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ESD protection diode -constant current?

Coper

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I am designing a device that has an external connector connected ,+24V, +5V, GND and several digital input.
for simplicity, I used an optocoupler in the digital inputs, although it is not a fully isolated solution
see picture S1 is out of device , and it can be anything

For ESD protection, I added ESD protection diodes PESD5V0S2U .
Because there is also + 24V on the same connector and the manuals are not read, it is only a matter of time when someone connects + 24V to this input.

The current is conducted by PESD5V0S2U, but nowhere can I find what is the maximum constant current that this ESD diode can conduct permanently?
Alternatively, how better to design the protection of the digital input, or the entire input?
On R2 is ARM MCU GPIO


ESD.jpg
 

betwixt

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I would make R1 much bigger, maybe 100K or more. Connect it to the base of a PNP transistor instead of D1A/LED1.
Connect the transistor collector to ground and emitter through a 1K resistor to LED1.

The new R1 will protect against nasty voltages and the C-B transistor junction will work as your ESD protection. If noise pick-up is a problem at the wiring from S1, add a 1M resistor from the transistor base to 5V to make sure it is held in non-conducting mode until S1 is closed.

As this isn't an isolating solution you could easily replace U1 with a second transistor to get the same output.

Brian.
 

FvM

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It's possible to design a different input circuit as suggested. You should be however able to figure the continuous current rating out yourself. Diode power dissipation with 24V input is about 7V*17mA. Thermal resistance isn't given in the datasheet but can copied from similar SOT-363 diodes. 120 mW should be no problem.
 

Coper

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I also considered the solution, see the picture, it is very similar to the one proposed .
R4 in in colector because , her is connect GPIO from MCU.

What do I dislike about it?

1. It will turn on a current higher than 9uA , that is, if the insulation condition is below 200k ohm is Switch on.
2. ESD protection, hard to say, but I've already fixed a few such inputs, they had a destroyed transistor.

Therefore, I would also use ESD protection and probably a parallel capacitor in the base of the transistor .


esd2.jpg
 

betwixt

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I would place R4 in the emitter and directly ground the collector. You could add an anti-parallel signal diode across collector and base so effectively have two diodes wired opposite ways around, that should stop any ESD quite effectively.

Brian.
 

KlausST

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Hi,

and the manuals are not read, it is only a matter of time when someone connects + 24V to this input.
Don't feel responsible for other's mistakes.
I think it's counter productive to make a system mor complex, more expensive more probe to assembling errors and part failure..to avoid all possible mistreating.

In my eyes to optocoupler makes no sense (it is no protection device)
A BJT is more rugged, thus I'm with the BJT circuit.

Go for high impedance input to lower the current for the protection devices.
But don't make it too high impedance to avoid noise catch up and thus generating malfunction.

Klaus
 

std_match

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Is this a professional design for mass production?
In that case I would not approve that the voltage rating of R1 is exceeded because of ESD and other high-voltage pulses.
You then need some component before R1 to limit the voltage to the rating of R1.
This can be a capacitor (capacitance calculated from the ESD human body model or the pulse generator specification).
It can also be a transient protection diode or a VDR, but the standoff voltage must be high enough to allow for "expected" short circuits (normally to the highest expected supply voltage).
 

carpenter

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if it is to have a really breakdown resistance to ESD, including induced voltage from long cables during a storm, a robust and comprehensive solution is needed
One example, including the arrangement on the PCB.

input_protection_v2.pngISOLATION.jpg
 

Coper

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74LVC14 have not defined output level for high impedance input.
if we connect a button to the input, for example, we have several options
1. Use internal pul up resistor and external input grounded
2. Use internal pul down resistor and external input connect to high level voltage
3. Use external ( at the button ) pul up resistor and external input grounded or vice versa
4. Use in a button with two contacts and 0 and 1 define on "hard" level.

Option one (two) is the most economical, but poor insulation of the conductor can lead to false positivity
Option three, requires 3 wires and even here there may be a problem with the insulation of the conductor. When using resistors, a lot depends on the size of the resistance, I consider 1M Ohm to be nonsense, and even with a resistance of 100k, the currents are in the order of micro Amps. Maybe 10k
Option 4 is probably the best condition, but not all buttons allow it.

Can anyone evaluate the benefits of which use is already risky?

The described connection theoretically allows the use as a scary signal other than 0 / 5V voltage level
For example 0/24V 24V flows through R2, D2 and is clamped on ZD1.
Would this be considered a normal operating condition for this connection or is such use possible but faulty?
 

KlausST

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Hi,

74LVC14 have not defined output level for high impedance input.
if we connect a button to the input, for example, we have several options
1) this is invalid, because there is no internal pull up
2) this is invalid, because there is no internal pull down
3) is valid
4) not recommended, because usual switches are "break before make" and thus cause intermedate floating signal, which should be avoided.

Option three, requires 3 wires
What does this mean? At the switch? No. At the resistor? No. At the IC? No....where else?
--> I see this as the useful method. You are most flexible because you may choose the resistor value on your own. You are free to adjust it to the switch requirements, wiring length, noise pickup....

In either case: consider switch contact bouncing.
* it may be no problem at all
* it may be removed by hardware
* it may be removed by software

You are free to input any voltage as long as:
* it results in valid levels at the logic IC input
* it does not violate any device specification
I did not check all devices' specifications, but 0V/24V should work..on a first sight.

Btw: I disagree with some informations at schematic of post#8. At least they are questionable.

Klaus
 

Coper

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) this is invalid, because there is no internal pull up
2) this is invalid, because there is no internal pull down
I put it inappropriately, I mean internal inside the device ie add a resistor to input 7414
 

Coper

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Btw: I disagree with some informations at schematic of post#8. At least they are questionable.
Can you clarify that?

It just hit me
C1 to 1kV , but D1 is only 50V. However, this may be related to the strength of the dielectric and the rate of voltage rise at the storm-induced voltage .
D2,D3 Here's the question, isn't it better to use a Transient Voltage Suppresssor Diodes ?
 

KlausST

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Hi,

C1 to 1kV , but D1 is only 50V.
Yes. This is one....
But it's even worse. C1 is in parallel to ZD1, which limits the voltage to +4.3V and -0.7V ....far from 1kV

*******
* R2 fusible type information
* R2 package size
* TTL/ CMOS information
...

Klaus
 

carpenter

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OK
for the introduction, it is not my desig, I put it here as an example of the complexity of the problem, an example with the arrangement on the PCB. That is, I do not need to defend him , but

Why is next to the 4V zener diode 1kV capacitor ?
Because of speed , Zener diode is semiconductor device and have some reaction time typical <3ns.
if the rising edge of the voltage pulse is fast enough, ZD can be destroyed before it starts working,
Solution? Use AVALANCHE TVS with reaction time about 50ps or use capactor for blunting edge .
C is of 1V order for the same reason plus it is about the quality of the dielectric and dV / dt. Of course it is a compromise price + size / function .

TVS/Zener Theory and Design Considerations

R2 size + fusible. Again, it is a compromise
larger dimensions = less likely voltage to flow out of resistance
 

KlausST

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Hi,

Why is next to the 4V zener diode 1kV capacitor ?
Because of speed , Zener diode is semiconductor device and have some reaction time typical <3ns.
Makes no sense to me. Two inductors plus a series resistor limit the current and the rise rate. At least for the first nanoseconds.
That's why they are there.
How can a 100nF capacitor be charged to several volts within those nanoseconds?
How much current is necessary for this?

Does X2K and Y2K mean "safety capacitor"? Makes no sense.

Klaus
 

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