Fast optocoupler shutoff - Page 1
Forum section: electronics / projects

author: matthuszagh
author_rank: Frequent Contributor
timestamp: January 27, 2023, 11:55:52 pm
content: I'm designing an over/under-voltage protection circuit that needs to have really fast shutoff (ideally in the low 10s of ns). The downstream circuit is connected/disconnected with a bidirectional mosfet switch whose gate is driven by an optocoupler (probably the VO1263). The optocoupler is needed to isolate the gate from the comparator circuitry. Otherwise the gate-source and gate-drain ratings would be greatly exceeded (input can go to +/-100V). I've attached an image showing this part of the circuit.My question is: how can I make this shutoff as fast as possible? The turn-on time isn't critical (within reason, anything in the ms range, or even up to a second is fine). Here's what I've come up with so far. I'm using a P-channel JFET to switch in and discharge the charge stored on the MOSFET gates (the capacitance is significant, as these MOSFETs are designed for high VDS and low Rds(on)). This is discussed in the VO1263 and VOM1271 datasheets. Additionally, I'm planning to add a bleed resistor (unlabeled resistor in the schematic) to hasten the turn on time of Q4. And, I can use JFET with a low on-state channel resistance, or parallel several. There's a tradeoff here, since I don't want the JFET gate capacitance too high. I'll use the smallest possible bleed resistor such that the optocoupler can still bring the MOSFET gate voltage high enough (optocoupler short-circuit current is in the 10s of uA). To be able to make this bleed resistor smaller, I'm planning to parallel a number of these optocouplers (I think 10, but that's TBD). Another thing I'll do is to make the JFET gate threshold voltage as high as possible so it kicks in as quickly as possible when the gate starts to discharge.Are there any other options? Have I overlooked anything in my various optimizations discussed above?

author: moffy
author_rank: Super Contributor
timestamp: January 28, 2023, 03:42:04 am
content: If you need 10s of ns shutdown then you will need a dedicated separate high speed shut down circuit. The photovoltaic coupler is orders of magnitude too slow, you would need something like a fast digital isolator to get the response you need.

author: matthuszagh
author_rank: Frequent Contributor
timestamp: January 28, 2023, 03:55:25 am
content: I don't understand how that would work. As far as I know, all digital isolators require power rails on the output side. That will exceed the MOSFET Vgd/Vgs rating.

author: matthuszagh
author_rank: Frequent Contributor
timestamp: January 28, 2023, 03:59:40 am
content: What is it precisely that makes the optocoupler shutdown slow? If the photodiode stops generating current quickly after the LED is shutoff, then it's just that the load capacitance has to discharge through the parallel resistance. The VOM1271 uses a parallel 200pF, 10Mohm load to measure shutoff time. I can use a much lower parallel resistance than 10M (probably 30k) and the JFET gate only has 5.5p of capacitance.

author: matthuszagh
author_rank: Frequent Contributor
timestamp: January 28, 2023, 04:03:49 am
content: I'll try to setup some measurements to actually verify this, but I looked up latencies of LEDs and photodiodes. LEDs apparently have latencies of single-digit to 10s of ns and photodiode latencies are in the picosecond range. So, I'm maybe a bit out of my budget when I consider other contributions to shutoff, but not by much. I don't see why I'd be orders of magnitude off.

author: moffy
author_rank: Super Contributor
timestamp: January 28, 2023, 04:18:39 am
content: Quote from: matthuszagh on January 28, 2023, 03:59:40 amThe datasheet:https://www.vishay.com/docs/84639/vo1263aa.pdfon Page 2 "Switching Characteristics" quotes a "Turn off time" of 472us. When the LED stops shining there is still a lot of conducting charges in the photovoltaic diode that need to be swept out. Because you want these charges to last a long time for efficiency sake the device is very slooow. And yes for a digital isolator you would need a supply on both sides, but if you want ns response that is part of the cost.

author: matthuszagh
author_rank: Frequent Contributor
timestamp: January 28, 2023, 04:45:16 am
content: Quote from: moffy on January 28, 2023, 04:18:39 amOk, but that's with a 1M, 15p load. Isn't that going to slow the turn-off time? The VOM1271, which has an integrated fast turnoff (I think this is just a p-channel jfet and resistor) is more than an order of magnitude faster, indicating at least some of the turn-off time isn't just optocoupler delay.Quote from: moffy on January 28, 2023, 04:18:39 amHm, ok this I don't know much about. I'll try to find out more. If you have any resources I can read that would be much appreciated!I think I'm going to setup a measurement circuit like the one attached. The load resistor will be small enough that the optocoupler still sources its full short-circuit current, but large enough to be clearly visible on the scope. The opamp is just there to isolate the test circuit from the scope input and attached cable capacitance. You might well be right, but I want to be sure of how much of this delay comes from the actual optocoupler and not the specific load conditions used for the datasheet values. Oh and I neglected to show the scope measuring the input signal too, but of course it will to measure the delay between input and output turn-off.

author: moffy
author_rank: Super Contributor
timestamp: January 28, 2023, 05:02:48 am
content: The turn off time measurement has an If of 20ma, or it is discharging at a constant 20ma, not the 1M/15p.

author: matthuszagh
author_rank: Frequent Contributor
timestamp: January 28, 2023, 05:05:54 am
content: Quote from: moffy on January 28, 2023, 05:02:48 amIF is the LED forward current. The short-circuit output current is in the range of 10s of uA. 20mA doesn't mean the output is discharging at that current.

author: matthuszagh
author_rank: Frequent Contributor
timestamp: January 28, 2023, 05:14:17 am
content: Here's a scope capture of the VOM1271 turn-off into a scope input (1M||10p, plus cable capacitance). This is not the circuit I proposed above, which I have not yet had the time to setup. This isn't an ideal measurement, and the time scale is too coarse, but clearly the photodiode stops generating current pretty quickly after the input current shuts off. This looks to me like the photodiode shuts off and the 10p cap (plus cable) start discharging through the 1M resistor. That drop off is, I think, the internal JFET kicking in.

author: moffy
author_rank: Super Contributor
timestamp: January 28, 2023, 05:18:43 am
content: Quote from: matthuszagh on January 28, 2023, 05:05:54 amGood point.P.S. The gate capacitance of the MOSFETS could be many nF.

author: Marco
author_rank: Super Contributor
timestamp: January 28, 2023, 05:49:51 am
content: Your bleed resistor will need to be upwards of 220k, even 10p is a lot then.I'd look at using a 6n135 just for turn off (the Avago datasheets have some fast linear circuits you could adapt).

author: matthuszagh
author_rank: Frequent Contributor
timestamp: January 28, 2023, 02:02:27 pm
content: Quote from: matthuszagh on January 28, 2023, 05:05:54 amThe MOSFET gates will discharge through the JFET channel. When the JFET turns on, it has a channel resistance of 85ohms (at least the one I'm currently looking at). There are JFETs with even lower channel resistance, but generally you trade off gate capacitance. The bleed resistor does not need to discharge the MOSFET gates because the 2M resistor isolates them. It only needs to discharge the JFET gate, so the total relevant capacitance across the photodiode is still just 5.5p.

author: matthuszagh
author_rank: Frequent Contributor
timestamp: January 28, 2023, 02:09:28 pm
content: Quote from: Marco on January 28, 2023, 05:49:51 amNo, I don't think so. This is the point of using multiple optocouplers in parallel. I should be able to divide the bleed resistance by the number of parallel optocouplers. I was planning to use something like 30k bleed resistance and 10 optos (5 of the VO1263s). That should turn on the JFET in under 50ns (ignoring any optocoupler delay). That's still a bit slower than I'd like, so I may add some more in parallel.Quote from: Marco on January 28, 2023, 05:49:51 amUnfortunately the 6n135 requires a power supply at the output (all fast shutoff optos do, I think). That will exceed the MOSFET Vgd/Vgs ratings. I'll take a look at the Avago datasheets.

author: Zero999
author_rank: Super Contributor
timestamp: January 28, 2023, 02:36:59 pm
content: How much current?You won't be able to turn it off in 10ns. Use a gas discharge tube or transient voltage suppressor, as well as switching off the MOSFET.

author: Kleinstein
author_rank: Super Contributor
timestamp: January 28, 2023, 03:02:14 pm
content: Photodiode / PV cells are slow to get good efficiency. In PV mode I would expect a silicon photodiode to take some 5-50 µs decay time. That is just the minority carrier lifetime in relatively clean silicon. The PV couplers may already include an extra circuit to turn of a bit faster on the scope this is seen as the fast drop after some 40 µs.Normal optocouplers with a phototransistor output can be resonable fast on turn on, especially when driven hard. It can still need quite some time to discharge the capacitance and maybe drive another transistor as emitter follower. Discharging the PV coupler may take some time, they may well act like a capacitance (e.g. 50 pF range) a bit similar to the storage time or reverse recovery of a very slow didoe.The 6N135 chould get away from the supply of the PV coupler still the turn off time is more in the 500 ns range, maybe 100 ns with high drive current.

author: matthuszagh
author_rank: Frequent Contributor
timestamp: January 28, 2023, 05:30:40 pm
content: Quote from: Zero999 on January 28, 2023, 02:36:59 pmYeah I was planning to put a TVS diode downstream. I may also put a small resistor in series with the downstream circuitry (0V-biased parallel n-channel JFETs) to provide some additional current limiting, though I can't get away with that much series resistance. A suitable TVS diode will be able to handle the current, assuming I get the MOSFET to switch off in time. I do want to limit the amount of time I short out the upstream device though.

author: matthuszagh
author_rank: Frequent Contributor
timestamp: January 28, 2023, 05:33:41 pm
content: Quote from: Kleinstein on January 28, 2023, 03:02:14 pmThe 6N135 looks really helpful here. Thanks for the suggestion!

author: Doctorandus_P
author_rank: Super Contributor
timestamp: January 28, 2023, 06:22:00 pm
content: 6n135 is rated for 1Mb/s, while the 6n137 is rated for 10Mb/s, also if you go to fiber optic modules such SFP (Small Formfactor Pluggable), they readily go to 10Gb/s but those are already getting expensive. You can go quicker, but that gets more expensive exponentially still.https://en.wikipedia.org/wiki/Small_Form-factor_PluggableAnd for speed, I would probably go for a dedicated gate driver IC. That would need a separate (isolated) power supply, but that can be a simple EUR4 sil module.And are you sure this is not anhttps://en.wikipedia.org/wiki/XY_problemWhy do you need that speed in the first place? Maybe it's a better option to add some kind of inductor or filtering on the "other side" to lower di/dt.

author: matthuszagh
author_rank: Frequent Contributor
timestamp: January 28, 2023, 07:02:47 pm
content: Quote from: Doctorandus_P on January 28, 2023, 06:22:00 pmAh yeah of course, I can just create an isolated supply. That would simplify this tremendously and then I can use proper, high-speed optocouplers.The question of how much speed I actually need is a bit tough to answer. I'm trying to protect a sensitive, downstream preamplifier. The first stage is some n-channel JFETs in parallel biased to 0V. I need to be able to handle hotplugging power supplies (DUT) up to +/- 100V to the input. There are at least two parts to this challenge: protecting the JFETs and not shorting out the DUT for too long. Eventually I'll do some surge testing of these JFETs, but otherwise its difficult to get data on how much power they can actually dissipate for very short periods (ns to us durations). Anyway, generating an isolated supply should make this whole thing a lot easier. Thanks for the suggestion!I will add some additional downstream protection, including some small series resistance and a TVS diode. If the TVS can't handle the current duration, I may also add a limiter like the BAV99, we'll see. As for a series inductor, I'm not really a fan of that solution. That limits the max voltage, but prolongs the duration, which could still blow up these JFETs. I looked at ferrite beads too, but they were also too inductive at applicable frequencies.

author: Kleinstein
author_rank: Super Contributor
timestamp: January 28, 2023, 07:25:18 pm
content: The inductor will prolong the pulse, but after some time the JFETs will turn off and than the current will go down quite fast. The tricky point could be ESD, as this gives a quite high voltage (e.g. kV range) and thus still a quite fast rise in current. With a more moderate voltage the inductor can be a good solution to limit the very fast part. 1 kV and 1 mH would give 1 A/µs for the speed that the current can go up - though only until the inductor saturates, which may not be very long. It also helps with EMI.There is also the option to add a local current limit to the MOSFET part. So a series resistor between the MOSFETs and BJTs to short out the gate to source of the FETs. With some 60 ohm and some 600 mV for the transistors to turn on one would be in the 10 mA range. This may be acceptable for the amplifier and DUT side for a short time until the optocoupler part turns on (some 1-10 µs range) to reduce the current even more.Some Keithley meters like K2001, K2000 use such a protection. As a simple part they use an optocoupler with AC drive (2 LEDs) also for the clamping part. So a rather direct feedback loop with no many extra parts. The steady state courent would than be at the PV drive current divited by the CTR (around 10-100%), so in the 10-100 µA range, which is normally OK.

author: matthuszagh
author_rank: Frequent Contributor
timestamp: January 28, 2023, 08:19:21 pm
content: Quote from: Kleinstein on January 28, 2023, 07:25:18 pmI looked into a BJT current limit, but unfortunately the thermal noise from the series resistor is too much. Even an order of magnitude less than this would dominate the preamp input noise. I may accept some degradation in input noise to add a protective series resistor, but 60ohm is too much.

author: Marco
author_rank: Super Contributor
timestamp: January 28, 2023, 09:49:13 pm
content: Depending on the duty cycle you don't really need an additional power supply, the VO1263 is a power supply.

author: Kleinstein
author_rank: Super Contributor
timestamp: January 28, 2023, 10:11:54 pm
content: If 60 Ohm is way to high this means a really low noise amplifier.One may consider having something like MOVs at the input to catch ESD or other high voltage peaks well higher than the normal DUT and than rely on an inductor (possibly physically relatively large, like a flyback transformer) to slow down the rise in current, before the protection can kick in. For a short time the LED in the OK should be OK for maybe 100 mA. So with lets say 100 V and 1 mH one would have 100 A/ms and thus 1 µs to reach 100 mA. With higher current to the LED the turn on time is expected to go down. So even "normal speed" optocouplers like TLP290 can turn on in less than 500 ns.A fast rising voltage would also need extra current to discharge the gate. So one may want fast switching fets for not too high a voltage.A problem could be that the clamp voltage from the LEDs may reach 1.5 or maybe even 2 V at higher current.

author: Zero999
author_rank: Super Contributor
timestamp: January 28, 2023, 10:45:41 pm
content: Quote from: matthuszagh on January 28, 2023, 05:30:40 pmYou still haven't said how many amps it needs to pass.How about an SCR in series with a depletion mode MOSFET? The SCR and MOSFET can be coupled to the drivers via pulse transformers. The SCR will need a short pulse to turn on and the MOSFET a pulse to turn it off for long enough for the SCR to latch off.

author: matthuszagh
author_rank: Frequent Contributor
timestamp: January 29, 2023, 12:38:22 am
content: Quote from: Zero999 on January 28, 2023, 10:45:41 pmWhen the input blocking capacitor is charged and the bidirectional MOSFET switch is closed the current will be miniscule. nV or uV into the high JFET input impedance. When a power supply is turned on or off (or hotplugged), the current could be quite large while the blocking capacitor charges or discharges. I don't know exactly how much, but this should be able to handle any level of current while the 10u blocking capacitor charges. When the MOSFET switch opens, the capacitor will (dis)charge much more slowly through a 100k load resistor to minimize loading on the DUT. The low MOSFET Rds(on) is more about limiting thermal noise contribution from the switch (though I could probably go up to about 1ohm total and that would be fine). It's also a byproduct of using high power / voltage MOSFETs.Quote from: Zero999 on January 28, 2023, 10:45:41 pmHm, I'm not sure I totally follow. Is this an SCR across the line or in series with it? Either way, the input can positive or negative and the switch needs to conduct both ways. Also, I don't want anything shorting out the input for any appreciable amount of time.

author: matthuszagh
author_rank: Frequent Contributor
timestamp: January 29, 2023, 12:56:57 am
content: Quote from: Kleinstein on January 28, 2023, 10:11:54 pmI was planning on using an SP00R6 or similar as high speed protection until the switch kicks in. If the switch is fast enough, I think this combination should be enough to avoid damaging the preamp. Though this is purely speculation still and I plan to test it. I'll reconsider inductors too, to limit current into the preamp.But, with an isolated supply and the isolators that permits, it should be possible to kick in the switch very quickly and I wonder how much additional protection I'll need (though I'll definitely still keep the TVS). For example, I'm looking at the ISO7710. That gets me back into the 10ns range, though a bit more when I factor in the comparators, charging and discharging the MOSFET gates, etc. The TVS clamps around 2V. It seems unlikely that a <50ns pulse at this voltage would damage a JFET.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: January 29, 2023, 01:36:53 am
content: Quote from: matthuszagh on January 28, 2023, 07:02:47 pmFinally some information!Please show the system connection (how, and what, they are hotplugged to, in whatever combinations are relevant) and the amplifier schematics.Tim

author: matthuszagh
author_rank: Frequent Contributor
timestamp: January 29, 2023, 02:19:14 am
content: Quote from: T3sl4co1l on January 29, 2023, 01:36:53 amI've attached a block diagram. The input will not always be a high voltage power supply, but it can be and that's the worst case in terms of damage protection. The purpose of this preamp is to amplify very weak low-frequency noise (<1Hz to 1MHz) in the presence of a potentially large DC offset.I've shown an LED+photodiode in the diagram but I'll actually use something like the ISO7710 and an isolated supply.The switched in load resistor is 100k, as already mentioned. The exact value isn't super critical, but it needs to be large enough not to load down low current, high voltage power supplies.Both blocking caps are 10u high-voltage WIMA MKP 4.I have not finished designing the preamp, but for the sake of damage protection the input stage will look like that. The series resistor may or may not be used, but if used it will be very low valued (a few ohms, max).

author: T3sl4co1l
author_rank: Super Contributor
timestamp: January 29, 2023, 02:36:22 am
content: Zero source resistance?No filtering?What about a startup delay on hot-plugging to do the discharge automatically?Tim

author: matthuszagh
author_rank: Frequent Contributor
timestamp: January 29, 2023, 02:46:24 am
content: Quote from: T3sl4co1l on January 29, 2023, 02:36:22 amFrom the power supply? Yes, it's a normal (real) power supply. But there's no stipulation on what power supply can be plugged in other than the peak value stay within the voltage range. Typically it will have bypass caps at the output and all power supplies have some source resistance... But, since any supply in that voltage range can be plugged in I don't want to rely on source resistance as part of the protection circuit.Quote from: T3sl4co1l on January 29, 2023, 02:36:22 amI'm not sure how this would work. Can you elaborate? This is not a fixed circuit. This is not always plugged into a power supply. This is a protection circuit for a preamplifier that can be used for many different things as long as the application stays within the voltage limits. One example is measuring the noise of a power supply. Another is measuring the noise of voltage references. Hot plugging and unplugging are supported. So is turning on the supply after it's plugged in. I don't know how a startup delay would address all of these. Maybe this modified diagram makes it more clear.

author: matthuszagh
author_rank: Frequent Contributor
timestamp: January 29, 2023, 03:21:01 am
content: Quote from: matthuszagh on January 29, 2023, 02:46:24 amOh, I think I misunderstood your question. I now think you were asking whether I will add any series resistance or filtering to the input. Probably not. I'm not eager to increase the equivalent input noise of the preamp unless I need to. As it is, even a few ohms will surpass the equivalent input noise. I also don't want to filter the noise of the DUT since that's precisely what I'm trying to measure.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: January 29, 2023, 05:04:24 am
content: Quote from: matthuszagh on January 29, 2023, 02:46:24 amBetween the supply and preamp I mean.1MHz bandwidth seems to imply you can put a filter in there, and not need to clamp things for at least some hundreds of nanoseconds. And the filter will necessarily introduce impedance, making the inrush current well-defined. But it also needs to be terminated and a zero-impedance source doesn't exactly work with that (termination can be on the amp side, but the source impedance still preferably needs equalization -- a constant resistance filter is best).Also if the source is always zero ohms, a JFET doesn't seem the lowest noise option; a nice thick BJT should be able to do better than a JFET array. Lowest noise may not be the top priority however, so that's fine.QuoteWell your first claim was hotplugging so if that's the only thing you're worried about, that could be worked around in many ways. You didn't say it needed to do anything else, so we're free to assume anything and everything until said otherwise...If it's not the only thing, maybe you could make a hotpluggable interface module and only use it to test supplies with that? More steps, but makes things more flexible / foolproof.No idea; we're not telepathic here, after all.Tim

author: matthuszagh
author_rank: Frequent Contributor
timestamp: January 29, 2023, 01:45:56 pm
content: Quote from: T3sl4co1l on January 29, 2023, 05:04:24 amYeah good point. The preamp should achieve a noise floor of about 300pV/rtHz and needs to measure noise down to about 2nV/rtHz for power supplies and lower for e.g., transistor noise measurements. That doesn't leave a ton of margin for additional noise sources as I'd like to stay far enough below the DUT noise to maintain accuracy. There are other tricks I'll emply here; mainly, two preamps and cross correlation but that imposes its own challenges and has its own drawbacks so I'd like to get the preamp noise as low as possible first.The protection circuit will be able to be switched out for certain measurements such as for 0-DC sources where it's really not needed. Such a protection circuit is probably also overkill for biasing arrangements used in transistor noise measurements. Anyway, maybe there's some headroom for adding a filter with a small series resistance if this protection circuit will be exclusively used for power supply and voltage reference measurements. I'll investigate; thanks for the idea.Quote from: T3sl4co1l on January 29, 2023, 05:04:24 amThis isn't necessarily true. While the DUT might be low impedance, the blocking cap at the lower frequency range of operation isn't. I specified <1Hz as the cutoff, which I appreciate is a bit vague (this should go down to at least 100mHz, preferably all the way down to 1mHz). But, using 1Hz as the frequency that 10u blocking cap looks like a 16k impedance. The BJT current noise through that cap will swamp the equivalent input noise. JFETs do much better in this regard. And, while you're right that JFETs don't do quite as well as BJTs for true low impedance sources, they can do remarkably well and are really not all that far off the performance of BJTs from what I've seen. I could, of course, increase the blocking capacitance, but this quickly becomes impractical.Quote from: T3sl4co1l on January 29, 2023, 05:04:24 amI apologize for being vague - that really wasn't my goal. And I do appreciate the suggestions I've received.As for hotplugging modules, I guess I just don't know enough about those so I'll spend some more time looking into it. The commercial hotplugging modules I've seen have all been for a single polarity supply and for a limited voltage range (makes sense, given the typical application). I suppose the principles underlying these modules could be used to make a custom one. But, looking at the ADM1177 for instance, doesn't make me think that my current topology is wrong. They use a single external MOSFET whereas I need a bidirectional switch. I monitor the voltage and trigger the shutoff with high-speed comparators, whereas they use a sense resistor and ADC. I like my approach better, since it's simpler, should be faster and doesn't require adding a sense resistor.

author: peter-h
author_rank: Super Contributor
timestamp: January 29, 2023, 01:57:02 pm
content: A photodiodeoptoisolator will be fast. A phototransistorisolator has a much higher gain but is much slower - largely due to Miller Effect (which is why the better optos have the base coming out on a pin so you can decouple it.Hot plugging needs to be done carefully because even a 100nF capacitor will short out the supply rail for some nanoseconds or more and crash everything upstream

author: Zero999
author_rank: Super Contributor
timestamp: January 29, 2023, 05:47:47 pm
content: Quote from: matthuszagh on January 29, 2023, 12:38:22 amWhat input blocking capacitor?What are you protecting?My post assumed you're protecting a circuit against over-voltage on the DC supply rail, but now it appears you're talking about an AC signal.Post a full schematic.

author: matthuszagh
author_rank: Frequent Contributor
timestamp: January 29, 2023, 06:10:04 pm
content: Quote from: Zero999 on January 29, 2023, 05:47:47 pmThis information has already been provided. Please see my recent explanations to T3sl4co1l and the associated block diagram. I haven't posted a full schematic yet since I'm changing the isolated MOSFET drive portion.This isnota DC supply rail over-voltage protection circuit. It's a protection circuit for the input of a very low noise preamp that can be exposed to inputs with DC voltage anywhere in the range of -100 to +100V. The preamp 1st stage is a 0-bias n-channel JFET in a common source configuration (or a variation on that, but for damage protection that's accurate).

author: matthuszagh
author_rank: Frequent Contributor
timestamp: January 29, 2023, 06:21:46 pm
content: Quote from: peter-h on January 29, 2023, 01:57:02 pmNoted, thanks Peter. I'm currently looking at the ISO7710, which is not actually an opto - it uses isolation capacitors instead. But it's quite fast: around 10ns.Quote from: peter-h on January 29, 2023, 01:57:02 pmIs a 10-50ns short really that bad? I figured the bypass caps would be ok supplying this. But if that's really very bad for the power supply attached to the input I'll need to rethink things.

author: matthuszagh
author_rank: Frequent Contributor
timestamp: January 29, 2023, 08:48:30 pm
content: There's an additional difficulty with the isolated power supply in that these switching power supplies are really noisy. For example, I've been looking at the DCP020505 to power the output side of the ISO7710. I'm worried about that making its way into the signal path.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: January 29, 2023, 11:11:06 pm
content: So what's the isolation about then? It's all common ground...Tim

author: shapirus
author_rank: Super Contributor
timestamp: January 29, 2023, 11:26:04 pm
content: Quote from: matthuszagh on January 29, 2023, 08:48:30 pmWhat is the supply current of the (sub)circuit that the isolated power supply will be powering? If it's not too high, it might be quite possible to utilize heavy RC filtering of its output with a large value resistor and still get sufficient voltage after the filter to power the circuit. Or, use a 5V->12V isolated converter, put an RC filter (multi-stage if necessary) at its output, then use a 5V linear regulator to stabilize voltage. Use proper board layout to minimize possible noise coupling.

author: matthuszagh
author_rank: Frequent Contributor
timestamp: January 29, 2023, 11:50:31 pm
content: Quote from: T3sl4co1l on January 29, 2023, 11:11:06 pmThe drain of the first MOSFET in the bidirectional switch (Q5 in the schematic attached, or first post) can go to +/- 100V. If I don't isolate the gate drive circuitry, the drain-gate voltage can greatly exceed the rating (typically 20 or 30V).It seems I've caused a lot of confusion by not showing the full protection circuit schematics. I've attached a preliminary version of this schematic. This still uses the old optocoupler method, whereas I'm now considering an ISO7710 and DCP02 isolated supply. Remember, too, that "out" is connected to the preamp, which looks like an n-channel JFET in a common source configuration with 0-bias (low clamp voltage TVS diode across it too). "in" is connected to whatever DUT is being tested. That may or may not be a high (or very negative) power supply, but that's the worst case from a damage protection scenario. I'm sorry for the confusion I've caused and I appreciate all the assistance I've gotten.

author: matthuszagh
author_rank: Frequent Contributor
timestamp: January 30, 2023, 12:16:53 am
content: Quote from: shapirus on January 29, 2023, 11:26:04 pmGood idea! The ISO7710 specifies a typical 1.7 mA current draw, so very minimal and there's definitely headroom for RC filtering. I'd been planning to use a high-PSRR linear regulator (probably an LT3042), but adding RC filtering seems like an easy way to get much more noise rejection.To further reduce noise, I was going to use the pi filter mentioned here:https://e2e.ti.com/support/power-management-group/power-management/f/power-management-forum/143862/dcp020515d-output-cap-to-reduce-output-voltage-rippleand the Y-cap across the negative supplies. If I go with an LT3042, the datasheet is pretty particular about proximity to the input switcher. Basically, they say to keep it about an inch or a little more away and get rid of the input bypass cap in order to maximize PSRR. This may take a little trial and error to get right. I was also planning to put these as far away from everything else as possible on the PCB and keep other conductors away to minimize capacitive coupling.I'm a little worried about noise being injected back into the input of the switcher and making its way into more sensitive circuitry that way. I don't know how much of a concern that normally is. I suppose if that is a problem I could add a linear regulator right at the input too. Unfortunately the DCP02 has a minimum load current and so RC filtering the input might be a little tricker.I'd love to use the original optocoupler (VO1263) as the power supply, but unfortunately I'd need way too many to provide enough current.

author: matthuszagh
author_rank: Frequent Contributor
timestamp: January 30, 2023, 01:20:09 am
content: Looking at this more, it seems there are better isolator choices. In particular, dedicated isolated gate drivers. These can source/sink enough current on their own and I could get rid of the P-channel JFET fast shutoff. And they still have low quiescent currents so I can still use RC filtering to suppress noise from the isolated switcher.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: January 30, 2023, 03:58:25 am
content: You're still locking yourself into one limited solution... What's wrong with:- Series limiting resistor e.g. 10R into bidirectional TVS; TVS capacitance can be used as part of filter capacitance to set bandwidth.- Current limiting depletion MOS; has low resistance at signal levels, zeners or clamp diodes protect amp.- Plain old polarity protection circuit like you'd use on a battery, chain two complementary ones and bias them to +V/-V so they pass from (-V + Vgs(th)) to (+V - Vgs(th)) and shut off outside there.- Classic, biased diode-bridge current limiter; bias can be from HV BJTs between supply rails.Tim

author: Siwastaja
author_rank: Super Contributor
timestamp: January 30, 2023, 09:25:31 am
content: Quote from: matthuszagh on January 28, 2023, 05:30:40 pmYou realize a TVS has a lot of capacitance which continues supplying the load?For the idea of turning a load off in nanoseconds, it's absolutely vital to know what the load is. You may need to modify the load, for example short any capacitors in it. Large short circuit currents in nanoseconds then mean high dI/dt, which means problems with inductance.Unless it's top secret, tell us what you are actually doing.

author: inse
author_rank: Super Contributor
timestamp: January 30, 2023, 10:08:00 am
content: What about a dedicated optocoupler to turn off the power MOSes fast?

author: shapirus
author_rank: Super Contributor
timestamp: January 30, 2023, 11:35:13 am
content: Quote from: matthuszagh on January 30, 2023, 12:16:53 amCheckhttps://www.analog.com/media/en/training-seminars/tutorials/MT-031.pdf.You may find some applicable ideas on the isolation of noisy components from sensitive ones there.

author: matthuszagh
author_rank: Frequent Contributor
timestamp: January 30, 2023, 01:36:48 pm
content: Quote from: Siwastaja on January 30, 2023, 09:25:31 amNo, it doesn't. I'm planning to use a SP00R6, which has 0.2p capacitance. Just because some TVS diodes have a lot of capacitance does not mean that all do.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: January 30, 2023, 03:44:23 pm
content: But that's rated for 2A (8/20us) and 12kV (ESD, contact). Can you limit currents accordingly?With 50 ohms in series, you'd pretty much be done. Not quite, as 10uF * 50R = 500us not 20us, but a bigger diode could be used trivially without affecting bandwidth.I don't get the series resistance at the gate, when it can do much more upstream of the transient protection. This isn't rocket science, it's basically solved passively this way.Tim

author: matthuszagh
author_rank: Frequent Contributor
timestamp: January 30, 2023, 06:10:05 pm
content: Quote from: T3sl4co1l on January 30, 2023, 03:44:23 pmI think I could with the protection scheme outlined above. I don't really want anything where I'm relying on the TVS too much because that means I'm over-exposing the DUT to a short condition. But, yes, that TVS will only work if I can limit current enough. If I can't, I'll have to choose something else.Quote from: T3sl4co1l on January 30, 2023, 03:44:23 pmI'm not at all against resistance upstream instead of at the gate. But 50ohms (as stated) is way too much. 10ohms is also too much. 5 ohms would be acceptable but that's pretty much at the upper limit.But, even if we could use 50R, it's not obvious to me that would totally protect the JFET. Even the TVS I proposed, which has a very low clamp voltage, still clamps above where the JFET gate conducts. Clamp diodes, too, will leave a lot of current for the JFET gate. And, these solutions create a much longer transient than my solution, which is about 25,000 times faster.Your language makes it sound like this whole thing is glaringly obvious, but that solution (a) fails to meet the noise criteria, (b) probably fails to meet the criteria for minimum disturbance to the DUT, and (c) might also fail the damage protection criteria. If it passes the damage protection criteria it's certainly not obvious to me and I'd like to see something that convinces me that it really does.

author: matthuszagh
author_rank: Frequent Contributor
timestamp: January 30, 2023, 06:15:39 pm
content: Quote from: T3sl4co1l on January 30, 2023, 03:58:25 amLimited in what sense? Sure, there are challenges associated with it, like mitigating noise coupling from the switcher. But, limited suggests to me that it fails to handle certain input conditions, which I don't think it does. Please clarify. Still, I'm more than happy to explore alternatives especially if they're simpler than this solution, which is getting quite complicated.Quote from: T3sl4co1l on January 30, 2023, 03:58:25 am10R adds more thermal noise than I'd like. More importantly, though, this exposes the DUT to a prolonged short and provides significantly less protection to the JFET input, which will now be exposed to elevated voltages and lots of current input for 100s of us. I would expect a JFET gate to be able to handle elevated currents for 10s of ns, but 100s of us seems to be pushing it (again, I will damage test the JFETs to actually make sure they can handle the 10s of ns).Quote from: T3sl4co1l on January 30, 2023, 03:58:25 amThis is a really neat technique that I wasn't aware of. Thanks for pointing it out!First addressing zeners and clamps: the JFET gate-source is just a pn junction. This doesn't leave a lot of room for devices to kick in before the JFET gate conducts appreciably. Zeners have soft knees, so anything that steals enough current away from the JFET will likely conduct too much at levels where it shouldn't. This greatly lowers the input impedance of the JFET amp and imposes unrealistic demands on the capacitance of the input blocking capacitor. Clamp diodes have similar issues. They provide some protection, but they're not adequate on their own - they only steal some of the current away from the JFET and so the input still needs to be shutoff. Schottky's do a much better job of stealing current from the JFET gate but lower the input impedance too much.As for the depletion MOS, I'm still exploring this, but I think the fundamental difficulty is that the steady state gate current needs to be kept quite low. I've heard 10mA per JFET thrown around. I don't know if that's accurate - I still need to test this, but let's go with that for now. I'll be paralleling these JFETs, so assuming the current is equally distributed between them that eases things a bit. Let's say 16 JFETs. Total steady state current then needs to be limited to 160mA. The boundary between conduction and non-conduction for depletion MOSFETs isn't very sharp. So, to adequately limit current will require enough series resistance between the MOSFET channel and any external resistors. I've looked at the datasheets for a few depletion MOSFETs and the total series resistance for those would be too high.I think using depletion MOSFETs could still be very useful as additional protection by imposing a higher current limit (several amps). This should limit the max current during large current transient events (eg plugging or unplugging a power supply).Quote from: T3sl4co1l on January 30, 2023, 03:58:25 amI'm intrigued by this idea, and hadn't thought of it. There are challenges with it, though, relating to the fact that MOSFETs don't have a sharp boundary between conduction and non-conduction. Additionally, threshold voltage varies from unit to unit and is temperature-dependent. The JFET forward drop is also, of course, temp dependent, but not necessarily in the same way. This makes biasing quite delicate, which I don't like. I'm not convinced it can't work yet, but it's tricky. I measured Rds vs vgs for a SiHD240N60E and I'm not sure that's sharp enough. That's just one FET though, I can look at others, perhaps with lower thresholds.Code:| vgs | Rds ||-----+------|| 4.0 | 2.2k || 4.1 | 1.0k || 4.2 | 959 || 4.3 | 472 || 4.4 | 149 || 4.5 | 66.1 || 4.6 | 32.7 || 4.7 | 23.9 || 4.8 | 10.5 || 4.9 | 5.59 || 5.0 | 6.20 || 5.1 | 2.05 || 5.2 | 1.38 || 5.3 | 0.99 || 5.4 | 0.75 || 5.5 | 0.62 |Quote from: T3sl4co1l on January 30, 2023, 03:58:25 amI'm not familiar with this. Can you explain further or point me to a resource that explains it?There's an additional consideration, which is that I still need to switch in a resistor during an over or undervoltage condition. The lowpass frequency cutoff is low enough that the capacitor would take too long to charge or discharge. This resistor can't always be in because it lowers the input impedance too much. This means that I still need something to switch this in. The requirements on this aren't too tough (and a simple window comparator would do), but I guess my point is that it still needs to be here, so none of the above solutions are sufficient in and of themselves.

author: matthuszagh
author_rank: Frequent Contributor
timestamp: January 30, 2023, 06:47:50 pm
content: Quote from: inse on January 30, 2023, 10:08:00 amThis is my current solution, or at least very similar to it (see messages above). Rather than optocouplers though I plan to use an isolated gate driver (eg UCC21530). The main difficulty with this solution, in my opinion, is the required isolated power supply, which is horrendously noisy. I think it will be possible to resolve this in an adequate way, but that remains to be seen.Optocouplers that don't require an isolated supply are what I investigated initially. A lot of people pointed out that won't be fast enough. I'll still measure how fast I can shut it off by placing multiple in parallel and adding a bleed resistor (see first couple messages), but I'm willing to accept this is unlikely to work in the way I originally hoped.

author: inse
author_rank: Super Contributor
timestamp: January 30, 2023, 07:20:09 pm
content: I thought it like this: one photovoltaic isolators as now for turning the MOSFETs on and a fast optocoupler to short the gate to source.Only needs a inverter on the primary side

author: matthuszagh
author_rank: Frequent Contributor
timestamp: January 30, 2023, 07:48:56 pm
content: Quote from: inse on January 30, 2023, 07:20:09 pmCan you recommend a fast optocoupler to short the gate to source? I haven't found anything suitable.

author: shapirus
author_rank: Super Contributor
timestamp: January 30, 2023, 07:58:33 pm
content: Quote from: matthuszagh on January 30, 2023, 07:48:56 pmWhen I looked for something to isolate an SPI communication path, the best I found was 6N137: approx. 100 ns, give or take, propagation plus fall time.I too wonder if faster optocouplers exist, but I doubt it. There's probably a good reason why faster isolators are based on magnetic or capacitive coupling principles rather than optical.p.s. at this point it might be worth it to do some experiments and see what you can achieve in hardware. Sometimes one can find out that something previously rejected can be quite suitable after all.

author: matthuszagh
author_rank: Frequent Contributor
timestamp: January 30, 2023, 08:12:24 pm
content: Quote from: inse on January 30, 2023, 07:20:09 pmActually, I'm not totally sure I follow what you're envisioning here. All fast optos I've seen require a supply on the output side. Are you suggesting that I use the photodiode output (from the turn-on opto) to power the fast opto, which, when triggered, shorts that to ground? Does this work?

author: matthuszagh
author_rank: Frequent Contributor
timestamp: January 30, 2023, 08:16:57 pm
content: Quote from: shapirus on January 30, 2023, 07:58:33 pmInteresting, that delay for the 6N137 is a bit slower than I would have thought from the typical numbers on the datasheet. Did you have a fair amount of load capacitance in your case?As for doing the actual experiments, yeah I totally agree. I'm still waiting on a few things before I can but then I will. Not that I'm at all hopeful, but I'm still curious how fast I can shut off the VO1263 by throwing a bunch in parellel and adding a bleed resistor.

author: shapirus
author_rank: Super Contributor
timestamp: January 30, 2023, 08:30:16 pm
content: Quote from: matthuszagh on January 30, 2023, 08:16:57 pmI really have no idea how much of a capacitance there was. The optocouplers were driving GPIO pins of a single-board computer in one direction and CS/MOSI/CLK pins of an ADC chip (MCP3204) in the other. It worked at 2 MHz, but I abandoned this idea because of the extra power consumption of the LEDs and the pull-up resistors on the output side, not to mention the number of discrete components necessary. All of this was replaced by a single ADuM2401 chip designed specifically for this job, but your case appears to be different from isolating digital lines.(also, the 100 ns that I mentioned are directly from the datasheet, only I used max, not typical -- I did not actually measure it.)Regarding the use of an optocoupler to flush the gate charge, I think something like the following was assumed:LED turns on -> transistor opens -> gate capacitor discharges.(this is actually an interesting idea, I never tried it, but I did experience the problem of speeding up the turn-off of a transistor, let us know how it goes if you decide to try it.)

author: inse
author_rank: Super Contributor
timestamp: January 30, 2023, 10:02:44 pm
content: Yeah sorry, maybe the idea wasn’t thought to the end..I didn’t expect that there were no optocouplers with faster turn-on time available without secondary supply.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: January 31, 2023, 02:29:11 am
content: Quote from: matthuszagh on January 30, 2023, 06:15:39 pmMy first objection is more semantic than practical: it's suggestive of "if-then" procedural thinking, which doesn't map well onto analog circuits. Embrace the continuum! If voltage goes outside of bounds, don't switch it off (what if switching it off causes voltage to rise further? or reverse so it oscillates?), just increase current, diverting it away from your load.The practical consequences come when considering marginal cases. What if the input is extremely noisy (some volts of AC)? Well, on a practical note, your amplifier is useless (driven to clipping), so maybe it doesn't matter. But randomly activating (and at a high rate) that switch logic won't do any favors for signal distortion, or time dependency of the response.A simple clipping filter has the benefit that it's not time dependent. It's there only when needed and goes away seamlessly. Distortion increases in an orderly fashion as signal level approaches the threshold. The onset of clipping is easily recognized in the response.The linearized version might be interesting to ponder, but isn't going to be easy to pull off (you need to synchronize the transfer curves of two switches, potentially while level shifting). That is, suppose the switches were adjustable resistors instead; or some kind of V(I) characteristic curve anyway, since we're using transistors here.Notice also you're making a negative resistance input characteristic: as voltage rises, current falls . This might not be well-behaved say with a reactive source. Or maybe it's balanced by the shunt switch, but again, it needs to track properly. And it needs to have the same response rate (control theory is involved too!), lest it have an impedance peak that causes ringing or oscillation (between one or the other "switch" and the source, or any other combination thereof).Maybe those are equally good reasons to reject the continuum, and stick to a logical method instead. I don't know.QuoteOkay but you're already hot-plugging the thing, what did you expect was going to......nevermind.(Can't you just plug the thing in,thenturn it on? Normal startup transient / soft-start cycle applies?)QuoteAlso can always add source degeneration (reduces gain without increasing noise), or bypass the gate resistor with a capacitor (another old school oscilloscope protection feature, does have a long recovery tail though). (Again, could use depletion MOS for faster recovery while limiting current, I suppose.)QuoteWell, what's it going to be biased at? Surely not Idss (Vgs=0) giving minimal dynamic range. You're already planning on basically a back-to-back diode (SP00R6) (actually I wonder if it's anti-series back diodes* or something, given they show a series diagram internally?) so biasing down a volt or two should be trivial.*A variant of tunnel diodes, with especially low reverse voltage. Oh but wait, normal forward voltage. Or hm, could tunnel diodes be fabricated with just enough negative resistance to compensate for the forward junction? No that wouldn't make sense, there'd be a shelf in the response. Hmm, maybe they are just punch-through diodes, just tuned for a very low voltage? Dunno. Anyway, laughable that they show anti-series zeners as the diagram, they are absolutely anything but.Also mind, the gate current flows into the channel, which has significant resistance. It's no rectifier. Actually it's notoriously poor: consider the pico-ampere diodes, which as I understand it are basically diode-strapped JFETs ala 2N4117. (Oh, in fact that's exactly their description(!):https://www.interfet.com/pad/) Your parallel array will be "stiffer" of course, but Rds(on) will be no less a factor in its response.So I don't see a need to be picky about a volt here or there.QuoteI would expect ratings to be driven by thermal and electromigration limits, or something to that effect. 10mA is the DC figure. I would expect pulsed currents of, oh I don't know, 10x for 10s of µs at least, being acceptable. Compare LED pulse ratings, those can have quite small junctions for example. I mean, those are in entirely different material too, so take this for the salt-grained hand-wave it is -- but perhaps your testing will find this out.Mind to do a good number of cycles, like, thousands, or millions even (should be easy enough to set it going in the background and accumulate that kind of time?), on a reasonable range (preferably a dozen parts or more, from various production dates and manufacturers if possible?), and check the small-signal parameters afterwards. Then re-run at more aggressive settings (higher Ipk, pulse width) until failures are observed.I suppose noise would be the most critical and least predictable variable here? I don't have a feel for what it should do. Want to say it depends on surface states and spooky stuff like that, which shouldn't be affected by forward bias? Leakage feels more likely to be affected, to me. Or Vbr, or gm or Vgs(off), those are pretty sensitive and something like spot heating or ionic motion (pushed by high current density?) might do it.Or just outright failure first (short/open), which seems most likely a fatigue or electromigration sort of mode I think.Will be interesting to know, independently; post that in a new thread even, if you could.QuoteWell so what... think of it this way: you're going from ~100V to 2-4V. That's a hell of an improvement, eh?Maybe a multi-stage process is needed; but you're already set on that [series and shunt switches] so it's more a matter of how than whatQuoteSure,Any current sink/source will do, this is just shown with R+D bias for simplicity. Can get closer to the rails with current mirrors (but watch for Early effect!!), or use even just bias resistors if you don't need the currents to cancel out over signal voltages (which is to say, loading the input) (which, really, with a sub-volt range, that should be fine regardless).Or somewhat higher supplies and generous emitter degeneration, if you need low current noise. This is the subtraction of current sources after all, their noise will be uncorrelated.QuoteOkay but what do you need a high impedance for when you're measuring fractional ohm sources?.....nevermind again.Oh, or is that so you can save on C in the first place? But that increases noise, at the low end. Because it's the -- a capacitance is lossless, but the implicit resistance that is always present alongside it, making that filter cutoff -- that has noise, and the noise is effectively filtered by it and there you go. It's identical* whether you have 1 ohm and 1mF or 1kohm and 1uF. The only difference is the transition region between those impedances and frequencies, and whatever the high-frequency characteristic should be (apparently very low impedance, to keep voltage noise down?).*In noise power, but voltage scales with impedance of course, so it's better to use large values here.Tim

author: matthuszagh
author_rank: Frequent Contributor
timestamp: February 01, 2023, 03:22:40 am
content: Thanks for the detailed response Tim - I appreciate you taking the time to explain all of this. And I will share the JFET stress testing results when I perform them.I spent a little time simulating the depletion MOSFET current limiter solution (schematic attached), and I'm starting to like it (thanks Tim!). The BSP149 MOS has 1.7ohm typical (3.5 max) at vgs=0. That's low enough in terms of noise performance and should limit the current to 400mA. Most of that can be diverted through a BAV99 limiter. I'll probably use 2 to limit the current through the JFET gate even further. That decreases the input impedance a bit and adds some capacitance, but it's still acceptable. The ferrite bead dampens the initial current peak. This configuration leaves 80mA current to flow through the JFET gates. With 16 in parallel, I think that should be low enough. I don't know how evenly that current will be distributed between the JFETs, but if it's a problem I can probably add source resistors before each JFET gate (assuming the JFET resistance at the limited voltage is sufficiently low). Another technique I could use if the current is too high is to add a small resistor to the MOS source terminals. Based on the simulation, just 1ohm at each source drops the total current almost in half.This solution draws a bit more current from the load and for a little longer, but I think that's probably ok. It's also not subject to the initial current surge, even if that was only for 10s of ns.Obviously, this solution has a huge advantage in that it's much simpler. There are only a few components and they're cheap. And no power supplies are needed. This should also work up to nearly +/- 200 V, which is the MOSFET breakdown voltage, though of course that needs testing.