How a precision DMM have a high input impedance in the Gohm Rang?

[David Hess]

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?

A varactor diode bridge type of operational amplifier could 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.

[JS]

Having a 1.1K high voltage in a DMM .....

That 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.

Or 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

[behzad63]

Hi friends
i 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.html
Do you have any circuit to Implementation it for DMM?
Thanks a lot

[David Hess]

Hi friends
i 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.html
Do you have any circuit to Implementation it for DMM?
Thanks a lot

Current 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.

[behzad63]

Hi Friends
Thanks 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 

[bktemp]

The output will not be very smooth unless you have a suitable LC or RC output filter after the rectifiers but then it should be good enough unless you need really low output noise.

Actually 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

[David Hess]

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.

[ali_asadzadeh]

Does this simple boost single transistor works?

https://learn.adafruit.com/diy-boost-calc/the-calculator

[bktemp]

Does this simple boost single transistor works?

A 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

[ali_asadzadeh]

Thanks bktemp,Sound a better Idea

[behzad63]

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?

[David Hess]

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?

Not 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.

[ali_asadzadeh]

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.

[wine+dine]

Having a 1.1K high voltage in a DMM .....

That 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.

Not 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.

[David Hess]

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.

Move 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.

[Kleinstein]

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.

[ali_asadzadeh]

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!?

[David Hess]

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!?

They 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?

[JS]

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!?

  For 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

[zlymex]

Having a 1.1K high voltage in a DMM .....

That 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.

Not 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.

1. >10G means exclude 10G
2. 100nA is still consider large for many high impedance 1kV source
3. not only the voltage coefficient, but the T.C. and aging rate of 10G resistors is worse than a 10M.

[behzad63]

Hi Friends
I 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?

[Kleinstein]

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.

[behzad63]

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

[Kleinstein]

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.

[David Hess]

Some DMMs (e.g. Keithly2000) even use a bootstrapped OP based buffer even for the low voltage ranges.

This 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.