Circuit explanation

[ElectricGuy]

Hi;

I just reversed engineered a circuit ( attached ), but i don't understand why are the 2 diodes that way.
I mean, if D1 was connected to ground, that's ok. If D2 was connected between the Cathode of D1 and VCC, that's also ok, but the two anodes are connected, and one cathode goes to the micro line, and the other to Vcc.
Can somebody explain to me?
Thanks

[PA0PBZ]

Looks like they clamp the input to the uC to + and - (6.4+0.7)V. I wonder what micro that is...

[Belgarion]

But it's referenced to 5V and not ground, so if my thinking is correct it should be clamping to (5+6.4+0.7) = +12.1V and (5-6.4-0.7) = -2.1V.

[ElectricGuy]

Looks like they clamp the input to the uC to + and - (6.4+0.7)V. I wonder what micro that is...

It is a PIC micro.

[Shadetreeprops]

So im no expert by any means what so ever, im as new as you. but the way im looking at it. Without the dioeds, the caps would not properly charge, casue how electricty likes the path of least resistance, as much as water...so that is the only reason i can see for the dioeds to be that way. but i would not take my word for it, casue i am absolutely new to all if it myself.

[suicidaleggroll]

So im no expert by any means what so ever, im as new as you. but the way im looking at it. Without the dioeds, the caps would not properly charge, casue how electricty likes the path of least resistance, as much as water...so that is the only reason i can see for the dioeds to be that way. but i would not take my word for it, casue i am absolutely new to all if it myself.

You've got it backwards.  The diodes aren't there to help the caps, the caps are there to help the diodes.  Those are TVS diodes, designed for ESD protection.  The caps help to slow the transient down...absorb some of the current to give the diodes time to open up and short it out.  At least as far as I understand it.

[ElectricGuy]

So im no expert by any means what so ever, im as new as you. but the way im looking at it. Without the dioeds, the caps would not properly charge, casue how electricty likes the path of least resistance, as much as water...so that is the only reason i can see for the dioeds to be that way. but i would not take my word for it, casue i am absolutely new to all if it myself.

You've got it backwards.  The diodes aren't there to help the caps, the caps are there to help the diodes.  Those are TVS diodes, designed for ESD protection.  The caps help to slow the transient down...absorb some of the current to give the diodes time to open up and short it out.  At least as far as I understand it.

Believe me, they are not backwards. That's why i post it here, i checked, re-checked, and ask other person to checked as well. The circuit is exaclty like i posted.

[suicidaleggroll]

So im no expert by any means what so ever, im as new as you. but the way im looking at it. Without the dioeds, the caps would not properly charge, casue how electricty likes the path of least resistance, as much as water...so that is the only reason i can see for the dioeds to be that way. but i would not take my word for it, casue i am absolutely new to all if it myself.

You've got it backwards.  The diodes aren't there to help the caps, the caps are there to help the diodes.  Those are TVS diodes, designed for ESD protection.  The caps help to slow the transient down...absorb some of the current to give the diodes time to open up and short it out.  At least as far as I understand it.

Believe me, they are not backwards. That's why i post it here, i checked, re-checked, and ask other person to checked as well. The circuit is exaclty like i posted.

I never said you did, my post was not directed toward you.  I was talking to Shadetreeprops who suggested that the only reason the diodes were there were to charge the caps.

[danadak]

Circuit just does not seem to be drawn right.

Spice simulation shows it does not look like it functions in any meaningful way.


Regards, Dana.

[IanB]

Hi;

I just reversed engineered a circuit ( attached ), but i don't understand why are the 2 diodes that way.
I mean, if D1 was connected to ground, that's ok. If D2 was connected between the Cathode of D1 and VCC, that's also ok, but the two anodes are connected, and one cathode goes to the micro line, and the other to Vcc.
Can somebody explain to me?
Thanks

If the voltage goes too high then D1 will conduct and stop the voltage going higher. If the voltage goes too low then D2 will conduct and stop the voltage going lower. Between them the two diodes will keep the voltage close to the expected value. If the voltage is in range then no current flows through the diodes.

The capacitors are there to absorb the energy from voltage spikes and help the diodes do their job. The whole thing is an input protection circuit to protect the micro from ugly spikes on the power supply.

[ElectricGuy]

If the voltage goes too high then D1 will conduct and stop the voltage going higher. If the voltage goes too low then D2 will conduct and stop the voltage going lower. Between them the two diodes will keep the voltage close to the expected value. If the voltage is in range then no current flows through the diodes.

The capacitors are there to absorb the energy from voltage spikes and help the diodes do their job. The whole thing is an input protection circuit to protect the micro from ugly spikes on the power supply.

Thanks IanB;

I get the first one. If the voltage is above 6.4V the diode starts conduct. But since D2 is conected to 5V i really don't know how when the voltage is more negative will protect the circuit.

Circuit just does not seem to be drawn right.

Spice simulation shows it does not look like it functions in any meaningful way.

Regards, Dana.

Like i told before, it was cheched, and re-checked, by me and other person. It was re-checked, because it does not  look right also for me.

I never said you did, my post was not directed toward you.  I was talking to Shadetreeprops who suggested that the only reason the diodes were there were to charge the caps.

sorry, i thought it was for me.

[suicidaleggroll]

I agree with Belgarion, it should limit the input voltage between about -2 to +12v.  Kind of an odd range considering it's feeding an MCU.

[ElectricGuy]

I agree with Belgarion, it should limit the input voltage between about -2 to +12v.  Kind of an odd range considering it's feeding an MCU.

And since is a 5V Pic micro, i really don't get it 

[Godzil]

Do you have the value of the capacitors?

[ElectricGuy]

Do you have the value of the capacitors?

It's on the schematic, 100nF ( first post )

[IanB]

I get the first one. If the voltage is above 6.4V the diode starts conduct. But since D2 is conected to 5V i really don't know how when the voltage is more negative will protect the circuit.

If the incoming voltage goes below about -2 V then D2 will conduct and prevent the voltage going any lower. Maybe that is the design intent?

[ElectricGuy]

I get the first one. If the voltage is above 6.4V the diode starts conduct. But since D2 is conected to 5V i really don't know how when the voltage is more negative will protect the circuit.

If the incoming voltage goes below about -2 V then D2 will conduct and prevent the voltage going any lower. Maybe that is the design intent?

I get your point, but if the PIC Micro only works between 0 to 5V, with +-0.5V tolerance, don't you think that -2V is over the specs?
In my opinion the desing intent is to protect the IOs, i just don't think it is done well, and i'm talking about a comercial product for the medical industry.

[IanB]

I get your point, but if the PIC Micro only works between 0 to 5V, with +-0.5V tolerance, don't you think that -2V is over the specs?
In my opinion the desing intent is to protect the IOs, i just don't think it is done well, and i'm talking about a comercial product for the medical industry.

But what is the line being protected? Is it an input signal? I don't think you have ever said what its function is?

[ElectricGuy]

I get your point, but if the PIC Micro only works between 0 to 5V, with +-0.5V tolerance, don't you think that -2V is over the specs?
In my opinion the desing intent is to protect the IOs, i just don't think it is done well, and i'm talking about a comercial product for the medical industry.

But what is the line being protected? Is it an input signal? I don't think you have ever said what its function is?

Sorry, yes, its an input signal line. It runs a +-200m cable lenght.

[Shadetreeprops]

So im no expert by any means what so ever, im as new as you. but the way im looking at it. Without the dioeds, the caps would not properly charge, casue how electricty likes the path of least resistance, as much as water...so that is the only reason i can see for the dioeds to be that way. but i would not take my word for it, casue i am absolutely new to all if it myself.

You've got it backwards.  The diodes aren't there to help the caps, the caps are there to help the diodes.  Those are TVS diodes, designed for ESD protection.  The caps help to slow the transient down...absorb some of the current to give the diodes time to open up and short it out.  At least as far as I understand it.

Believe me, they are not backwards. That's why i post it here, i checked, re-checked, and ask other person to checked as well. The circuit is exaclty like i posted.

And i thank you for that correction. i am still trying to get all the diffrent symobls, i did not notice the crook at the end of the dioed referenced. thank you for letting me know that too. i will make a note of it..

I never said you did, my post was not directed toward you.  I was talking to Shadetreeprops who suggested that the only reason the diodes were there were to charge the caps.