Reading and understanding this schematic [Discussion]  - Page 1
Forum section: electronics / projects

author: XaviPacheco
author_rank: Regular Contributor
timestamp: June 14, 2019, 01:23:33 pm
content: See attachments Power1 and Power2. I'm doing some reverse engineering and I'm trying to understand the functionality. The two schematics are located in different PC boards that are interconnected. As you can see, for example, the +27V point comes out from Power1 schematic and connects with Power2 schematic where it's feeding a LM317 regulator. I'm not showing the whole circuitry. There's another control circuit which processes a signal coming from a current transformer. Additional information: the input line to line voltage ranges from 220 to 480 VAC. So based on the schematics:1) How are the +27 and -13 voltages generated? I'm kind of confused as it seems like in Power1 schematic the reference is NOT zero volts. I know there should be a trick with the zener diodes.2) How is the FET turned on? Is it through the +V point which connects with -13V through the SCR in Power2 schematic?3) At the drain of the FET, there is a freewheeling diode for the coil and a stack of three TVS diodes in series (not sure why three).4) At last, I want to understand better the setup at the source of the FETI'm just asking for few hints that help me interpret the circuit in a better way.What improvements would you make to this circuit?

author: Psi
author_rank: Super Contributor
timestamp: June 14, 2019, 01:58:03 pm
content: For the LM317 the 13V zener adds 13V, canceling out the -13V to zero.Next the 12V zener raises the output by 12V.Then you have the internal 1.25V reference of the LM317.Hence the final output voltage is 12+1.25 = 13.25V with respect to what he is calling GND.Note: This designer has chosen to GND a mid point in the output voltage in order to get the -13V. He could just as well have called that point GND.

author: magic
author_rank: Super Contributor
timestamp: June 14, 2019, 02:16:15 pm
content: M1 is turned on by R1 charging its gate towards positive rail.R2+R3 and D2 implement current limiting - if current reaches 15~20mA, there is almost 10V loss across R2+R3 and D2 discharges M1 gate to a voltage barely above M1 source, limiting further current flow through M1.D1 limits M1 gate to 43V above the negative rail, so for M1 to conduct at all, about 40V above the negative rail must be at its source. Since the negative rail is regulated by zener D3 to -13, that makes about 27V at the output of M1 if there is no load current across R3.The SCR appears to be a crowbar which shorts M1 output to ground to shutdown downstream circuitry.Not sure what's the point of the choke.

author: XaviPacheco
author_rank: Regular Contributor
timestamp: June 14, 2019, 02:30:00 pm
content: D3 in schematic 1 is the freewheeling diode for the coil I think. This coil should be activated when commanded by the control circuit (that I'm not showing) to trip something. The control circuit is being powered by the 13.25V coming out of the LM317 and uses both references -13V, and 0V. I was trying to see what's the origin of signal the FET needs to get activated. How can this circuit be improved? Also, what's the point of the three TVS diodes in series? The reverse breakdown voltage is about 900V.

author: XaviPacheco
author_rank: Regular Contributor
timestamp: June 14, 2019, 02:35:11 pm
content: Quote from: magic on June 14, 2019, 02:16:15 pmSo in the first circuit the 0V reference is brought down to -13V by the second circuit?

author: magic
author_rank: Super Contributor
timestamp: June 14, 2019, 10:36:27 pm
content: FET is permanently pulled up by R1, there is no signal to disable it. Unless you mean that "the coil" is not an inductor between the indicated points but a relay which switches power on and off to the drain of the FET. Then nobody can tell you what controls that relay because you didn't draw itYes, the second circuit regulates 13V above the first's negative rail and ties it to ground, therefore forcing the first's negative rail to -13V. The frst looks like a secondary of a transformer so it "floats" and adapts to whatever voltage you connect it to because it has no electrical connection to ground on its own.

author: XaviPacheco
author_rank: Regular Contributor
timestamp: June 17, 2019, 02:13:56 am
content: Quote from: magic on June 14, 2019, 10:36:27 pmThe coil attached to the indicated points is a 115 VDC trip coil. I swear I have drawn the whole circuit around the FET though. It doesn't make sense the trip coil is always active. So according to the schematic I attached the FET is always active as well as the coil?

author: floobydust
author_rank: Super Contributor
timestamp: June 17, 2019, 03:10:35 am
content: I think the designer is using the coil as a power resistor to drop the HV down to power the LM317 and other circuitry.It's hard to analyze a circuit given bits and pieces. What does the mystery coil do? Ohms?Where does +V go? Where does the LM317 output go? If you redraw the circuit on one page it's much easier.If the MOSFET is on a big heatsink, then it's doing some constant-current control, instead of on/off switching.Multiple TVS diodes is the only way to up the voltage and make a 900V clamp. It's needed to protect the MOSFET from mains transient overvoltages. The MOV clamps up to 1.65kV with lightning hits.

author: magic
author_rank: Super Contributor
timestamp: June 17, 2019, 07:15:32 am
content: I'm not sure what a "trip coil" is and how it works, apparently some sort of overcurrent device?If so, then it is possible that when this device trips, power is disconnected from the FET's drain and it cannot supply current to the other board even if it wanted to.However, the gate of the FET is permanently wired to positive rail through R1. The FET is always ON unless one of the zener diodes limits it. D1 comes into play if there is more than ~40V on the output, D2 if there is more than 15~20mA of output current. Yes, I calculated the current limit wrong in my previous post, corrected it now.

author: XaviPacheco
author_rank: Regular Contributor
timestamp: June 17, 2019, 02:23:23 pm
content: I have gathered some additional info. The device is an earth leakage protector that detects any current imbalance and sends a trip signal to other device. That's the circuit I'm not showing which is being fed by the LM317. I just drew the power supply part that seems pretty odd to me. I don't understand why the designer chose -13V as reference for the power board. I still don't get how the +27V is generated with the zeners and the FET. I thought the +V was some kind of signal that turned on/off the FET, but I think I'm wrong.Thanks for your inputs, I know it's kind of hard to understand this all without enough details. I will try to gather more info.

author: XaviPacheco
author_rank: Regular Contributor
timestamp: June 17, 2019, 05:06:28 pm
content: Just a brief question that came to my mind now. As the capacitor after the rectifier is just 22 nF in the first schematic, does this mean the inrush current is harmless, right?

author: magic
author_rank: Super Contributor
timestamp: June 17, 2019, 10:36:52 pm
content: The FET is a "source follower". Its source voltage follows gate voltage (minus a few volts) because if source voltage falls, source-gate voltage increases and the FET turns on.Its gate is held 43V above negative rail by R1/D1 so you get about 40V at source.The power board doesn't care whether its output is -13V/27V or 0V/40V. It would work exactly the same.

author: XaviPacheco
author_rank: Regular Contributor
timestamp: June 18, 2019, 01:40:58 pm
content: Quote from: magic on June 17, 2019, 10:36:52 pmThank you. At last, for me to understand better, I would like to prove these value by doing some quick math like KVL/KCL. How did you calculate the 15~20 mA? Maybe if I simulate it I can be more clear. Someone told me that in order to switch the FET, the V+ point should be grounded, which is what the SCR does in the second schematic, it connects +V to -13V. When the coil is activated, the main line power (480 VAC) is removed, and the circuit shuts down.I would happy if I'm able to prove (mathematically) the +27V. Yes, I know you already told me, but if I write some equations to show it, I would be more clear.

author: dmills
author_rank: Super Contributor
timestamp: June 19, 2019, 10:15:05 am
content: When you short the V+ and -13V point together the mosfet current rises (because the load in the source is now ~5R plus a diode drop or two), this causes the coil current to rise until either it is limited by the 5.1R resistor robbing Vgs or by the coil resistance.The trip coil will be designed to not release the trip when only the idle current is flowing but to cause the breaker to drop when this much larger current flows.This sort of thing is very typical of highly cost optimised design.Regards, Dan.

author: XaviPacheco
author_rank: Regular Contributor
timestamp: June 19, 2019, 12:10:07 pm
content: Quote from: dmills on June 19, 2019, 10:15:05 amWhy is this a highly cost optimized design? What are other things take into account in optimized designs?