issue because the resulting voltage would be very low due to the built-in voltage divider resistor in the regulator.

4). The diode connecting the output of the LM340 to the input. I do not understand the purpose of this diode. As I understand it, current always flows from high voltage to lower voltage. The output of the LM340 should always be a lower voltage than the input, so no current should ever flow through that diode. Even if the LM340 fails short, the voltage should be the same...so how could any current even pass through this diode? Even in a reverse polarity situation, the optional diode (dotted line) should form a short circuit, so again I cannot see how any current could ever pass through the diode.

Read the datasheet for the LM340 carefully--they explain it.  If your output has any significant capacitance (in the devices typically) and the power goes out, those capacitors may try to 'backfeed' through the regulator and it can't tolerate that.  The diode bypasses the regulator and allows that current to flow through it instead.  This is a very common setup with LM-style regulators.

Also if I connected a diode in series, I would have to use a higher voltage power supply, since 7.5V is just enough for the voltage regulator, if I put a diode in front of the voltage regulator then it will drop the voltage too much for the voltage regulator to work.

The actual dropout voltage of the LM340 at 300mA is probably more like 1.3-1.4V, so you'd probably get away with a diode.  Using a Schottky would help, as mentioned.  However, the regulator will regulate a bit better with a higher input voltage, so if you have very dynamic loads perhaps a 9V suppy plus an appropriate resistor would be better.  Don't use multiple diodes for voltage dropping, they don't have the desired current tapering effect in case of overload or failure of the regulator.

Your circuit looks just like the crowbar circuit without the thyristor and capacitors.

Is my understanding correct that the difference between your circuit and a crowbar circuit is that, in the event of a short voltage spike, the crowbar will "lock in" and short until the fuse is blown, while your circuit will simply "clamp" the voltage to 5V?

In over-voltage condition the zener diode will begin to allow current through, right? But only so much current until the voltage drops back down to 5V? Maybe the current is all used up for heating the components and won't be enough to blow the fuse??

Actually, the major difference I see between your design and crowbar is that in a crowbar circuit, over-voltage will create an actual short, but in your circuit, over-voltage will allow current to flow through the voltage regulator, the zener diode, and the fuse.

Since the current still has to go through the voltage regulator, it's not a short and won't actually blow the fuse?

If the LM340 fails and becomes a dead short, the fuse will indeed blow. But if the LM340 fails and becomes a resistor, then the fuse won't blow...what will happen in that situation? I guess the components in the "protection circuit" (zener, voltage regulator, and fuse) will just slowly get warmer until something fails?

Yes, the zener clamps as opposed to shorting.  That's a lot handier in case of transients and if there is a failure but insufficient current to blow the fuse, it will simply continue to function, albeit with a somewhat higher output voltage.  Which solution you choose depends on the desired behavior.  There is an MDL-3/10-R fuse that is guaranteed to blow at 500mA (but never at 300mA) but you still could have a fault condition in between there.  You might add a voltage indicator circuit that tells  you your output is low, good or high.  However much effort you put into making something bulletproof or foolproof, there's always something that you didn't anticipate that will trip you up.  Personally, I think the original circut I drew, without the optional diode and with a 22µF input capacitor, all powered by your 7.5V supply, will be robust enough that you really ought to worry about other things.

If you want more robust reverse polarity protection without any risk of fuse-blowing, you could also put in a series diode, but that would really be serving a different function.  A series diode alone may not help with transients in certain situations. If your devices are extraordinarily sensitive to reverse current and limiting the reverse voltage to one diode drop (0.7V or so) isn't good enough, you could use Schottky diodes instead.  The leakage of the Schottky diode would not be an issue because the resulting voltage would be very low due to the built-in voltage divider resistor in the regulator.

Sorry, let me make sure I understand what you're saying. Are you talking about adding an additional diode connected in series in addition to the optional (dotted line) diode which is connected in parallel in your schematic?

Here's what I think you're suggesting:



Let's say D1 is a Schottky diode with 30uA of reverse leakage current.

In a reverse polarity situation, that means we'll have 30uA of current flowing through the circuit in reverse, right?

When the voltage is reversed, D2 and zener diode will be the lowest resistance components in the circuit, so that means all of the reverse current will be flowing through them, and no current is going through the rest of the circuit (voltage regulator and PIR sensors)? Is my understanding correct? What kind of numbers are we talking about here?

The above circuit should fully protect against both reverse polarity as well as transients, right?

If your devices are extraordinarily sensitive to reverse current and limiting the reverse voltage to one diode drop (0.7V or so) isn't good enough, you could use Schottky diodes instead.  The leakage of the Schottky diode would not be an issue because the resulting voltage would be very low due to the built-in voltage divider resistor in the regulator.

Sorry, I don't understand this. Are you talking about diode connected in series or in parallel? Am I understanding correctly that the diode D2 will limit the reverse voltage to 0.7V by conducting current until the voltage has dropped to 0.7V, but you are suggesting to use a Schottky diode here in order to drop the voltage even further, down to say 0.3V? But this reverse voltage of 0.3V will be dropped more by the voltage divider resistor in the regulator? I searched online and couldn't find any information about the voltage divider in the regulator...


EDIT: Wait, thinking more about this, in the case of reverse polarity, D1 will have a resistance in the mega-Ohms, which means the path through D2 and the path through the voltage regulator will be of similar resistance overall, so the current flow through all these paths should be roughly equal...which means the micro-amps will be flowing through all of the paths! So my reasoning is wrong...my components will still suffer the micro-amps of reverse current!

[ejeffrey]

Noise is not usually a big concern for linear voltage regulators and regulator noise is often only a small part of supply rail noise.  Any other electronics on the same power supply will be injecting noise onto the supply rail from their normal operation.  So it's normal to design circuits to have good power supply rejection (PSRR).  There are exceptions but I doubt a PIR sensor is especially sensitive.  Ripple from switching  power supplies can be much more objectionable due to typically higher amplitude and the spikiness that can be hard to filter.

[CatalinaWOW]

I do suspect that you are overthinking the noise problem.  Even if your sensors are sensitive to supply noise it is unlikely that series voltage regulators will be the best, or even successful approach.

As a beginner this will be hard to sort out.  It will involved understanding how noise affects the sensor, frequency characteristics of the noise and other factors. 

I would recommend hooking your sensors up and trying them.  If they do what you expect them to you are done.  If not you can dig into why not.

Some obvious experiments would be trying different supply voltages.  I would not expect your sensors to show any sensitivity to changes within their recommended supply range (probably something like 4.75 to 5.25 V).  Running off a battery pack that supplies in this range would give one of the lowest possible supply noises. 

Much of your learning will come from actually building and operating your circuits

[Gyro]

Yes, you are probably overthinking the noise a bit. As your sensors are such low supply current, you could probably get around any over-sensitivity with individual RC (or LC) filters to buffer them from the regulator and whatever else you have on the 5V rail.

Off hand, I don't know any quieter (automotive) alternatives to the LM2940.

[bdunham7]

Let's say D1 is a Schottky diode with 30uA of reverse leakage current.

In a reverse polarity situation, that means we'll have 30uA of current flowing through the circuit in reverse, right?

When the voltage is reversed, D2 and zener diode will be the lowest resistance components in the circuit, so that means all of the reverse current will be flowing through them, and no current is going through the rest of the circuit (voltage regulator and PIR sensors)? Is my understanding correct? What kind of numbers are we talking about here?

The above circuit should fully protect against both reverse polarity as well as transients, right?

Yes, that's the idea.  As far as numbers, you can learn quite a bit by carefully reading the actual datasheets of the various devices.   Leakage current, forward voltages, etc are all quite variable and interdependent.  If you had two typical 30V rated Schottkys, say 1N5818, then with an applied reverse voltage the leakage through D1 (at room temperature) would be ~3µA and if all of that went through D2, the voltage across it and the regulator input would be no more than 0.25V.  We'd need more specific information on your PIR sensors, but I'm doubtful that a reverse 250mV would harm them.  Reverse current can't flow through them without enough voltage to push it.

https://www.mouser.com/datasheet/2/389/1n5817-1848842.pdf

Sorry, I don't understand this. Are you talking about diode connected in series or in parallel? Am I understanding correctly that the diode D2 will limit the reverse voltage to 0.7V by conducting current until the voltage has dropped to 0.7V, but you are suggesting to use a Schottky diode here in order to drop the voltage even further, down to say 0.3V? But this reverse voltage of 0.3V will be dropped more by the voltage divider resistor in the regulator? I searched online and couldn't find any information about the voltage divider in the regulator...


EDIT: Wait, thinking more about this, in the case of reverse polarity, D1 will have a resistance in the mega-Ohms, which means the path through D2 and the path through the voltage regulator will be of similar resistance overall, so the current flow through all these paths should be roughly equal...which means the micro-amps will be flowing through all of the paths! So my reasoning is wrong...my components will still suffer the micro-amps of reverse current!

Again, vital information is found in the actual PDF datasheets and you can learn a lot by reading them.  Look at an application circuit for the LM317 regulator and you'll see two resistors, one from the output to the adjust pin and the other from the adjust pin to ground.  In series, these form a resistor going from output to ground.  Integrated single-voltage devices like the LM7805 and LM340 work the exact same way, except that those two resistors are internal to the regulator.  There is a block diagram (simplified schematic) of the internal structure of the LM340 in the datasheet and it shows those resistors and they add up to about 3k.  If you have 3µA of leakage current and it all goes through that resistor, you'll have at most 9mV of reverse voltage. 

https://www.ti.com/lit/ds/symlink/lm340.pdf?

Anyway, this is a good way to learn some basic electronics but it may not be the best way to power your sensors because you haven't addressed some other issues that are more likely to cause you grief than any regulator noise or failure.  The main issues that are likely to damage your devices are transients and ESD that result from using multiple things that plug in or are otherwise connected and disconnected.  There's also a good possibility that your wall wart may have signficant common-mode AC leakage.  I just grabbed two random DC wall warts on my bench and found that one had 300mVAC and the other over 10VAC when measured with a 10M DMM from the output to an earth ground.  I've seen them as high as 90-100VAC.  You don't specify any grounding and I'd suggest that you plan on grounding the negative supply rail to something instead of leaving it floating, especially if you are using a mains-derived power source.

The best ways to safely power your devices with low noise would be a commercial linear lab PSU with a 5V output, a dedicated linear supply that you build ourself or a 12V SLA battery with a 15R or so resistor in series.  Or, perhaps the wall wart route is good enough.

[David Hess]

Linear regulator noise is hardly ever a consideration, and improving on it is not trivial because most of the noise comes from the bandgap reference.

The easiest way to improve the noise is to use an adjustable regulator and bypass the adjustment pin to ground, like a 317, or I guess the Texas Instruments TL1963A-Q1.  Some fixed voltage regulators have an extra pin just to bypass the reference but I do not know that any of these meet your other requirements.

[seqcst]

For reverse polarity protection, I guess a diode connected in series would be the simplest and most straightforward solution, so I searched online for some diodes to protect against reverse polarity and found only two that seemed suitable:

1). PMEG120G20ELR-Q - This one is discontinued, but I cannot find any suitable replacements for it. It has 2A average forward current, 120V reverse voltage, 30nA max reverse current at 25C, 11ns reverse recovery time, and the datasheet says it is qualified according to AEC-Q101 and recommended for use in automotive application, and under Applications it explicitly lists Reverse polarity protection. So I really want to get this SiGe rectifier.

2). 1N4149 - This one is not discontinued, but it has only 500mA average forward current, 100V reverse voltage, 25nA reverse leakage at 20V at 25C, and only 4ns reverse recovery time. But the datasheet does not say that it is suitable for use as reverse polarity protection.

Based on the datasheets alone, it seems that the PMEG120G20ELR-Q is better than the 1N4149 in every way - the datasheet even explicitly recommends it for reverse polarity protection.

But why is nobody recommending the 1N4149 for reverse polarity protection? It seems to have good stats (only 4ns recovery time!) other than the 500mA current limit.

Should I use the PMEG120G20ELR-Q for reverse polarity protection? It's even rated for automotive use.  But why is it being discontinued? I can't find any replacement for it. I looked on the Nexperia website and couldn't find anything that had similar stats.

[bdunham7]

But why is nobody recommending the 1N4149 for reverse polarity protection? It seems to have good stats (only 4ns recovery time!) other than the 500mA current limit.

Because an ordinary 1N4001 is probably more than good enough for your application and those other parameters just don't matter in a circuit like this.  That and there's not much reason to use it over the common-as-dirt 1N4148.  Say you use a 22µF input capacitor--what difference (and under what circumstances) would it make if the recovery time were 4µs vs 4ns?  You have to read the whole datasheet--the 1N4149 is a small signal diode and has a V_f of >1.1V @ 300mA.  There's no need to go hunting down unicorns to make a very straightforward circuit that most of us would make out of parts from the junk drawer.  I suggest you start a junk drawer if you don't already have one!

[BrokenYugo]

Use cheap common "jellybean" parts whenever possible, for a reverse polarity protection diode that would be a basic rectifier/power diode, 1N4007 (1A 1000V) being the most common 1 amp example, for overkill use 1N5408, the common 3 amp 1kV diode.

[seqcst]

Let's say D1 is a Schottky diode with 30uA of reverse leakage current.

In a reverse polarity situation, that means we'll have 30uA of current flowing through the circuit in reverse, right?

When the voltage is reversed, D2 and zener diode will be the lowest resistance components in the circuit, so that means all of the reverse current will be flowing through them, and no current is going through the rest of the circuit (voltage regulator and PIR sensors)? Is my understanding correct? What kind of numbers are we talking about here?

The above circuit should fully protect against both reverse polarity as well as transients, right?

Yes, that's the idea.  As far as numbers, you can learn quite a bit by carefully reading the actual datasheets of the various devices.   Leakage current, forward voltages, etc are all quite variable and interdependent.  If you had two typical 30V rated Schottkys, say 1N5818, then with an applied reverse voltage the leakage through D1 (at room temperature) would be ~3µA and if all of that went through D2, the voltage across it and the regulator input would be no more than 0.25V.  We'd need more specific information on your PIR sensors, but I'm doubtful that a reverse 250mV would harm them.  Reverse current can't flow through them without enough voltage to push it.

I am really confused by this. I thought current will always flow unless the resistance is infinite. Surely the resistance of a PIR sensor is not infinite, therefore even at 0.25V there should be at least some current flowing across the PIR sensor.

If I understand correctly, the ratio of the resistances between the two branches determines the ratio of the currents going through the circuits.

So if a diode has 20 times smaller resistance than the alternative branch, then current will flow through it in a 20:1 ratio, so if there are 21uA of current in total, then 20uA will flow through the diode and 1uA will flow through the alternative branch.

But 1uA is potentially still enough to damage my PIR sensor, right? Since it only draws 1uA in normal operation. So that's why I wanted a diode with only 25nA reverse leakage current.



I'm trying to figure out if, in a reverse polarity situation, the reverse current will really "all" flow through the diode, or if the ratio of the resistances is low enough that a significant amount of those 30uA of reverse leakage current will reverse-flow through other parts of the circuit.

But to figure out the resistance ratio, I need to know the resistance of my PIR sensor in reverse polarity situation. If the sensor normally draws 1uA at 5V, does that mean that it has 5 megaOhms of resistance? Do electronic components have fixed internal resistance?

The PIR sensors I'm using are the EKMB and EKMC PIR sensors from panasonic such as these ones:

EKMB (1uA):
https://industrial.panasonic.com/cdbs/www-data/pdf/EWA0000/bltn_eng_ekmb110611_ast-ind-247372.pdf

EKMC (170uA):
https://industrial.panasonic.com/cdbs/www-data/pdf/EWA0000/bltn_eng_ekmc160611_ast-ind-247327.pdf

I can't find the resistance listed in the PIR sensor datasheet, but it says:

6) When wiring the product, always use shielded cables and minimize the wiring length to prevent
noise disturbances.

7) The inner circuit board could be destroyed by a voltage surge. Use of surge absorption elements
is highly recommended.

Surge resistance ： below the power supply voltage value indicated in the maximum rated
values section.

Please use a stabilized power supply. Power supply noise can cause operating errors.

Noise resistance ： ±20V or less (Square waves with a width of 50ns or 1μs)

To reduce the effect of power supply noise, install a capacitor on the sensor’s power supply pin.

Does the requirement to use shielded cables mean that I can't use the standard breadboard jumper wires?

[MarkT]

Any regulator can fail short circuit.  That's just a fact of life with semiconductors.

I think the best approach is a buffered zener shunt regulator - crowbar's are destructive for no good reason here, a zener that's bolstered with a power transistor is pretty solid, can be made indefinite-short-circuit proof, and can be paralleled for redundancy.  And often a beefy zener is enough, again easy to parallel for redundancy.  And if it fails short-circuit your load is protected.  A TVS diode on the input can provide extra protection from high energy spikes.

[Terry Bites]

I know that indefinite short and thermal protection claims ofter aren't worth the toilet paper they're written on.
Nothing's foolproof, ever. Linear regs are more (much more) prone to OV failure than switchers. Non-isolated buckers are an invation to OV hell though.
The death of hard pushed power transistors in isolated DC-DC converters leads to no output.

There's the risks you can live with and the one's you can't.

A regulator is unlkely to just go short because it does a boring job. Dont blame the reg, it's just as reliable/ unrelaible as any other component.
Do some careful analysis of power source and load issues that drag the regulator outside it's SOA. Design conservatively. That heat needs to be managed.
Protecting agains the vagueries of the bargain regulator is pointless when a premium device might save the day.

Crowbar's aren't seen too often these days, smsps have had a lot of those OV problems designed out.
A crowbar needs careful design or it will fail to fire or fire too early. I'd use a real crowbar protection ic MC3423.
Dont entrust your crown jewels to a zener voltage!

[Manul]

I am really confused by this. I thought current will always flow unless the resistance is infinite. Surely the resistance of a PIR sensor is not infinite, therefore even at 0.25V there should be at least some current flowing across the PIR sensor.

If I understand correctly, the ratio of the resistances between the two branches determines the ratio of the currents going through the circuits.

So if a diode has 20 times smaller resistance than the alternative branch, then current will flow through it in a 20:1 ratio, so if there are 21uA of current in total, then 20uA will flow through the diode and 1uA will flow through the alternative branch.

But 1uA is potentially still enough to damage my PIR sensor, right? Since it only draws 1uA in normal operation. So that's why I wanted a diode with only 25nA reverse leakage current.

Man... Stop worrying about microamps and nanoamps. I understand your pursuit of perfectionism, because I have it too, but you need to draw a line when it becomes irrational. Your sensor will not be damaged from 0.25V of reverse voltage. With small compliance voltage, it is safe to inject hundreds of uA of current, both reverse and forward to almost any circuit. For instance, when doing VI signature analysis, I inject even 1000uA of current into various nodes of high-tech devices, some worth as much as a house. And never ever damaged anything. All will be good.

[bdunham7]

But to figure out the resistance ratio, I need to know the resistance of my PIR sensor in reverse polarity situation. If the sensor normally draws 1uA at 5V, does that mean that it has 5 megaOhms of resistance? Do electronic components have fixed internal resistance?

No, electrical components do not have fixed internal resistance.  Even if some current does flow, it is really unlikely to cause any harm if the power involved is in the nanowatt range.

Does the requirement to use shielded cables mean that I can't use the standard breadboard jumper wires?

So this is why I warned you that you might be worrying about the wrong things.  First, one of those datasheets lists a maximum supply voltage of 4.5VDC, so how are you handling that?  If you are using the other one, it says 7.0VDC and that seems like the basic circuit above is perfectly adequate.  However, note that it specifies a maximum of 200V static electricity and shielded cables.  This is very, very low and indicates that these things are delicate and sensitive to transients.  A breadboard or breadboard jumpers will give you pretty bad results.  Wall-wart leakage may be an issue, as I mentioned earlier.  You need soldered twisted pairs inside a shield or coax or something like that.  And ground the negative supply rail properly.

[CatalinaWOW]

The data sheets for your sensors provide an excellent learning opportunity for new engineers and students.  These data sheets are written with input from salesmen, lawyers and even engineers in some cases.  The result is something that must be read with a grain of salt and much thought.

The takeaways I got from reading these sheets (and understanding how these sensors generally work) are these.

1.  You should not apply a steady reverse voltage above 0.3 volts for a "long" time.  The implication from the data sheet is that excess current will flow and cause essentially thermal damage to the device.  "Long" is ill defined and depends on how far the 0.3 volts is exceeded.  It shouldn't be hard to avoid this type of damage to your part, just be careful during fabrication and make sure that any cables which can be disconnected have polarized connections (or use a rectifier bridge to make the connection reversible at the cost of a diode drop in voltage).

2.  The 30 second stabilization time on turn on implies that your voltage supply needs to be pretty stable over this time scale.  Given the low current draw of the sensor the simplest way to achieve this would be the zener supplies suggested by others. 

3.  The test conditions, range and mask requirements imply that the sensor is tuned to respond to signals changing at very low frequency, only a few Hz.  Noise at higher frequencies will not be terribly important as indicated by the allowance for up to 20 V P-P for pulses of less than microsecond duration.

4.  Most of the applications cautions are best practices or CYA from the legal department, not dire indications of an actual likelihood of device destruction if the guidelines are not followed.  Use a little thought.  If your hobby lab is adjacent to the hospital CAT scanner follow the advice about shielded wire and twisted pairs to the letter.  If you are operating in a normal housing unit far from heavy loads like the heater or a super fast kettle you can be more relaxed.

5.  I suspect this will work reasonably well if hooked up exactly as the diagram in the spec sheet shows.  The two biggest issues are whether the 100 microamp output limit of the circuit will provide enough base drive for the load transistor, and whether the load current turning on will cause enough change in the supply to bother the sensor.  Only testing the device will tell how sensitive it is, the data sheet is no help.  Or just use separate regulators for the sensor and load (or put the load on the wall wart and use a single regulator for the sensor.

6.  One warning they have is real.  This class of sensor does need a stable environment for best results.  If the HVAC cycles frequently, changing temperature by a few degrees over a couple of minutes, or if cars with warm motors are driven into your testing space or if a south facing window gives you a moving sunny spot or any number of other thermal events can cause false alarms or reduced sensitivity (range).

7.  These parts are static sensitive and ESD precautions should be followed.  This might be a warning about damaging the circuitry, but it is equally likely that charge accumulating on portions of the sensitive element will semi-permanently change the operating characteristics and thus ruin the sensitivity.

[EPAIII]

5.5 or 6 V Zener diode rated for twice the current rating of your wall wart. And call it a day.

[Terry Bites]

The Zener will melt pretty quickly if it's Pd rating is exceded (P=Id*Vz). That puts you back to square one with your precious parts aflame.
Ay, caramba! Even if the Zener can take the pain the trashy wall wart might not.
Magic smoke looms on the horizon either way.

An alternative to a crowbar is fast load disconnect. www.eevblog.com/forum/beginners/trying-to-work-out-how-overprotection-circuit-works-with-a-tl431-mosfet-design/msg4940407/#msg4940407  See www.ti.com/lit/an/snva681a/snva681a.pdf You dont need to reset it or put a new fuse in after tthe fault has cleared either. Happy days.

TPS2400 any one?

[seqcst]

But to figure out the resistance ratio, I need to know the resistance of my PIR sensor in reverse polarity situation. If the sensor normally draws 1uA at 5V, does that mean that it has 5 megaOhms of resistance? Do electronic components have fixed internal resistance?

No, electrical components do not have fixed internal resistance.  Even if some current does flow, it is really unlikely to cause any harm if the power involved is in the nanowatt range.

Does the requirement to use shielded cables mean that I can't use the standard breadboard jumper wires?

So this is why I warned you that you might be worrying about the wrong things.  First, one of those datasheets lists a maximum supply voltage of 4.5VDC, so how are you handling that?

I was thinking of using a Raspberry Pi Pico to provide 3.3V to the EKMBs.

The Pico has a RT6150B buck-boost converter with a current limit of 800mA and a power dissipation of 2-3W, which should be enough so that it doesn't overheat.

But now I think about it, the RT6150B might fail too? If it fails short, then it will pass the 5V USB power directly to the EKMB which can only tolerate 4.5V.

And what if the RT6150B fails in such a way that it drives up the voltage very high? Maybe I should put a shunt zener diode to prevent over-voltage here too? I guess I should also add a fuse.

I also could not find information on the RT6150B datasheet about its noise and voltage regulation.

If you are using the other one, it says 7.0VDC and that seems like the basic circuit above is perfectly adequate.  However, note that it specifies a maximum of 200V static electricity and shielded cables.  This is very, very low and indicates that these things are delicate and sensitive to transients.  A breadboard or breadboard jumpers will give you pretty bad results.  Wall-wart leakage may be an issue, as I mentioned earlier.  You need soldered twisted pairs inside a shield or coax or something like that.  And ground the negative supply rail properly.

I found this YouTUbe video:

In that video, it looks like the guy (Andreas Spiess) is just using a regular ESP32 with some regular jumper cables to attach to the PaPIRs. It looks like the jumper cables are attached via some heat-shrink wrap? Did he solder the pin to the exposed jumper wire first and then use heat-shrink tubing to wrap it?

Anyway, that video is what gave me the idea that it is probably okay to connect a PaPIR to directly to an Arduino Uno or a Pi Pico via some jumper cables, but then I got worried about the voltage regulator failing on the Arduino Uno and that's what made me ask about all this.

If using coaxial jumper wires, I would have to ground the shielding, right? Seems really complicated for a jumper wire that will be at most 20cm long.

Is grounding something just means connecting it to -ve? I'm thinking of connecting my power supply to the breadboard using one of those barrel jack screw terminal things - I'm not sure what the rating is on these things, can never find a datasheet, but most places say they're okay for 1-2A which should be plenty for my needs. Anyway, the screw terminal has a +ve and -ve coming out of it, so grounding something just means connecting it to the -ve, right?

[seqcst]

Wow, this looks complicated.

Does noise, line regulation, and load regulation not apply to shunt zener regulators?

Sorry if I wasn't clear--I'm only proposing the shunt as a protection system.  Using only a shunt regulator in this case would be more complicated.

I'm going to be running this from mains power. I'm literally going to just plug the wall wart directly into my mains power supply.

Mains power is what goes into your wall wart, what comes out is not unless there is a terrible failure inside of it.  Mains power has fault currents up to thousands of amperes, your wall wart probably has a fault current of 1 or 2 amperes.  The fuse and SCR being discussed are capable of working with mains power directly which is much more difficult.

If transients exceed the maximum rated input voltage of the device, or reach more than 0.8 V below ground and
have sufficient energy, they will damage the regulator. The solution is to use a large input capacitor, a series
input breakdown diode, a choke, a transient suppressor or a combination of these.

The zener and the other diode in the circuit I've attached will provide some protection against those startup transients, but if you are concerned about them you can add another regular 1N4001 in front of the regulator for complete reverse polarity protection.  The values of the capacitor and resistor will depend on your power source.

I've been trying to find a suitable fuse for that circuit and haven't been able to.

Main problem is that I want to make sure the fuse blows up before the zener diode does, but the 400mA fuses I've found all have HUGE voltage drop (around 1V) and takes like 5 seconds to blow even at 200% current.

I found some other fuses such as this one: 3414.0115.26

But that one has an almost 100mV voltage drop too.

I was going to ask "why can't I just connect the fuse in parallel?" then I realized, if the fuse blows, then all the current will go through the other path, so that would be stupid.

The problem is that I want to put the fuse after the voltage regulator, because I want to power my device from an Arduino, and the 5V output from the Arduino is already a regulated 5V, so if I put a fuse there then it drop the voltage to like 4.9V.

And also I'm worried that the zener fail before the fuse does...but then I heard that when a zener fails, it fails short, so that would actually be okay.

[CatalinaWOW]

If your goal is general learning you are doing very well.  You are understanding the crux of why fuses are not effective at protecting semiconductors.  In general fuses prevent fires, they don't do as well protecting equipment.  Those pesky details (time to blow, voltage drop, etc.) which are the real behavior of components, as compared to the conceptual ideal (opens instantly and completely when current rating is exceeded) are much of the essence of engineering.

It may take a while learning about all of the pieces of your circuit before you start learning about the IR sensor.

[seqcst]

I found this StackExchange answer that seems to answer my question: https://electronics.stackexchange.com/a/85099/367680

According to the datasheet for the 1N5339B-TP (the zener diode that I'm about to buy), it looks like it can withstand 6A for 1000ms.

If the fuse will blow in only 0.2s at 300% of its rated current, which is around 1.5A, then the fuse ought to blow before the zener diode does.

So that seems like that should solve the problem.

From looking at different datasheets, it looks like there is a trade-off between voltage drop and melting time.

It seems that the higher the voltage drop, the faster the fuse can blow.

I guess that makes sense since you need power in order to blow the fuse?

Should I use a TVS diode or a zener diode, or both in parallel?

[bdunham7]

The problem is that I want to put the fuse after the voltage regulator, because I want to power my device from an Arduino, and the 5V output from the Arduino is already a regulated 5V, so if I put a fuse there then it drop the voltage to like 4.9V.

And also I'm worried that the zener fail before the fuse does...but then I heard that when a zener fails, it fails short, so that would actually be okay.

After the regulator would be an entirely different problem and would leave you searching for some magic fuse that probably doesn't exist.  If you want protection like that, you'll need a low value current sense resistor and an active circuit of some kind. 

If you know the characteristics of the supply and choose the correct input resistor, the circuit as I've drawn it--with the fuse I've specified--will work and the zener will not be in any danger. 

I looked up the sensors you are using and it appears that they are $11?  If so, you may be overthinking this a bit...

[CatalinaWOW]

I found this StackExchange answer that seems to answer my question: https://electronics.stackexchange.com/a/85099/367680

According to the datasheet for the 1N5339B-TP (the zener diode that I'm about to buy), it looks like it can withstand 6A for 1000ms.

If the fuse will blow in only 0.2s at 300% of its rated current, which is around 1.5A, then the fuse ought to blow before the zener diode does.

So that seems like that should solve the problem.

From looking at different datasheets, it looks like there is a trade-off between voltage drop and melting time.

It seems that the higher the voltage drop, the faster the fuse can blow.

I guess that makes sense since you need power in order to blow the fuse?

Should I use a TVS diode or a zener diode, or both in parallel?

We are all trying to nudge you into two things.  1.  Figuring out how things work.  2.  Picking your battles.  You can't stick your thumb in every hole in the dike.

Look a little closer at your proposed zener clamp solution.  See the graph of zener voltage vs current.  The graph stops at the maximum allowed current of 1 amp, but is rising more and more rapidly with current.  The data sheet I looked at did not have a curve for your selected 5.1 volt zener so can you really satisfy yourself that the voltage across the zener will remain below the limit for your sensor at 6 amps? 

If you can, it indicates that you are not making a data based decision, and really need to re-evaluate why you are searching for this certitude that you will never over voltage your sensors. 

If not, the only path forward I can see is to buy a large quantity of the zener diodes and perform testing to see how they behave under the proposed test conditions. 

Or you could relax your requirements in another direction.  For example putting a few ohms of resistance in series with the output of your regulator (upstream of the zener).  Say 30 ohms.   Now if your regulator shorts the maximum current is on the order of 300 mA.  This hurts load regulation, but your sensors draw a maximum of about 300 microamps.  The real variation in this current will be far smaller, but assume that it bounces from zero to the max all the time.  The change in supply voltage would be about 9 millivolts, which could be further reduced with a capacitor across the load.

Have you decided on a voltage regulation spec?

[tooki]

I think you’re way, way overthinking this. Can the regulator on an Arduino or ESP32 board fail? Sure. Is it common? No, not at all, unless you abuse the regulator by significantly overloading it for an extended period of time. A cheap PIR sensor isn’t going to depend on exotic voltage regulation, it’ll be just fine with a regular one.

[kevin.gibbs]

Why don't you buy a 5-volt stabilized power supply? IR sensors consume little power; you can additionally put an RC filter on each.

[DavidAlfa]

A semiconductor can fail in so many ways depending on the stress/damage taken, I've never seen specs were it guaranteed no shorts, but I haven't seen so much either.
But:
- Add proper input filtering and protection against spikes/trasients.
- Don't rely on cheap chinese parts.
- The only true protection against shorts between input/output is to use a crowbar, so there's absolutely no way the output can be overvolted and damaged.
- Given the low current, just use a small resettable fuse for protection against input/output short to gnd.