High-side 450V, micro-amp sensing - Page 1
Forum section: electronics / beginners

author: sdouble
author_rank: Frequent Contributor
timestamp: December 10, 2020, 11:36:50 pm
content: Dear all,I am currently after a reasonnable scheme for sensing the current sucked by silicon detectors (400 to 450 V biasing). nA resolution would be great, 10 nA would make it. 10 uA range.I never did that. would you use an optocoupler to isolate the amp located in the high side from gnd ?thanks for your help

author: fourtytwo42
author_rank: Super Contributor
timestamp: December 11, 2020, 09:25:40 am
content: A LEM module with many turns wrapped around it ?Not much info to go on.

author: Le_Bassiste
author_rank: Frequent Contributor
timestamp: December 11, 2020, 01:54:39 pm
content: Quote from: fourtytwo42 on December 11, 2020, 09:25:40 amLEM module at (say) 10 A full scale and TO's 10 µA max. range gives 1000000 turns. what else info do you need?

author: KrudyZ
author_rank: Frequent Contributor
timestamp: December 11, 2020, 04:12:36 pm
content: Quote from: sdouble on December 10, 2020, 11:36:50 pmAssuming you need DC current measurements, I would suggest using the standard shunt resistor method. This allows a low bias current front end OPAMP.Then feed this into one of the isolated sigma delta ADCs you can get from TI or Analog Devices. Pay attention to the leakage current your DC/DC converter adds, so make sure you make the common the side of the sense resistor that doesn't care, i.e. don't have the leakage current flow through the sense resistor.

author: jmelson
author_rank: Super Contributor
timestamp: December 11, 2020, 05:22:36 pm
content: Quote from: sdouble on December 10, 2020, 11:36:50 pmCan you isolate the ground side of the power supply? If the bias supply is fully floating, then it will be MUCH easier to measure the current on the ground side. This is typically how it is done on such bias supplies.If you REALLY need to have it on the HV side, you need a transformer-coupled power supply for the measurement circuit, a sense resistor, an instrumentation amp and a serial ADC that can send the reading across an opto-coupler. WAY more complicated to do it that way.Jon

author: fourtytwo42
author_rank: Super Contributor
timestamp: December 11, 2020, 10:04:40 pm
content: Quote from: Le_Bassiste on December 11, 2020, 01:54:39 pmAnd your contribution was ?I made the suggestion as the first thing that came off the top of my head as a potentially cheap solution not knowing what the range of current was, for example what any surge might be. Try to be a bit more considerate when slinging mud

author: Wolfgang
author_rank: Super Contributor
timestamp: December 11, 2020, 10:40:55 pm
content: Quote from: sdouble on December 10, 2020, 11:36:50 pmHow precise ?

author: Doctorandus_P
author_rank: Super Contributor
timestamp: December 13, 2020, 09:13:44 pm
content: My first approach would be to use a microcontroller with a suitable frontend, and then push the digitized signal through some isolation barrier.

author: sdouble
author_rank: Frequent Contributor
timestamp: December 13, 2020, 10:55:04 pm
content: dear all,thanks for the feedbacks.They really help. No way to measure on the low side due to the fact that the ground will not float.I thought about using the shunt resistor approach and the LTC6102 as shown page 23 of the datasheet : 2 couples of mosfets and diodes would make it. The power consumption seems a bit high but the dynamic range is probably ok : 80 dB seems reachable.Any thought ?

author: exmadscientist
author_rank: Frequent Contributor
timestamp: December 14, 2020, 03:42:36 am
content: The detailed design and whether it is a proper fit for your requirements is up to you, but that architecture is very useful and a completely viable choice.If you haven't come across it already, take a look atLTC AN105: Current Sense Circuit Collection. It does try to sell you a lot of LTC products, as these things do. Just remember that many of the same circuits are possible without the expensive single-source ICs, if you are willing to trade off some accuracy or overall complexity.

author: David Hess
author_rank: Super Contributor
timestamp: December 15, 2020, 06:40:33 pm
content: The high voltage supply is unlikely to provide enough current to drive an optocoupler and even if it did, the return current for the optocoupler to ground defeats the purpose of using it because if that is acceptable, then there are better ways; see below. The non-linearity of an optocoupler will also limit accuracy.The problem with return current can be solved with a floating low voltage supply to power the high side circuit, but the non-linearity of the optocoupler will still limit accuracy.A high side current sense amplifier using a micropower low input bias current operational amplifier is feasible. One or more high voltage p-channel MOSFETs are used to cascode the output allowing it to drive a ground referred load. Usually a separate ground connection is required to power the circuit however with some cleverness, the connection to ground can both power the high side sense circuit and act as the signal return by subtracting the operating current on the ground side.I would consider a part like the LTC2066 because it draws 7.5 microamps and has a noise of 1.5 microvolts however the LTC2063 draws 1.4 microamps for 4.6 microvolts of noise if that is acceptable and a lower supply current is required. The LT1494 is another candidate. For the transistors, there are some small 500 volt p-channel MOSFETs available so given the voltage, I might use two in series which will require basing their gates to suitable levels with a resistor divider but the same resistor divider can provide power to the circuit.

author: Renate
author_rank: Super Contributor
timestamp: December 15, 2020, 07:55:30 pm
content: Quote from: David Hess on December 15, 2020, 06:40:33 pmSay what? Who is advocating for pushing analog though an optocoupler?I go to stuff I know.Use a TI INA226 high side current sensor.No, it doesn't work for 450 V.Make a 5 V supply down from the 450 V.Then you have current sense with 5 V common mode.Send it into a µP. Dare I recommend an AVR8? Maybe an ATTiny85?Pump it into an optocoupler. I'll try to live with the non-linearity.

author: David Hess
author_rank: Super Contributor
timestamp: December 16, 2020, 10:07:26 am
content: Quote from: Renate on December 15, 2020, 07:55:30 pmObviously what I said does not apply for digital data but using a digital isolation barrier with optocouplers has the other problem I identified in trying to power the optocouplers and the optocoupler loads. The linear method I suggested without an optocoupler can work within 20 microamps and even 10 microamps with some care.

author: Renate
author_rank: Super Contributor
timestamp: December 16, 2020, 12:15:20 pm
content: Well, a TI INA226 wouldn't work so hot as its input bias of 10 µA would have to be considered.How fast does this circuit have to be?How protected does it have to be from flash over or shorts?Would a low-tech current mirror with two (high voltage) transistors work?

author: David Hess
author_rank: Super Contributor
timestamp: December 16, 2020, 08:53:10 pm
content: Quote from: Renate on December 16, 2020, 12:15:20 pmThat is essentially what I suggested but to get the needed precision, an operational amplifier is used to correct the offset voltage between the current mirror transistors and once that is done, a slightly different circuit is used with resistors replacing the transistors.A current mirror, with an operational amplifier to correct offset, would be suitable in an application which requires a very high dynamic range but can tolerate less accuracy. What is really going on is that the first transistor takes the logarithm of the current and then the second transistor takes the antilogarithm producing the original current or some fraction of it. The temperature compensation that a logarithm or antilogarithm amplifier requires is not necessary in this case because the transistors operate at the same temperature and compensate each other. (1)If in this application a current of up to 10 microamps with 1 picoamp resolution was required, then a transistor current mirror would be more suitable.(1) This circuit shows up when you have an NPN logarithm amplifier which requires a negative input current, and you want to use a positive input current so you mirror it. You cannot do this with a normal inverting amplifier because it lacks the dynamic range of the logarithm amplifier. Thedatasheet for the Burr-Brown LOG101briefly mentions it.

author: sdouble
author_rank: Frequent Contributor
timestamp: December 16, 2020, 09:08:48 pm
content: Dear David, dear Renate,thanks again for your suggestions.Power consumption could be an issue, this is clear. My current design sucks 0.3mA from the HV source. Not terribly high, but no negligible either.I'll look at the LTC2066 option in the next days.

author: Renate
author_rank: Super Contributor
timestamp: December 17, 2020, 12:30:43 am
content: You probably have your 450 VDC already.If you want to get good retro-credibility use an old TV vacuum tube transformer.Those had the high voltage secondary, but also the 5 VAC secondary for the filaments of your 5U4GB.The filaments ran raised at the full B+ voltage.I wonder if the 5 VAC secondaries were better insulated than the 6.3 VAC secondaries?

author: sdouble
author_rank: Frequent Contributor
timestamp: December 17, 2020, 02:19:55 am
content: Until now, I just have LTSpice simulation.The boards and the associated enclosure are supposed to come in January.Design phase first and JLCPCB will work in early January !

author: dcarr
author_rank: Regular Contributor
timestamp: December 17, 2020, 02:34:17 am
content: If this is a one-time or shorter term measurement, consider using a few AA batteries for the floating power supply. Heck if the current consumption is low, they may even be feasible "in production".

author: sdouble
author_rank: Frequent Contributor
timestamp: December 17, 2020, 07:54:26 am
content: Not Really : 12 Volts DC generates my 450 V. 100-150 mA sucked out of the 12V.

author: Le_Bassiste
author_rank: Frequent Contributor
timestamp: December 17, 2020, 08:41:45 am
content: Quote from: fourtytwo42 on December 11, 2020, 10:04:40 pm1. my contribution was the emphasis on the fact that for a given LEM module, a million turns are not practicable. so, saved all the other readers some time following the path of a brainfart.2. asking as to what else info you would actually require to come up with a viable solution wasn't "slinging mud", it was asking for a competent answer.3. the current range was stated by TO in post #1

author: fcb
author_rank: Super Contributor
timestamp: December 17, 2020, 10:42:25 am
content: So just starting from your original posting.You want to measure the current used by a silicon detector driven by a 450V rail. You want to do this on the high-side and at nA or better resolution.1. What bandwidth?2. What are you doing with the measurement (is it ending up in a microcontroller or computer?), is it part of control loop?3. What is the minimum current and the maximum current? 10nA to 10uA in 1nA steps? Accuracy? Linearity? Do you need 1nA at 10uA? Can you cope with range switching?4. Is there a budget? Is it a one off, or are you making lots? Does cost matter?You might for example do the sensing and sampling using a battery powered (isolated) circuit floating on your 450V rail and then send down digits over a piece of plastic fibre optic (POF). Or perhaps just mulitply up the current sensed and use a current mirror to transfer this to 0V. Or use a logarthymic converter and an optocoupler...

author: sdouble
author_rank: Frequent Contributor
timestamp: December 17, 2020, 11:02:36 am
content: Thanks for your interest.My first post was a bit short maybe. I do need to power a set of multistripped silicon detectors. A total of 16 detectors. They will be powered individually. I expect the leakage current to increase during irradiation startin from about 10-50 nA to 10 - 50 uA. the idea is to follow the leakage current increase. I just want to check that the leakage current is low (say below 40-50nA) at the beginning of the experiment and do not saturate before 50uA. THe 50uA upper limit does not need to be measured with a very high accuracy : a few percents is enough.Bandwith request is of 1kHz only. Nothing fancy here.about 20 units need to be produced. I am not much budget limited. I do not assume this device to cost a fortune anyhow.At first, the output will be connected to a 10 Bit scope . THen, we'll move for a full integration into the data acquisition slow control system through a ADC.Quote from: fcb on December 17, 2020, 10:42:25 am

author: fcb
author_rank: Super Contributor
timestamp: December 17, 2020, 11:57:45 am
content: Quote from: sdouble on December 17, 2020, 11:02:36 amFrom what you describe I would build the circuit like so:Range switching would ease the development of the ADC and allow you possibly to just use the 12bit one in alot of micro's, or get a decent 12-16bit one from the usual suspects.Isolation of the supply is probably simple for what you want, and you might be able to get away with one of those little RECOM style 1-2W modules.If you sync all the units together, then you could probably create a giant OR with all the received data and just end up with one 1Mbit RS232 line.

author: sdouble
author_rank: Frequent Contributor
timestamp: December 17, 2020, 12:20:03 pm
content: thanksMy source of HV is already defined. It's an integrated module from XP POWER : CA05P

author: fcb
author_rank: Super Contributor
timestamp: December 17, 2020, 12:23:47 pm
content: Quote from: sdouble on December 17, 2020, 12:20:03 pmThe txfmr/isolate PSU on my dwg isn't for your HV - it's to drive the isolated electronics that measure your current.

author: mikeselectricstuff
author_rank: Super Contributor
timestamp: December 17, 2020, 12:38:24 pm
content: Quote from: fcb on December 17, 2020, 12:23:47 pmWhy not just use an off-the-shelf isolated DC-DC converter?

author: fcb
author_rank: Super Contributor
timestamp: December 17, 2020, 01:13:55 pm
content: I did suggest you could also use one of those isolated modules. As the OP is already using a module to generate the 450V then high-frequency coupling is unlikely to be a factor. Those little 1-2W DCDC blocks have isolation at low frequency (i.e. 50/60Hz), but tends to drop off rapidly (interwinding capacitance etc..) as the frequency goes up.Good example of the lengths you have to goto sometimes when building isolated supplies for nA/nV measurements in one of daves teardowns of a Tek/Keithley DMM)

author: mikerj
author_rank: Super Contributor
timestamp: December 17, 2020, 01:36:04 pm
content: Quote from: fcb on December 17, 2020, 11:57:45 amI don't see any DC connection between the HV supply and the LV supply, apart from maybe through ESD diodes in the op-amp if it has any...

author: fcb
author_rank: Super Contributor
timestamp: December 17, 2020, 01:49:14 pm
content: My sketch was more of a 'pointer' rather than a fully worked design! 0VISO, +VISO, -VISO are the isolated supply. I've added some red ink to help:

author: mikerj
author_rank: Super Contributor
timestamp: December 18, 2020, 05:19:10 pm
content: What ensures that +Viso and -Viso are sitting symmetrically around the 450V rail? There is no DC connection between them, just some decoupling caps. The non-inverting input to the op-amp will be at 450v wrt ground, but the voltage between the non-inverting input and the op-amp supply rails is undefined as the supply rails are isolated and not DC referenced to anything.

author: Renate
author_rank: Super Contributor
timestamp: December 18, 2020, 07:13:22 pm
content: I presume that they forgot to draw in the CT of the secondary?

author: fcb
author_rank: Super Contributor
timestamp: December 18, 2020, 07:50:05 pm
content: Yup, draw a centre-tap. It was literally a two minute sketch.. E&OE