PV inverter - Voltage reading circuit - Page 1
Forum section: electronics / repair

author: webgiorgio
author_rank: Regular Contributor
timestamp: January 08, 2019, 11:40:21 pm
content: Hi,I am repairing a 3 KW Power One inverter for grid connected photovoltaic.I get "error E031" when the inverter connects to the grid.This error means that the two AC voltage measurements do not match.It is typically caused by a bad relay, but replacing all of them didn't fix the problem.In the block diagram you can see that there are two AC measuring points, one on the grid side, and one on the inverter side.Soon after the inverter closes all the relays, the two voltages are compared, and if a mismatch is found, the inverter disconnects from the grid.So, I am investigating the voltage reading circuit.The voltage reading on the grid side is ok, because I see correct values in the inverter display when it is preparing to connect to the grid.I was looking for a voltage divider, but instead I found 4x 10 Mohm resistors connected as shown in attached schematic.The end of the resistors go into a flat cable to a board with 3 DSPs.What kind of circuit is it?It looks like the input of a high-voltage differential probe for oscilloscope (just learned it from EEVblog #932).With a 50:1 isolated probe and oscilloscope I found 300 mVpp on one channel, and about 100 mVpp on the other one (captured with the scope in "single trigger"). The 10 Mohm resistors are within specs.

author: coromonadalix
author_rank: Super Contributor
timestamp: January 09, 2019, 12:43:22 am
content: Have you tested all the 10 megs ohms resistors ? maybe a short, or bad solder ??An grid inverter always check the line vs solar panel / battery packs etc .. and decide to switch its output on the conditions given ...not necessarly an high voltage diff probe, but an high voltage divider

author: boB
author_rank: Frequent Contributor
timestamp: January 09, 2019, 01:16:14 am
content: So, you measured 300mV on the AC line and 100mV on the other side of the relay at the same time while the relay was ENGAGED ? TWO channels isolated ?The relay is probably engaged for a very short time before the error turns the relay off but maybe enough time for a reading. Or not.If the DSP has a differential input then the resistors may go there... Otherwise, there may be a differential op-amp circuit where each of these pairs of 40 Meg Ohm resistor strings connect to and then the single output of that op-amp goes to the A/D input of the processor.But if the relay trips too fast, you might not be able to take a good reading.I'd check the resistors too like coromonadalix mentioned.

author: webgiorgio
author_rank: Regular Contributor
timestamp: January 09, 2019, 10:43:16 am
content: The 10 Mohm resistors are within specifications.The solder joints seems ok.I only have one differential probe for the Oscilloscope.You see here the voltages I measured at 3 positions in the schematic, indicated with black arrows.The green means that all the relays are closed (parallel between grid and inverter).In the top left figure you see that the inverter gets the grid voltage for about 1.2 seconds, then the relays open (orange arrow) and the voltage decay (some filter capacitors discharge).Bottom right is the differential voltage of the grid side. Note the amplitude.Bottom left is the differential voltage of the inverter side. There are some oscillations when the relays switch to do the ground isolation measurement. Then all the relays close, but the amplitude of the differential voltage looks smaller than for the grid side. Hence the inverter disconnects the parallel.Sorry for taking photos of the digital scope, I have some problems with the USB-drive (when I plug it in, the scope becomes unresponsive)Also, the resolution is quite bad because I am using a 50:1 probe to measure a small (300 mV) signal.Update:The input impedance of the measuring circuit is 1.06 MOhm for the good differential channel, and 3.88 MOhm for the faulty channel. mhhh looks like some resistor is disconnected somewhere.

author: webgiorgio
author_rank: Regular Contributor
timestamp: January 09, 2019, 01:39:03 pm
content: So, I have looked at the control board, see attachment.I guess one 16 bit DSP is for the DC/DC, one for the inverter, and the 8 bit Atmega is for the relays and measurements.I guess that all those op-amps are signal conditioning of the two AC voltages, 2 AC currents, 2 DC currents, and 2 DC voltages, and differential current transformer (blue one).I see some 3.3 Mohm resistors near op-amps...Damn, how to do without schematic with this level of complexity?

author: Inverted18650
author_rank: Frequent Contributor
timestamp: January 10, 2019, 02:03:37 am
content: Is this one of the eBay Aruroa grid-tied inverters?

author: webgiorgio
author_rank: Regular Contributor
timestamp: January 10, 2019, 02:52:13 pm
content: it is a Power One Aurora PVI 3.0.

author: boB
author_rank: Frequent Contributor
timestamp: January 10, 2019, 04:33:40 pm
content: Yes, I see the difference in your scope pictures. Not sure why you would see a difference at the end of thoseresistor chains ? I can only wonder if what those resistors go into, (diff op amp ?) is bad on one measurement side ?We have a Power One Aurora at our shop but not sure if it's the PVI 3.0 ?You know...... One thing that comes to mind that might be your problem might be a small capacitor shorted.Differential op-amp circuits will have so many pF to GND and so many pF negative feedback. If one of thoseare open or shorted, relative to the other, it could drag down the level you see at 50 Hz when you look at itwith your scope. Probably best to trace those MegOhm resistors to where you can and see if that makessense.

author: webgiorgio
author_rank: Regular Contributor
timestamp: January 10, 2019, 11:09:25 pm
content: The resistance is different on the measurement side (with the 10 MOhm resistors disconnected).I traced the pins to two op-amps, two different part numbers! What the hell? what is the point in doing so? make one accurate measurement, and one crap just for reference?It might explain why I measure different input resistance (1 Mohm and the other is 4 MOhm).One is OPA270, the other is TI ezx6111 (I can't find the datasheet of this, the part number is quite well readable though. 14 pin).https://www.analog.com/media/en/technical-documentation/data-sheets/op270.pdfThere is 10k resistor in series with each input of the op-amp.A negative feedback of 75k with parallel C.A 75 k resistor with parallel C from the non-inverting input to 1.65V (thanks to the designer for the label!). It looks like 3.3V/2.What kind of measurement technique is this? does the input resistance of the op-amp plays a role in attenuating the voltage? (together with the 20 Mohm external).All the C in the area seems not shorted.With the inverter off, if I feed let's say 200 mVp-p with a signal generator into the 10k resistors, then measure the voltage with the scope across the op-amp input...Does it make sense? (anyway I don't have a signal generator here :/ )

author: boB
author_rank: Frequent Contributor
timestamp: January 11, 2019, 04:20:09 am
content: I wonder if both op-amps share the same GND and 3.3V ? Not that that necessarily should make a difference, but...But you're getting that error that says that they DO measure different.But yeah, a schematic are pretty darn handy !Maybe when you're done fixing this thing, there WILL be at least a partial schematic ?Is there two auxiliary power supplies does it look like ?

author: Inverted18650
author_rank: Frequent Contributor
timestamp: January 11, 2019, 05:41:00 am
content: I would bet its a relay issue. Try to bypass the relays or bridge the connections around them. Based on everything i have read today, that always seem to be the err031 problem. I read a lot about it trying to figure this out, as I am really into Inverter repairs as well.

author: boB
author_rank: Frequent Contributor
timestamp: January 11, 2019, 05:42:38 am
content: Quote from: Inverted18650 on January 11, 2019, 05:41:00 amI thought he had already tried that. Maybe not ? That would have been a good first thing to try.

author: fzabkar
author_rank: Super Contributor
timestamp: January 11, 2019, 09:04:53 am
content: IIUC, the non-inverting input is biased at a DC level of +1.65V, ie midway between ground and Vcc. This means that the inverting input will also sit at this same DC bias, if the op amp is operating within its linear region. Therefore the op amp configuration is that of an inverting amplifier with a gain of …Av = -Rf / Rin = - 75K / (10M + 10M + 10K) = -75 / 20010Now the AC input is 240Vrms, so the output of the op amp is …Vsense = 240 x 75 / 20010 = 0.9Vrms = 1.27VpThis means that the output of the op amp should swing by +/- 1.27V on either side of the DC bias level of 1.65V, ie 0.4V - 2.9V.

author: webgiorgio
author_rank: Regular Contributor
timestamp: January 11, 2019, 06:31:39 pm
content: @Inverted18650 First thing I did was replacing all the relays with new one. One of the old relays had a burned via, which I repaired.QuoteI can't find the datasheet of the 14 pin op-amp to identify the gnd and + pins(the marking looks like "ST 24I EZX 6111")However I could figure out which is the output thanks to the negative feedback resistor 75 k.The other 75 k is to the same 1.65V pad of the OPA270.I am working on the schematic and I will share as soon as is decent.I would like to measure the output of the op-amps without using my isolated probe 1:50, but using two normal 1:10 probes, so that I can take simultaneous measurements.What if I float my scope, connect the GND clip to the 1.65V node, and the probes to the output of each op-amp?So that I can verify the values offered by fzabkar!

author: Inverted18650
author_rank: Frequent Contributor
timestamp: January 11, 2019, 07:27:14 pm
content: Quote from: webgiorgio on January 11, 2019, 06:31:39 pmSorry mate, I did see you already said you replaced them all...I am tryingt o find the datasheet of the ezx6111 part now. Not much luck,TI does make a level shift IC but not in the 14-pin SOIC package that has EZX in the part number....I'll keep looking

author: fzabkar
author_rank: Super Contributor
timestamp: January 11, 2019, 07:55:02 pm
content: The AC voltage at the junction of the 10M and 10K resistors should be …10K / (10K + 10M + 10M) x 240Vrms = 120mVrms

author: webgiorgio
author_rank: Regular Contributor
timestamp: January 11, 2019, 11:52:02 pm
content: Sorry, it is ST, not Texas Instruments!The marking has the ST logo, then "24I", and line below is "EZX6111". That's so strange that it does not come up anywhere. Attached photo is the most left chip.The 8 pin ishttp://www.ti.com/lit/ds/symlink/opa2277-ep.pdfMy scope measurement on the L-N voltage between the 10k-10M node is showing 3 divisions times 250 mVpp (see earlier post).250 mV * 3 /2 /sqrt(2) = 265 mV rms.

author: fzabkar
author_rank: Super Contributor
timestamp: January 12, 2019, 12:07:27 am
content: The voltage at the junction of the two 10M resistors should be half the AC output. Is it?An input voltage of 265mV rms (750mVpp) would drive the op amp output to each of its supply rails., ie it would clip. :-?AISI, Vout and V+/V- should all be biased at 1.65VDC. I suspect that something is amiss at the op amp's V- pin.

author: webgiorgio
author_rank: Regular Contributor
timestamp: January 12, 2019, 12:37:41 am
content: Not half because of the 10 k resistors (and the op-amp need to be not saturated).If the V+ equals V-, then it is like we have the 10 k connected together.The voltage divider looks like 10M 10M 10k - 10k 10M 10M.In theory the voltage I should see on the scope across the 20 k is:V_LN*2*sqrt(2) * (20k / 20020 k) = 650 mVppI measure about 750 mVpp, so, it is correct (on the V_LN channel)On the bad channel I see less, probably because the bad O-Amp is not producing an output that makes V- equal to V+. A bad op-amp or something else.

author: fzabkar
author_rank: Super Contributor
timestamp: January 12, 2019, 01:23:31 am
content: I mistakenly thought that each of the L & N inputs were going to separate op amps. Now, IIUC, the two inputs are being sensed by a single op amp configured as a differential amp. If so, then the gain should be the same as I calculated previously, but you should be measuring the differential input between the two 10K-10M junctions and then comparing this against the L-to-N reading (Sorry, I see you have already done this). Am I right to assume that the op amp's 0V reference is floating midway between the L & N potentials?

author: fzabkar
author_rank: Super Contributor
timestamp: January 12, 2019, 01:49:16 am
content: Quote from: webgiorgio on January 12, 2019, 12:37:41 amShouldn't that be …V_LN*2*sqrt(2) * (20k / 40040 k) = 325 mVpp

author: webgiorgio
author_rank: Regular Contributor
timestamp: January 12, 2019, 08:39:00 am
content: Quote from: fzabkar on January 12, 2019, 01:49:16 amyou are right, I was mistaken. Then the O-amp is not doing its job to keep V+ equal V- ! However the voltage indicated in the display of the inverter for V_LN is correct. mmmhThe "unkown" ST 24I EZX 6111 is used for the V_LN which is measured correctly and (if my guess on the pinout is correct) its input resistance is 1.13 Mohm. I call it U1.Above it, there are U2 and U3 that are OPA2277U K.U3 is used to measure V_L1-L2 (which is measured wrongly).I measured the input resistance on U2 and U3, and found it different! Same chip, different input resistance.U2: 312 kohmU3: 4.12 MohmShouldn't they be equal? I know I am measuring in circuit, but...The datasheed say 100 Mohm 3 pF.I expected something like 400 k (like the OPA270 from Analog (not on the board) that is also a precision op-amp).U2 has differential input configuration, with two 10k resistors, and the feedback is 12k (instead of 75 k).I promise you the schematic (so far) today

author: webgiorgio
author_rank: Regular Contributor
timestamp: January 12, 2019, 03:05:11 pm
content: Schematic so far attached.While checking the various connection with the continuity test, I also measured resistors (because is more comfortable than reading the small numbers).Two resistors don't match the markings!There are several 10kohm, but there are two that measure 75 k, despite being marked like other resistorsStill I can't figure out much of the lowest chip (U1), but at least I found out the supply pins thanks to the pinout I've got for U2 and U3.

author: fzabkar
author_rank: Super Contributor
timestamp: January 12, 2019, 08:14:36 pm
content: FWIW, here is a quad op-amp with a similar marking.TS924ID, STMicroelectronics, RRIO current quad op-amp, SO14, marking 924I:https://www.st.com/resource/en/datasheet/ts924.pdfNote that its Vcc+ and Vcc- pins are swapped when compared with your circuit diagram.

author: webgiorgio
author_rank: Regular Contributor
timestamp: February 20, 2019, 06:21:23 pm
content: Thanks.So, I replaced those two resistors with 10k resistors. The old resistors read 100k measured out of the circuit.The grid voltage (L - N) is still measured correctly, with U1.The inverter voltage (L1 - L2) is still measured wrong, with U3. (Hence I get E031 error).So, the problem might be U3 (it cost 8€).Before going trough the effort of de-soldering it, I would like to see its output.I want to see the voltage at:- the output of U1, top left pin (I guess)- the output of U3, top left pin (I have the pinout)I could float the scope, and attach the ground clip to the 1.65 V pad.

author: wek
author_rank: Frequent Contributor
timestamp: December 08, 2024, 10:02:02 am
content: 24I is probably TSH24.JW

author: dustooff
author_rank: Contributor
timestamp: December 09, 2024, 11:11:39 am
content: While we have this Lazarus on the table,Quote from: webgiorgio on January 12, 2019, 03:05:11 pmthe markings are alike but not exactly the same10k SMD code 01C100k 01Dwe can now return this thread to 2019.