Can I auto-zero a sub-microvolt DC signal with a relay? (null detector)

[dietert1]

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

[Echo88]

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/#msg2593941
The obsolete PVA1054/3054 for example state 200nV offsetvoltage: http://www.irf.com/product-info/datasheets/data/pva33.pdf

During 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?

[dietert1]

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

[Marco]

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.

[dietert1]

Where does the gate charge come from? I think for charge injection you need some external drive with a common ground.

Regards, Dieter

[Marco]

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.

[dietert1]

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, Dieter

PS: 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.

[Kleinstein]

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 leakage

Charge 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.

[dietert1]

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

[BrianHG]

Where does the gate charge come from? I think for charge injection you need some external drive with a common ground.

Regards, Dieter

The 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.

[dietert1]

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

[David Hess]

The 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.

Charge 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.

[splin]

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.

Do 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?

[David Hess]

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.

Do 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?

All 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.

[Echo88]

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).

[dietert1]

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 measurement
LED continuosly on: Photoeffect, thermoelectric effect
LED continuosly on, then off: thermoelectric effect only (temperature gradients decay after ten seconds or a minute)
LED continuously switching: Charge injection, Photoeffect and thermoelectric effect

Regards, Dieter

[Marco]

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.

[Echo88]

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.

[Kleinstein]

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.

[dietert1]

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

[Kleinstein]

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.

[dietert1]

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

[dietert1]

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

[David Hess]

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.

They 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.

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.

I 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.

[Echo88]

Can you give a link to the mentioned HP-forum-post Dieter?