High bandwidth, isolated differential probe. - Page 1
Forum section: electronics / projects

author: Mandragora
author_rank: Contributor
timestamp: September 05, 2015, 05:42:19 pm
content: Hello everyone, I want to consult with You my idea for high bandwidth, isolated differential probe.Most of the ideas that I've met were basing on instrumental amplifier and voltage divider (100:1 or 1000:1). Many of them don't allow measurements of low voltage (less than 1V). The reason is that in design with instrumental amplifier DUT and scope aren't really separated.After searching a solution to measure etc. 10mV voltage drop across high side transistor (100V from ground) working at 500kHz I found suitable optocoupler.Main problem in linear (analog) optocouplers is bandwidth, most of them got smething about 100kHz, only few are close to 1Mhz which is useless when i need to see 500kHz a square wave. Fortunately there is one exception, the holy HCPL-4562, linear optocoupler (0,25%) with 17MHz bandwidth (-1db at 6MHZ, -1,85db at 10MHz). Best part is price, something about 3$.I want to make a probe using it as a main part. Below is a circuit showing basic operation (without any protection etc, it will be disused later).The input part:I need bipolar input but OPt can transfer only unipolar signals. It is recommended in data sheet to apply 10mA quiescent current to input diode in that case. After few calculations I came up with small table. +-10V on input will be reflected in +-4mA current change.Amplifier acts as V/I converter.+10V input voltage (buffered by amplifier) +23,75V (bias voltage) giving 33,75V across 2,5k resistor which is 13,5mA.0V input voltage (buffered by amplifier) +23,75V (bias voltage) giving 23,75V across 2,5k resistor which is 9,5mA.-10V input voltage (buffered by amplifier) +23,75V (bias voltage) giving 13,75V across 2,5k resistor which is 5,5mA.The output:Now we need to transfer current back to voltage. I/V converter is a transimpedance amplifier (many circuits can be fount in the net). With 390k resistor output voltage will vary from 6,864V (-10V input) to 16,848V (+10Vinput). When we will add bias voltage +11,856V to the ground of a scope we will be able to get +-5V signal.In summary this basic idea allow to accurately measure +-10V, 1000V over common ground.==================================================Now I am writing more practical part which will include part selection, input protection, supply voltage etc.I'm open for any suggestions, if I've made mistake, please point it out and lets find solution together.RegardsMandragora

author: cat87
author_rank: Regular Contributor
timestamp: September 07, 2015, 07:19:54 am
content: Very nice and interesting projectI'm curious what the CMRR of this probe would be.

author: tggzzz
author_rank: Super Contributor
timestamp: September 07, 2015, 08:48:39 am
content: A 400kHz COTS device:http://www.analog.com/en/products/interface-isolation/isolation/standard-digital-isolators/adum4190.html

author: physicboy
author_rank: Newbie
timestamp: December 05, 2015, 03:35:16 pm
content: Hi, I am currently also working on the same problem voltage probe with 4562I am thinking about making isolated voltage probe to probing my smps project.do you already implement it?why do you inteded to use opamp instead of the bjt circuit from the datasheet?what is the opamp you gonna choose for this project?AD8651?Please let me know, I would be very happy if we can collaborate and share data between each other.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: December 05, 2015, 10:51:42 pm
content: I expect that, without extra isolation channels for feedback or self calibration, you'll have a hard time reading a useful signal out of that circuit. Optos are notorious for tempco, drift and aging, so you'll always be trimming it for offset and gain.Tim

author: MagicSmoker
author_rank: Super Contributor
timestamp: December 05, 2015, 11:19:12 pm
content: If I were to roll my own diff probe the approach I would try first is a dual channel delta-sigma converter using cheap pulse transformers for isolation. The so-called linear optocoupler you are using isn't really all that linear ; you really need two matched photodiodes in the package so that one can be used for feedback. A couple of examples are HCNR200 and LOC111. Best bandwidth you'll get with either is around 300kHz, IIRC. (Correction - the HCNR200 is supposedly good to 1MHz; the LOC111 tops out around 200kHz).

author: free_electron
author_rank: Super Contributor
timestamp: December 05, 2015, 11:47:11 pm
content: you'll never get this to be linear without optical feedback.take the base circuit used in a agilent 3631 power supply . they use another optocoupler to get the thing to be linear.but, here is another question : since it is a diff probe : why do you need isolation ?

author: Mechatrommer
author_rank: Super Contributor
timestamp: December 06, 2015, 04:10:26 am
content: yup why isolated? if its due to high voltage input, there is no easy KV (even mains) rated on the left opamp U? anyway...

author: MagicSmoker
author_rank: Super Contributor
timestamp: December 06, 2015, 04:46:37 pm
content: Quote from: free_electron on December 05, 2015, 11:47:11 pmLOL... I dashed off a reply without even considering the obvious here... Good catch!

author: DanielS
author_rank: Frequent Contributor
timestamp: December 06, 2015, 05:22:24 pm
content: Quote from: free_electron on December 05, 2015, 11:47:11 pmBecause OP wants to measure a 10mV signal that's at 100V common-mode voltage and differential amplifiers often have a worse offset from common-mode voltage than the signal of interest. Also, the signal of interest is 80dB below common-mode voltage, which is near the limit of many opamps' CMRR.