Maker Pro
Maker Pro

Circuit Protection For Inputs In Hostile Environment

mpopeang

Aug 5, 2017
17
Joined
Aug 5, 2017
Messages
17
Hi

Is there any modern IC that can be used as a protection circuit for bad inputs ?
I am trying to see any working solutions available instead of using a custom one using diodes and resistors so i can minimize space on a board.
The protection should handle -7V - +12 V on each input pin.

Thanks a lot
Marian
 

(*steve*)

¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd
Moderator
Jan 21, 2010
25,510
Joined
Jan 21, 2010
Messages
25,510
It depends on what you're trying to protect the input from, the nature of the input, the impedance of the usual source, and the requirements for handling fast edges.

For example, what you need is vastly different if you're trying to protect against direct lightning strike vs a line level output connected to a phono input.

Protection circuits can have an impact on the signal or the input impedance, so if these are important, you need to take it into account. (The signal fidelity may not be so important for digital inputs as it is for analog inputs)
 

mpopeang

Aug 5, 2017
17
Joined
Aug 5, 2017
Messages
17
I am trying to protect the signal lines that go to a microcontroller (or some ttl logc) that will work at 5V but can be under short/high voltage etc. If i can protect the signals between -7V up 12V is very good. Thats how wrong that circuit can end up with.

And i am trying to minimize the number of components. I've seen solutions that use BAV99 that can protect 1 signal line in concordance with 100ohm resistor. Using network resistors can be fine. I'll need 5 of these. (8x5 = 40)
I have at least 40 signal lines which needs protection. Older solutions used in fluke 9010A pod for example uses some custom made IC. (like 53-4066T -i can't find these anywhere) and can connect 8 signals per IC. And for 40 lines you use 5 of them.
This was 30+ years ago. I am trying to see newer solutions.(hoping there are :) )

If i use the BAV99(has 2 clamped diodes) , i'll need 40 of these on the board.
If i use BAV99BRW( has 2 pairs of 2 clamped diodes), i'll still need 20 of these. Plus they are smd and i might have issue soldering these.

Any IC with many inputs to do this job should do it. But one IC that is still produced. The ones that i found are not produced anymore.

Any suggestions ?

thanks a lot.
 

(*steve*)

¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd
Moderator
Jan 21, 2010
25,510
Joined
Jan 21, 2010
Messages
25,510
The inputs already have their own diode protection. For minimal parts count you could just add a simple series resistance that is sufficient to limit the current through this protection.

There are packages that contain multiple resistors, so you may only need one device per 4 or 8 inputs.
 

Harald Kapp

Moderator
Moderator
Nov 17, 2011
13,700
Joined
Nov 17, 2011
Messages
13,700
Check USB port protection chips (e.g. here). A single small chip protects 2 lines against ESD. Series resistors need to be added if the source of overvoltage has a low impedance.

I'm not aware of any single chip that protects more than these 2 lines.
 

kellys_eye

Jun 25, 2010
6,514
Joined
Jun 25, 2010
Messages
6,514
There are also three-terminal signal conditioning components often used on I/O line - they look like little 'blobs' and comprise a simple LC filter.

Darn.... can't find a quick link to one.... typical :rolleyes:
 

mpopeang

Aug 5, 2017
17
Joined
Aug 5, 2017
Messages
17
I think i'll go with BAV99 if i can't find anything else. The BAV99BRW is much smaller and not sure how much harder to solder.
 

Harald Kapp

Moderator
Moderator
Nov 17, 2011
13,700
Joined
Nov 17, 2011
Messages
13,700
Not a bad choice, but: the BAV 99 is good for low currents only, 130 mA to 215 mA depending on the version you use. You'll need current limiting resistors to prevent destruction of this diode by a low impedance overvoltage source.
The series resistor needs to drop 7 V (from -7 V to GND or from +12 V to +5 V). 7 V/100 mA = 70 Ω, therefore a 100 Ω resistor is a suitable choice.
You'll have to take into account the input current of your circuit. This resistor will drop some voltage even under regular coperating conditions due to this input current. The resulting voltage, after accounting for the voltage drop, needs to high enough to be detected as a logic high signal by your circuit.
In case your input circuit draws very little current, you can increase the series resistor's value to reduce the burden on the protection diodes, The math is up to you.

You may also want to consider the reaction of your circuit to such an overvoltage/undervoltage event, You may want to filter the input signal to suppress short spikes. The series resistor is a good starting point. Add a capacitor after the resistor from input to ground and you have a low pass filter which will suppress high frequency disturbances and short spikes. The backside is a slow edge of te signal after the filter. A Schmitt-Trigger input characteristic is what the doc recommends to restore the input to a clean signal.
 

Arouse1973

Adam
Dec 18, 2013
5,178
Joined
Dec 18, 2013
Messages
5,178
The diodes you are suggesting may not work properly. They may give the impression they are working. This is why, if the inputs you are protecting have internal protection diodes as most IC's have now you may find the maximum input to this device is only +/- 0.3 Volts above or below the supplies.

This means using an external diode which does not have a lower forward voltage than the internal diodes will cause the internal diodes to switch first in a transient condition. The external diodes comes in too late to help.

You can use the internal diodes and a series resistor to prevent latch up but I personally don't use this method. There is a reason for using the correct transient protection device.

You will have to specify the nature of this transient pulse because some transient protection devices are too slow in some cases.

Thanks
Adam
 

(*steve*)

¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd
Moderator
Jan 21, 2010
25,510
Joined
Jan 21, 2010
Messages
25,510
The diodes you are suggesting may not work properly.

+1

If you're going the pair of diodes approach, Schottky diodes are better, and a resistor before and after them will limit the current through them as well as ensuring most goes through the external diodes.

One of the other problems with using diodes this way is that they can shunt excess voltage to the power rails. This can cause the power rails to rise in voltage, and depending on the total energy you might preserve the input but kill the chip via excess voltage anyway. Maybe this means you need to place a zener diode across your power rails to crowbar any excess voltage.
 

mpopeang

Aug 5, 2017
17
Joined
Aug 5, 2017
Messages
17
I am trying basically to replace those old IC custom made ( 53-4066T ) and from the schematic they are using the 2 diodes with an 100 ohm resistor to protect 1 signal line. I assume the other problems you have rised up are already solved by the overall circuit. The power lines to the diodes are coming from another IC (9000A-8076 ? which i dont have any idea how to replace it) which shuts down the power once there are fluctuations from the diodes.
It works on -5V and +5 V and the other sense inputs are connected to a series resistor from input voltage of 5V (which can be bad and assumed in between -7V - +12 V). These resistors are from a package similar to a SIP but with individual specific values : 2275 ohm, 250 ohm, 1512 ohm, 333 ohm, 2997 ohm (called 54-4066T).
They seem to work to drop voltage on specific values so that the IC can cut of the power off. The output of the IC is a power fail signal which controls the power on the board.
The power lines to diodes from it are 4.3V and 0.7V. (and backed up by pull up and pul down resistors).
 
Top