Circuit explanation - Page 1
Forum section: electronics / beginners

author: ElectricGuy
author_rank: Regular Contributor
timestamp: March 04, 2016, 08:31:45 pm
content: 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

author: PA0PBZ
author_rank: Super Contributor
timestamp: March 04, 2016, 09:18:51 pm
content: Looks like they clamp the input to the uC to + and - (6.4+0.7)V. I wonder what micro that is...

author: Belgarion
author_rank: Unknown
timestamp: March 04, 2016, 10:09:20 pm
content: 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.

author: ElectricGuy
author_rank: Regular Contributor
timestamp: March 04, 2016, 10:18:57 pm
content: Quote from: PA0PBZ on March 04, 2016, 09:18:51 pmIt is a PIC micro.

author: Shadetreeprops
author_rank: Regular Contributor
timestamp: March 04, 2016, 10:32:29 pm
content: 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.

author: suicidaleggroll
author_rank: Super Contributor
timestamp: March 07, 2016, 03:52:06 pm
content: Quote from: Shadetreeprops on March 04, 2016, 10:32:29 pmYou'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.

author: ElectricGuy
author_rank: Regular Contributor
timestamp: March 07, 2016, 04:18:28 pm
content: Quote from: suicidaleggroll on March 07, 2016, 03:52:06 pmBelieve 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.

author: suicidaleggroll
author_rank: Super Contributor
timestamp: March 07, 2016, 04:21:46 pm
content: Quote from: ElectricGuy on March 07, 2016, 04:18:28 pmI 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.

author: danadak
author_rank: Super Contributor
timestamp: March 08, 2016, 12:34:23 am
content: 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.

author: IanB
author_rank: Super Contributor
timestamp: March 08, 2016, 12:46:07 am
content: Quote from: ElectricGuy on March 04, 2016, 08:31:45 pmIf 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.

author: ElectricGuy
author_rank: Regular Contributor
timestamp: March 08, 2016, 01:06:06 am
content: Quote from: IanB on March 08, 2016, 12:46:07 amThanks 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.Quote from: danadak on March 08, 2016, 12:34:23 amLike 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.Quote from: suicidaleggroll on March 07, 2016, 04:21:46 pmsorry, i thought it was for me.

author: suicidaleggroll
author_rank: Super Contributor
timestamp: March 08, 2016, 01:44:08 am
content: 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.

author: ElectricGuy
author_rank: Regular Contributor
timestamp: March 08, 2016, 01:47:57 am
content: Quote from: suicidaleggroll on March 08, 2016, 01:44:08 amAnd since is a 5V Pic micro, i really don't get it

author: Godzil
author_rank: Frequent Contributor
timestamp: March 08, 2016, 02:17:45 am
content: Do you have the value of the capacitors?

author: ElectricGuy
author_rank: Regular Contributor
timestamp: March 08, 2016, 02:20:47 am
content: Quote from: Godzil on March 08, 2016, 02:17:45 amIt's on the schematic, 100nF ( first post )

author: IanB
author_rank: Super Contributor
timestamp: March 08, 2016, 02:29:56 am
content: Quote from: ElectricGuy on March 08, 2016, 01:06:06 amIf 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?

author: ElectricGuy
author_rank: Regular Contributor
timestamp: March 08, 2016, 10:06:18 am
content: Quote from: IanB on March 08, 2016, 02:29:56 amI 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.

author: IanB
author_rank: Super Contributor
timestamp: March 08, 2016, 10:32:34 am
content: Quote from: ElectricGuy on March 08, 2016, 10:06:18 amBut what is the line being protected? Is it an input signal? I don't think you have ever said what its function is?

author: ElectricGuy
author_rank: Regular Contributor
timestamp: March 08, 2016, 10:36:12 am
content: Quote from: IanB on March 08, 2016, 10:32:34 amSorry, yes, its an input signal line. It runs a +-200m cable lenght.

author: Shadetreeprops
author_rank: Regular Contributor
timestamp: March 08, 2016, 12:58:52 pm
content: Quote from: suicidaleggroll on March 07, 2016, 04:21:46 pm