How a precision DMM have a high input impedance in the Gohm Rang? - Page 1
Forum section: electronics / general

author: ali_asadzadeh
author_rank: Super Contributor
timestamp: September 30, 2016, 01:07:29 pm
content: Hi,I always wanted to know how a precision DMM have a high input impedance? don't they use 10M ohm resistor dividers?

author: Gyro
author_rank: Super Contributor
timestamp: September 30, 2016, 01:17:37 pm
content: Try this thread as a starting pointhttps://www.eevblog.com/forum/beginners/input-protection-for-high-impedance-voltmeter/msg992838/

author: David Hess
author_rank: Super Contributor
timestamp: October 02, 2016, 06:11:12 pm
content: If the input signal is within the common mode range of the input buffer, typically up to about +/-5 volts maximum, then nothing special needs to be done and this is why many high performance voltmeters support gigohm input resistance at low input voltages.For a larger input voltage range, the input buffer is bootstrapped to follow the input voltage. This can provide a gigohm input resistance over an input range of hundreds of volts which is common with electrometers. Bootstrapping also reducing errors caused by changes in operating point just like operational amplifiers can be bootstrapped to improve common mode rejection ratio which would otherwise limit performance of non-inverting amplifier configurations.

author: zlymex
author_rank: Frequent Contributor
timestamp: October 03, 2016, 01:27:19 am
content: I made myself a front-end buffer allowing 5V to 1000V DC signal to be measured at ultra-high input resistance of multi-Tero Ohms.The error(the in-out difference) is mainly duo to the Vos of the opamp, and the leakage current is mainly duo to the protection diodes and Ib of the opamp.If this buffer is built into a desktop multimeter and supply with -1015V and +1015V, the input resistance of all the DCV ranges will be very high too.

author: ali_asadzadeh
author_rank: Super Contributor
timestamp: October 03, 2016, 06:09:01 am
content: Thanks for sharing,I can not see your schematics completely, would you attach the full circuit, and Also it would be very appropriated if you explain your circuit.Thanks in advance.

author: behzad63
author_rank: Contributor
timestamp: October 03, 2016, 08:53:52 am
content: HII had read about bootstrap for DMM high impedance.do you have any article or schematic to explain that? Do you know any solution for designing DMM high impedance?Thanks

author: zlymex
author_rank: Frequent Contributor
timestamp: October 03, 2016, 04:40:28 pm
content: Here is the complete schematics. This is the one ideally working at 1000V to 2500V. If the input is 100V to 800V, it can be simplified. I have an old thread at 38hot discussing the high voltage related topic:http://bbs.38hot.net/thread-11133-1-1.htmlThere are two tricks involved, one is the boostrape supplies of opamp that make them float. This idea is known for sometimes and can be found in some existing circuit such as Linear AN67 page 58. But here I use transistors instead of MOSFET because of the low dropout. Also, I use a current source(consisting of D1, R21 and Q21) to provide constant working current and high dynamic for the float.The second trick is the cascade of transistors allowing high working voltage than single transistor. There are three current paths, one is the supply path for the opamp(D1, Q26-Q29), two is the constant current(R21, Q21-Q24,D2,,,), three is the start-up and balancing voltage distribution among the cascade transistors(R22-R29).Q5, Q6 and R10 is a bidirectional current limiter(1.5mA).I actually use 600V rated transistor complimentary pair of 2SC3632 and A1413 in my trial PCB. Another option would be 900V rated 2SC4030 and 2SA1967 allowing 2500V to be buffered. In theory, the number of series connected pairs may extend as many as you like, to increase the max. working voltage.Replace the opamp with low-Ib one to achieve higher input impedance.

author: bktemp
author_rank: Super Contributor
timestamp: October 03, 2016, 08:27:07 pm
content: Quote from: zlymex on October 03, 2016, 04:40:28 pmThanks for the schematic. It is an interesting solution using the opamp supply current for biasing the current source.I built a similar circuit a while ago. Basically it worked, but I experienced a couple of problems:The circuit was quite unstable. The culprit was the input protection diode. The parasitic capacitance fed the output signal back into the high impedance input, making the circuit oscillate somewhere between 100-500kHz.I simulated your circuit using different transistors and a generic opamp model and it also shows a significant spike (+30dB) in the ac response at about 200kHz. Depending on the diodes used I can see the circuit oscillating after a voltage step. Without the 4 input protection diodes there is almost no spike in the ac response and no overshoot/ringing after a voltage step.Did you see a similar behaviour in your circuit or do you have any idea how to solve this problem (maybe the opamp is critical for stability)?

author: Gyro
author_rank: Super Contributor
timestamp: October 03, 2016, 09:03:21 pm
content: Quote from: behzad63 on October 03, 2016, 08:53:52 amHi behzad63, Welcome.You'll find some references in the thread I linked in reply #1...https://www.eevblog.com/forum/beginners/input-protection-for-high-impedance-voltmeter/msg992838/

author: David Hess
author_rank: Super Contributor
timestamp: October 03, 2016, 10:19:31 pm
content: Quote from: DiligentMinds.com on October 03, 2016, 08:30:41 pmMaybe but the coupling to the non-inverting input can be halved by doubling up on the input diodes by using pairs in series which will halve the capacitance. The LT1012 shown has internal back to back diodes across the inputs anyway.

author: zlymex
author_rank: Frequent Contributor
timestamp: October 04, 2016, 03:43:03 am
content: Quote from: bktemp on October 03, 2016, 08:27:07 pmI did experience unstable problem when I first assembled the trial PCB and use LMC6062 for the opamp(no oscillation for LT1012 though), and then I realized that the feedback capacitor(C2) is missing(not shown in my partial diagram) so I soldered an 1nF capacitor at the back of the PCB.I also found out that large current(of the opamp, and of the constant current) helps to stabilize so I modified R11 and R21 to 5.1k on the PCB(and to 2.2k in the new circuit).Q1 and Q2 may be omitted for LT1012 and reduce the capacitance by more than half because they are zero voltage biased. I used these because I'm trying other opamps as well that may not have the internal protection. These protection diodes are the bc junctions of small signal transistor which have small capacitor of about 4pF when reverse biased at 2V.Attached is a new circuit for 7V-500V. If up to 1000V is required, a cascade has to be used(add two pair of transistors).

author: bktemp
author_rank: Super Contributor
timestamp: October 04, 2016, 06:45:34 am
content: Using a LT1012 the overshoot is much lower than when using a generic opamp. Looking deeper into the datasheet, LT1012 seems to have a large phase margin of 70°. This seems to be really important here.Increasing the current seems not to help with unstable opamps. Also increasing or shorting C2 has no effect, because it does almost nothing in the unity gain configuration (except helping with driving capacitive loads, but that is not the problem here).Except from changing the opamp and selecting one with a large phase margin or using a heavily overcompensated one, the only other working solution I have found is adding a small capacitor at the input to ground. 1pF seems so be a low value that is always there due to parasitic capacitance to ground, but using guard traces reduces this capacitance to almost zero. Using guard traces actually increases the positive feedback capacitance, making the circuit even more unstable.The capacitance to ground can also be dangerous (especially when using a shielded cable with some 100pF of capacitance), because if left floating the opamp may drift to one of the supply rails charging this capacitance to some 100 or 1000V and delivering quite some charge when connected to a target at a different voltage.

author: ali_asadzadeh
author_rank: Super Contributor
timestamp: October 04, 2016, 07:46:40 am
content: zlymex why did you use big transistors? couldn't you use small transistors like MMBTA92LT3G or similar?

author: zlymex
author_rank: Frequent Contributor
timestamp: October 04, 2016, 11:41:01 am
content: Quote from: ali_asadzadeh on October 04, 2016, 07:46:40 amMMBTA92LT3G only withstand 300V, big transistors usually rate more such as 600V for 2SC3632-Z, 900V for 2SC4030 and 2000V for 2SC4913. It takes space/gap for high voltage anyway. I was aiming at 1kV to 2.5kV for a start at that time and I had already got many 2SC3632-Z and 2SA1413 at hand too.Of course, if only 200V is required(like Keithley 6514), MMBTA92LT3G or similar can be used.Another factor in choosing these transistors is not too small DC gain(hFE) at low Ic from which is often suffers for high voltage power devices.

author: ali_asadzadeh
author_rank: Super Contributor
timestamp: October 05, 2016, 10:07:58 am
content: How do you generate the High voltage for example 1.1kV for the circuit? Also can we apply AC voltage to the input? what's the maximum frequency limit of the circuit? how can we calculate it?

author: zlymex
author_rank: Frequent Contributor
timestamp: October 05, 2016, 03:28:02 pm
content: Quote from: ali_asadzadeh on October 05, 2016, 10:07:58 amI have three high voltage power supplies. As for DIY, a high voltage transformer plus a regular will do, the current consumption is only a few mA making things easier.From the simulation, the follower can handle AC voltage up to a reasonably high frequency, and with high input impedance as well because of the bootstrap, provided the follower is powered by negative voltage as well. The maximum frequency limit of the circuit is determined IMO by the phase shift and gain variation that you can tolerate, that can both be seen from the simulation. However, the frequency limit is also bounded by the slew rate which may be obtained by large signal step-response test.

author: VintageNut
author_rank: Frequent Contributor
timestamp: October 06, 2016, 02:07:19 pm
content: For the OP, I recently read a Fluke discussion of using a DMM for a null meter. The DMM can be used if the input bias current is 50pA or less for a 40k ohm source resistance. ( If I recall correctly)To measure this, place a short into the DMM and measure voltage on the most sensitive range. Perform a REL or record the voltage if REL is not supported. Then replace the short with a 40k resistor and note the voltage measured by the DMM. If you performed a REL the input bias current is Vmeasured/40k. I did this for my Keithley 2000 and the bias current is such that the input resistance is >10G as specified in the datasheet.If you do not have a REL, you have to subtract the initial measurement with the short in place from the 40K voltage measurement. Delta V / 40k is your input bias current.

author: Kleinstein
author_rank: Super Contributor
timestamp: October 06, 2016, 04:08:20 pm
content: A 40 K resistor might be rather low to measure the input bias current. The more usual value would be 1 M or 10 M.

author: David Hess
author_rank: Super Contributor
timestamp: October 06, 2016, 04:58:23 pm
content: Quote from: Kleinstein on October 06, 2016, 04:08:20 pmOr higher if your meter is that good.I have occasionally used my meters which have a standard 10 megohm input resistance to measure leakage of diodes and such with the meter measuring DC volts and the 10 megohm input resistance acting as the current shunt.

author: Gyro
author_rank: Super Contributor
timestamp: October 06, 2016, 05:26:52 pm
content: Datron specify (or specified) 10M in parallel with low leakage 100nF for bias current evaluation / adjustment.

author: VintageNut
author_rank: Frequent Contributor
timestamp: October 07, 2016, 12:19:33 am
content: Ok, I can see using a larger resistor to use more of the voltage range. I just tried a 10M on a DMM7510. -2mV measured yields 20pA yielding about 50G input impedance. The error is going to be on the order of 0.2% (10M / 50G). Not any kind of error to care about if all you are looking for is 2 significant digits to characterize the input impedance.

author: Kleinstein
author_rank: Super Contributor
timestamp: October 07, 2016, 08:47:13 am
content: The input is not that well described with just a simple resistor. There definitely is a capacitance at the input too and the bias current is often more important than the the "resistance". It is more like having a small bias current (e.g. in the 10-100 pA range) that slightly depends on the voltage. The voltage dependence of the bias current is what is describe by input resistance. However this does not have to be constant slope and could very well get nonlinear (especially higher currents) when you get close to the min or max voltage.The bias can also very much depend on the temperature of the DMM. Higher temperatures increase semiconductor leakage and on the other side might reduce surface leakage. There are usually currents in both directions that compensate, so it can change in both directions.

author: ali_asadzadeh
author_rank: Super Contributor
timestamp: October 07, 2016, 11:22:52 am
content: Having a 1.1K high voltage in a DMM is a practical thing? or is there some other way of achieving high input impedance for higher voltages? if the 1.1KV supply is the way to go doesn't it affect the precision of the circuit? I mean coupling the high voltages to the front end or reference path?

author: Kleinstein
author_rank: Super Contributor
timestamp: October 07, 2016, 01:01:57 pm
content: There are no real alternatives to having a kind of buffer amplifier for the high voltage if you really want high impedance at high voltages. There is one variant of driving the negative side terminal. But this is rather similar and also needs the high supply voltage, though it could save on the current.The main backdraw is, that the high voltage amplifier needs a significant amount of power. It could also be a safety concern. Otherwise a high voltage is not a big deal for the input stage. There are a few electrometer amplifiers that do it, but general purpose DMMs have high impedance only for lower voltages like +-12 V or +-20 V.If one does not need high resolution, there are static electrometer type solutions, like a field mill. However noise and drift is much higher - but it can work to 10 kV or more.

author: zlymex
author_rank: Frequent Contributor
timestamp: October 07, 2016, 01:22:48 pm
content: Quote from: ali_asadzadeh on October 07, 2016, 11:22:52 amThat seems to me is the only option if one wants to measure 1kVDC precisely and draw very little current(like >10G Ohm). I don't think a 9.9G:100M resistive divider is practical in this case. Further more, if 1000VAC is to be measured in this manner, a dual supply of +-1500V is needed.Having a 1000V in a meter is not a hard thing if the power consumption is low. I have two meters(one is Chinese hand held, the other is HP 4329A) both with internal 1000V source.

author: David Hess
author_rank: Super Contributor
timestamp: October 07, 2016, 02:56:57 pm
content: Quote from: ali_asadzadeh on October 07, 2016, 11:22:52 amAvaractor diode bridge type of operational amplifiercould be used. Before low input current FETs became available, they had the lowest input current (100pA max and 5pA to 10pA typical but it could be adjusted lower if higher Vos is acceptable) for solid state designs. An added bonus is that they have transformer isolated input circuits so with the right transformers, they can have 100s or 1000s of volts of common mode input range. It still needs a high voltage output stage to drive the other side of the bridge but that is easier to do than bootstrapping the whole amplifier.

author: JS
author_rank: Frequent Contributor
timestamp: October 07, 2016, 11:08:46 pm
content: Quote from: zlymex on October 07, 2016, 01:22:48 pmOr one and two floating amplifiers, if dealing with 3kV becomes a problem, just 1k5 could do it. Note that is very common DMM with 1KV DC range usually quote for about 700V AC. As the limitation is usually insulation is expected the peak value to be the same as the DC range, in which case the PS should be the same as the DC range, accounting for dual tracking amps, PS should be floating, which may needs to be accounted for.JS

author: behzad63
author_rank: Contributor
timestamp: October 08, 2016, 04:52:22 pm
content: Hi friendsi research about high impedance buffer and amplifier ,one of ways to achieve that use current positive feedback.pls see this link:http://www.analog.com/library/analogDialogue/Anniversary/22.htmlDo you have any circuit to Implementation it for DMM?Thanks a lot

author: David Hess
author_rank: Super Contributor
timestamp: October 08, 2016, 10:50:53 pm
content: Quote from: behzad63 on October 08, 2016, 04:52:22 pmCurrent feedback amplifiers like the bipolar one shown do not have a low input bias current. In theory you could make one using a JFET or MOSFET input stage but I have never seen such a thing and it would not have the low input bias current and inherent precision of an amplifier which uses a differential input pair.

author: behzad63
author_rank: Contributor
timestamp: October 09, 2016, 08:55:24 am
content: Hi FriendsThanks for your reply.I want to know how do i design safe 1500 v /10 ma power supply for bootstrap circuit ? and do you any idea for precision OHMmter?Thanks

author: bktemp
author_rank: Super Contributor
timestamp: October 09, 2016, 05:43:59 pm
content: Quote from: evb149 on October 09, 2016, 04:53:37 pmActually the voltage will be rather clean even without much filtering, because the royer type converters use an LC resonant circuit, therefore it generates a pretty clean sine wave with no noisy current spikes.Here is an app note on how to generate high voltages with low noise:http://cds.linear.com/docs/en/application-note/AN118fb.pdf

author: David Hess
author_rank: Super Contributor
timestamp: October 09, 2016, 09:46:07 pm
content: I would have linked that application note if you had not.I was going to suggest an inverter like that shown in figure 16 which is very similar to several low noise high voltage oscilloscope power supplies. It only requires one custom wound transformer which is simpler than the transformer required for a Royer converter.

author: ali_asadzadeh
author_rank: Super Contributor
timestamp: October 10, 2016, 06:15:56 am
content: Does this simple boost single transistor works?https://learn.adafruit.com/diy-boost-calc/the-calculator

author: bktemp
author_rank: Super Contributor
timestamp: October 10, 2016, 06:33:53 am
content: Quote from: ali_asadzadeh on October 10, 2016, 06:15:56 amA boost converter with such a high voltage ratio is inefficient because it both has to handly a high peak current and high peak voltage. If you want to go that way, you should at least use a voltage multiplier with about 4 stages: For 1100V the stepup needs to generate only 275Vpeak, the multiplier adds this voltage 4 times, generating 1100V. If you add another 4 stages with the diodes reversed, you also get -1100V.https://en.wikipedia.org/wiki/Voltage_multiplier

author: ali_asadzadeh
author_rank: Super Contributor
timestamp: October 10, 2016, 06:44:00 am
content: Thanks bktemp,Sound a better Idea

author: behzad63
author_rank: Contributor
timestamp: October 13, 2016, 01:26:49 pm
content: Actually I don not get the basic idea of boot-strapping in here? I have made some simulations in LT-Spice but the input voltage does not follow the output voltage exactly? Am'I missing something?

author: David Hess
author_rank: Super Contributor
timestamp: October 13, 2016, 02:40:48 pm
content: Quote from: behzad63 on October 13, 2016, 01:26:49 pmNot all operational amplifier macromodels include dynamic supply current. If the output current does not draw from the supply connections, then modeling the bootstrap circuit will not work.

author: ali_asadzadeh
author_rank: Super Contributor
timestamp: October 13, 2016, 03:49:27 pm
content: The next question is how they remove the fusible resistor from the normal mode impedance(10M ohm mode) , I believe they have way higher temp coefficient from the precision resistor ladder for example see the attached circuit, R1 is a fusible resistor and R2-R5 are the division ladder.

author: wine+dine
author_rank: Contributor
timestamp: October 13, 2016, 04:10:48 pm
content: Quote from: zlymex on October 07, 2016, 01:22:48 pmNot at all - you can use a 10 GOhm input resistor if that is all you need, and measure the current.For 1 kV that is 100 nA, plenty to work with. Your nanoamperemeter can be a simple transimpedance amplifier. For 1 V out that's a 10 M feedback resistor around a low-bias, low-offset opamp.Downside is the voltage coefficient of the 10G resistor.

author: David Hess
author_rank: Super Contributor
timestamp: October 13, 2016, 04:11:50 pm
content: Quote from: ali_asadzadeh on October 13, 2016, 03:49:27 pmMove it and any shunt protection to the output side of the divider between the divider and high input resistance buffer. The divider itself does not require protection and always presents a 10M input resistance to the input.

author: Kleinstein
author_rank: Super Contributor
timestamp: October 13, 2016, 04:25:31 pm
content: If you use the high voltage amplifier to get the > 10 G input impedance even for the high voltage, the protection is on the input side of the amplifier and the divider is on the output side of the amplifier.With the more normal input with 10 M input resistance for higher voltages, the divider itself usually needs no extra protection. Any protection would be behind the divider and in the extra path for low voltage measurements.

author: ali_asadzadeh
author_rank: Super Contributor
timestamp: October 13, 2016, 04:34:09 pm
content: Thanks,Kleinstein you mean moving the fusible and moves to the branches of the divider? if so why all the DMM schematic protection circuits are similar to the schematic that I provided!?

author: David Hess
author_rank: Super Contributor
timestamp: October 13, 2016, 05:13:50 pm
content: Quote from: ali_asadzadeh on October 13, 2016, 04:34:09 pmThey are not for the reason you identified; the series resistor would screw up the accuracy of the divider. All of the DMM schematics I have show the input protection resistor between the divider and input buffer where it has no effect.Some really cheap meters like the ones without a constant 10M input resistance might do it the way you show.Do you have some links to example schematics?

author: JS
author_rank: Frequent Contributor
timestamp: October 13, 2016, 05:17:26 pm
content: Quote from: ali_asadzadeh on October 13, 2016, 04:34:09 pmFor 3.5 digits with 10M input you can live with 1k drifting as much as you want, 100% drift won't change one LSB. Those meters are the ones on the battle field which may encounter problems big enough for that fuse to mean something. If you take into account normal mode operation and a reasonable temperature range you can live with one or two digits better, that resistor won't self heat in any operational (non faulty) condition.Lab meters in the other extreme (the ones with 10G? inputs) shouldn't be exposed to that kind of abuse, the divider can stand spikes much better than the circuit after the divider, which is protected after the division, in a lower impedance place.A spark gap at the input helps to deal with big spikes, maybe a less problematic but fast enough fuse could be used to be blown if the spark gap sparks.JS

author: zlymex
author_rank: Frequent Contributor
timestamp: October 14, 2016, 02:35:39 pm
content: Quote from: wine+dine on October 13, 2016, 04:10:48 pm1. >10G means exclude 10G2. 100nA is still consider large for many high impedance 1kV source3. not only the voltage coefficient, but the T.C. and aging rate of 10G resistors is worse than a 10M.

author: behzad63
author_rank: Contributor
timestamp: October 14, 2016, 03:01:44 pm
content: Hi FriendsI saw this schematic in the Ti application note ,can we use this buffer for high impedance DMM?if we replace FET and PNP with High voltage FET and PNP its possible?and output follow input correctly?

author: Kleinstein
author_rank: Super Contributor
timestamp: October 14, 2016, 03:58:40 pm
content: You can replace the PNP with a high voltage p-channel MOSFET. For the JFET is might get a little more complicated, but there is the option to add an extra high voltage MOSFET (e.g. depletion mode one) in a kind of cascode / bootstraping way. Bootstraping also helps to get a higher input impedance and less thermal drift.This simple circuit has a limited loop gain, so it might not be that accurate. At least there is quite some DC offset and also quite some offset drift. A more accurate input stage would be more like is difference stage with a pair of fets.The input impedance depends on the parts, the layout and the additional protection circuit you need.

author: behzad63
author_rank: Contributor
timestamp: October 15, 2016, 09:53:30 am
content: Dear zlymex thanks for sharing your time and knowledge with us, I have some basic questions regarding your bootstrap circuit.First of all can I power your circuit with +1100V and -1100V power supply so that I could measure AC voltages?2- what is the purpose of Q1-Q4? I think they are acting like normal diodes, why didn’t you use a normal diode instead?3-what are R6-R9 doing in your circuit? Doesn’t they generate some voltage drop for us, because we should connect the output of this circuit to the 10Mohm voltage divider, also for current limitation you have used Q5,Q6 as a J-FET can we replace them with simple BJT or MOSFET.4- why did you use R1-R5 in series with the op-amp? What are they doing?5- you mentioned the input should be 50V-1000V, so what happens to lower input voltages? For example, the mV and uV input voltages? Also what would affect the circuit performance, I mean which part are critical for achieving the Best lowest temp co-efficient?Thanks in advance

author: Kleinstein
author_rank: Super Contributor
timestamp: October 15, 2016, 12:21:33 pm
content: The maximum voltage is set by the high voltage transistors.Q1-Q4 are used as diodes and one could use low leakage diodes as well.R6-R9 (at the output of the OP) together with the capacitor are used to isolate the OP from a capacitive load. Drop at the resistors is compensated by the OP. The JFETs in the current protection part could be replaced with depletion mode MOSFETs as well. BJT would need a different circuit, as this simple circuit needs depletion mode operation. One might even get away with just a resistor at the OPs output.R1-R5 are for overvoltge protection, the series connection is for a higher peak voltage. An important case to deal with a fast transients, as the circuit by itself has a limited speed to follow fast transients. One might need more protection here, depending on the application.The DC performance is set mainly by the OP. The resistor chains and JFETs might contribute a little via thermal EMF. Leakage from the protection circuit could also contribute a little. The circuit would would also work as a buffer for low voltages. Some DMMs (e.g. Keithly2000) even use a bootstrapped OP based buffer even for the low voltage ranges.

author: David Hess
author_rank: Super Contributor
timestamp: October 15, 2016, 01:04:59 pm
content: Quote from: Kleinstein on October 15, 2016, 12:21:33 pmThis may be to increase the operational amplifier's CMRR (common mode rejection ratio) which would otherwise compromise linearity. This will especially be a problem with JFET and CMOS input operational amplifiers needed for the lowest input bias current because they have a lower CMRR than bipolar operational amplifiers unless they use chopper stabilization which has its own disadvantages.

author: zlymex
author_rank: Frequent Contributor
timestamp: October 16, 2016, 06:28:26 am
content: Quote from: behzad63 on October 15, 2016, 09:53:30 amKleinstein explained almost everything, here are something I'd like to add.1 - Yes, the circuit is designed to use at low frequency AC as well, 60Hz is fine, probably can achieve good result at even higher frequency.2 - Those transistors only have pA leakage, but normal diodes such as 1N4148 have nA leakage.3 - R6-R9 also function as short time short circuit protection.4 - It works fine for small voltages like -10V to +10V, provided it is powered by +15V and -15V.As for the best performance, it depends on whether you need a very small voltage error or very high input impedance. For former, use a precision opamp and small R1 to R7. For later, use opamp with very small bias current and also pay attention to leakage of Q1 to Q4 and guarding.

author: wine+dine
author_rank: Contributor
timestamp: October 16, 2016, 09:42:46 am
content: Quote from: zlymex on October 14, 2016, 02:35:39 pmWhat a really helpful reply, thanks.

author: alireza7
author_rank: Regular Contributor
timestamp: September 18, 2017, 09:00:03 am
content: Quote from: zlymex on October 04, 2016, 03:43:03 amcan you explain the function of q1 q2 q3 and q4 in this circuit?

author: d-smes
author_rank: Regular Contributor
timestamp: September 18, 2017, 10:11:55 am
content: Quote from: alireza7 on September 18, 2017, 09:00:03 amReread replies 51 and 53:Q1-Q4 are used as diodes and one could use low leakage diodes as well.Those transistors only have pA leakage, but normal diodes such as 1N4148 have nA leakage.

author: David Hess
author_rank: Super Contributor
timestamp: September 18, 2017, 11:17:54 am
content: Quote from: d-smes on September 18, 2017, 10:11:55 amI wonder why they included Q1 and Q2 though since the LT1012 shown has these built in.

author: d-smes
author_rank: Regular Contributor
timestamp: September 18, 2017, 04:26:55 pm
content: Quote from: David Hess on September 18, 2017, 11:17:54 amIn an earlier post, zlymex explained that he tried a few different op-amps and not all had built-in protection diodes. Another post said that if protection diodes are built-in, eliminate Q1 and Q2 to lower input capacitance and yield improved phase margin.

author: agaelema
author_rank: Contributor
timestamp: December 27, 2017, 04:40:25 am
content: Where I can find the values of leakage (B-E and B-C) of this bjt and others like 2n3904, 2n2222, etc. Normally datasheet not show this parameters.

author: Kleinstein
author_rank: Super Contributor
timestamp: December 27, 2017, 11:49:31 am
content: Only very few datasheets show leakage for BJTs used as a diode. One can get a first idea from looking at the C-E Leakage with open base, that is more often in the DS: this is higher by about the transistors gain. The pA range currents mentioned are measured values - so not guarantied but typical numbers measured on sample transistors by others. Leakage current is also scattering between samples - so they might need special selected / checked parts.In the better DMMs the protection diodes are usually operated at a very low voltage (e.g. mV range) and thus the leakage is low even for quite normal diodes.

author: David Hess
author_rank: Super Contributor
timestamp: December 28, 2017, 05:19:45 am
content: Quote from: Kleinstein on December 27, 2017, 11:49:31 amJust make sure your "normal" diode is not a gold doped switching diode which is almost all of the small signal ones. They have considerable leakage even at millivolts and present a low impedance at zero volts which just gets worse at higher temperatures. A 2N3904 or 2N2222 is a great way to avoid this problem and cheaper than a real non-switching small signal diode.

author: floobydust
author_rank: Super Contributor
timestamp: December 28, 2017, 06:56:49 pm
content: What about theBAV199? Dual diode and 3pA typicalbut ugh 5,000pA max.terrible spreadIs a diode-connected transistor's spread better for leakage?

author: Kleinstein
author_rank: Super Contributor
timestamp: December 28, 2017, 09:14:20 pm
content: The spread is the difference form testes value (5 nA) to typical value ( 3 pA). Testing for very low leakage currents is expensive as it takes time to settle. Only very few parts would come close to the 5 nA limit (e.g. more than 500 pA).With transistors there are often not even tested specs for leakage, thus typical values only. Not all transistors are equal, especially fast ones can show higher leakage. The actual spread can be similar.In many circuits the critical diodes would be used with a very small voltage across, not the rather high voltage for the leakage specs.The BAV199 is a good choice. The expected leakage at comparable conditions is expected to be lower for the diodes.

author: MisterDiodes
author_rank: Frequent Contributor
timestamp: December 29, 2017, 02:56:18 am
content: ...Or instead of trying to make transistors work with unknown specs - just use a real purpose-built sub-pA leakage (at low voltage) dual protection diode - now you've got a spec for leakage current.These are a based on Jfets, work very well & the cans are light-tight:http://www.linearsystems.com/product-search-result.html?type=products&partnumber=id101Another handy tip - if you're needing a low leakage diode and trying to use something encapsulated in glass (or some plastics) be aware that light hitting the junction will be converted into higher leakage current flow.

author: David Hess
author_rank: Super Contributor
timestamp: December 29, 2017, 03:45:05 am
content: Quote from: floobydust on December 28, 2017, 06:56:49 pmThe BAV199 has the same problem with requiring a test for low leakage and costs more than a transistor. The cheapest way to get a tested low leakage specifications is to use a JFET. It is too bad that Linear Systems' parts do not have better availability.I always found sample testing sufficient for transistors from the same lot.

author: MisterDiodes
author_rank: Frequent Contributor
timestamp: December 29, 2017, 05:08:37 am
content: Quote from: David Hess on December 29, 2017, 03:45:05 amHuh? We've never had a problem ordering - it's easiest to order direct from LS and generally they have many hundreds on hand - at least that's been the case when we order. It's a popular part.

author: floobydust
author_rank: Super Contributor
timestamp: December 29, 2017, 05:40:12 am
content: I've had no problems ordering from Linear Systems, aside from a lighter wallet. You can order directly from Linear Systems California sales or theirdistributorssuch asTrendsetter Electronicsin Texas, it was painless.I don't know how you'd make a 10Gohm input resistance multimeter and not have clamp leakage current in the way.I did not see protective clamps in the Keithley 2000, 2001 multimeters (but don't have official schematics) and wondered if this is why JFETS are failing in them, from (test lead) ESD?There's no ESD or impulse rating on those LS ID101 lovely gold legs diodes, just a "20mA max."

author: Kleinstein
author_rank: Super Contributor
timestamp: December 29, 2017, 04:15:07 pm
content: The meters do have clamping inside.In the Keithley2000 they use LEDs inside an optocoupler towards bootstrapped zener diodes. A problem however might be that the current limiting relies on MOSFETs and a rather small resistor. So over-voltage with enough power might go through to sensitive parts. For too high a voltage at the input, there is a spark gap, but this might be slow sometimes. It's also parasitic effects (series inductance) that can cause ESD to reach point one does not expect.

author: Alex Nikitin
author_rank: Super Contributor
timestamp: December 30, 2017, 02:29:09 am
content: I'vemeasured a number of BAV199 diodesand some makes of these are the best choice for low leakage protection IMHO. The spread from good makes is surprisingly small. You need to keep things very clean though.CheersAlex