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?