JFET as replacement for analog switch like DG444 and question about RDS(on)flat - Page 1
Forum section: electronics / beginners

author: julian1
author_rank: Frequent Contributor
timestamp: March 14, 2017, 08:30:42 pm
content: I am interested in using an analog switch to switch various shunt resistors, for current measurement ranges.An analog switch like the DG444 has a typical RDS(on) resistance of around 50ohms.However, it also has a variation of around 9ohms over the range of +VE to -VE, which will influence current measurement when used in series with a precision resistor and limit choice to higher-valued sense resistors. And 9 ohms may well be a best case, for Maximhttps://datasheets.maximintegrated.com/en/ds/DG444-DG445.pdf.Neither Vishay or Intersil quote the parameter.There are other analog switches with lower RDS(on) that also have flatter responses - eg. DG1411, but they are less common, have higher leakage currents (1-2nA versus 80pA), and come in more difficult to use packages (tssop versus soic).As an alternative I am wondering if a JFET might be used as a bidirectional analog switch?. Something like a J111http://www.mouser.com/ds/2/149/J111-889320.pdfThe J111 specs suggest 30ohm with 0V VGS, and leakage is only 1nA when VGS is -10V. I have no experience with JFETs but understand they can be used bidirectionally. My question is - how RDS(on) of the JFET would be characterized if the D or S was varied over a range of say +10V to -10V?

author: Kleinstein
author_rank: Super Contributor
timestamp: March 14, 2017, 08:43:05 pm
content: To have a JFET turned on, one usually sets the gate - source voltage to zero, so one often needs a kind of sensing amplifier. After that, there will be essentially no variation of the on resistance with the DC voltage of drain and source. It is more in the off mode, that the negative gate voltage is changing and thus the leakage can depend on the voltages.With the CMOS switched much of the R_on variations happen more close to the supply rails. At the typical low voltage used at shunts this is much smaller. One might even be able to use MOSFETs, if voltages are below about 200 mV or with two back to back MOSFETs in series.However there still is a variation of the on resistance with temperature. So usually one designs the a current sensing circuit to have separate paths / switched for current and voltage sensing. So even if the switch resistance is changing it would not make a big difference.Dave has in interesting video one current sensing on his new DMM on how to switch shunts and still get a low burden voltage.

author: Benta
author_rank: Super Contributor
timestamp: March 14, 2017, 08:54:45 pm
content: QuoteThis is totally misunderstood. A JFET is a voltage controlled current source, and with the gate voltage set to 0 it will act as a current regulator, which in an ideal case would have infinite "on resistance".

author: Kleinstein
author_rank: Super Contributor
timestamp: March 14, 2017, 08:59:54 pm
content: Quote from: Benta on March 14, 2017, 08:54:45 pmIn the low drain to source voltage (e.g. less than 100 mV) range a JFET acts like a resistor controlled by the gate voltage. It is only at higher voltages the the current can not increase further and thus the JFET will act as a current limiter.

author: BrianHG
author_rank: Super Contributor
timestamp: March 14, 2017, 09:04:49 pm
content: Use an optomosfet with AC output support. Some exist with on impedance less than 0.06 ohm and are good to +/-6 amps.(This is 3 orders of magnitude compared to any of your solutions, this will give you the best accuracy unless you go to power mosfets with a bunch of gate driving support voltages and circuitry.)Check this part number > CPC1907BYou wont find this kind of isolation from your control supply voltage with discrete components with ZERO leakage current.The down side to the part I listed is that when off, it will leak 1ua. But you can always use a weaker optomosfet with a higher on resistance, but, a lower leakage current.

author: julian1
author_rank: Frequent Contributor
timestamp: March 14, 2017, 09:12:49 pm
content: To clarify a couple of points. This is for higher valued current sense resistors eg. > 1k.For lower valued shunts (higher-current) two back-to-back series n-channel mosfets or relays would be desirable for their current handling ability.I am interested in bidirectional current sense - eg. a positive or negative drop across the shunt and switch with reference to GND, and with a +-10V D to S sense range.

author: Benta
author_rank: Super Contributor
timestamp: March 14, 2017, 09:23:44 pm
content: We're talking switching in different shunt resistors for current measurement, right?You can eliminate the errors complete by using two switches per resistor.On switch for connecting the relevant shunt resistor to the input socket, and one switch connecting the shunt resistor to your sense amplifier.The first switch is outside the measuring loop (like a test-lead resistance).The second will (hopefully) look into a high impedance amplifier, so on-resistance is negligible.

author: BrianHG
author_rank: Super Contributor
timestamp: March 14, 2017, 09:24:01 pm
content: These are a little pricey as switches, but they go to 2.5 ohms when on:https://www.maximintegrated.com/en/products/analog/analog-switches-multiplexers/MAX4669.htmlRun the 2 in parallel for 1.25ohm on, however, you wont get the blind supply isolation/impedance at any A to B impedance of the optomosfets who's gates are optically driven.

author: julian1
author_rank: Frequent Contributor
timestamp: March 14, 2017, 09:35:17 pm
content: Quote from: Benta on March 14, 2017, 09:23:44 pmYes, I have considered that and am still considering it. The trade-off is extra circuit complexity and I am aware of schematics for range switching that do it with one switch - using different strategies depending on the specific range and leakage control needed.But, in a similar vein - I wonder about connecting two analog switches- in series on just one side of the shunt. I think analog swiches like the dg444 use a p-channel and n-channel in parallel, which is what gives their non flat response when swept between +VE and -VE. But it might be possible to arrange two of them in series to cancel some of this effect.

author: Benta
author_rank: Super Contributor
timestamp: March 14, 2017, 09:45:54 pm
content: QuoteSo that's not extra complexity?My suggestion is a sure-fire way to solve this, but you'd rather speculate on things that are not in the datasheet?

author: Kleinstein
author_rank: Super Contributor
timestamp: March 15, 2017, 05:37:35 pm
content: For current shunts one can avoid to include drop on the switches in the measured voltage. One does not even need a second switch, just measure the voltage through the higher shunts. For very low currents (e.g. < 10 µA) one could consider a trans-impedance amplifier instead of a simple shunt.For shunt use the voltage level seen by the shunt is usually rather constant and often in the middle. So this is more like the sweet spot of R_on vs V_DC curve. Using a N and P MOSFET in parallel is already made to compensate voltage dependence, but near the supply rails just one of them is active and thus the not so good performance.Besides voltage dependence there is also temperature dependence, so one should avoid having the switch resistance in the critical path, except if it is really small.Can you show us an approximate schematics on what is planed ?

author: Benta
author_rank: Super Contributor
timestamp: March 15, 2017, 06:48:07 pm
content: QuoteDo you mean series connecting the shunts and then tapping into the shunt column from the input side?Trying to visualize it, it's a nice idea. I like it. So you'd have the highest current shunt to ground, the next highest on top of that, and the third on top of that again?And with a high impedance amplifier connected to the top of the shunt column, you can measure from there directly, and the switch impedances are completely eliminated?Very nice, good idea. It should work.

author: Kleinstein
author_rank: Super Contributor
timestamp: March 15, 2017, 07:19:15 pm
content: The shunts in series is the standard circuit in multi-meters. The switches are for current steering only.If works well up to about 100K for the highest shunt (thus about 2 µA current range) - for a much higher resistor the noise from that resistor adds significant to the noise.The other point to take care of is leakage of MOSFET switches - so it might be worth using a relay for the very high currents. A < 1 µA range is tricky with the leakage current of a 50 A MOSFET in the circuit.

author: julian1
author_rank: Frequent Contributor
timestamp: March 16, 2017, 09:33:06 am
content: QuoteSomething like this,I like the idea of having the switches in series with the resistors for simplicity - we get a single-ended output voltage referenced to AGND.The disadvantage is that the circuits are sensitive to any non-linearity in the switch over the voltage range.But - almost all the voltage drop is across the resistors and not the switch. So we're really only concerned with very small voltage differences - as you already noted.Additionally, having all the switches with one side tied to gnd means we get a common S (eg for Jfets) which makes it easy to drive G with respect to GND and not some floating/changing value.I am considering dual n-mosfets with sources tied, or relays for lower value resistors (eg. 1-100ohm).higher value resistors I am considering jfets or analog switches.The additional bank relay r1 is to isolate high-leakage devices like mosfets - where leakage is on the order of 1uA.Jfets and common analog switches have leakage on the order of 1nA.Some analog switches are available with much lower leakage - eg. pA range - and they may be useful for higher valued resistors (eg > 1M) although their RDS(on) is > 1k.I am considering using an addition bank-like relay on the higher valued resistor bank. A limitation of signal relays that I've found - such as the Omron GS6 is a minimum current of 10uA - so that would preclude their use for shunts over 1M.

author: BrianHG
author_rank: Super Contributor
timestamp: March 16, 2017, 10:29:37 am
content: Q: You do require voltage support below the GND, correct?Q: What supply voltages are available to you?

author: julian1
author_rank: Frequent Contributor
timestamp: March 16, 2017, 11:27:48 am
content: QuoteYes. And I'm unsure of the implications of that for Jfets. I *think* that for symmetrical jfets (like j111) they should conduct current equally since S and D can be swapped. But when reverse biased (so to speak) VGS will be slightly less than the 0V on or -10V off VG to GND, because D is now tied to GND rather than S and there is some small additional voltage drop across the device. If that difference is linear it shouldn't matter.Quotenot constrained.

author: BrianHG
author_rank: Super Contributor
timestamp: March 16, 2017, 04:07:26 pm
content: Quote from: BrianHG on March 14, 2017, 09:24:01 pmI would recommend my last recommended switch with a +/- 20v supply. Your on resistance will be 1 ohm per switch and the off leakage current will be 0.01na at room temp.

author: Kleinstein
author_rank: Super Contributor
timestamp: March 16, 2017, 05:05:38 pm
content: One can still get a voltage referenced to a common GND with the shunts in series connection. Especially for the low ranges (R < 100 Ohms), the switch directly in series is not practical - It would at least need a second switch to also switch the voltage sensing path. For the low resistors there is usually little downside on having the shunts in series. It only gets a problem if the second lowest one is a shunt with 4 wire connections.Usually the On resistance of FETs has a temperature coefficient in the 7000 ppm/K range - so the switch resistance needs to be much smaller part of the shunt resistance.Another point to look at is overload protection. This is usually something like 2 or 3 diodes in series to limit the voltage at the shunts. This could be a problem for the very low ranges.Even if the switches are not directly towards GND, the voltage at the shunts is usually small (e.g. < 200 mV) and thus does not make a big difference for JFET / MOSFET working as a switch. However gate leakage could be a problem with a +200 mV gate voltage to a JFET, so gate drive might need to be adjusted.For the symmetrical JFETs D and S are really exchangeable - so calling them drain and source is at will. When off it does not matter if the gate voltage is at -10 V or -15 V. If on, the DS voltage is very small.For shunts one usually does not need +-20 V rated switches. Something like the +-5 V or +-10 V rated ones are usually good enough. Very low on resistance switches usually also have more leakage.

author: julian1
author_rank: Frequent Contributor
timestamp: March 16, 2017, 06:56:23 pm
content: Quote from: BrianHG on March 16, 2017, 04:07:26 pmThanks. I agree, that is a superior part for maximizing the range between on resistance and off leakage. Also soic and with a pinout that would support low pcb leakage if that became a factor to optimize. I am not price sensitive. It looks like they might be getting the good switching spec by trading off band-width - which is perfectly fine for constant DC application.

author: julian1
author_rank: Frequent Contributor
timestamp: March 16, 2017, 07:19:40 pm
content: QuoteGood point, I need to think about this.

author: Benta
author_rank: Super Contributor
timestamp: March 16, 2017, 07:42:59 pm
content: This is the basic "serial shunt" Kleinstein suggested as a quick sketch from me. It's a very clean design and completely eliminates switch resistance in the measurement.It will work with bipolar signals.You'll need an opamp with high input impedance, but hey, those are thirteen a dozen nowadays.And you can refine it with protection circuitry, relays for higher currents etc. This is just the principle.

author: djnz
author_rank: Regular Contributor
timestamp: March 17, 2017, 01:10:01 am
content: Do you need to make one or two units incorporating this, or do you want to use it in a manufactured product made at higher volumes? I am guessing you'll use high accuracy, high precision resistors, and having a mux or a FET in series will defeat the point of that a little / add significant cost of human time for calibration.

author: julian1
author_rank: Frequent Contributor
timestamp: March 17, 2017, 06:21:05 am
content: Quote from: Benta on March 16, 2017, 07:42:59 pmThanks for the sketch, I wasn't quite understanding the in-series topology and I agree it's very elegant.There might be a practicality given that IN (and therefore S) would be moving over the entire sense-range making gate control more difficult for active devices. But this would not be an issue for relays.

author: Kleinstein
author_rank: Super Contributor
timestamp: March 17, 2017, 08:29:44 am
content: Due to the normally small (e.g. less than 300 mV) voltage at shunts the variable voltage at the shunts is not a big problem - not even when using JFETs as switches. Over-current protection usually limits the voltage to something like 1-1.5 V.The main advantage of a relay it that they have low leakage even for high current versions. Still the contact resistance can be a problem when in series with a low ohms shunt, but it does not matter if only in the current path.The on resistance of FET switches is also temperature dependent with a rather high TC on the order of 7000 ppm/K. So one should really avoid having a FET as part of the shunt. As leakage goes up for low on resistance switches going very low in R_on is also limited.

author: Benta
author_rank: Super Contributor
timestamp: March 17, 2017, 08:46:10 am
content: Quote from: julian1 on March 17, 2017, 06:21:05 amThat's no problem at all. If you check out the Siliconix/Vishay switches, they almost all have a +/- 15 V analog range and TTL compatible control inputs.

author: mfratus2001
author_rank: Contributor
timestamp: March 17, 2017, 03:37:25 pm
content: The current domain is hard to maintain with accuracy, since all analog switches are going to be variable.Instead, why don't you operate in the voltage domain for your selectors, and use a voltage-to-current conversion? That way the on-resistance is a minor factor.I just can't see switching a 1-Ohm resistance in with any analog switch (even a big MOSFET) and having repeatability. Otherwise, use relays.

author: julian1
author_rank: Frequent Contributor
timestamp: March 17, 2017, 10:21:01 pm
content: Quote from: mfratus2001 on March 17, 2017, 03:37:25 pmA trans-impedance amplifier as Kleinstein mentioned? I had to wikipedia it - but was thinking about whether it would be possible to use a jfet input op-amp with an output diode (ideal rectifier) and shunt-valued resistor feedback (tia), and then use the non-inverting input (0V or -10V) to control which tia op-amp was active. But I'm not sure it can be done without some additional switching complexity.QuoteAgreed.And once there's one relay in the path - then iterating ranges will always be slowed down by the mechanical action of the relay. I'm still thinking it may be simplest (and economically equivalent to analog-switches) to just use relays for all ranges up to 10uA. 10uA being the minimal permissible current to ensure good relay contact connection according to the datasheets for Omron signal-relays.

author: Kleinstein
author_rank: Super Contributor
timestamp: March 18, 2017, 10:09:59 am
content: The trans-impedance amplifier is attractive for low to very low currents. It gets more difficult for more than about 1 mA, as the OP has to drive the measured current and at a higher voltage drop self heating of the resistor might get a problem. The advantage is that in an TIA the drop on the resistor can be higher (like 1 V or even 10 V) this way offset errors of the OP are not that important and one can thus use low bias JFET of CMOS types. The main range to use an TIA is if you want to measure with better than pA resolution - so more like below 1 µA full scale.For the large currents, the chain of shunts is the usual way, possibly using a fully separate shunt for very high currents. The usually higher bias current of low drift / offset OPs somewhat limits the use of the shunt scheme to very low currents. So ranges of less than about 10 µA might need a compromise on the OP.I am a little confused with the resistor and current values used in the first circuit plan - usually the voltage at shunts is low - more like 20 mV - 400 mV. It is only in circuit more like an TIA that compensate for the drop, that a higher burden is more common. At high currents, self heating is strong argument for a low drop. What is the application ?As the switch is not in the voltage reading path, the switches are not that critical with respect to on-resistance. So there is no big problem in also switching 10 A through MOSFETs (or JFETs if you are willing to spend the money) - it is leakage that might be a problem. So for leakage reasons a relay might be better here - still leakage of the protection circuit remains. As the switches are only for directing the current, they could even work for the very low currents below the 10 µA limit from the relay specs - we don't really care about a little higher drop that might occur at such low currents. However, FET switches are usually cheaper and lower power at such low currents.

author: julian1
author_rank: Frequent Contributor
timestamp: March 19, 2017, 11:12:18 am
content: Quote from: Kleinstein on March 18, 2017, 10:09:59 amA current source. The resistors are part of the regulation loop, so burden voltage is not an issue. I was aiming for 1A max output - not 10A - but you're still correct about self-heating which I have somehow managed to overlook. The higher current ranges will need to use resistors with a lower voltage-drop and amplification rather than a buffer to align with other ranges.QuoteI've seen schematics for older current sources with a 1G sense resistor switched by a relay (10nA range) where the relay is part of the measurement- using what I think is a Coto relay part. But I can't find any relays that match that kind of minimum permissible current.I suspect that if one can periodically (eg. on instrument startup) push a higher current through the contacts then that would be sufficient to clean them of any oxides/debris.Alternatively perhaps any added resistance - from the low current being outside spec - is simply dwarfed by the size of the G-ohm range shunt resistor.

author: Kleinstein
author_rank: Super Contributor
timestamp: March 19, 2017, 12:04:49 pm
content: For an active current source over-current protection is less of an issue and the possibly high drop is allowed if self heating is considered. It's also possible to have current in one direction only. I am not sure about the very small currents - the power part might limit the useful range due to leakage too.For operation with very low currents there were special mercury whetted relays, that work well and very reproducible, but are not available anymore for normal use. AFAIK a normal relay contact at very low current might give an extra voltage drop of a few µV. So it should not be a problem in the current path, but it might be a problem in the voltage path. Relays with gold contacts are already quite good - typical performance can also be much better than guaranteed.For most of the ranges the series connection of the shunts should be fine. Otherwise one could use separate switching for the current (e.g. relay or MOSFET) and the voltage sensing (CMOS switch chip). It is much less trouble than having the switch resistance as part of the shunt, even if a cheap CD4052 or similar is used.

author: julian1
author_rank: Frequent Contributor
timestamp: March 19, 2017, 12:10:19 pm
content: QuoteIf we organize for the higher current ranges - to use lower voltage-drops to limit self-heating (eg 1V rather than 10V). Then that makes controlling the n FET gates much simpler. eg. Just drive it +10V above GND, rather than +10V above a S that might be +-10V wrt GND.I think that removes my hesitation about gate control complexity. So the in-series shunt approach with the switch out of the measurement is an excellent fit for higher current ranges.

author: Kleinstein
author_rank: Super Contributor
timestamp: March 19, 2017, 01:14:07 pm
content: For an 1 A current, the drop on the shunt should not be much higher than 200 mV. This already gives 200 mW and thus should use a resistor that is good for something like 2 W or more to limit the self heating. Even for the lower currents, one should get away with less than 1 V - one can still use the same amplifier.With MOSFETs one can just apply something like +10 V to turn them on. With JFETs one might still need to take into account the approximate voltage levels, as here on is at 0 V and already +400 mV might give considerable leakage to gate. This makes JFETs less convenient in many applications.For current in both directions one might need something like 2 MOSFETs in series (different direction) because of the substrate diodes. It usually still works just driving the gates.For low currents a MOSFET with separate substrate or a CMOS switch might be a better choice.

author: KubaSO
author_rank: Contributor
timestamp: January 14, 2024, 01:47:18 am
content: Quote from: julian1 on March 14, 2017, 08:30:42 pmJ111..113 and 2N4391..4393 were designed specifically for use as analog switches. They are good enough to be used as switches in 4+ digit double-slope DMMs, as a point of reference.Quote from: julian1 on March 14, 2017, 08:30:42 pmIt doesn't change at all, since these parts are meant to be turned on by setting the gate voltage to be the source voltage. So the RDS(on) is not absolute-voltage dependent. It varies slightly with the channel current, and around the saturation current the incremental channel resistance goes very high, i.e. you get a current source, not a resistor.Say the source of the JFET is connected to the output of an op-amp. It is a low impedance-driven node. The other side - drain - is connected to a high impedance load, like say an integrator input.To turn the switch on, a resistor - say 100kΩ - permanently connects gate to the low-impedance source.To turn the switch off, a current sink is connected to the gate, using a bipolar transistor as a switch, for example. A current is sunk from the low impedance source-side driver, through the 100kΩ GS resistor, establishing a negative gate-source voltage, i.e. the gate is lower than the source.Since the gate-channel capacitances are in single pF, the charge injection is minuscule.As long as the source nodes are driven from a low-impedance source, the JFETs are excellent switches. In that particular application, they outperform low-end integrated MOS switches. Many early DMMs used JFETs as switches, with much success.hXn5D.png(9.63 kB. 600x373 - viewed 194 times.)

author: the_cake_is_a_lie
author_rank: Contributor
timestamp: January 16, 2024, 04:55:10 am
content: Douglas Self has pages devoted to switching with JFET and CMOS 4016 and 4066 gates in chapter 21 of Small Signal Audio Design. I think half a page is fair use.Quote from: Small Signal Audio DesignHe shows example circuits next. No PCBs or circuit simulation links to download. I bought J111 and J112 but also J109 and P-channel J176 that he doesn't mention. Not sure why he doesn't mention J109. I'd appreciate it if anyone can answer if J109 isn't legit to use in switching given it has the lowest resistance.I'm not smart enough to make tables. Maybe someone can make a pic but I'm going to act like it's 2000s GameFAQs:N-ChVgs(min), Vgs(max), Rds(on)ΩJ109:......-2...........-6..............12J111:......-3...........-10............30J112:......-1...........-5..............50J113:......-0.5........-3..............100P-ChJ175:.......+3.........+6.............125J176:.......+1.........+4.............250J177:.......+0.8......+2.5...........300The scam of JFETs, besides their price, is the Vgsspread being greater than MOSFETs.