Circuit still sends voltage surge through when plugging into Power Supply - Page 1
Forum section: electronics / beginners

author: Falcon69
author_rank: Super Contributor
timestamp: December 19, 2014, 06:24:11 pm
content: So, I went with this circuit, as suggested per data sheet for the LM2596s-ADJ chip.as you can see from the circuit, I added the enable pin circuit which was suppose to be a slow on start that I thought would prevent all my LED's in the circuit from turning on for a split second when the circuit board was plugged in.FYI, once the initial plug in (and that voltage surge), is done, the entire circuit works as it should! SO I know this is a plug-in problem.Would changing C3 to a higher value solve this problem?Thanks in advance,Jason

author: Seekonk
author_rank: Super Contributor
timestamp: December 19, 2014, 08:16:09 pm
content: Don't you mean C4? .1uF seems like a very short period.

author: Falcon69
author_rank: Super Contributor
timestamp: December 19, 2014, 08:22:15 pm
content: okayI think I have a 1.0uF I could try. Or would I need to go MORE than that?Oh, I think the 1.0uF i have is only 16volts. I would probably need at least a 35volt one, just in case the input voltage is 24volts?I also have a 4.7uF, it's an 0805, but I think I can still make it work on the 0603 pad on the board. Just for testing anyway.

author: Falcon69
author_rank: Super Contributor
timestamp: December 19, 2014, 09:42:48 pm
content: Well, I tried the 1.0uF I had, It didn't change anything.So, Before I place another order to buy components, what are your suggestions, anybody, as to what size I would need to change the C4 to?I was reading on the data sheet, that changing C4 to be careful, because a too high value will cause the LM2596 to turn on and off at random.

author: Seekonk
author_rank: Super Contributor
timestamp: December 19, 2014, 10:17:48 pm
content: For a longer delay I would add a transistor or a FET. That's getting complicated and I'm not sure I see a need for a delay anyway.

author: Falcon69
author_rank: Super Contributor
timestamp: December 19, 2014, 10:31:00 pm
content: There probably is not a need. I was just concerned with the initial plugging in of the circuit board. For some reason, a spike happens and flashes all the LED's on for a split second. (even ones that should not be on, which of all are controlled via the logic gates...LED's current sinks or sources through the logic gates).Does this initial spike hurt the components?This circuit board will be turned on and off, based on what the other items are doing within the control box for a machine. For example, each time the machine turns on and off, so will this circuit board.Do you think it will still send this voltage spike, if and when the power inputs will be switched on and off by a separate control box switch (which turns off the machine), each time it is turned on and off?

author: Falcon69
author_rank: Super Contributor
timestamp: December 19, 2014, 11:18:30 pm
content: well, I fond some almunium caps, so I just placed them parrallel across that 0603 0.1uF cap I had.a 33.1uF is too long of a delaya 22.1uF is too long of a delaya 10.1uF is too long of a delay (but i think should work just fine. It's about a 2 second delay before circuit turns on).a 4.7.1uF doesn't workI think maybe a 6.8uF might work, but I don't have one to testSo, I think I will order up some 10uF 35v ceramic capacitors and replace that with the C4 in the schematic.N/M, they don't make them in an 0603 package. Damn. hmmm, any ideas? I already got 14 of these boards made. Would like to try and use the boards I have. I can redesign the next batch for the 0805 size 10uF 35v

author: Falcon69
author_rank: Super Contributor
timestamp: December 19, 2014, 11:21:11 pm
content: would i be safe with a 25volt for C4? The max input for this circuit is 24volt. I've just been using a value rated for at least 10volts over for the capacitors.I ask, because, i found a 10v 0603 XR5 package in 25volt I could use for these 14 boards.

author: kxenos
author_rank: Frequent Contributor
timestamp: December 19, 2014, 11:31:26 pm
content: Q1 is reversed.

author: Falcon69
author_rank: Super Contributor
timestamp: December 19, 2014, 11:35:11 pm
content: ya, but for this, I'm not asking about that. I merely drew it wrong in the schematic, but it is the correct way on the circuit board. The main schematic for the entire board, it is drawn correctly. Just not in the picture I grabbed.

author: Falcon69
author_rank: Super Contributor
timestamp: December 26, 2014, 05:42:53 am
content: well, I got in the 10uF 25volt ceramic capacitors to replace C4.It didn't work. the LED's still get a current spike when the board is initially plugged in and flash for a split second. I'm worried that by doing this, it is going to damage the components. The board will be turned on and off via a switch, which turns off other boards on the machine. So, everytime the machine is shut down, so will this circuit. And then, when it is turned back on, it will have a voltage spike again.It seemed the Aluminum Cap worked (i had a thru hole one I tried earlier). Why would the aluminum work, and not the ceramic? I would have to completely redesign the board to make the aluminum one work. Don't really want to do that. I cant afford another order for boards.Any suggestions? Would changing the 47k resistors to another value help and go back to a 0.1uF ceramic cap, as in the data sheet?

author: Falcon69
author_rank: Super Contributor
timestamp: December 26, 2014, 06:36:19 am
content: well, I just tried it again with the 10uF aluminum Cap.I must have overlooked this last time i tried it.On initial plug in, there's a spike that lights all the LED's, then no LED's are lit until the cap charges and then circuit operates as normal.So, How do you solve the voltage spike problem then?Apparently, the suggested circuit as drawn above and in the data sheet is correct, but, I still get a voltage spike on plug in.I'm thinking that maybe it can't be fixed? Since on plug in, the logic chips are like, "what do we do?" as the inputs would be floating at intitial plugging in of the circuit, and that is why the LED's flash for a brief second, until the logic chips realize the inputs are no longer floating? Is this normal operation for them? I just don't want the circuit to get damage when an on/off switch is hit.

author: Circlotron
author_rank: Super Contributor
timestamp: December 26, 2014, 07:24:44 am
content: Quote from: kxenos on December 19, 2014, 11:31:26 pmIt looks like it is, but in this cct it is the right way around. When normal polarity power is applied the body diode of the mosfet conducts and also the mosfet turns on, helping the body diode conduct. But yeah, the current is in fact going the reverse direction through the mosfet, but mosfets can do that. If you reverse the supply polarity, the mosfet stays switched off and it has "normal" polarity across the source-drain.

author: Falcon69
author_rank: Super Contributor
timestamp: December 26, 2014, 07:28:50 am
content: Ya, it is a reverse Polarity circuit, using a P-Channel mosFET.I noticed too, that day, after I made that comment that it IS in the correct orientation.

author: Circlotron
author_rank: Super Contributor
timestamp: December 26, 2014, 07:58:57 am
content: Oh yeah, P-channel. Whoops!

author: Falcon69
author_rank: Super Contributor
timestamp: December 26, 2014, 08:04:21 am
content: I was just thinking, and this would actually be a GREAT idea if it works.Pin 5 of the LM2596S-ADJ voltage regulator is an ON/OFF control for the voltage regulator. Normally, it would be connected to ground, or, if you need a delay start (in my case, that doesn't seem to function right for my application), it would be connected as in the schematic above.My thought was, what if a simple SPST switch was placed between pin 5 and ground. If the switch was Open, would that NOT allow the voltage regulator to turn on? If it was closed, that would allow it to turn on?Is there any problems with doing this?That would be perfect, because using this method, the circuit board could be switch on and off, based on the position of the switch. That could be wired to a relay, that could be activated by the main control panel for the machine which turns on all electronics of the machine.What are your thoughts on this? ANyone have experience with the LM2596 and has done this before with Pin 5?Thanks in advance.

author: tautech
author_rank: Super Contributor
timestamp: December 26, 2014, 08:30:25 am
content: Lower the value of your R4, R2 in the datasheet.This will delay the startup further.

author: Falcon69
author_rank: Super Contributor
timestamp: December 26, 2014, 08:38:39 am
content: Thank You tautech, I will try that, however, I am not sure that IS the problem.with the circuit just as it is in the data sheet, but changing that 0.1uF cap to a 10uF, I get the delay, however, this is how the circuit works....Plug in Circuit (connect the walwart)LED's flash for a fraction of a secondThen nothing, no LED's, no circuit power for like 2 seconds (well, the output of the voltage regulator to rest of circuit)Now circuit turns on as normal, no flash of the LED'sThat is what is happening, so they delay start circuit is working, however, somehow, a bit of current is getting through, turning on the logic gates and supplying power to the LED's for that fraction of a second when the walwart is plugged in.I'm not exactly understanding what is happening.

author: tautech
author_rank: Super Contributor
timestamp: December 26, 2014, 08:54:41 am
content: Time to get the scope out and see what and why it is happening.Little else will tell you.Monitor the pins that matter and study the datasheet.Set the trigger and use single.Then you will find it.

author: Falcon69
author_rank: Super Contributor
timestamp: December 26, 2014, 08:59:04 am
content: I think I found my answer searching the net.The LM2596S-ADJ has a voltage threshold of 1.3V on PIN 5 with a maximum voltage of 25v.THe LM2596S-ADJ will default to the 'ON' state if PIN 5 is left floating or connected to ground.If a voltage source above 1.3v is connected to the LM2596S-ADJ PIN 5, then the voltage regulator will revert to an 'OFF' state.If I understood what I read, then connecting a resistor (i'm thinking a 100k) from the walwart supply to pin 5, and then to the in of a switch and out to ground from there, by turning the switch on and off it should shut the voltage regulator down.Seeing how my circuit will not see more than 24 volts, I should be okay with a 100K ohm resistor?FYI, I don't have a scope, nor do I know how to use one....YET. Can't afford one right now anyway.

author: Falcon69
author_rank: Super Contributor
timestamp: December 26, 2014, 09:08:12 pm
content: Still having trouble with this.I had the circuit like the first pic below. It was according to the Data Sheet. I tried changing the C4 to 10uF and didn't help. It gave the delay, but, upon plug in, the LED's still flashed for a fraction of a second, then it did the delay start for the voltage regulator, and then all the circuit turned on correctly.Now, I tried the circuit this way (second pic) and this way it still does the same thing, but without the delay start. Although, I think I will go with this idea, because it allows the whole circuit to be turned off via a standard SPST switch.Now, I've included the debounce circuit I have. The output of the inverter goes to every logic gate that is supplying power to the red and green LED's I have. The green LED's are always on (unless one of the hall sensors triggers it, switching it to a red LED) and when the tactile switch in the debounce circuit is pressed, all the LED's light, this is a feature I added so that you can see if all the LED's are turning on as they should, or if one has burned out, etc.So, I am wandering if maybe changing a value in the debounce circuit will solve the voltage spike that happens on plugging in of the wallwart to the circuit board, or flipping the rocker switch for pin 5 of the voltage regulator?

author: Christopher
author_rank: Frequent Contributor
timestamp: December 26, 2014, 10:46:19 pm
content: The on off pin literally only turns the device on or off. There is no soft start.You can either change to a chip which does include a soft start pin, which allows you to ramp up the voltage over say 100 ms to stop the device from folding back due to current limit, or add in an external soft start to the feedback pin as shown herehttp://m.electronicdesign.com/site-files/electronicdesign.com/files/archive/electronicdesign.com/files/29/6384/figure_01.gif

author: Falcon69
author_rank: Super Contributor
timestamp: December 26, 2014, 11:25:42 pm
content: okay, don't understand that one. that's a completely different chip than the LM2596.So it is the feedback pin of the LM2596 that needs to have a circuit implemented onto it to have the delayed start up?Do you have a schematic I could look at that is specific to the LM2596 or similar? That would be very helpful to me.Thanks

author: Christopher
author_rank: Frequent Contributor
timestamp: December 26, 2014, 11:37:08 pm
content: The way your chip works is increasing/decreasing the frequency of the pulses of current through the inductor. It keeps changing this until the voltage on the feedback pin reaches a specific voltage then keeps doing what it can to the output till the vfb is constant.You can put a 100 or 1u or 10u ceramic cap 6v rated+ right across r1 to see what I mean. But do not keep this across there forever as this will reduce the transient response and is a bad idea.U1 in the circuit I linked does this in a more controlled way at startup

author: Falcon69
author_rank: Super Contributor
timestamp: December 26, 2014, 11:48:27 pm
content: okay. If I leave the circuit as is, will it damage the components? If it doesn't, I'll just leave as is, as It's really not a big deal I guess for the function of the circuit.

author: Christopher
author_rank: Frequent Contributor
timestamp: December 26, 2014, 11:52:53 pm
content: If the output is being switched to 24v as it turns on them clamps to 5v then it's not good at all... suppose the only way you'll tell is with a scope or dmm in peak but that might not pick it up.

author: Falcon69
author_rank: Super Contributor
timestamp: December 27, 2014, 12:00:49 am
content: I guess I don't understand. I'm trying to follow in what that schematic is doing, but I'm just confused. My chip does not have Sense + or sense -.I don't want to change the lm2596, as I've already bought alot of them.

author: tautech
author_rank: Super Contributor
timestamp: December 27, 2014, 12:07:41 am
content: You are really just pissing in the dark without a scope.Nearly anything with 2 channels will show what you want to see.Get something, anything and explore another world.

author: Falcon69
author_rank: Super Contributor
timestamp: December 27, 2014, 12:28:55 am
content: What about using an P-Channel Mosfet or PNP transistor on the output of the LM2596, like right after the 1uH inductor (L2) on my schematic? With a cap and resistor on the gate? That way, the cap will take time to charge to reach the voltage, turning on the transistor. That would give the LM2596 enough time to properly switch on. That would use a lot less components than the circuit you had posted Christopher.

author: Christopher
author_rank: Frequent Contributor
timestamp: December 27, 2014, 12:31:14 am
content: I'll do a couple of simulations in the morning for you but if you could post an image of your pcb that'd be good. Doubt it's a layout problem as these chips are pretty resilient.

author: Falcon69
author_rank: Super Contributor
timestamp: December 27, 2014, 12:35:23 am
content: I would appreciate that Christopher.I will send you a pic of the board.I was also thinking maybe something like this using a low side transistor or mosfet to switch it on.Would something like this work? I can adjust the regulator to put out 6 volts, this should offset the voltage drop across the transistor.

author: Christopher
author_rank: Frequent Contributor
timestamp: December 27, 2014, 12:57:23 am
content: Can you take the circuit back to normal, remove r3 and short the on pin to ground then see what happens?Then again but removing the cap between the feedback and switching pins

author: Falcon69
author_rank: Super Contributor
timestamp: December 27, 2014, 01:44:29 am
content: Yes, I had done that. It still does it. I made pin 5 directly connected to ground.

author: Andy Watson
author_rank: Super Contributor
timestamp: December 27, 2014, 01:48:28 am
content: Quote from: Falcon69 on December 19, 2014, 06:24:11 pmWhat is controlling the LEDs? Is it the same logic that is being powered by this supply circuit? If it is, then the problem is happening when the +5V becomes active. As has been pointed out, the enable pin is an On/Off switch - not a slow start. No amount of delaying the start-up is going to stop the LEDs flashing when the power supply output is activated. Need to be looking elsewhere to cure the problem.

author: Falcon69
author_rank: Super Contributor
timestamp: December 27, 2014, 02:15:30 am
content: yes, that is what we are trying to track down now Andy. I was hoping for an easy fix, but doesn't seem so.

author: Andy Watson
author_rank: Super Contributor
timestamp: December 27, 2014, 02:38:41 am
content: Ah, I must have missed a bit of the threadIn that case could you give some more details about the LED drive/control? Are you looking for a positive or negative enable signal ?

author: Falcon69
author_rank: Super Contributor
timestamp: December 27, 2014, 02:52:23 am
content: i need a positive 5.3 volts from the LM2596. I could just use a 5v walwart instead of this headache, but I'm trying to offset the voltage drop from length of wire that goes to some hall effect sensor switches.As said before, the whole circuit works fine, except for when initially plugging in the board. All the LED's flash for a fraction of a second. I'm trying to stop that from happening.If I understood Christopher right, The LM2596 is supplying on the output at plug-in the full supply voltage of the input for a fraction of a second, i.e. 24 volts if it was the supply amount, until the voltage regulator fully switches on, and supplies the 5.3volts.That could prove fatal to the logic gates and inverters on this board and on the switches.

author: Alex30
author_rank: Regular Contributor
timestamp: December 27, 2014, 02:56:47 am
content: Falcon, what voltage are you running your tests at when you plug it into your power supply?

author: Falcon69
author_rank: Super Contributor
timestamp: December 27, 2014, 02:58:09 am
content: right now, i'm using a 12volt walwart plugged into it. But, the circuit needs to be able to handle between 7-24volts input for the voltage regulator.

author: Alex30
author_rank: Regular Contributor
timestamp: December 27, 2014, 03:07:30 am
content: This might seem dumb as I don't fully understand the inner workings of the LM2596 (but then again the slow startup circuit should work in my opinion), would you try removing C4 entirely and short circuit R3 to the voltage supply of the LM2596? In this configuration the reg should never switch on, even partially, if it does then no amount of fiddling with pin 5 will work. I hope we can fix this as I will be using this for some of my work.

author: Falcon69
author_rank: Super Contributor
timestamp: December 27, 2014, 03:16:02 am
content: no, the regulator will not turn on then, as the voltage then supplied to the lm2596's pin 5 will be above its 1.3v threshold. In order for the lm2596 to turn on, pin 5 must have a voltage below 1.3v.

author: Alex30
author_rank: Regular Contributor
timestamp: December 27, 2014, 03:33:31 am
content: Exactly, but there are some internal capacitors within the lm2596 - just not sure how it's set up and how it impacts this slow startup circuit. My point being that using the slow startup circuit given in the datasheet should work, given that C4 is a short circuit in initial startup. I'm trying to prove the configuration in the datasheet will never work by asking you to change C4 to a short circuit.

author: Falcon69
author_rank: Super Contributor
timestamp: December 27, 2014, 03:37:30 am
content: As others have enlightened me, the delay start is just that, a delayed start, meaning, it will delay the turn on of the LM2596. However, it seems the problem I am having is not that, and needs to be fixed using some kind of a Soft Start circuit on the output of the LM2596. This is where I am confused, but happy to learn.I am trying to search for 'LM2596 Soft Start' circuit schematics, but I'm running out of luck. I am a visual person, and I need to see stuff drawn in order to understand it.

author: Alex30
author_rank: Regular Contributor
timestamp: December 27, 2014, 03:50:09 am
content: Oh I see. Is undervoltage an issue for any micro controllers or other chips you have in your circuit?

author: Falcon69
author_rank: Super Contributor
timestamp: December 27, 2014, 03:51:24 am
content: all the chips I have (logic and inverters) operate on 2-6 volts.

author: Alex30
author_rank: Regular Contributor
timestamp: December 27, 2014, 04:18:05 am
content: Well the simplest slow startup application may be of use here. But I'm not 100% sure if this is the best thing to do here as I am only a hobbyist. Simply a power transistor (perhaps a darlington) with a cap at the gate as shown on the picture. The turn on time can be altered by changing the RC time base to your needs. My thinking would be to use this before the lm2596 then use the under voltage lockout circuit given in the datasheet (attached) to suit your needs.Things to look out for:If you have lots of changing loads in your circuit you could cause oscillation by using smaller capacitor valuesVoltage drop over the power transistor could increase your power requirementPower rating of capacitor

author: Falcon69
author_rank: Super Contributor
timestamp: December 27, 2014, 04:22:57 am
content: I was thinking about the same thing, but using a mosFET instead, and AFTER the inductors. That would be the trace going out to the rest of the circuit.The problem is the inductor is messing with the feedback for a fraction of a second when the board is plugged in using the walwart or another supply voltage.If I can somehow delay the current/voltage from going to the rest of the circuit (after the inductors) until the voltage regulator turns on completely, maybe a 1/2 a second delay, then I think everything will work just fine.

author: Alex30
author_rank: Regular Contributor
timestamp: December 27, 2014, 04:29:03 am
content: Well it shouldn't be a problem if you set up the lockout to happen at 2.5 volts as there should only be 2.5 volts available on the power rail anyway until the slow start up is fully initialized.

author: Falcon69
author_rank: Super Contributor
timestamp: December 27, 2014, 04:36:40 am
content: i think you are confused. The input with the lm2596 will always be 7-24v (depending on which walwart is used). There is no undervoltage here.I need to control the output somehow, so the current and voltage is delayed until the lm2596 turns fully on, thus, eliminating the problem of the surge current and voltage at plug-in.

author: Alex30
author_rank: Regular Contributor
timestamp: December 27, 2014, 04:43:08 am
content: No, my suggestion is to put the slow startup circuit on the Vin side of the LM2596. The voltage at Vin therefore will slowly ramp up. So use the lockout pictured above to ensure that the LM2596 does not turn on until the voltage at vin > 2.5V. At this point, even if the feedback isn't working properly and tries to draw tens of volts, there is only ~2.5V available to the lm2596 anyway, so you won't see any high voltage transients when the lm2596 switches on.

author: Falcon69
author_rank: Super Contributor
timestamp: December 27, 2014, 04:48:11 am
content: i see a 12v zener. How's that work with variable input? i.e. 7-24volts.and both the circuits you shown, go BEFORE the IN pin of the LM2596?

author: Alex30
author_rank: Regular Contributor
timestamp: December 27, 2014, 05:03:21 am
content: Yes so my suggestion is to slow start the input (Vin) of the lm2596 and use the low voltage lockout to make the lm2596 turn on at ~2.5V. Then the voltage will slowly continue to rise until the slow start has fully turned on.As shown in the datasheethttp://www.ti.com/lit/ds/symlink/lm2596.pdf:UNDERVOLTAGE LOCKOUTSome applications require the regulator to remain off until the input voltage reaches a predetermined voltage. Anundervoltage lockout feature applied to a buck regulator is shown in Figure 34, while Figure 35 and Figure 36applies the same feature to an inverting circuit. The circuit in Figure 35 features a constant threshold voltage forturn on and turn off (zener voltage plus approximately one volt). If hysteresis is needed, the circuit in Figure 36has a turn ON voltage which is different than the turn OFF voltage. The amount of hysteresis is approximatelyequal to the value of the output voltage. If zener voltages greater than 25V are used, an additional 47 k? resistoris needed from the ON /OFF pin to the ground pin to stay within the 25V maximum limit of the ON /OFF pin.So try a 1.8-2V zener, or even up 4V. Then when it turns on you won't see the output of the lm2596 jumping any higher than the voltage on the Vin side.

author: Alex30
author_rank: Regular Contributor
timestamp: December 27, 2014, 05:17:29 am
content: So in the picture shown, I have put a red box on areas where the lm2529 may turn on and then drawn a line at the point at which the lm2596 is set to turn on due to the zener diode (this one is a 4.7V diode). At this point Vin is about 5-6V. The switching regulator can only theoretically supply a maximum of the voltage at Vin. As you can see it's impossible to see higher voltages than ~6V at the output until the voltage at Vin is higher.

author: Falcon69
author_rank: Super Contributor
timestamp: December 27, 2014, 05:36:40 am
content: what is the supply voltage? did you try it with 7 volt input and 24 volt input? Does it work as in the picture above you just tested?

author: Falcon69
author_rank: Super Contributor
timestamp: December 27, 2014, 05:46:25 am
content: is the Cincapacitor no longer needed with this circuit? Before, I was using a 220uF aluminum. Ti Data sheet shows a 680uF for it for the lockout circuit, however, this is not with the added soft start circuit you added.

author: Alex30
author_rank: Regular Contributor
timestamp: December 27, 2014, 05:56:14 am
content: Quote from: Falcon69 on December 27, 2014, 05:36:40 amI just chose 16V for that one but I have attached one showing 7V input and 24V input to allow you to get your head around it. I have not properly designed the power transistor or the RC circuit so you will have to work that part out. But as you can see by doing this method, your effective maximum Vin is going to be your input voltage (power adapter) - 0.7V roughly as you have a voltage drop on the transistor. But you should still be able to provide a 5.25V output at 6.3V Vin (7V supply).Quote from: Falcon69 on December 27, 2014, 05:46:25 amYou certainly do. the schematics given in the datasheet often only relate to the part that is being discussed. I recommend you go through the LM2596 Series Buck Regulator Design Procedure section in the datasheet to make sure your component values are correct if you haven't already.I think using this method should be fine as long as you aren't pulling lots of current, otherwise the power transistor will cook.

author: Falcon69
author_rank: Super Contributor
timestamp: December 27, 2014, 06:02:50 am
content: with the on/off (pin 5) now occupied, any suggestions where a switch can be used to turn off the circuit. This circuit board will be inside a control panel. As the control panel and machine are turned off, this board, as well as other circuit boards in the control panel, need to be turned on and off at the same time.Any suggestions as to where? I'd hate to use a big Solid State relay to do this.

author: Alex30
author_rank: Regular Contributor
timestamp: December 27, 2014, 06:16:02 am
content: If the switch has to go near the on\off pin just put a switch between D1 and R3. When it is switched on (Short circuit) the supply will be on, when it is off (open circuit) Q2 will be driven low by R3 and Q2 will switch off, then the on\off pin will be driven high by R4, turning off the circuit.EDITSorry don't do that. As the slow turn on is dependent on when the supply turns on, so your switch must be before the slow turn on circuit.

author: Falcon69
author_rank: Super Contributor
timestamp: December 27, 2014, 06:27:31 am
content: I don't think that will work? That would leave C1 fully charged, thus, when the switch is switched to "ON" position, the current/voltage will flow, without a soft start, then I am back to this same problem.What do you think of coming off the junction of the 50ohm resistor and C1, put a 47K resistor and sink to ground? Then placing the switch between the 50ohm resistor and the supply voltage? On simulation, it seems the switch is only about 3.5mA, so, a it should be a problem for the control box switch. The 47k resistor would be so that the cap drains fast, ready for the circuit to turn on again.

author: Falcon69
author_rank: Super Contributor
timestamp: December 27, 2014, 06:53:14 am
content: so, something like this I was thinking.....

author: Christopher
author_rank: Frequent Contributor
timestamp: December 27, 2014, 11:22:15 am
content: They really should supply a spice model for this chip!Are you sure the problem is with the buck regulator and not the logic at the output?Can you bypass the buck with a 5v wall art and see if the problem still occurs?

author: Falcon69
author_rank: Super Contributor
timestamp: December 27, 2014, 06:08:43 pm
content: I will try that now.EDIT: Interesting. It does it still. Is it possible that I am still getting feedback from the inductors?The only thing I can think of is that debounce circuit. I have an LED that is on that circuit, and that too is flashing for a fraction of a second, which is only suppose to turn on if the switch is pressed.

author: Andy Watson
author_rank: Super Contributor
timestamp: December 27, 2014, 06:29:24 pm
content: What precautions have you taken in the design of your LED control circuit to ensure that it powers-on in to a known state?

author: Falcon69
author_rank: Super Contributor
timestamp: December 27, 2014, 06:31:07 pm
content: well, with the debounce circuit, the inverter is floating, unless the switch is hit, then it goes to ground. WHen switch is open, the inverter is floating until power is applied, then it goes to power, for the input. So, upon plug-in, it is floating

author: Andy Watson
author_rank: Super Contributor
timestamp: December 27, 2014, 06:36:20 pm
content: I think I will need to see more of the circuit to understand what your are trying to achieve with the inverter.Where does the inverter get its power from?

author: Falcon69
author_rank: Super Contributor
timestamp: December 27, 2014, 06:52:07 pm
content: This is the debounce circuit currently. ("Debounce Circuit")I have the older boards, that have no debounce circuit (no inverter either), it is just a tactile switch, 1k resistor, and a capacitor ("Debounce Circuit 3") and this one does not have the LED flash when using it. However, I can not use that, as I DO NEED a debounce circuit; the LED's all act funny when the tactile switch is pressed.SO, I've at least isolated the problem to the Debounce Circuit now. Now, any ideas how to fix it?EDIT: that should be a 0.1uF capacitor. Mistake there.

author: Falcon69
author_rank: Super Contributor
timestamp: December 27, 2014, 07:33:12 pm
content: Maybe I can do this.It would only add 1 more component (the other 0.1uF cap, the inverter comes in a dual chip, the same size), And this would allow the inverter to default to ground and only have power supplied to the input WHEN the switch is pressed.I think if I don't have the Schmitt Trigger of the Inverters, I would have a problem with switch debounce still, otherwise I would just eliminate the inverters.What do you think?

author: eetech00
author_rank: Regular Contributor
timestamp: December 27, 2014, 09:23:26 pm
content: HiIn your debounce circuit (attached)...Try changing R6 from 1M to 82kSee if that helps.But I would have used these values:R6 = 82kR5 = 18kC4 = 1uf, 25vthat would give enough debounce time to cover just about any crummy switch.eT

author: Falcon69
author_rank: Super Contributor
timestamp: December 27, 2014, 10:13:16 pm
content: The Debounce circuit works, it's just that when plugging in the voltage source (i.e. 12volt walwart), a voltage spike happens and the debounce circuit causes the LED's to Flash.That's why I was thinking that if I take away the power supply to the input, as in the schematic with the two inverters, it should eliminate that problem, I hope. It should make the inverter's input default to a low state as the only connection it would have is ground.

author: eetech00
author_rank: Regular Contributor
timestamp: December 27, 2014, 11:22:09 pm
content: Quote from: Falcon69 on December 27, 2014, 10:13:16 pmDuring and after power up, the inverter's output should remain low...right?I think the voltage isn't rising fast enough at the input to the Schmitt inverter.Basically, the Schmitt is initializing faster than its input, so you see a momentary flash.You don't need another circuit...Try changing the 1M resistor to something much lower...like 82k (or even 100k)eT

author: Falcon69
author_rank: Super Contributor
timestamp: December 27, 2014, 11:33:08 pm
content: Okay, I will try that right now.EDIT: Nope, changing to 100k did not fix it. I will try 47K, I don't have anything between that.EDIT AGAIN: Nope, changing to 47k or 10k did not fix it either.Thanks for the suggestion eetech00, it was worth a try.

author: eetech00
author_rank: Regular Contributor
timestamp: December 28, 2014, 12:17:07 am
content: Quote from: Falcon69 on December 27, 2014, 11:33:08 pmOk...but I still think something similar what causing the momentary flash. Somewhere there is a input that should be initialized high, to keep an output low, is not initializing fast enough.Using your debounce circuit and values, I was able to duplicate the flash. Changing the resistor value fixed it which makes perfect sense. So this same issue is probably happening somewhere else in your circuits.eT

author: Falcon69
author_rank: Super Contributor
timestamp: December 28, 2014, 12:23:15 am
content: that's what I thought as well, but, this single LED is based of that Debounce circuit, and it too is flashing. i.e. OR Gate A input tied to inverter's out, and OR Gate B input tied to ground, LED and resistor on output of OR Gate. That LED is flashing, along with all the other LED's in circuit controlled by other Logic Gates.

author: eetech00
author_rank: Regular Contributor
timestamp: December 28, 2014, 04:03:50 am
content: Quote from: Falcon69 on December 28, 2014, 12:23:15 amTry this...just to test the theory..If you can...connect the input of the inverter directly to +5 volts...with nothing else connected to the inverter input. Connected this way, there is nothing slowing down the voltage rise on the inverter input. Then connect/disconnect supply power while watching the LED. The inverter output should never go high and the LED should never light or flash.eT