Will this mosfet circuit work? - Page 1
Forum section: electronics / beginners

author: liquibyte
author_rank: Frequent Contributor
timestamp: October 08, 2014, 12:38:17 am
content: I have been looking for a solution to a problem I have and after coming up with the circuit in the picture I was reading down the new threads and came uponthisdiscussion which kind of touches on a few concerns I have about VGS and such.I've got an over voltage situation happening and was wondering if I could just put this circuit in line before the power supply circuit and have it delay the turn on time to the power supply by around 850ms or so and eliminate the 60+ volts I'm getting now due to the transient surge. I've been looking into solutions and found thisapp noteand while trying to calculate the values in figures 5a and 5b came across the attached circuit and thenthis calculator / equation. I thought it might be on the simple side but since I prefer a simple solution I thought I should ask before attempting it.

author: Falcon69
author_rank: Super Contributor
timestamp: October 08, 2014, 12:40:37 am
content: correct me if I am wrong, but I think most mosFETs can only handle +- 20 volts on the gate. That means You'll need to place a 30volt zener or so between the gate and source. Anode to the gate.Also, if the mosFET doesn't have one (most do), you may also want to put a Diode in there. Anode connected to the Source, Cathode to the Drain.

author: liquibyte
author_rank: Frequent Contributor
timestamp: October 08, 2014, 01:03:00 am
content: Quote from: Falcon69 on October 08, 2014, 12:40:37 amI actually have a couple of 30V zeners, so that's good to know. I did know about the diode and just mocked this up quick so I could visualize the circuit a bit better. I guess I could have chosen a part that had one on the schematic but then I wouldn't have had that issue verified. What I'm worried about is if this will keep power off for the time calculated or am I just wasting time with something so simple. I've never worked with mosfets before other than using one once for an electronic load and I didn't come up with that circuit, I just verified the mosfet I used was close enough to the one they used because I didn't have the exact part.Edit: How's this one?

author: Falcon69
author_rank: Super Contributor
timestamp: October 08, 2014, 01:19:53 am
content: now you switch to a p channel mosfet. You need to switch the drain and source on it. and move the diode to the other side and connect the anode between the resistor and the gate. I am by no means an expert, in fact a beginner myself, but I think the resistor needs to be connect to ground instead, and maybe the cap connected to the supply?

author: Dave
author_rank: Super Contributor
timestamp: October 08, 2014, 01:32:07 am
content: Quote from: Falcon69 on October 08, 2014, 12:40:37 amMOSFETs don't actually have separate diodes in the package. The construction of the transistor itself makes a PN junction between the bulk (which is in most cases connected to the source) and the drain. See the picture attached below.As for the circuit above, you are going to get a exponential ramp on the output, which in its steady state is going to be a couple of volts lower than the supply voltage. If the losses don't bother you, you are going to be fine.As Falcon69 suggested, add a zener between gate and source (15V or so). You aren't going to have a problem when your circuit will be powered, but you will have a problem when you cut the power. The potential of your drain and source will suddenly drop, while the capacitor on the gate will still be fully charged, holding the potential of the gate high.

author: Falcon69
author_rank: Super Contributor
timestamp: October 08, 2014, 01:39:12 am
content: Dave is correct, I just simulated it, and a 15v zener will work when the supply voltage is 43.7v? Wont that put ~28 volts to the gate? could blow the mosFET won't it?Here's the pic of the simulation. Dave is right, when the supply is removed, the cap holds power and slowly drains as the LED dims to nothing.

author: Dave
author_rank: Super Contributor
timestamp: October 08, 2014, 01:45:07 am
content: You might want to put that zener between source and gate, not drain and gate. Also, that cap should be in parallel with the zener and you should keep the resistor the way you have it now. Add another resistor (higher value) in parallel with that capacitor to drain it when the circuit is powered off.

author: Falcon69
author_rank: Super Contributor
timestamp: October 08, 2014, 01:53:51 am
content: fixed

author: Dave
author_rank: Super Contributor
timestamp: October 08, 2014, 02:01:09 am
content: nope

author: Falcon69
author_rank: Super Contributor
timestamp: October 08, 2014, 02:02:32 am
content: This simulation program is fun, but the way they have their mosFETs drawn confuses me.Here's another one.

author: rs20
author_rank: Super Contributor
timestamp: October 08, 2014, 02:09:01 am
content: Quote from: Falcon69 on October 08, 2014, 02:02:32 amYou've got a diode straight from input to output there, that's not going to stop any overvoltage...

author: Falcon69
author_rank: Super Contributor
timestamp: October 08, 2014, 02:12:04 am
content: Sorry, don't mean to hijack your thread liquibyteI'm confused, it looks the same as this circuit.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: October 08, 2014, 02:15:30 am
content: Quote from: Falcon69 on October 08, 2014, 02:12:04 amThis is a reverse protection circuit, not overvoltage.You're looking for something that looks more like a linear regulator or LDO.You can't use an N-channel MOSFET here because you lose at least Vgs(th) (and it's not "low dropout). An NPN emitter follower might be marginal (> 0.7V drop).You can also get purpose made protection chips that include drive circuits to use N or P channel MOSFETs correctly (without losing supply voltage in the process).Tim

author: Falcon69
author_rank: Super Contributor
timestamp: October 08, 2014, 02:18:39 am
content: That's why I said the circuit simulator is confusing. That's a P-Channel Mosfet in the circuit I drew.And yes, it's a reverse Polarity Protection Circuit. But, it does work in simulation. remove the supply voltage, and the cap keeps the LED lit for a time being. I think that is what liquibyte was after?

author: liquibyte
author_rank: Frequent Contributor
timestamp: October 08, 2014, 02:22:04 am
content: Quote from: T3sl4co1l on October 08, 2014, 02:15:30 amThe other guy working on this solution came up with the circuit below. Is this the kind of thing you're talking about?I don't really need low dropout because, as I said, the op amps are being powered by the full 43.7 volts and they are rated at 44 so it's always seemed like I'm too close to the tolerance for the part anyway.

author: liquibyte
author_rank: Frequent Contributor
timestamp: October 08, 2014, 02:27:11 am
content: Quote from: Falcon69 on October 08, 2014, 02:18:39 amNot really, no. I've got a transient surge of around 61-62 volts happening at power on that I'm trying to delay. I figure maybe I can use a mosfet as a delay until the transient passes. If there's a few volts drop as a by product then I've actually solved a couple of issues.

author: Falcon69
author_rank: Super Contributor
timestamp: October 08, 2014, 02:31:10 am
content: Oh! Can't you just put a TVS diode in there to suck up the transient voltage? I think TVS diodes reset once the transient is passed and circuit should operate as normal then.

author: liquibyte
author_rank: Frequent Contributor
timestamp: October 08, 2014, 02:46:02 am
content: Quote from: Falcon69 on October 08, 2014, 02:31:10 amThe problem is that the transient changes at the output of the power supply based on the voltage setting so what value am I shooting for? I've measured it at different voltage settings but If I clamp it at 43.7V then it will still allow more than I want when I've set the supply at a lower voltage wouldn't it? I'll give you guys a run down of basically what I'm seeing. The first voltage is where the power supply is set at after things settle down. The second voltage is what I measure as the spike if I turn off the power and let the filter caps drain and then turn it on again.30V - 36.7V25V - 33V20V - 29V15V - 23.5V12V - 19.7V10V - 16.2V5V - 7.95V3.3V - 5.49V2V - 3.6V1.5V - 3.2V1.2V - 3V1V - 2.7V.8V - 2.2V.5V - 1.6V.2V - .9V.1V - .36VMy thought was to delay the power to the supply circuitry until any transient has passed and the problem would go away.

author: Falcon69
author_rank: Super Contributor
timestamp: October 08, 2014, 02:56:34 am
content: so you are looking for something that would connect the circuit to the power supply after a set time amount, say 5 seconds for example? Dunno, as your power to supply the 'something' would also undergo those transients as well, unless you had an external power supply for that. Maybe a button battery, 555 timer, and some caps/resistors?

author: liquibyte
author_rank: Frequent Contributor
timestamp: October 08, 2014, 03:03:59 am
content: I've looked into using a relay as a soft start on the primaries of the transformers using a separate 12V supply but this adds quite a bit in complexity. What I want to do with this is connect it between the filter caps and the input to the power supply so that it delays turn on for just under a second until the filter caps have had the time to fully charge after power on. Basically it's a wedge in solution because the boards are made and the supply (dual actually) works fine but for this one issue.

author: rs20
author_rank: Super Contributor
timestamp: October 08, 2014, 03:29:41 am
content: I'm so confused by this thread. If you want the switch to turn on x seconds after power is applied, that is straightforward to do with a PFET, resistor and cap (and maybe a zener if your PFET can't withstand enough Vgs, although I'd prefer to just get a strong enough MOSFET if possible). You were all going down that route before, what happened?

author: liquibyte
author_rank: Frequent Contributor
timestamp: October 08, 2014, 03:49:28 am
content: Quote from: rs20 on October 08, 2014, 03:29:41 amI'm confused by this. Why would a PFET work better than an NFET? The only PFET I saw washereand I thought using one would require a low side configuration. I thought using an NFET was where this was going by adding the zener to control VGS which I wasn't really aware of for the most part because I've never really used mosfets before. The rest of the calculations were done but I wasn't sure if they were right because I've never tried this before and thought I'd ask those that know more than I do. The app note I linked had a filter cap configuration for dealing with inrush current and while I was doing research on the calculations for that I stumbled on this circuit and because it seemed simpler I was curious if it would work.

author: Falcon69
author_rank: Super Contributor
timestamp: October 08, 2014, 03:52:51 am
content: PFET are normally used as a High Side Switch. NFETs are for Low Side switching.

author: Falcon69
author_rank: Super Contributor
timestamp: October 08, 2014, 03:56:04 am
content: Oh, I also want to say liquibyte. NEVER use Google images for circuits. I did that once, and screwed up an order of circuit boards. The picture I was going off of, had the source and drain switched. Really messed up the entire circuit. However, if I had gone to the link where the picture was, and read the thread, I would have known it was wrong, as it was mentioned in the thread. Lesson learned.

author: rs20
author_rank: Super Contributor
timestamp: October 08, 2014, 04:46:37 am
content: Quote from: liquibyte on October 08, 2014, 03:49:28 amThe process goes like this:1. Draw a PFET on your piece of paper. Make sure the body diode is pointing against the normal flow of current.2. (Optional) add a zener from G to S, to prevent Vgs from getting too high.3. We want the MOSFET to stay switched off initially; capacitors tend to hold voltage steady. Since the capacitor will be discharged initially, it'll have zero volts across it. We want the FET to be off initially, Vgs = 0 volts. Therefore, we should connect the capacitor from G to S (in parallel with the optional zener).4. We need the FET to eventually turn on. A PFET is turned on by a negative Vgs, S is fixed at the positive supply, so this is great because we just have to pull the gate to ground to turn it on (if you used an NFET, this is where you would get stuck). Therefore, we connect a resistor from the gate down to ground.Design done, with simple reasoning only. Now, the only issue is that the FET might turn on sort of gradually, and might burn a lot of power while it's only partially turned on. If you want to avoid that, you'll need some more active components.Quote from: Falcon69 on October 08, 2014, 03:56:04 amI think surely the lesson here is, understand and test your circuits rather than blindly accepting schematics from ANY source. If you got circuit boards fabricated without ever testing or thinking "What's the VGS on this FET here?", then you're orders of magnitude short of due diligence. Google images is just fine (wonderful, even), as a source of inspiration.

author: rs20
author_rank: Super Contributor
timestamp: October 08, 2014, 05:13:27 am
content: Quote from: Falcon69 on October 08, 2014, 02:31:10 amThis is also a good suggestion, although whether it's actually a suitable solution depends on how strong that 60V spike is. If the power supply is happy to provide that 60V at 100A, your TVS is going to have trouble dissipating 4000W, I suggest. I'm not sure how likely this is to be an issue in practice. (Often you use a TVS along with a fuse, so that when that lightning strike hits, the fuse sacrifices itself.)TVS diodes aren't stateful devices, so you're right that they "reset" once the transient passes. (Pedantic: But I wouldn't use the word "reset", because that implies that the TVS "triggered" or "latched" into a conducting state. They don't, TVS diodes are just fast-reacting (sometimes bi-directional) zener diodes with a particular V/I characteristic.)

author: liquibyte
author_rank: Frequent Contributor
timestamp: October 08, 2014, 09:28:50 pm
content: I think I'm going to ditch the mosfet idea in favor of the real world circuit that redwire built and tested. I've done a spice sim, and please don't laugh at my ignorance with it because I know almost nothing about LTSpice, and am not even sure I did the thing right. In the real world circuit that has been built, R1 is 39K, R2 is 4.7K and there's an R3 in between R2 and C1 that's 2.2K, C1 is 47uF, and Q1 is a TIP141. I'm not even sure that the way I have things would actually work but the real results with the different parts are not too far off from what I have here.Edit: forgot the output pic.

author: revilo951
author_rank: Newbie
timestamp: October 09, 2014, 01:18:29 am
content: Quote from: Falcon69 on October 08, 2014, 01:53:51 amWhat simulation program is this!?

author: liquibyte
author_rank: Frequent Contributor
timestamp: October 09, 2014, 01:27:15 am
content: Quote from: revilo951 on October 09, 2014, 01:18:29 amhttp://www.falstad.com/circuit/It's actually kind of handy too.

author: rs20
author_rank: Super Contributor
timestamp: October 09, 2014, 02:18:41 am
content: Quote from: liquibyte on October 08, 2014, 09:28:50 pmThat circuit will work fine too. The only disadvantage with using a BJT is that it will drop 0.7V (and get quite hot if your circuit consumes several amps). You can't see this in your simulation because your load is a tiny capacitor; put a resistor in there to simulate a real load. The circuit I describedherewill waste less power in the transistor, and will consume (theoretically) zero quiescent current because the MOSFET doesn't require bias current like the BJT does. I'll edit this post with a sim later.

author: liquibyte
author_rank: Frequent Contributor
timestamp: October 09, 2014, 05:15:25 am
content: I've been trying all night to get the spice simulation working on the circuit you described and I just can't get it to behave. I've got the input spiking but no matter what I do, the gate source voltage does too and then so does the output because the mosfet is conducting.

author: rs20
author_rank: Super Contributor
timestamp: October 09, 2014, 01:55:38 pm
content: Your pulse source isn't working properly. To be fair, it seems to be a bit of a bug with LTSpice maybe, since you asked for all sources to start at 0V, but your pulse source isn't doing that (EDIT: No, not a bug at all -- the startup option startsDCsupplies at zero, it makes no promises about other supplies, like pulse). Hence, the capacitor is pre-charged to 18V, and the switch is on the whole time. ( Reasonably easy to methodically debug, first step was to ask "why is the FET switched on already? Why is Vgs nonzero the whole time? Shouldn't the capacitor start at zero? It's not starting at zero! Shouldn't the supply start at zero? It's not! etc" )Use .tran 2 startup, I can't wrap my head around this logarithmic time axis thing (the left hand half of the graph is full of the random details of how LTSpice ramps up the voltage sources, it's distracting).But more importantly, change the pulse source to PULSE(0 18.3 0 35m 810m 1m).(Bolded because this is the one take-away sentence that fixes your problem). Get rid of the top resistor, a FET doesn't require bias current. With these changes, the FET turns on after only 0.4s, since the Vgs threshold is orders of magnitude less than supply, but it's easy to tune that by making the resistor bigger. I've got it working here, but it's 1am here so I can't be bothered with screenshots right now

author: liquibyte
author_rank: Frequent Contributor
timestamp: October 09, 2014, 03:14:09 pm
content: I have a hard time wrapping my head around spice for some reason. 0 18.3 0 35 810 1 did the trick along with changing R1 to 180K to keep the delay until after the pulse settled. The gate is still seeing the pulse according the the simulation though, wouldn't that be an issue that would destroy the mosfet in the real world?

author: rs20
author_rank: Super Contributor
timestamp: October 09, 2014, 09:28:22 pm
content: Quote from: liquibyte on October 09, 2014, 03:14:09 pmYou should be measuring Vgs (Vg - Vs), not Vg. (Google how to measure relative voltages). The gate could be at a million volts with respect to ground, and it would be just fine as long as the source is within 20 volts of a million volts.

author: Dave
author_rank: Super Contributor
timestamp: October 09, 2014, 09:43:40 pm
content: You can plot the difference in LTspice by subtracting one function from the other: V(n00_) - V(n00_). You right-click on an existing trace name and edit the function there or hit Ctrl-A in the graph window. You can really do all sorts of weird and wonderful math.

author: liquibyte
author_rank: Frequent Contributor
timestamp: October 09, 2014, 09:57:00 pm
content: Quote from: rs20 on October 09, 2014, 09:28:22 pmI got it, thanks. You've been really helpful the last couple of days, I appreciate you're patience.

author: rs20
author_rank: Super Contributor
timestamp: October 09, 2014, 09:58:15 pm
content: Quote from: Dave on October 09, 2014, 09:43:40 pmOr you can just press the mouse down on the gate node of the schematic, and drag over and release on the source node. No need to manually type formulae if you just want a relative voltage.

author: liquibyte
author_rank: Frequent Contributor
timestamp: October 09, 2014, 10:10:17 pm
content: Quote from: rs20 on October 09, 2014, 09:58:15 pmThat's the way I looked it up. Other than some of the math going way over my head, I've been learning a lot using it. This is going to be one useful program going forward.

author: Dave
author_rank: Super Contributor
timestamp: October 09, 2014, 10:38:03 pm
content: Quote from: rs20 on October 09, 2014, 09:58:15 pmThanks for the tip, that does save some time.

author: liquibyte
author_rank: Frequent Contributor
timestamp: October 09, 2014, 10:50:16 pm
content: Quote from: Dave on October 09, 2014, 10:38:03 pmI found thisguidethat had a few other useful tips in it along with that one.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: October 10, 2014, 01:02:04 am
content: D1 is backwards relative to M1.In the MOSFET symbol itself: you see the little triangle 'pointing at' the middle line segment? That's the contact to the substrate. The junction dot says it's shorted to the source terminal (part of the channel). The substrate and channel form a PN junction. What the triangle / arrow is telling you is, there's a PN junction here, and it conducts (conventional / hole current flow) in this direction. In other words, the intrinsic body diode.Every Single Symbolthat shows an 'antiparallel diode' around the MOS structure isredundant: that diode isalready there, people just willfully ignore it.So please don't make the same mistake: there is never any reason to put an independent diode across a MOSFET in a simulation. It's already present and accounted for.Definitelydon't fool yourself by accidentally putting one in backwards and hoping in vain that, by putting that diode there, you can somehow tell the circuit which way you intend current to flow!Tim

author: liquibyte
author_rank: Frequent Contributor
timestamp: October 10, 2014, 03:16:28 am
content: Quote from: T3sl4co1l on October 10, 2014, 01:02:04 amThanks for the advice. I will definitely remove it from everything I'm working with. I think that was explicitly mentioned earlier in the thread too. Can you tell me if I'm doing the current part of this simulation correctly? I'm not sure how this works but the plot looks right when I do it like the following.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: October 10, 2014, 04:46:17 am
content: Is V2 the PULSE and I1 a true ideal DC current source?You don't need to put DC voltage in series with a PULSE, though you can and it doesn't hurt anything. You can set the ON and OFF voltages to arbitrary levels, including a DC bias, so it's redundant.What hurts is, the voltage sources DO NOT MATTER, because when SPICE says "constant current", it means constant. You can put the ass end of the current source to any other node connected by voltage sources, and it makes no difference. Doesn't matter if the voltage across the current source is positive or negative, large or small, current is current.So, all that you're seeing is a capacitor charging, into a diode-strapped MOSFET (not really, but sort of), into a load resistor, basically. It's not at all representative of a real circuit, and isn't showing you what you want it to.Remember, SPICE doesn't know what your intent is. (Silly and obvious, but also easily forgotten!) It's just ideal models slapped together. It's entirely your responsibility to make a model that's realistic. Usually, you should be experienced enough toalready know what's going to happen before you even hit 'simulate', which makes it somewhat redundant, but this should give you some perspective: unless you're doing something really "out there", you never use SPICE to determine functional performance ("does it do what I think it should", let alone "I don't know what to think"!), only to tweak the parameters and optimize the circuit to what you need.Which... doesn't help you any, I know, because that's like saying "you're dumb, give up and read a book", and you're just trying to figure things out. Just be very weary that, if you're trying to learn based on what SPICE gives you... it's only as good as what you're putting into it. Nothing written in SPICE is absolute, definitely not if you're unsure about your model.(Also... Comic Sans? Reeeeally?...)Philosophy aside, I think what you wanted to be going for is, a voltage source (PULSE alone, or PULSE plus DC -- or whatever, any voltage source combination is valid) with a series resistor, corresponding to either the short circuit capacity of the power source you're looking to model, or to the rated current at some voltage drop. Two examples: a 50V 2A (nominal) supply might drop to 45V at full load, in other words, a (50 - 45) = 5V drop, which looks like 5V/2A = 2.5 ohms. (Evidently, the short circuit current would be 50V / 2.5 ohm = 20A.) Another example, maybe you know your supply's short circuit current is 50A, in which case 50V/50A = 1 ohm. This is a Thevenin voltage source.You can also do it by using a current source (PULSE, DC or whatever, any combination in parallel) in parallel with the same resistor. This is a Norton current source. The two are related by having the same resistor (one series, the other parallel), and I = V/R relating the two sources. Don't mind that the current source appears to be delivering short-circuit power into its resistor: that's virtual watts, you're just using that for its behavior at the terminals.Tim

author: liquibyte
author_rank: Frequent Contributor
timestamp: October 10, 2014, 05:14:45 am
content: I'm going to do more studying and come back to this as it relates to what I'm trying to accomplish. I've read about trying to model real world results and how much of a pain it is to do in spice and ultimately futile to boot but I think it is helping me understand a few things regardless. I'll try my best to figure out how to set things up with as accurate a model as can be done just for the practice I suppose. Either way, I do appreciate the help. I like to read your posts even if I don't always understand everything in them because, no matter what, I always come away with something new learned and that's always a good thing.As for the font, it was set at Arial but I'm running this through wine so I guess I don't have my fonts set up right there. I tried changing it to Courier, Times, and Verdana but it stayed the same regardless. Finally I tried Tahoma and that I guess I have set up right, so from now on I promise not to post crappy fonts if I can help it.To be honest, I don't think I have Comic Sans installed so I have no idea how it was rendering that way.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: October 10, 2014, 05:57:34 am
content: ThanksWeird... actually on closer inspection, it looks like the graph is labeled with something normal like MS Sans, and the .TRAN label is some normal, kinda tilted sans, I don't remember which. But the schematic labels are all kinds of goofy, not Comic come to think of it, but something just as random. Go figure!Tim

author: liquibyte
author_rank: Frequent Contributor
timestamp: October 10, 2014, 06:50:44 am
content: Quote from: T3sl4co1l on October 10, 2014, 05:57:34 amI have no idea why it's so wonky but at least one of the options worked right. I could swear I had MS fonts installed but it turns out I didn't. I do now so that's not going to happen again. I should see if I can move my DejaVu fonts over and have them picked up by the program because I like the way they look.Istartedreading and then found things likethisthat is probably going to make my head hurt. I don't want to turn into a voltnut, or spicenut if you will, but my OCD is liable to take over at some point and I'm going to have to model a mosfet or an op amp for the fun of it. At this point I'm just glad I got the theoretical working. I'll have to see if I can find a PFET to try the circuit out and if it works I'll be one happy camper. The problems I've had with this power supply would probably have made most people quit or at least move on to something else but it's helping my learning experience so I'd say it can't be all bad. I'm determined to fix it or, if it can't be, at least learn enough as to why and then just design something new.

author: rs20
author_rank: Super Contributor
timestamp: October 10, 2014, 10:26:32 am
content: Quote from: T3sl4co1l on October 10, 2014, 01:02:04 amNo it's not, it's a PFET and in a PFET the diode arrow points towards the Source. (Assuming you're talking about the diagram in #30? Or did someone go back and retroactively change a diagram?). I agree that the diode need not be explicitly added to the diagram though, of course. I believe the original circuit in #30, without the current source added, is a reasonable model (modulo the unnecessary but harmless body diode)?Quote from: T3sl4co1l on October 10, 2014, 04:46:17 amIn this case, he wants the voltage to start at 0 (which, AFAICT, isnotenforced by using ".tran xxx startup"), then rise to a high value, then fall to a stable nominal voltage. Since the pulse settings only have 2 voltage values available, it can't be done with a single pulse source. His solution, a pulse atop a DC source (that respects startup) is a bit hacky, but its as good as any other method I can think of.

author: liquibyte
author_rank: Frequent Contributor
timestamp: October 10, 2014, 02:33:28 pm
content: Quote from: rs20 on October 10, 2014, 10:26:32 amI haven't gone back and changed any of them, I promise.Quote from: rs20 on October 10, 2014, 10:26:32 amThat's exactly what I was after. I wanted a spike that lasted for ~300ms or so and then went away allowing a final steady voltage of 43.7V which is what I measure after rectification and filtering. I needed to be able to tune the resistor and cap value based on where that pulse ended. I'm still not done simulating because I want to, as Tim said, get the models as accurate as possible as a learning exercise. Is there a better way to simulate a transient surge that I may not be aware of? Leaving the accuracy of the part models aside, I need to be able to simulate to 30V @ 3A with a startup transient of around 64V. I haven't measured the effect on the current but, to be honest, I don't think I have the right equipment for that. I only own one multimeter and it's not all that great but seems accurate enough as kind of verified by my latest acquisition of a 465B. The bad part about all this is that I can't really see the transient with the scope with any accuracy because it's happening to fast and I can't store it.One thing I've noticed is that after I added the current with a better load to the simulation, the spike doesn't behave as it did before. I still get a spike but it's not rising up to the 64V I wanted, it's only topping out at 44V. I think it may be because the load I have on it is R=limit(10,V(n002)**2/90,10). I'm still new to this so I'm absolutely sure I'm doing it wrong.Quick edit: I've also tried R=limit(14.56667,V(n001)**2/131.1,14.56667) which seems to give better results as far as stability goes but adding the 18V on top of 43.7V isn't doing what I would expect. I have to explicitly state 64V in V2 now.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: October 10, 2014, 07:59:30 pm
content: Quote from: rs20 on October 10, 2014, 10:26:32 amHmm, thought I was looking at #30. Perhaps I was a bit hasty, it does look the right way around!QuoteI'd say, with the current source replaced with a suitably chosen resistor. That'll be more representative of a typical power source.QuoteYeah, PULSE has four states in order: initial, slope, final, slope, <repeats>. You need more than one source if you need more than two static states, and an overshoot pulse requires three. You could also do a PWL source (a table of points and times describes the voltage), or build one yourself out of discretes (which has the advantage of potentially being representative of what's actually making the pulse in the first place, and the disadvantage of being hard / tricky / tedious to design and calculate the parameters of).Tim

author: T3sl4co1l
author_rank: Super Contributor
timestamp: October 10, 2014, 08:01:24 pm
content: Quote from: liquibyte on October 10, 2014, 02:33:28 pmWhat is this, actually? A load dump or something..??What are the characteristics of this pulse?QuoteLike I said, because your source is entirely constant current, the response is dominated by the bulk capacitors and load resistor, and the VPULSE does nothing at all.Tim

author: liquibyte
author_rank: Frequent Contributor
timestamp: October 10, 2014, 09:17:43 pm
content: Quote from: T3sl4co1l on October 10, 2014, 08:01:24 pmI'm trying to delay the power into the power supply circuit past the filter cap by the time between power on and the filter caps being charged. What's happening is that I'm getting about 63 volts at power on when the supply is set at 30 volts. The surge at the output is actually dependent on the setting of the voltage pot. I posted some of the numbers I'm seeinghere. I've tested fully loaded and unloaded and get basically the same results. I don't have a DSO, just analog so I can't really get images of it. The attached pic is the latest revision I've done of it with this mosfet circuit included.Quote from: T3sl4co1l on October 10, 2014, 08:01:24 pmI guess I'll have to live with the limitations of the simulator. I have been doing more reading and have figured out variable loads etc. I'll keep studying to learn the ins and outs of voltage and current sources.

author: rs20
author_rank: Super Contributor
timestamp: October 10, 2014, 11:25:24 pm
content: Quote from: liquibyte on October 10, 2014, 09:17:43 pmTo clarify, what T3sl4co1l is pointing out here is that if you put an ideal current source in series with anything (including, for instance, a pulse voltage source), then you just have a current source. In a similar way to how if you have an ideal voltage source in parallel with anything, you can just ignore the parallel components. This is not a limitation of simulation, this is just how it is.

author: liquibyte
author_rank: Frequent Contributor
timestamp: October 11, 2014, 12:45:11 am
content: I'm not sure if I understand completely but I've been playing around with a few things and I feel like I'm getting a more accurate representation of how I think the spike is behaving in the real world. It seems like the circuit should work for the intended purpose so I'll be getting a couple of PFET's to test out with on my supplies and see how it goes compared to how I think it should. As long as the startup delay is longer than the worse case transient scenario I should be good. If everything works out, I'll call this project done and button up the case and start using it to help building the next one.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: October 11, 2014, 02:43:34 am
content: And putting a current source in parallel with a voltage source gets you a voltage source again...Just use a voltage source and series resistor.So, you have a power supply, which exhibits nasty overshoot when turned on/off? That sounds like an out of spec issue, or a bad design issue, or a compensation issue. Fix that instead.Tim

author: liquibyte
author_rank: Frequent Contributor
timestamp: October 11, 2014, 03:10:36 am
content: Quote from: T3sl4co1l on October 11, 2014, 02:43:34 amHeh, I just took out the current source and I'm getting the same neat results. I feel dumb now.Quote from: T3sl4co1l on October 11, 2014, 02:43:34 amI don't even know where to begin to track the problem down at. I always thought it was just because of the 15000uF cap being discharged at power up. Looking back, I should have picked a different project but I've got so much into it now that I don't want to just give up. Getting help with this or even what to look for has been difficult. Q3 in the above schematic took care of the shutdown transient. I got a little help with that here by David Hess mentioning an HP service manual that had that kind of scheme in it. The part was actually in the original design but was redesigned without it. To be honest, I'm not sure why it was taken out at all since it actually did something useful.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: October 11, 2014, 04:10:45 am
content: Don't suppose you have a manual (with schematics) for the power supply in question?

author: liquibyte
author_rank: Frequent Contributor
timestamp: October 11, 2014, 04:37:26 am
content: Quote from: T3sl4co1l on October 11, 2014, 04:10:45 amNope, this is an old hobby build that was redesigned using modern parts and values that didn't fail due to being overloaded. Apparently the original had component failure issues but a couple of things were pulled out of it during the redesign that may have been needed. I'm attaching the pics of the old and new version with part lists from each. This is about all there is of this thing other than the work I've done with it in Eagle.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: October 11, 2014, 05:53:59 am
content: Oh yeah, back in August, I think I remember that thing...Try... different values for C6 and C9, and with resistors in series with them. Like, 100p-1n and 10k-1M. You probably don't need C6 at all, when you have a resistor in series with C9.You have to do some sort of transient testing to prove out the selection -- if you have a DSO, you can do this by just shorting a load resistor across the output momentarily, and triggering on the little transient blip the output voltage goes through. Use AC coupling so you can zoom in on the disturbance. If you don't have a DSO, you can still do this, but it'll be kind of hard to see... instead, you can switch the resistor automatically, using something like a MOSFET driven by a 555 or signal generator (at a low enough frequency to see things -- figure it out based on how long the transient is... might be 10Hz to 10kHz). Also, you can try with or without the final output capacitor -- it's not always the best thing for a power supply (they all do it, but that doesn't mean they have *reason* to do it!), and it can mask spooky behaviors like this, that need to be dealt with in a basic fashion.Also, does Q1 even do anything? I guess it was added or changed during/after that discussion, but it seems to me, the -1.3V supply will come up as quickly as the main supply, so that Q1 is basically just asking the question: is the supply over 136V (give or take temperature)? Which seems like a really odd thing to do, disabling the output (forcibly, not in a nice way -- it's shunting the op-amp output!) when the voltage is already burning up all the transistors and op-amps. (That's 1.2V for the supply, 0.6V for the transistor Vbe, and 12k into 160 ohms is a 0.013 voltage divider ratio, or for 1.8V from the divider, an input of 136.8V.)Another approach, which isn't applicable here, but can be used if you have OTAs (e.g., LM13700) or a discrete circuit: bring up bias slowly, so the amplifiers begin to amplify gradually. While coming up, they'll settle on the correct output voltage, while still having too little gain to cause overshoot.Tim

author: liquibyte
author_rank: Frequent Contributor
timestamp: October 11, 2014, 11:27:55 am
content: Quote from: T3sl4co1l on October 11, 2014, 05:53:59 amI've got pictures of some testing that the other guy did that's working on this too. Just as an FYI here, his circuit with the TIP141 didn't do so great under load which is why I've kept pursuing this avenue. He's new to scopes I think so I'm not sure if these will help or not. No DSO here unfortunately, but I did just pick up a Tek 465B so I have two channels to muck about with. I'll try the trick you suggested as well. I have a few 555's but not a real sig gen, just one I built out of a 555 oddly enough. It's a simple circuit but it's allowed me to verify that my scope does indeed work.Quote from: T3sl4co1l on October 11, 2014, 05:53:59 amWe had a discussion in the other thread about that. Apparently it serves the same function that theTek PS-503uses it for. It seems to work as you can see in the screenshots. Of course it may be causing other issues besides but I'll jump that hurdle when I get there. I think D10 might be needed too, it seemed to have helped with a noise issue or something I believe.Quote from: T3sl4co1l on October 11, 2014, 05:53:59 amI had to look up OTA's (Operational Transconductance Amplifier if any other newbies are looking). I'll keep trying to find the problem based on your advice but I thought what I was doing with this mosfet circuit was essentially a solution in the same ballpark. Allow the transient to pass before applying power to the supply. It's hackish but I guess it would work. Like I said though, I'll keep working on it and see if I can figure out where the problem(s) is(are)

author: T3sl4co1l
author_rank: Super Contributor
timestamp: October 12, 2014, 02:36:38 am
content: 100ns/div shots? Those aren't startup transients, those are RFI from the switch contacts closing! Absolutely nothing to do with the power supply itself.If those are causing problems, try some RFI methods:- C or R+C across the switch contact(s), usually 1n to 10n (must be X1/X2 "across the line" rated) and 0-100 ohmsFiltering applied after the switch:- Y1 ("line to ground") caps from AC line hot/neutral to ground (chassis ground), typically 1-4.7n- Ferrite bead (run both H/N together through a ferrite bead, or make multiple turns through), proceeded and/or followed by caps- Common mode choke (in series with H/N, observe winding polarity -- take apart an SMPS to see how it's done, or just hack the entire filter circuit out for use), esp. in combination with the Y1 caps- Or, same things on the output side (use caps rated for the isolation voltage -- if your output is grounded or SELV (safe, person-touchable), you don't *have* to use Y1 type here, unless you'll sometimes be connecting the output side to AC line stuff again).- Or, if this is periodic, the rectifier diodes can be doing the same thing as the switch. An R+C across each diode can help that (usually on the order of 0.01-0.1uF + 10 ohm).Tim

author: liquibyte
author_rank: Frequent Contributor
timestamp: October 12, 2014, 03:02:29 am
content: I just wanted to be able to give you what I thought were the relevant pics. I'll agree that the first two are set at a really small V/div setting and I couldn't figure out why he was looking at things at that setting but the one that concerns me is the third pic. I've measured this effect at 10V/div and am seeing an overshoot from the steady rectified and filtered voltage of 43.7V up to ~63V which is what I think is pulling the output at 30V up by 6 or 7 volts at turn on. If this is a real situation and I'm looking at it right, it's going to eventually damage the op amps right? This is why I thought that a slow start or soft start circuit might be needed. I'm fairly sure it's not just RFI because I can see the trace rise by that amount and I can also see it with my multimeter so it must be happening over a longer time period.I've got a shot from mine, but just at operating conditions and not power on. Like I said, I only have a 465B but it's a real treat compared to the Conar 255 I was using. This screenshot is 30V @ 3A fully loaded at 2mV/div @ 20ms/div at the output.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: October 12, 2014, 03:43:35 am
content: Oh, the blue trace I guess? Those waveforms are pretty crusty anyway, lots of ripple or HF interference or something.6-7V out of 30V doesn't sound like a big deal. A power supply, being what it is, shouldn't have any overshoot and should be tuned for overdamped response instead, but it's not terrible.Why would it be bad for the op-amps? They're the things making it in the first place...Are you saying the overshoot is on the internal supply, not the output? I'm really confused about what voltages and parts you're concerned about.I remember from the thread, there was concern about op-amps rated for enough voltage, which you'll have a hard time finding regardless. Your only major alternatives there are:1. Replace the transformer so it runs at lower voltage (24 or even 18VAC?), adjust output voltage scale for consistency2. Replace op-amps with discretes3. Use op-amps, but make a low voltage supply for them to run on, and level translate with discretes(1) changes the output range which might be undesirable (i.e., sometimes, you simply have to have 30V from it), and (2) and (3) basically mean ripping everything out and starting over. Which I would kind of rather have, just because it's a somewhat janky design to begin with, but it's a lot of work.Tim

author: liquibyte
author_rank: Frequent Contributor
timestamp: October 12, 2014, 11:11:34 pm
content: Quote from: T3sl4co1l on October 12, 2014, 03:43:35 amI'm getting 43.7 volts at pin 4 of U1 and U2 which is right at the upper limit of the 2141's tolerance. If I get a +20 volt spike on top of that voltage at power on, I'm worried that I'm going to damage those two op amps. U3 is behind the 10V zener due to the negative supply. What I think is happening is that the 63 volt spike is pulling the output up by the 6-7 volts @ 30V I see there.Quote from: T3sl4co1l on October 12, 2014, 03:43:35 amIf U3 is behind a 10V zener, why could't I do that for the other two so that I drop a few volts there? One of the things suggested in the other thread was that we really need an easy to build supply for hobbyists. I see discussions come up all the time about which supply to build and the two options most talked about are variable voltage regulator supplies and this one, or one if its variations. Now that I'm this deep into it, I will have to agree that it's somewhat janky. I think my best bet would be to use this mosfet circuit on what I have now and then work on a redesign that doesn't suffer from these issues. I'm far from qualified for this but I think I may give it a go.

author: liquibyte
author_rank: Frequent Contributor
timestamp: October 13, 2014, 01:34:16 am
content: I asked redwire to get some screenshots of the voltages across pin 4 and pin 7 of U1 and U2. He said that the "Second try was a bit shaky as it was just holding against pin 7 while trying to plug in the transformer" for the fourth picture. I think this should clearly show what kind of voltages we're getting into the power pins of the op amps.Edit: I meant to add that this is on power up.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: October 13, 2014, 01:48:26 am
content: I don't see overshoot? Or at least not much.

author: liquibyte
author_rank: Frequent Contributor
timestamp: October 13, 2014, 02:22:15 am
content: Do you think that the 2N3055's could be at fault here? 2N3055's have a Vceo of 60V, will that break down and allow a voltage across it if it's a bit more than that? If so, would a faster transistor help in this, say 4, 6 or 8MHz and perhaps a bigger Vceo like aMJ15022?

author: T3sl4co1l
author_rank: Super Contributor
timestamp: October 13, 2014, 03:59:06 am
content: I'd be more worried about the op-amps. Though I'dneveruse 2N3055 out of principle, anyway...Vceo only matters on supply to output voltage.Tim

author: liquibyte
author_rank: Frequent Contributor
timestamp: October 13, 2014, 04:55:32 am
content: I think I'll go with this mosfet circuit then and see how it goes. I'll try and get to the bottom of the issue and see if it can eventually be solved.