Why will BAV99 not clamp to Vcc?

[Azagon]

I am trying to get a the input protection circuit right for protecting I/Os from outside signals (which are up to 30v). For this, I am trying to use BAV99, that is connected to ground on the one side and Vcc to the other (Vcc being 3.3v). But when I connect the external voltage, I get 3.75 at the BAV99 junction. Now I (sort of) understand that this is Vcc + the drop on the diode. But what I don't understand is, given this this is a standard way of protection, how can it ever be proper if it is always Vcc + 0.4? That will always push it very close to the micro controller's limits? (my circuit is attached)

 

[DemonicSandwich]

You could always place a 3.0-3.3v zener diode from the input to ground.

The zener will clamp for forward voltage to 3.3v or so and (being a diode) will clamp voltages that go below ground so you can ditch the BVA99.

 

[Azagon]

You could always place a 3.0-3.3v zener diode from the input to ground.

The zener will clamp for forward voltage to 3.3v or so and (being a diode) will clamp voltages that go below ground so you can ditch the BVA99.

Even a 3.3v zener has a forward voltage. For example, a BZX84C3V3 which is a "3.3v" zener, actually will clamp anywhere from 2.3v to 4.2v depending on the current (which depends on the voltage). So I feel this is not a good solution either.

 

[ericgibbs]

Even a 3.3v zener has a forward voltage. For example, a BZX84C3V3 which is a "3.3v" zener, actually will clamp anywhere from 2.3v to 4.2v depending on the current (which depends on the voltage). So I feel this is not a good solution either.

hi,
On your MCU type there could be already inbuilt clamping diodes on the port pins.???

Whats the MCU type.?

 

[MikeMl]

The MCU is almost certainly Silicon, while your external protection diodes are Schottky. The Si protection diode inside the MCU would have a forward conduction voltage of ~0.6V, while your external Schottky is ~0.4V, ergo, it you try to force the input pin high, the Schottky conducts, steering the current away from the MCU input.

 

[Nigel Goodwin]

I am trying to get a the input protection circuit right for protecting I/Os from outside signals (which are up to 30v). For this, I am trying to use BAV99, that is connected to ground on the one side and Vcc to the other (Vcc being 3.3v). But when I connect the external voltage, I get 3.75 at the BAV99 junction. Now I (sort of) understand that this is Vcc + the drop on the diode. But what I don't understand is, given this this is a standard way of protection, how can it ever be proper if it is always Vcc + 0.4? That will always push it very close to the micro controller's limits? (my circuit is attached)

It's perfectly fine, it's all designed to work at those levels,

 

[audioguru]

With an input as high as 30V then the current through the 1k series input resistor will be about 26mA which is too much current for the input protection diodes of a microcontroller. Increase the value of the 1k series input resistor then use the built-in protection diodes.

 

[MikeMl]

With an input as high as 30V then the current through the 1k series input resistor will be about 26mA which is too much current for the input protection diodes of a microcontroller. Increase the value of the 1k series input resistor then use the built-in protection diodes.

Most CMOS devices are still prone to latch-up. It is extremely poor design practice to rely on the "protection" diodes inside a CMOS device, especially if you are going to be injecting 10s of mA into the input. If you are expecting transient voltages far in excess of the CMOS device's Vdd (or less than Vss), using external clamping diodes is a much better way to go.

 

[OldTechie]

What are the static voltages at the signal input and the P1.3 connection? Is P1.3 an input or an output port pin?
Does either connection have a pullup to Vcc?

Dave M

 

[Nigel Goodwin]

Most CMOS devices are still prone to latch-up. It is extremely poor design practice to rely on the "protection" diodes inside a CMOS device, especially if you are going to be injecting 10s of mA into the input. If you are expecting transient voltages far in excess of the CMOS device's Vss (or less than Vss), using external clamping diodes is a much better way to go.

You're thinking CMOS logic chips, we're thinking micro-controllers, where they are specifically designed to operate in that way (at least PIC's are).

As suggested by AG though, his series resistor is much too low.

 

[MikeMl]

You're thinking CMOS logic chips, we're thinking micro-controllers, where they are specifically designed to operate in that way (at least PIC's are)....

Nigel,

I suggest you go and read the "Absolute Maximum Maximum Ratings" section of a PIC Data Sheet. For example: Read the line "Voltage on all other pins with respect to Vss"

 

[crutschow]

You're thinking CMOS logic chips, we're thinking micro-controllers, where they are specifically designed to operate in that way (at least PIC's are).

And what type of transistors and fabrication process do you think are used to make micro-controllers?

 

[Nigel Goodwin]

Nigel,

I suggest you go and read the "Absolute Maximum Maximum Ratings" section of a PIC Data Sheet. For example: Read the line "Voltage on all other pins with respect to Vss"

I suggest you go and read the MANY MicroChip application notes that use them - the protection diodes are there specifically for that reason, and the chips are designed to be use in that way.

 

[Nigel Goodwin]

And what type of transistors and fabrication process do you think are used to make micro-controllers?

What has that got to do with it? - the claim was that the chips 'lock up', PIC's (and other MicroControllers) don't - as simple as that. If CMOS logic chips lock up, then it's a design failure of the chip, NOT of the fabrication process.

 

[dougy83]

With an input as high as 30V then the current through the 1k series input resistor will be about 26mA which is too much current for the input protection diodes of a microcontroller. Increase the value of the 1k series input resistor then use the built-in protection diodes.

Seems like a little mistake here... current through the 1k resistor shouldn't exceed (3.75-3.3)/1k = 450uA (but will generally be less due to the Vf or the internal clamp diodes).

The configuration shown by the OP is likely fine.

 

[MikeMl]

Seems like a little mistake here... current through the 1k resistor shouldn't exceed (3.75-3.3)/1k = 450uA (but will generally be less due to the Vf or the internal clamp diodes).

The configuration shown by the OP is likely fine.

The OP stated that the input end of the series resistor can go as high as 30V, which means that without the external protection diode network, the current would be (30-(3.3+0.7))/1000 = 26mA!

 

[MikeMl]

I suggest you go and read the MANY MicroChip application notes that use them - the protection diodes are there specifically for that reason, and the chips are designed to be use in that way.

Since you are so certain, you go find the MicroChip App Note and post an excerpt here.

ps: I went and did your homework:

Here is what MicroChip says:

Anybody that relies on the internal ESD protection diodes to act as a clamp on any CMOS IC is extremely naive.

 

[Nigel Goodwin]

Since you are so certain, you go find the MicroChip App Note and post an excerpt here.

ps: I went and did your homework:

Here is what MicroChip says:

Anybody that relies on the internal ESD protection diodes to act as a clamp on any CMOS IC is extremely naive.

You mean you're done YOUR homework, ignoring all the MicroChip application notes (such as AN521) which you didn't like.

So presumably you think MicroChip advising their users to do this is "extremely naive"?, and that the (presumably millions of) commercial devices doing this all over the world don't actually work?.

 

[crutschow]

What has that got to do with it? - the claim was that the chips 'lock up', PIC's (and other MicroControllers) don't - as simple as that. If CMOS logic chips lock up, then it's a design failure of the chip, NOT of the fabrication process.

What's that got to do with it, is that all CMOS devices (expect for specialized military devices and some analog switches) use a similar fabrication process and design. Microprocessors are no exception to that. They all have some degree of susceptibility to latch-up. Now they may have minimized that tendency in the newest processes but that's not necessarily unique to microprocessors.

 

[audioguru]

They got rid of Cmos latch-up when they discontinued CD4xxxA ICs and introduced CD4xxxB devices about 40 years ago.

I am suprised the two Micro-Chip articles do not agree with each other.