Can I auto-zero a sub-microvolt DC signal with a relay? (null detector) - Page 1
Forum section: electronics / general

author: ckocagil
author_rank: Contributor
timestamp: July 01, 2019, 08:21:03 pm
content: I've studied a few DIY null detector designs so far, and I got to a point where I'm really wondering if a scheme like the following could work:Take an SPDT electromagnetic relay and have it switch at something like 10 hz. Connect the relay such that its output goes into the non-inverting input of the opamp, while the input switches between the signal and ground. Have a decent ADC measure the amplified signal, get it into a microcontroller which would then report the difference between the high and low half cycles.If this works, I would have the best of all worlds: low noise, sub-microvolt input offset and thermal EMF, low bias current (I'd use a JFET input opamp). It likely won't, which is why I'm posting it here so one of you can point out the glaring error.

author: Kleinstein
author_rank: Super Contributor
timestamp: July 01, 2019, 08:45:47 pm
content: Mechanical relays tend to have problems with thermal EMF, as the coil needs quite some power for switching. At 10 Hz even a bistable relay needs quite some power.A second problem is the contact forming at very low current - not all contacts work that well at very small currents. Gold or mercury whetted contacts are the normal choice for similar jobs - though hardly at 10 Hz.The more suitable type of switching are electronic switches, like JFETs, MOSFETs or PhotoMos couplers.

author: thermistor-guy
author_rank: Frequent Contributor
timestamp: July 02, 2019, 04:06:57 am
content: Quote from: ckocagil on July 01, 2019, 08:21:03 pmNot sure about sub-microvolt, but the principle works. Leeds & Northrup used electromechanical choppers in its DC measurement and recording instruments 40 or so years ago. E.g. have a browse through:https://f01.justanswer.com/QnsDptPr/PF.pdf

author: TiN
author_rank: Super Contributor
timestamp: July 02, 2019, 04:17:34 am
content: Here is how Keithley does it:This is part of schematics fromKeithley 7168nanovolt scanner card, specified <30 nV offset per channel.

author: guenthert
author_rank: Frequent Contributor
timestamp: July 02, 2019, 04:16:07 pm
content: Well, not just in pre-FET times, but also in the 21st century relays have been used for that purpose:"Input signals are applied to the AVM-2000’s low-thermal EMF binding posts. From here the signals arerouted directly to the programmable input attenuator. Depending on the selected measurement range, theinput signal is applied directly to the input amplifier (100 nV – 1 mV ranges) via a matched set of lowthermalEMF polarity reversal relays, or is attenuated to be compliant with the input amplifier’s dynamicrange and then applied to the input amplifier (3mV – 1000V). On all ranges, all stages of the inputamplifier allow sufficient head room so peak noise does not cause limiting and very high loop gains ensurelow-distortion. This allows the AVM-2000 to make full use of the digital filtering technology to eliminateunwanted noise from the measured signals."[https://www.tegam.com/wp-content/uploads/2015/11/AVM-2000-Manual-Rev-D-for-firmware-2.1.14-or-later.pdf]

author: dietert1
author_rank: Super Contributor
timestamp: July 02, 2019, 04:39:33 pm
content: For the analysis of thin film resistor arrays i am using a relay scanner made of bistable relays with gold plated contacts. The coils of those relays are 180 Ohms and operate from 3 V. Current pulses are 5 msec, so there is no measurable thermal voltage as long as you operate a relay once per second or less (average coil power less than 0,5 mW).We have lots of VXI switch boxes built with reed relays. Those are specified with "thermal voltage < 4 uV".Regards, Dieter

author: ckocagil
author_rank: Contributor
timestamp: July 02, 2019, 07:24:29 pm
content: What worries me about JFETs is that I don't know which parameters (if any) should match. Do I need a matched pair? Do I need low R_on? Are the JFETs going to introduce any voltage offset (except the thermocouples)?Since I can't find much info online about solid-state switch selection for low DC voltages and I don't have any gear to run experiments with, I considered relays to be the more fool-proof way. Mercury wetted relays look harder to buy these days due to their mercury content. I'll look into gold plated relays.Quote from: TiN on July 02, 2019, 04:17:34 amGreat article, thank you! Too bad it doesn't give me the exact JFET used in the Keithley. It likely would be out of production or my budget anyway. Do you have any resources on JFET selection for this purpose?

author: Kleinstein
author_rank: Super Contributor
timestamp: July 02, 2019, 08:53:27 pm
content: A JFET should (no RF signal superimposed) not add any offset voltage except thermal EMF. The main parameters to look at are the R_on, leakage and maybe gate capacitance (as an approximation to get charge injection).A classical choice for switching is 2N4392 or a modern SOT23 equivalent. These are for relatively low R_on and reasonably low leakage.For a chopper there could be other better choices (especially lower voltage types) with less gate capacitance. A good resource on JFETs the older Fairchild (now ON) application note AN-6609.An important point is to keep thermal gradients small at the FETs.As the voltage is expected to be relatively small, one could also consider a modern CMOS switch chip (e.g. DG4053).Switching with 10 Hz sounds like relatively fast, especially for mechanical relays.

author: thermistor-guy
author_rank: Frequent Contributor
timestamp: July 02, 2019, 11:40:05 pm
content: Quote from: Kleinstein on July 02, 2019, 08:53:27 pmLeakage will rise exponentially with ambient temperature, so be conservative if designing with room temperature leakage specs. Any ESD damage to the JFET will increase the leakage, so I suggest characterizing the leakage of the finished assembly at the upper ambient temp.QuoteHistorical note: loop disconnect dialling for step-by-step relay-based telephone exchanges (1970s and earlier) used to use 10 pulses per second. The relay equipment of the time could detect and regenerate the pulses at that speed.https://en.wikipedia.org/wiki/Pulse_dialing

author: David Hess
author_rank: Super Contributor
timestamp: July 03, 2019, 03:44:42 am
content: Chopper stabilization can be done with relays but there are lots of options including:1. FETs, both MOSFETs and JFETs.2. Bipolar transistors intended for chopping applications and with some cleverness, even normal bipolar transistors.3. Photo resistors - these are one of the higher performance options.4. Photo JFETs - I do not know how these work.

author: Gyro
author_rank: Super Contributor
timestamp: July 03, 2019, 09:39:23 am
content: Quote from: Kleinstein on July 02, 2019, 08:53:27 pmA lot of (well at least, some) Clare mercury wetted relays were rated to >100Hz.Of course this depends a lot on whether this is a one-off experiment and probably the vagaries of ebay.

author: voltsandjolts
author_rank: Unknown
timestamp: July 03, 2019, 11:13:59 am
content: Why not use an auto-zero op-amp? Less than 1uV offset for less cost than any workable relay system.

author: Vtile
author_rank: Super Contributor
timestamp: July 03, 2019, 06:33:17 pm
content: Rotating mercury switch.

author: Marco
author_rank: Super Contributor
timestamp: July 03, 2019, 07:10:40 pm
content: Quote from: ckocagil on July 01, 2019, 08:21:03 pmFor the lowest bias currents you want to precharge the input(s) of the amplifier, even with fA leakage CMOS inputs chopping will pull a DC current.For IC designers mitigating that sort of thing just means throwing a few more square um at it, for you not so much.

author: Vgkid
author_rank: Super Contributor
timestamp: July 03, 2019, 09:01:08 pm
content: Quote from: Vtile on July 03, 2019, 06:33:17 pmI have seen that used in polarity reversal switches...

author: ckocagil
author_rank: Contributor
timestamp: July 23, 2019, 10:20:36 pm
content: I played with some switch designs based on JFETs. A very good candidate seems to be MMBF4391/2/3 in plastic package. But the spec says a maximum of 100p leakage current and the LTSpice simulations using the official ONSemi model show a leakage current around 20-30p for my case. The current is pulled from the device being measured.Is this too much current for a decent-ish nanovoltmeter that would affect my measurements?My gut said yes, so I looked for some analog switch ICs. Analog Devices has these switch/muxes with only a few picoamps of typical leakage:Their specs look incredible. It looks like I should be able to use these switches for a nanovoltmeter, perhaps even an electrometer/picoammeter. Am I overlooking something?

author: niner_007
author_rank: Frequent Contributor
timestamp: July 24, 2019, 12:17:25 am
content: COTO 3500 reed relays are rated at < 0.5uV EMF, but I believe you have to make a custom order, and they will be expensive, I expect around $60-$100 per relayA mechanical relay doesn't make sense if you are going to turn it on and off all the time

author: Kleinstein
author_rank: Super Contributor
timestamp: July 24, 2019, 07:08:38 am
content: Testing leakage at the sub 10 pA level is slow and thus many parts have quite a spread from the typical to the maximum specs for leakage.Using CMOS switches can be a good option. One point to watch for is the charge injection: this can also act similar to bias currents, especially if switching is relatively frequent. Charge injection also usually depends on the voltage level and the specs are sometimes confusing - especially 1st page number may be near best case. Low charge injection parts tend to use some compensation technique. So the current noise from variations in the charge injection can be higher than just looking at the charge injection and leakage numbers.In a chopper application the voltage over the switches is usually small, so that the leakage may well be smaller than under the test conditions for the switches (quite often with a relatively high voltage).

author: David Hess
author_rank: Super Contributor
timestamp: July 24, 2019, 03:48:55 pm
content: I wonder if anybody ever made a chopper or did low input current multiplexing with diode switches.

author: guenthert
author_rank: Frequent Contributor
timestamp: July 24, 2019, 04:25:18 pm
content: Quote from: David Hess on July 24, 2019, 03:48:55 pmI didn't make one, but ran some simulations in LTspice in the context of a precision PWM voltage source. It didn't seem to work well. I attributed the failure in the end to the small, but not negligible (if switched with high frequency) capacity of the diodes.

author: Kleinstein
author_rank: Super Contributor
timestamp: July 24, 2019, 04:39:32 pm
content: Diode switching is used in ring mixers. They are nice for RF, but I would not expect good performance at near DC due to thermal effects. The diode drop is some 2 mV/K, while thermal EMF from dissimilar metals is more like 20 µV/K. The diodes also have quite some power dissipation compared to CMOS switches.I used a ring mixer for essentially DC, and it worked better than expected. Still there was some low frequency noise in the µV range. I don't know the source, but temperature fluctuations are well possible. A more classical chopper was not an easy option at 50 MHz. Due to thermal effects I would also expect some residual temperature drift, possibly not even better than a classical BJT based amplifier.

author: iMo
author_rank: Super Contributor
timestamp: July 24, 2019, 05:22:46 pm
content: When talking mixers - long time back on the eve of modern SDR I used to use FST3253 switches as the mixer. A few ohm when on, pretty fast, 5V, no idea about other parameters, however.

author: Conrad Hoffman
author_rank: Super Contributor
timestamp: July 24, 2019, 08:30:05 pm
content: You can get reed relays (search on reed contact) at about ten for a couple dollars. Add your own coils for a cheap experiment. They should be very fast.

author: dietert1
author_rank: Super Contributor
timestamp: August 06, 2019, 08:22:21 pm
content: The proper technology is using opto fets as in later Fluke 845A null detector models, for example ON Semiconductor H11F1M. Those parts are supposed to deliver clean signals up to some KHz without charge any injection.Regards, Dieter

author: David Hess
author_rank: Super Contributor
timestamp: August 06, 2019, 09:56:35 pm
content: Quote from: dietert1 on August 06, 2019, 08:22:21 pmYou could also *make* photoFET using a part like the Linear Systems LS627 or one of the parts fromCrystalonicsand the emitter of your choice. I have seen some precision instruments that went this route with light pipes or just connecting the emitter and detector plastic tubing.

author: dietert1
author_rank: Super Contributor
timestamp: August 07, 2019, 08:45:58 pm
content: Yes, one disadvantage of photofet and LED in one package is a small thermal offset voltage in the switch circuit generated by the LED heat. So better keep them separate.Regards, Dieter

author: Echo88
author_rank: Frequent Contributor
timestamp: August 07, 2019, 09:20:38 pm
content: Interesting idea David! Need to try that and compare it with other OptoFETs like the H11F1/PVA3054.The H11F1 is no good example for a low offsetvoltage switch, as it develops ~200µV across its contacts like its shown in the following document and interestingly not shown in the available datasheet:https://www.eevblog.com/forum/metrology/wescor-psychrometric-microvoltmeter(-~keithley-155)/msg2593941/#msg2593941The obsolete PVA1054/3054 for example state 200nV offsetvoltage:http://www.irf.com/product-info/datasheets/data/pva33.pdfDuring my crude charge injection experiments ive seen the ~200µV across the H11F1 and couldnt confirm the "no charge injection"-claim. It behaved pretty much like the other tested Optofets PVA1054/PVA3054.Can anyone give good advice on better charge injection measurement techniques on such OptoFETs?Can you remember which instruments use the LED-lightpipe-OptoFET-solution you mentioned David?

author: dietert1
author_rank: Super Contributor
timestamp: August 07, 2019, 09:55:23 pm
content: What you saw is not charge injection but a small thermal voltage like it happens with relays. While the heater in a relay is the coil, in a photofet coupler it is the LED. Appearently some mW are enough to create gradients.The light pipes were in the original version of the Fluke 845A null detector. Later they used H11F1 photofets to replace LDRs, light pipes and neon bulbs.Regards, Dieter

author: Marco
author_rank: Super Contributor
timestamp: August 07, 2019, 10:09:56 pm
content: I don't see why a photofet wouldn't have charge injection, a voltage is developed across the gates to turn it on. When it turns off there will be imbalances in how the charge from the gates gets divided between the signal source and the amplifier. Meanwhile the FET channel will probably also see a little light leaking through while it's on, so it will start generating a photocurrent.

author: dietert1
author_rank: Super Contributor
timestamp: August 07, 2019, 10:42:04 pm
content: Where does the gate charge come from? I think for charge injection you need some external drive with a common ground.Regards, Dieter

author: Marco
author_rank: Super Contributor
timestamp: August 07, 2019, 11:20:02 pm
content: Lets say you make our own photofet from a N-MOSFET P-MOSFET, zener, resistor and photodiode. The charge on the output comes from the uneven distribution of the induced currents during turn off. They don't just go in a neat little circle between gates and sources.

author: dietert1
author_rank: Super Contributor
timestamp: August 08, 2019, 06:05:03 am
content: The mosfet relays i have seen are made of two FETs of the same polarity for the parasitic source-drain diodes being back to back. Then the drive is between gates and sources and the drains are the relay contacts. Since the drive has no external connection in a photo fet, i can't see how it injects charge to the load except from the source.The Fluke 845A null detector is designed as a regulator. As long as the regulator is near balance the switches turn on and off small currents at near zero voltage difference and the generated AC signal is almost null. Maybe this design also helps to avoid charge injection. Due to the different chopper technology the bias current of a 845A is about 700 times smaller than that of a modern autozero amplifier like the ADA4522.Regards, DieterPS: Another way of separating the light source and its heat from the switches may be a combination of a PVI1050N and small FETs of your choice.

author: Kleinstein
author_rank: Super Contributor
timestamp: August 08, 2019, 08:10:39 am
content: A large part of the much higher bias currents and current noise of the ADA4522 comes from the high chopper frequency, that is something like 3000 times higher. So the switches inside the AZ OPs are not that bad when it comes to charge injection.Besides charge injection there is also possible gate leakageCharge injection is a tricky effect and it does not need an outside connection. There is internal gate charge that on turning off has to flow to drain and source. If this flow is asymmetric there will be a charge appearing. This is superimposed with the normal capacitive coupling from the drive signal. I also don't see a reason why a photo-fets should intrinsically have no charge injection.A possibly offset source for monolithic photo-fets is from light leaking to the FET area and producing photocurrent there.Similar there could be parasitic photo-current even in the photo-resistive elements, e.g. neat the contacts or at inhomogeneities.Using an extra PV opto-coupler with external FETs can avoid the thermal coupling and light leakage. It also helps with gate leakage - this may be important with modern MOSFETs that usually contain ESD protection at the gate. However it does not help with charge injection.

author: dietert1
author_rank: Super Contributor
timestamp: August 09, 2019, 06:07:38 am
content: Sorry, the H11FXM datasheet states: "Accuracy and range are improved over conventional FET switches because the H11FXM has no charge injection from the control signal." Let's assume this is valid until proof of the contrary. They don't give an upper limit because the reason is in the principle, as i wrote before: There ist no electrical connection between drive signal and the switch.To prove the contrary you will need a valid measurement excluding thermoelectric effects.Regards, Dieter

author: BrianHG
author_rank: Super Contributor
timestamp: August 09, 2019, 06:27:37 am
content: Quote from: dietert1 on August 07, 2019, 10:42:04 pmThe gate charge in the optofet and optomosfets comes from the photovoltaic action. It's like there is a small solar cell tied between the gate and source of the fets which receives infrared light from the isolated LED.Internally, to rapidly turn off the mosfet, the gate does have a parallel load resistor to the source as well.Note that the optofets have a linear transfer function related to led current by design. They are also good for linear isolated resistor control. They also have less parallel capacitance than the optomosfets.The optomosfets don't usually do this with any real control. Once the LED reaches a specific current, the mosfet switches on like an avalanche. This is why these devices are also called optical solid state relays.The FET version has better high frequency (RF range) AC performance as the Mosfet versions have built in 2 back to back mosfets for AC signal applications. As the protection diode in each mosfet may snap on through the drain-source capacitance when the voltage first swings in the opposite way when the switch is open. (This is so minor and happens potentially only once due to a charge held by only a few pf. If you can detect this, you must be relying on impedances in the multi megaohm region at RF frequency.) There is no such effect in the FET style optocouplers.In other words, then H11FXM has no problems, but it's on resistance is high and varies with led current.Say 10ohm at 1ma led current, or, 1 ohm at 10ma led current.

author: dietert1
author_rank: Super Contributor
timestamp: August 09, 2019, 10:00:09 am
content: In the schematic of the Keithley nanovolt scanner posted above by TiN you can see several interesting details.Appearently they wanted to use the H11F1M but found it was not good enough due to thermoelectric voltages (thermal gradients in device connections by LED currents). So they separated the switches from the LEDs by adding another set of FET switches driven by batteries. Those FET switches and their connections are embedded in plastic covers filled with thermal grease. Another detail: High ohm resistors between the drivers and the switches to block RF.One thing i don't quite understand are the surge arrestors. Don't know if they can protect the FET switches.Regards, Dieter

author: David Hess
author_rank: Super Contributor
timestamp: August 09, 2019, 04:24:52 pm
content: Quote from: BrianHG on August 09, 2019, 06:27:37 amCharge injection is a common mode phenomenon. The only common mode coupling is magnetic and electrostatic when the switch is optically driven so there might be an effect if the emitter is physically close to the switch but that may be why some older designs physically separated them instead of using common optocoupler modules.Incidentally, the old way to do this in sampling applications was to use transformer coupling to drive a pair of bipolar transistors as shown below. I suspect this was done even after JFETs became available because of lower charge injection.

author: splin
author_rank: Frequent Contributor
timestamp: August 09, 2019, 06:26:22 pm
content: Quote from: David Hess on August 09, 2019, 04:24:52 pmDo you know how well this arrangement works over time and temperature? There will be some offset due to differences in Vce sat between the transistors and because the base currents won't match perfectly. Presumably the transistors will be closely thermally coupled, but i'd guess the offset drift could be quite high using discrete transistors?

author: David Hess
author_rank: Super Contributor
timestamp: August 09, 2019, 06:45:43 pm
content: Quote from: splin on August 09, 2019, 06:26:22 pmAll I can say is that it works reliably. That schematic is from the Tektronix 7T11 sampling sweep and in other places in the same instrument, they used JFETs and it is not clear to me why they chose one way versus another.I think you would just have to test and compare the methods.

author: Echo88
author_rank: Frequent Contributor
timestamp: August 11, 2019, 12:45:33 pm
content: I did some tests regarding the H11F1-offset voltage of about 200µV, to determine wether its of photoelectric or thermal EMF nature:Tests were done in an earthed aluminium case with pomona 3770-binding posts (and normal twisted copper wire/oropax-setup for low thermal emf errors) connected to the H11F1-output-leads, a BNC-connector connected to the H11F1-input-leads via current limiting resistor.A 34465A was used to measure the resulting voltage in the following mode: 100mV range, Autozero off, 10Meg Input-impedance.Tests with Autozero on and 10G-input impedance each added considerable error due to the charge pumping/bias-current-effects which is probably due to the capacitances on the JFET in the H11F1.The input leads were heated with a soldering iron at 300°C and a lighter, both methods proved not really suitable, since with the aluminium-case open and my hands at the lighter/soldering iron acting as antenna the introduced errors were to big to reliable determine the voltage resulting from the temperature at the inputs.With a 100R resistor connected across the H11F1-inputs, the earthed aluminium case closed, 4V supplied in reverse (to avoid turning the emitter in the H11F1 on) the resulting heating power from the resistor was ~160mW -> no resulting output voltage change measured during a test run of 1min.So it seems we can indeed conclude that the offset voltage of the H11F1 is produced by photoeffects instead of thermal EMF.I attached a simulation of different charge injection circuits, which contain common compensation measures on circuit 1 and 5 (out1, out5).

author: dietert1
author_rank: Super Contributor
timestamp: August 11, 2019, 02:24:33 pm
content: In the H11F1 datasheet they say there is no charge injection. Photoelectric effect and charge injection are not the same. Charge injection is a capacitive transfer, while photoelectric effect is from incident light. Ways to separate these effects:LED continuosly off: no effect, null measurementLED continuosly on: Photoeffect, thermoelectric effectLED continuosly on, then off: thermoelectric effect only (temperature gradients decay after ten seconds or a minute)LED continuously switching: Charge injection, Photoeffect and thermoelectric effectRegards, Dieter

author: Marco
author_rank: Super Contributor
timestamp: August 11, 2019, 03:03:09 pm
content: That depends on the level of symmetry. There's obviously next to no charge injection from the driver side. But there is a voltage created on the gates, which will cause displacement currents and charges, some of which will get trapped unevenly between source and amplifier rather than summing to zero again if the device isn't perfectly symmetrical.In that respect LDR is better, no currents generated, just resistance directly modulated.

author: Echo88
author_rank: Frequent Contributor
timestamp: August 11, 2019, 04:22:10 pm
content: The H11F1 produces continously about 200µV across its output, its not a charge injection effect.Worded myself not correctly in this regard a few posts ago.You can leave the LED on and load the H11F1-output with a resistor-decade and then disconnect, it will always get back to the 200µV offset voltage.Need to build a fA-buffer/amp from a LMC662 or equivalent to avoid the autozero/bias-current errors introduced with the direct DMM-measurement.Then well see if at least the PVAs show charge injection, since those dont suffer from the constant offset.

author: Kleinstein
author_rank: Super Contributor
timestamp: August 11, 2019, 05:20:26 pm
content: An LDR is also not necessary ideal. There can also be parasitic photoelectric effects near the contacts of from inhomogeneities. For an intrinsic semiconductor is takes little to disturb the bands. I don't know how ideal the CdS film is.Separating the thermal and photoelectric effect is via the timing. However off phase is only of limited use, as bias current from the DMM can disturb things. Only the charge injection part may need to turn of AZ and use rather fast conversions. With a 1 nF cap 1 pC of charge injection would be 1 mV - so not that very small, but still overwhelmed in leakage currents quite fast.The charge injection is an effect to change the readings just after turning off, especially with a small capacitance across.The Photoelectric effect is quite fast (e.g. some µs scale) and is the dominant effect just after turn on.The thermal effects are slow and need some 1 s or more to build up.So the thermal effect comes from comparing just after turn on and some 10 second or so later.200 µV of offset would be huge. As a good switch one would like to be in the < 200 nV range.

author: dietert1
author_rank: Super Contributor
timestamp: August 11, 2019, 06:14:47 pm
content: I tried to do some measurements, too. Made a circuit to simulate something like the Fluke 845A input stage. I made an 80 Hz clock generator do drive a PVI1050 and then a dual P-Mosfets (all parts we had). Those Mosfets are huge (82 milliOhm Rdson).I found they produced quite some charge injection when i routed the multiplexed signal to the scope. But that disappeared, when the load was a 33 pF capacitor instead - to simulate the FET input of the AC amplifier. To determine to what extent it disappeared, i put a 50 MOhm resistor across the 47nF input capacitor instead of the 1 MOhm i had before and my Fluke 8600 multimeter read 0,57 mV. This reading drifted by about +/- 30 uV. So my limit would be about 0,6 pA. Another factor ten lower would be interesting.Then we have a two channel low noise FET amplifier, but i have to see how i can bypass the input capacitors to make it into something like a Fluke 845A. Then the demultiplexer is still missing ..Regards, Dieter

author: Kleinstein
author_rank: Super Contributor
timestamp: August 11, 2019, 07:50:07 pm
content: 82 mohms p-channel MOSFETs are really huge. Smaller MOSFETs could make things a lot easier. For a chopper amplifier it would be more like an 2N7000 or similar - ideally even smaller.I don't know how effective the opto-coupler is in suppressing charge injection. It reduces the direct capacitive coupling, but not really the gate charge part. So it does not help very much. Well adjusted gate drive to small MOSFETs may be about as good.Especially if used at a low frequency a little charge injection is not that bad.There are still some LDRs available as NOS parts. Though more like cheap versions (so possibly more PV effect), one might still be able to test such parts with LED drive.

author: dietert1
author_rank: Super Contributor
timestamp: August 11, 2019, 08:56:22 pm
content: I wanted an experimental check of the conclusion in the H11F1 datasheet, that optical isolation inhibits charge injection. That's why i thought those FDC6306P dual fets were good for a test. Their characteristic charges are 0.8 nC to 4.2 nC in the datasheet. And i did not use pairs of mosfets as switches because all voltages are small - like some uV. (1 dual mosfet = 2 switches = 1 MUX. The switches are connected such that the sources are the two inputs (with 47 nF capacitors on them) and the two drains are the MUX output (with a 33 pF capacitor). Single mosfets don't have the charge injection compensation of mosfet pairs.Now, if you divide my 0,6 pA limit by 80 Hz this is a charge transfer of less than 7.5 fC per cycle, which is roughly 100 000 times below those datasheet numbers. So i think the idea is valid that optical isolation inhibits charge injection from the driver. And using an optical isolator separate from the switch also avoids heating and photovoltaic effect.LDRs seem to be extremely slow in turning off. I think for an LDR to go back to MegOhm after being illuminated it is slow. So the MUX in the Fluke 845A will definitely pass current from one input to the other until the feedback settles. My clock generator for the FETs needs a dead time of about 100 usec to avoid both fets being on at the same time.Regards, Dieter

author: dietert1
author_rank: Super Contributor
timestamp: August 13, 2019, 10:13:21 am
content: In another forum (HP equipment) i found some statements about the photovoltage of H11F1. Someone wrote that the two H11F1 to be used in a null detector must be matched carefully for similar photovoltage and that the parts needed aging before. The person said that out of 10 parts he got 3 good pairs of H11F1 photofets with less than 30 nV difference.Don't know how the photovoltage depends on temperature.Regards, Dieter

author: David Hess
author_rank: Super Contributor
timestamp: August 13, 2019, 04:06:51 pm
content: Quote from: Kleinstein on August 11, 2019, 07:50:07 pmThey use tiny low capacitance parts for this like JFETS and the 3N series MOSFETs. Long ago I tried using 2N7000s and they are just way too large. Even many JFETs are too large.QuoteI disagree. Most of the charge injection is a common mode effect between the control lead (gate) and source/drain. It is especially bad because it varies with voltage which leads to the various ways of compensating for it. An optically driven device simply starts out with much less because the common mode (gate) connection is only parasitic.

author: Echo88
author_rank: Frequent Contributor
timestamp: August 14, 2019, 11:21:17 am
content: Can you give a link to the mentioned HP-forum-post Dieter?

author: Squantor
author_rank: Contributor
timestamp: August 14, 2019, 12:16:44 pm
content: QuoteI was actually following that discussion a while ago when I was repairing my HP419A, here is the link:https://groups.io/g/HP-Agilent-Keysight-equipment/topic/24812737#68881They moved from yahoo groups to groups.ioSearching for hp419A or the 3400A you can read up on a lot of discussion replacing the photochopper with various other devices. Worth digging through that.

author: Echo88
author_rank: Frequent Contributor
timestamp: August 14, 2019, 02:04:14 pm
content: Thanks Squantor.Found the following:https://groups.io/g/HP-Agilent-Keysight-equipment/message/68873?p=,,,20,0,0,0::Created,,h11f1+30uV,20,2,0,24812737But its 30µV instead of 30nV and no mention of H11F1-aging. Maybe i missed the aging-part in some other post?

author: dietert1
author_rank: Super Contributor
timestamp: August 14, 2019, 05:33:02 pm
content: It's in a HP419A thread:https://groups.io/g/HP-Agilent-Keysight-equipment/message/75908?p=,,,20,0,0,0::Created,,H11F1,20,2,0,24814081Yes, there is a difference between nV and uV. My mistake - difficult to believe that Fluke put something with an offset voltage of several hundred uV into a uV null detector that is supposed to work down to 200 nV. Really, i still don't believe it.Regards, Dieter

author: SilverSolder
author_rank: Super Contributor
timestamp: August 14, 2019, 09:40:43 pm
content: As long as the offset is stable and consistent - it can probably be nulled out?

author: dietert1
author_rank: Super Contributor
timestamp: August 31, 2019, 09:44:39 am
content: I just tested a COTO 6213-1 LED driven mosfet switch we had (old parts from the 90s). The part shows an offset voltage of 0.32 uV. On state resistance is 5.5 Ohms. Off resistance as measured by HP 3456A is 5.35 GOhm.Another part one can order from RS would be Vishay LH1510 that has a spec: "Switch offset at 5 mA: 0.17 uV typical". Typical on state resistance is 12 Ohm.That means these opto mosfet switches do not generate "several hundred microvolts". A stable offset voltage of less than a microvolt would be something one may compensate in the input stage of a null meter.Regards, Dieter

author: Kleinstein
author_rank: Super Contributor
timestamp: August 31, 2019, 09:52:38 am
content: The offset of less than 1 µV would be acceptable for most cases, especially higher impedance use. Beside the photovoltaic, relatively fast part, there may be another slow thermal part, as the LED side will heat up the switch.However there could still be charge injection from the internal FET. It is no per se clear that charge injection is less than with a normal MOSFET or JFET. It is only the normal capacitance around the switch that is reduced, not that actual redistribution of the gate charge.