Prop Plug and LQFP (Prop 1)

13

Comments

  • @Brian Fairchild The 5v switcher is rated at 32 volts.

    You say you have a 24V AC Transformer. That's the on-load voltage with the mains at the nominal voltage, with a small load on it it's likely to rise to at least 10% above that, so 26.4V AC minimum. 26.4V AC is 37V peak-to-peak. Feed that into a bridge rectifier and you get 35V DC out. Now, when your mains goes to the upper limit, which it's perfectly allowed to do you can easily add another 6% on top of that, so 37.6V DC.

    Using any component over its maximum rated voltage is a sure fired way to kill it.

    Using any component near its maximum rated voltage and relying on everything staying within limits is a sure fired way to kill it.
  • @Brian Fairchild There is a 12v 1w zener between the bridge and the 5v switcher. That should cut the voltage to 12 volts. The new design should never come close to the max rating of the switcher. I uploaded a new schematic last night with the new layout.
  • There is a 12v 1w zener between the bridge and the 5v switcher. That should cut the voltage to 12 volts.

    Not with it as drawn it won't. You've show it the wrong way around; as drawn it will just be a diode. The schematic also shows it as a 5V1 part.

  • @Brian Fairchild

    Cluso99 stated:
    Better solution. Buy a 12V zener 1W through hole and wire this instead of your 1N4004s. The anode goes to the 24V side. This will drop 12V across it. My presumption here is the max current to the 5V reg is 80mA. When you cut the track, measure the current. Allow 50% safety so you want to see 40mA max. Else use two 5V 1W zeners in series. That will cover you safely for 100mA (160mA calculated). Dont forget you will need that 47uF tantalum across the 5V reg input after the zener(s) as you want the input to the 5V reg to be rock solid. A resistor could be used but then you need to know the range of current accurately for the line.

    He said the anode goes to 24 volts. Thats the way I have it. Is that not correct ?

  • Sorry yes it should be the cathode so reverse it, and 12V.
    Like Brian, I am still concerned about you transformer. You haven’t given us full info.

    You should try these mods on your current pcb as you must confirm these fixes will also fix your reset issue. These mods are just basic d3sign principles. We need to see a new circuit with the caps at their correct position.
  • @Cluso99

    The transformer is Honeywell AT140A1000 40Va, 120V Transformer - 60 Hz. From amazon.

    I posted the pcb last night with all the new caps in place.
  • As a tip, you should provide a link to specs ;)

    Anyway, the full details I found are lacking but say 24Vac. So there is already going to be some range here.
    24Vac with full wave rectification gives 1.414 * 24V = 34VDC less 2* 0.7 for bridge so the nominal output will be about 32.5VDC which is far greater than the 24VDC you've been saying.

    BTW A 24VDC power supply may have been a better choice and they often have 5V out as well.

    You will need to check your relay specs to see if they can handle 34VDC and likely more than 10% higher too.

  • Peter JakackiPeter Jakacki Posts: 9,719
    edited 2020-03-22 - 01:21:13
    Putting a diode or diodes in series with the power supply is an effective way to drop the input voltage. SInce they will drop around 0.8V each then if the board is drawing 1A, then each diode will be getting very hot as they dissipate 0.8V x 1A = 800mW. Trying to do the same with Zener diodes will only work if the total power dissipated by the diode is within its ratings but 12V at 200ma for instance would be a killer 2.4W and awfully inefficient, you may as well just use a linear reg with a huge heatsink in that case. Most of your problems stem from using an old style mains transformer when there is no need to these days. Typically I will just use a 19VDC "laptop" power pack or a 24VDC if I need 24V for solenoids and then feed that into my switcher etc. If you are simply driving relays then always use 12V coils and then you can have a 12V switcher. These things are available from you supermarket even.

    As for a zener + diode across coils they will help with faster release but remember that zeners appear as a normal diode in the forward direction but they are like really low voltage diodes that breakdown at a set reverse voltage, so for them to work you need them in reverse to the series diode, back to back.

    BTW, putting a zener in the input supply was my solution for a 48V product that we found was going to 60V or something when they were charging battery banks and there were some failures as the components weren't rated for that. But the unit didn't need too much current so a suitable Zener inserted in the supply line was a quick fix.
  • I haven't studied every posting on this subject. But why 24v on the on board relays? why not use 5v or 12v. Whatever the case use a seperate small power supply to drive the relay coils and your prop 3.3v regulator to operate the opto. If you must stay with 24v use a DC/DC converter from 24v to prop power. Such as a Cosel they have 5v, 12v, 24v, and 48v inputs with various outputs. single and dual supply.
  • Cluso99Cluso99 Posts: 16,674
    edited 2020-03-22 - 09:36:30
    I suggested zeners rather than diodes as the voltage drop needed is considerable. The string of diodes would be excessive IMHO.

    But the whole design needs re-examining. I don’t know the whole story of what’s required, just getting snippets of info.

    At the risk of repeating, why is C1 on the output and not the input of U1? My suggestion was to smooth the input following the the zener.
    What else is being powered by 5V and 3V3?

    My guess is you’re looking at 100-150mA tops but we do need to know for calculations.

    BTW the 5V and 3V3 tracks should be much thicker. What’s you track thickness? 0.012-0.020” is at least preferred, but depends on total current too.
  • DigitalBob wrote: »
    I haven't studied every posting on this subject. But why 24v on the on board relays? why not use 5v or 12v. Whatever the case use a seperate small power supply to drive the relay coils and your prop 3.3v regulator to operate the opto. If you must stay with 24v use a DC/DC converter from 24v to prop power. Such as a Cosel they have 5v, 12v, 24v, and 48v inputs with various outputs. single and dual supply.

    Because he has and needs a 24VAC supply in an enclosure where he wants to mount his board, and does not have space for another power supply.

    IMHO the first thing that needs to be done at this point is to verify that all the changes have created reasonable voltages for the board circuitry and then connect everything up to see if the problem is resolved or still there. If it is still there then disconnect the solenoid and see if the problem goes away or not. That will determine if the problem is with the solenoid or the controller board.
  • Cluso99 wrote: »
    Anyway, the full details I found are lacking but say 24Vac.

    The off-load figure is worse than that though, 27V ac. And I don't see much load on a 40VA device to pull that down much. A good design would allow the mains to go 10% above nominal so 29.7V ac.

    Full wave rectify that and it's (29.7 x 1.414) - (2 x 0.7) = 43.4V dc.
  • Here is the meat and potatoes:

    Drive a 120 Volt 1.4 Amp pump, a 24 VAC Gas valve, 2 temperature sensors and a float valve. Since the gas valve is 24 volt AC I started with a 24 volt AC power supply.

    Temperature sensors are DS18B20's

    I opted for 24 volt relays because I was already at 24 volts after the bridge. I am not hard set on 24 volt relays. Biggest problem is the control box size. There is a lot of hot liquid and huge gas burners so I want to minimize to one power cord.

    @Cluso99 5 volt and 3 volt traces are .31.

    Problem: Prop chip resets randomly when gas valve turns off. I don't have a scope so a deep dive is not possible but I believe the 5 volt switcher is shutting down.

    I found this on amazon: AC to DC Isolated Power Supply Module

    I think that would work and I can just remove the bridge and drop down to 12 volt relays, no diodes needed!

    Best of all its small enough to fit in the control box.

    I was also looking at adding a buck converter to the PCB.
  • I had the same problem switching a/c loads, solenoids, etc.. In the short term I put 500v 1uf metal film caps. across the contacts of the relay (not the coil) and grounding the solenoid. The prop board has to be totally isolated with its own power. All outside functions I/O , comm. must have an opto isolator.
    More recent I bought a arduino relay board with 8 relays. The arduino relay board had power for the relays and a seperate source for the opto. I can turn 60 amp. 3 phase contactors on and off with no issues (120v coil). Not all relays boards are the same, make sure it has isolation with no common ground.
    The cap. across the contacts was a definate winner in the short term, no issues with resets. Can't argue with success. If for some reason your worried about high a/c or rectified voltage just use a UA7824 regulator first and then drive the other regulators off of that.
  • Yes, there is a good chance the 5V switcher is shutting down. The input voltage is way over spec and worse, 24V is even above the recommended operating conditions (see the spec).

    Check your current to the 5V regulator under all conditions. I am sure you will see it is quite low. Then install that 12V 1W (or bigger if you can get one) and try that first. No use making another board if you cannot fix your current problem first.
  • For some reason this prop chip is more sensitive. I think the BS2 has less glitches in this type of application. Maybe because the BS2 chip is a PIC and is made by Microchip. That's my guess.
  • DigitalBob wrote: »
    For some reason this prop chip is more sensitive. I think the BS2 has less glitches in this type of application. Maybe because the BS2 chip is a PIC and is made by Microchip. That's my guess.

    I don't think that is so. I've built 100's of prop boards without problems. It just needs proper designs.

  • Perfect. Just make sure it'll deliver the necessary amount of current and you're done.

    These days, except for certain applications, it's easier to use an AC/DC switching power supply; they are often cheaper than a transformer.
  • So I was able to prove it was definitely dirty power. I did redesign of the PCB and dropped to 12 volts on the output side and much smaller relays. Also incorporated a buck converter and used a second inductor to provide really clean and consistent power. I was able to cycle the gas valve and the pump on and off very rapidly and the prop did not reset.

    I am getting 13.95 volts out of the buck converter that should be 12 volts. the voltage divider is using 10% resisters, I have ordered some 5% ones hopefully that will get me to 12. If not I can live with 13.95 volts.

    I have attached the schematic for the power supply. I didn't take a picture of the board before I installed it, but I have also attached a shot of the render from 3d viewer.
  • Great news :)

    You should be able to get 10K and 1K 1% resistors easily. However, it's just as likely that the 10% resistors are in fact accurate anyways as these days making resistors to a tight tolerance is easy.

    My preference would be to have a higher C12 - 1000uF would probably be an overkill tho but wouldn't hurt.
    C14 and C9 make 1100uF which I think is overkill. On power up it places a big surge current on the switcher which is nice to avoid. A single C14 may do, or a little larger.

    For future reference, L2 is to block transients, so any voltage drop/spike is resisted. Therefore, it would be better to power the 5V (and hence 3V3) from before L2 (or have its' own inductor too). What is needed is to prevent any spikes or droops getting onto the input to the 5V regulator.

    I note you didn't take my advice and also have a 100nF on the input and output of the 3V3 regulator. While many may not consider this necessary, I have found it improves noise on the power rail. 100nF is used to bypass high frequency noise, which is why we put them onto the power/ground at the input to every IC. The regulator is an IC too.

    Am I seeing this correctly... You have a series resistor with the reverse diode across the relay coils (eg R3 and D6) ???

    BTW just for reference, it's more common (and easier to see what the circuit is actually doing) to put the Vout above the FB pin even tho the zener and cap will cross the feedback line.
  • @Cluso99 I did take your advice on the caps, You said put a 10uf tantalum on input and output of 3v3 regulator and a 47uf tantalum on the switcher output.

    R3 and D6 is just an led and a 4.7k resistor to show when the relay is on.

    As for C9 It can really go. It was there from before when I had the larger 24 volt relays.
  • @Cluso99 I did take your advice on the caps, You said put a 10uf tantalum on input and output of 3v3 regulator and a 47uf tantalum on the switcher output.
    OK. Remember it for next time - also add a 100nF across the 47uF as it will improve the noise response. As the 3V3 is regulating down from a 5V regulator you have good isolation from any relay induced spikes on the 12V line.

    R3 and D6 is just an led and a 4.7k resistor to show when the relay is on.
    Can you draw the relay circuit? Just one is fine.

    Seems that C9 is the big blue cap on your pcb. I am a little confused as to how you have this hooked up. Looks like you have the 12V relay coil supply running between the relay contacts???

    Also, I don't see the snubber diodes???

    FWIW these days I use superbright leds because the price difference is minor and then design the resistor to give 1mA thru the LED. When doing the calculation, don't forget the LED voltage drop.
    As for C9 It can really go. It was there from before when I had the larger 24 volt relays.
    If it is the blue cap, then leave it in although it could be reduced to say 220uF. Presume its 35V or larger.

  • @Cluso99 Attached is the current schematic. I actually got rid of the whole shebang and went with a NUD3124 Relay Driver. It replaces all those parts for like 40 cents and reduces 4 components to 1.

    I don't know if the led is a super-bright, but its a 603 footprint and it is very bright.

    what type of cap should the 100nf be?
  • Cluso99Cluso99 Posts: 16,674
    edited 2020-04-04 - 04:17:02
    People still assume they should be running LEDs with 20-50mA but those days are well gone although many still waste way too much power in the LEDs. At night they are sooo bright because they are being driven hard. Choose a resistor that makes them visible in daylight without having to squint ;)

    The caps should be mono (MLCC) X5R or X7R and double the voltage they are likely to see. 50V and 35V are easy to get and little price difference. Of course you can get away with less but then you limit possible future use on left-overs.

    You need to keep the coil side of the relay separate to the contacts side - as much as you can get, so your driver, LED and 12V need to come out one end of the relay, and the contacts the other end. And there should be no ground plane in that section either. This is a critical isolation path!!!

    Since your're making further changes, you should get used to the proper way to use the I2C bus. A 10K pullup is meant to be on both SDA and SCL even tho the PropPlug does work without the SCL pullup. I use a 4x10K network which does the SDA/SCL/Reset and I have an SD card which I put a pullup on /CS even tho it's not necessary.

    You don't necessarily require both R16 and R17 as a single resistor would do.

    I am not sure about L2 as it's a monster. A value between 1uH and 10uH would be good. There are some nice Bourns or Murata SMT inductors. You need to know the current usage and allow for double if possible. You would be best to use a pair, one for the relays and one for the 5V regulator input. Are you powering anything off-board with the 12V or 5V?
  • My solution.
    1113 x 829 - 146K
  • Looks good feng. Cannot recall what is being switched without searching back - needs to be ac for those ss relays. With the higher volts being switched J1 should be 9 pin with an unused pin between the input and relays. I can’t recall specific isolation gaps but think it was 5mm which would still be compromised with an unused pin in the gap. Perhaps the pin can be removed and no pad for the pin either. Alternately an isolation slot can be milled in the pcb but again specifics regarding the connector.
  • I guess 110VAC. 1.15x110x1.4141 = 178V peek
    With 220VAC we get 1.15x220x1.4141 = 358V peek
    Current for the motor was 1.4A

    Voltage and minimum distance(mm) between conductors
    50V/0.3mm
    150V/0.6mm
    300V/1.2mm
    500V/1.8mm

    Standard terminalblocks
    Raster 5.08 Distance pad to pad 2.6mm
    Raster 3.81 Distance pad to pad 1.6mm
    Raster 3.50 Distance pad to pad 1.0mm
    Raster 2.54 Distance pad to pad 0.8mm
    Raster 2.50 Distance pad to pad 0.7mm

    5.08 is overkill. 3.81 is enough.
  • Cluso99Cluso99 Posts: 16,674
    edited 2020-04-04 - 11:36:30
    The standards call out IIRC 1500V for 110Vac and 2500V for 240Vac but it’s been 20 years since I designed for these specs so I could be wrong. I am sure that at least 5mm isolation would be required. I used to put 2mm slots in the pcb if 5mm could not be attained. Specs have no doubt also changed but I wouldn’t think the isolation would be any less.
  • I don't have the official docs at hand, but KiCAD has a table, they say it is from IPC 2221:
    for 250V outer layer, no coating up to 3050m min spacing 1.25mm
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