Hi all,

I have a ciruit which can operate anywhere from 3.3V up to 5V.
My plan was to have three input supply options:

1) Regulated 3.3V (12V max input supply)
2) Regulated 5V (12V max input supply)
3) Unregulated input supply - can be anywhere from 3.3V to 5V (ex: powered directly from an arduino supply etc)

Is there a way to clamp any voltages above 5V when the 3rd option is selected (hence preventing the circuit fromany damages) ?
Ideally this protective circuit should be small (ex: just few passive components).
I did some research and could only find zener (which is not good for this application) and a clamp circuit (Vcc not available as this is to be used for protecting the main supply and not a secondary input)

Any suggestions would be highly appreciated.

 

For an effective suggestion to be made we also need the current requirement for this power supply. E.g. 50mA vs 5A. Without this information it is guessing.

 

Use a TL431 and transistor to clamp above 5V.

A circuit diagram of your inputs would be appreciated .

 

A thyristor crowbar. When the voltage is above the threshold the thyristor is triggered and shorts the supply.
Simple circuit but not very precise threshold voltage:

More complicated but more accurate circuits:

 

Probably the best simple solution is to use an ultralow-dropout 5 V regulator. It would need to be chosen carefully to be sure that it would not consume too much power itself when the input voltage was too low to allow it to regulate. Some of the old designs would take a lot of current under such circumstances. Newer designs, especially those using a FET as the pass element, are much better in this regard. One limitation is that the maximum current from some such regulators is quite low - in the 50 to 100 mA range. You may be out of luck if you need a through-hole part.

 

Hello,

@ebp , would this circuit do as ultra low LDO regulator?



Bertus

 

Bertus, that is a truly impressively low dropout voltage for the current!

The problem that arises with all regulators that use a PNP device as the pass transistor is that the base current for it is sunk to "ground." This reduces efficiency, which often is of no particular concern. However, if the input supply voltage has dropped to the point where the regulator can maintain the output voltage, the base current will be increased to the maximum by the error amplifier. For something powered from a battery, this is a bad thing because as soon as regulation is lost the regulator itself will very rapidly complete the discharge of the battery.

Devices that use a FET for the pass device can achieve low dropout with more-or-less "zero" drive current for the FET. Other problems remain. For example, if the input voltage is abruptly raised from below to above the dropout threshold, you can get overshoot on the output voltage because it takes some time for the error amp to come "off the rail" and regain control (this isn't unique to LDO regulators and is an issue in all sorts of closed-loop control systems - the "large signal" behavior versus the "small signal" behavior).

Picking the best LDO regulator for a particular application can be rather time consuming because of the need to pay attention to a lot of the details in the datasheet.

 

Hi all,

Thanks for the replies. The circuit will consume around 50mA.
Did a quick schematic of the intended circuit:

 

@AlbertHall , I did some research about the circuit you highlighted, but the information is a bit vague.
Tried the circuit and it is not working (zener 5.1V, supply 8V and output 8V -- not shorting)
What is wrong with the circuit please?

 

Did you try this in a simulator or real life?
If it was in a simulator and the input 8V comes from a voltage source, then even if the thyristor triggers the input will still be 8V even though a large current will be flowing. Try a low resistance between the circuit and the source, say 1Ω.

 

Did you try this in a simulator or real life?
If it was in a simulator and the input 8V comes from a voltage source, then even if the thyristor triggers the input will still be 8V even though a large current will be flowing. Try a low resistance between the circuit and the source, say 1Ω.

Thanks a lot, it worked What I am noticing is that the circuit is only being triggered at 6V when using a 5.1V zener. Should I replace the zener with a 4.7V one, or is this a simulation issue as well?

 

Hello,

Thanks a lot, it worked What I am noticing is that the circuit is only being triggered at 6V when using a 5.1V zener. Should I replace the zener with a 4.7V one, or is this a simulation issue as well?

That would be normal, as the gate voltage of the thyristor must be taken in account.
The datasheet of the thyristor will tell you the value of the trigger gate voltage.

Bertus

 

The circuit at #9 is a latching crowbar - it works by short-circuiting the input supply when the SCR is triggered. At low power it isn't too bad, but at high power it is poor because it triggers the SCR with the lowest possible current. This can lead to hot-spotting in the SCR - only part of the total area turns on, overheats and potentially fails because of excessive current through the hot spots. Albert's circuits with the TL431 are much superior, not only because the trigger voltage is more accurate but also because the SCR gets a good strong current for the gate when it is first triggered.

However, the more significant issue is that once the SCR is turned on, it will stay on until the current through it falls below its "hold current." Hold current for an SCR rated at a few amperes can is usually no more than a few tens of milliamps. Depending on the effective source impedance, which includes any added series resistance, this would mean the source voltage would need to be reduced far below "normal" to get the SCR to turn off. For example, if the SCR's hold current were 20 mA and the source impedance were 100 ohms, the source voltage would need to be reduced to below 2 volt to get the SCR to turn off. 100 ohms would generally be considered to be a very high source impedance for a low-voltage power supply (resulting in a 100 mV drop per milliamp of load).

For simulation, instead of DC input, use a triangle wave that swings from 0 V to 8 V and back to zero. And look at currents - always look at currents, not just voltages.

Zener diodes below something in the about the 5 to 6 volt range are usually "true" zeners. The "knee" of the voltage versus current curve tends to be very rounded and moderate reverse current will flow well below the nominal reverse breakdown voltage. Above that voltage range, the devices are actually "avalanche diodes" and have sharper knees. Simple simulation models may not be very good matches for real devices unless they actually have part numbers corresponding to real devices. Again, a triangle (current) wave is useful for testing the model.

 

Hi again,

Thanks for the useful information.
I did some further experimental simulations and was comparing the performance of a crowbar circuit (left plot) vs a zener diode (right plot).
For the purpouse for protecting the circuit, what would be the ideal circuit to use, and why?

Blue: Input voltage
White: Output
Red: Threshold

 

A crowbar is normally used to intentionally short-circuit a power supply that has gone out of regulation and is putting out a voltage that could damage the powered circuit. The expectation is that if the crowbar has "deployed" the power supply will require repair. Sometimes the crowbar SCR will also fail, hopefully short-circuit so protection is maintained.

A zener will limit voltage, but it is not easy to get precise limiting and the amount of current a particular type can handle is limited. A more precise circuit is a "shunt voltage regulator." The TL431 is a shunt regulator or reference. The voltage is set by a pair of resistors and it acts much like a high-precision zener. Again, the current handling capability is limited and there must be a series resistor between it and the input supply to limit the current. There are ways of boosting the amount of current it can handle and I think you will find circuits in the datasheet or applications notes. What you must consider is that the series resistor will always produce a voltage drop even without the shunt regulator conducting. The value must be carefully evaluated to be sure that the drop is not excessive when the reg is not conducting, the current when it is conducting with the highest input voltage is acceptable and the power dissipation of both the resistor and the regulator will be OK.

Again, for the amount of current involved I suggest a 5 V series regulator designed for low dropout voltage and low "quiescent current."

 

@ebp , If I opt for the zener configuration (due to the small area required by this configurtion), the equation for calculating the series resistor is R = (Vin - Vout) / I_load

For a 5V input: R = (5V-5V)/0.08 = 0ohms
For a 3.3V input: R = (3.3V-5V)/0.08 = n/a
Assuming the max input can reach up to 12V: R = (12V-5V)/0.08 = 88ohms

Will an 88ohm resistor work for all scenarios (i.e. input of 3.3V, 5V and any voltage up to 12V) ?