how to get ground referenced amplified V difference from high side (250VDC) shunt?

--- oOPS...

It should look like this:

. FWB REG . +----+ +-----+ .120AC>-+ +--|~ +|---+---| |--+------------+------>Vcc . P||S | | |+ +--+--+ | | . R||E | | [BFC] | | | . I||C | | | | | | .120AC>-+ +--|~ -|-+-+------+ +--|----+ | . +----+ | | | | | ADC . | | Vcc [RFB] +---+---+ .250DC>------------+-+-[Rs]--+---+-|-\\ | | Vcc | . | | | >--+---|IN OUT1|-->ADOUT1 . +---------|---+-|+/ . . . | | GND | OUT8|-->ADOUT8 . | | | | GND | . | +--+ +---+---+ . [LOAD] | | . | +------------+ . | .250GND>---------------------+

-- JF

Reply to
John Fields
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Neat... I've tried to tackle a similar impedance/level translation problem some time ago and came up with a dozen lame designs and few good ones.

One design was based on the LTC6101 differential inputs across the I sense resistor.

D from BC

Reply to
D from BC

Yes, but that IC only goes to 100V and 100kHz, the design above can be easily extended to 1kV and goes beyond 1MHz.

Reply to
Winfield Hill

Two more problems: if the amp rails, it can cave in the power supply, maybe latching things up. Or it might turn off Q2 through the b-c junction of Q1.

A fet for Q1 helps. A resistor between Q1 and Q2 fixes the other problem.

It's almost appealing to go back to discretes, especially for a fast differential version.

Simple circuits can have so many hazards!

John

Reply to
John Larkin

--- Sure, but's that not what I was suggesting.

Drawing it this way:

. FWB REG . +----+ +-----+ .120AC>-+ +--|~ +|---+---| |--+------------+------>Vcc . P||S | | |+ +--+--+ | | . R||E | | [BFC] | | | . I||C | | | | +--|----+ | .120AC>-+ +--|~ -|-+-+------+ | | | | . +----+ | | Vcc [RFB] | ADC . | | | +---+---+ .250VDC>-----------+-+-[Rs]--+---+-|-\\ | | Vcc | . | | | >--+---|IN OUT1|-->ADOUT1 . +---------|---+-|+/ . . . | | GND | OUT8|-->ADOUT8 . | | | | GND | . | +--+ +---+---+ . [LOAD] | | . | +------------+ . | .GND>------------------------+ May make the concept easier to understand, which is that you can float everything on the 250VDC signal, using it as a pseudoground, and do your processing something like this:

250 + LVDC / +----------+----------+---------+ Vcc | | | | | [LVDC SUPPLY] | | | [R] | | |ADC |µC OPTO | | +---|-\\ +--+--+ +---+---+ +----+ | | | | >--+--|A D|--|I/O I/O|--|A C|-+->OUT 250VDC---+-----|-+-|+/ | +--+--+ +---+---+ | | | | | | | | | | | [RS] | +--+----|-----+---------+------|K E|-+->GND | | | \\ +----+ | +-----+--[RF]---+ PSEUDOGROUND | | | [RL] | | | GND>-----+---------------MAYBE-------------------------+

What does your system look like, anyway?

--- Only the one on the load side.

---

--- Here's what you're talking about:,

E1 E2 | R5 |

+250>--+--[SHUNT]--+--[LOAD]--+ | | | [R1] [R3] | | | | +-E3 +-E4 | | | | [R2] [R4] | | | | GND>---+-----------+----------+

and if we wanted to start putting some numbers in there, we'd start with the shunt and the load.

Since you're looking for low millivolts out of the shunt let's assume the entire 250V is dropped across the load. Then with your specified 10mA max current into the load, it'll look like:

E 250V R = --- = ------- = 25000 ohms I 0.01A

Now, assuming "low millivolts" means 10mV with 10mA through the shunt and the load, that makes the shunt resistance:

E 0.01V R = --- = ------- = 1.0 ohm I 0.01A

So your circuit now looks like this:

E1 E3 | R5 |

+250>--+--[1R]--+--[25kR]--+ | | | [R1] [R3] | | | | +-E2 +--E4 | | | | [R2] [R4] | | | | GND>---+--------+----------+

Now, assume you've got a rail-to-rail input opamp which you can drive from a 25V supply and which has a common mode range from 0V to the supply voltage. Then its inputs will have to be slightly below

25V and to get there you'll need to drop the 250V to --+--[1R]--+--[25kR]--+ | | | R1[226k] [226k]R3 | | | | E2--++--E4 | | | | R2[24k9] [24k9]R4 | | | | GND>---+--------+----------+

With the 250V at zero, E2 = E4 = 0V, and the output of your instrumentation amp would be 0V, ideally.

With the supply at 250V, though, we have for E2:

E1 * R2 250V * 24.9kR E2 = --------- = ---------------- = 24.81068 volts R1 + R2 226kR + 24.9kR

For E3, assuming the load takes 10mA and R3R4 takes 1mA, we have:

E3 = E1 - (I * R5) = 250V - (11mA * 1R) = 250V - 11mV = 249.989V

And for E4:

E3 * R4 249.989V * 24.9kR E4 = --------- = -------------------- = 24.80959 volts R3 + R4 226kR + 24.9kR

The difference between E2 and E4 is:

E5 = E2 - E4 = 24.81068V - 24.80959V = 0.00109 volts

so to get 5V out of your opamp with 0.00109 volts in, you'll need a gain of:

Eout 5.000V Av = ------ = ---------- ~ 4587 Ein 0.00109V

Tricky at 1MHz. ;)

Of course you could always increase the resistance of the shunt.

10 ohms would get you to a gain of ~ 459 and 100 ohms to ~ 46.

-- JF

Reply to
John Fields

Winfield Hill snipped-for-privacy@rowland.org posted to sci.electronics.design:

As seems to have been conventional for North America and most of Europe since the 1930's or so.

Reply to
JosephKK

--
Or,  if you wanted to do it my way:

Version 4
SHEET 1 1216 1252
WIRE -48 384 -224 384
WIRE 224 384 80 384
WIRE 432 384 304 384
WIRE 560 384 432 384
WIRE 752 384 640 384
WIRE -224 448 -224 384
WIRE 608 464 192 464
WIRE 608 496 608 464
WIRE 432 544 432 384
WIRE 576 544 432 544
WIRE -224 560 -224 528
WIRE -224 560 -288 560
WIRE 752 576 752 384
WIRE 752 576 720 576
WIRE 784 576 752 576
WIRE -224 592 -224 560
WIRE -48 592 -48 384
WIRE 80 592 80 384
WIRE 192 592 192 464
WIRE 432 592 432 544
WIRE 576 608 512 608
WIRE -224 704 -224 672
WIRE -48 704 -48 672
WIRE -48 704 -224 704
WIRE 80 704 80 672
WIRE 80 704 -48 704
WIRE 192 704 192 672
WIRE 192 704 80 704
WIRE 432 704 432 672
WIRE 432 704 192 704
WIRE 512 704 512 608
WIRE 512 704 432 704
WIRE 608 704 608 656
WIRE 608 704 512 704
WIRE 656 704 656 656
WIRE 656 704 608 704
WIRE 752 704 656 704
WIRE -224 784 -224 704
FLAG -224 784 0
FLAG 752 704 +250VDC
SYMBOL res 656 368 R90
WINDOW 0 -47 59 VBottom 0
WINDOW 3 -44 59 VTop 0
SYMATTR InstName R4
SYMATTR Value 499
SYMBOL voltage 80 688 R180
WINDOW 3 24 104 Invisible 0
WINDOW 123 0 0 Left 0
WINDOW 39 0 0 Left 0
WINDOW 0 15 113 Left 0
SYMATTR Value PULSE(0 250 1 3e-6)
SYMATTR InstName V3
SYMBOL res 208 400 R270
WINDOW 0 70 56 VTop 0
WINDOW 3 69 56 VBottom 0
SYMATTR InstName R9
SYMATTR Value 25k
SYMBOL res 448 688 R180
WINDOW 0 36 76 Left 0
WINDOW 3 46 46 Left 0
SYMATTR InstName R5
SYMATTR Value 1
SYMBOL voltage 192 576 R0
WINDOW 3 24 104 Invisible 0
WINDOW 123 0 0 Left 0
WINDOW 39 0 0 Left 0
WINDOW 0 -37 0 Left 0
SYMATTR Value 10
SYMATTR InstName V1
SYMBOL Opamps\\\\LT6230-10 608 576 R0
SYMATTR InstName U1
SYMBOL res -240 432 R0
SYMATTR InstName R1
SYMATTR Value 490
SYMBOL res -240 576 R0
SYMATTR InstName R2
SYMATTR Value 10
SYMBOL voltage -48 576 R0
WINDOW 3 24 104 Invisible 0
WINDOW 123 0 0 Left 0
WINDOW 39 0 0 Left 0
WINDOW 0 -39 -1 Left 0
SYMATTR Value PULSE(0 250 1 3e-6)
SYMATTR InstName V2
TEXT -56 736 Left 0 !.tran 2
TEXT 312 632 Left 0 ;SHUNT
TEXT 232 424 Left 0 ;LOAD
TEXT -496 560 Left 0 ;PERFECT dV/dT <
TEXT 784 576 Left 0 ;> REAL dV/dT


Probe the perfect edge and then the real edge and then zoom in to
see the details.


The chip goes for about $2 US.
Reply to
John Fields

The original post was about producing a ground referenced voltage proportional to load current with 8-bit accuracy and 1MHz bandwidth, so what has changed?

Deconvolution has little to do with the sampling theorem...

Reply to
Fred Bloggs

But please explain what mathematical transforms you had in mind, and what analog requirements could be relaxed, when you wrote

"This means a more intelligent approach would relax the complexity and performance requirements of the analog kluge and compensate with a deconvolution of the data."

John

Reply to
John Larkin
[snip]

Fred doesn't know, he just copied the verbiage from Wikipedia because it sounded intelligent ;-)

...Jim Thompson

--
|  James E.Thompson, P.E.                           |    mens     |
|  Analog Innovations, Inc.                         |     et      |
|  Analog/Mixed-Signal ASIC\'s and Discrete Systems  |    manus    |
|  Phoenix, Arizona            Voice:(480)460-2350  |             |
|  E-mail Address at Website     Fax:(480)460-2142  |  Brass Rat  |
|       http://www.analog-innovations.com           |    1962     |
             
         America: Land of the Free, Because of the Brave
Reply to
Jim Thompson

Yup. Sounded like word salad to me.

John

Reply to
John Larkin

John, I'll have to say, no, that's wrong. If you think it through carefully, you'll see those states either cannot occur, or they quick move on to a safe operating condition.

Reply to
Winfield

Actually, the 9.90k resistor leaves the bias cancellation in error by 1%. So, correcting this minor problem:

Fast high-voltage high-side current monitor, r2-c

10.0 +250V Iout ------+------+---/\\/\\----+----o ---> 0 to 10mA load | | | / '--+--------|---, \\ 200 | | | / 1% | 190 / | | | 1% \\ | +---------|----, / +---, | LT1783_|_ | | | | 7.5V zener Q1 e / -|--' | _|_/ | fast b ---< | | /_\\ _|_ PNP c \\__+|------+ | --- 0.1uF | | | | | Q2 e +--------|---+---+ mpsa92 b ------' | | c 470k \\ / 330k | Ib = 0.5 0.5W / \\ 0.5W | to 1.0mA \\ Ia / | | | 1k / ,------' gnd \\ | 10.0k / +---/\\/\\---, | 1.111k | __ | 0 to +5V out | ,--/\\/\\--+--|- \\ | for 0 to 10mA | gnd | >---+------ +--------------|+_/ | 1.00k '---/\\/\\/--- gnd
Reply to
Winfield

Apparently you have absolutely no clue of the analog GBW requirement of the circuit to maintain less than 1/2 LSB error at 8 bits out to 1MHz. Obviously you would want to keep things simple, something like a dominant pole low pass roll-off for that, then unless you have a DSP with FFT, you would use a time domain convolution to reconstruct the unfiltered data record of the current. Maybe this doesn't have to be done in real time, maybe the circuit can just acquire the data on both the current profile and the sensor response to a known stimulus, and this can be processed later, download the files to a PC or workstation for processing and display. I'm not real concerned about the OP and his imagined requirements, skill set, or anything else, including your opinion or ideas, you're so dull and boring...

Reply to
Fred Bloggs

What neither you nor I have a clue of is whether the op expects 1/2 lsb at 1 MHz, or whether 1 MHz might be an acceptable -3 dB point. In the real world (hello!) the latter would be the likely situation. Especially seeing that the latter is totally out of the league of this situation.

You can't computationally replace bits once they are lost. The key to accuracy here is doing the analog stuff (and digitization) right, not massaging the data after it's trashed. Yes, analog circuit design bores you.

What doesn't bore you?

John

Reply to
John Larkin

On a sunny day (Wed, 12 Dec 2007 07:23:08 -0800) it happened John Larkin wrote in :

I have thought of a third way to do this. Say you use a 74HC4054 CMOS switch on the high side, that switches at about 10MHz between both ends of the shunt, and apply that 10MHz carrier, with amplitude exactly equal to the voltage drop over the shunt, via a 1:1 transformer to an opamp precision peak rectifier on the ground level. I am still thinking what to use to swith the 4053, a 555 is not fast enough. UJT?

Reply to
Jan Panteltje

On a sunny day (Wed, 12 Dec 2007 15:46:24 GMT) it happened Jan Panteltje wrote in :

I have got it, integrated xtal osc module.

Reply to
Jan Panteltje

The high-side stuff could be completely unpowered: one transformer sending a carrier square wave up, to switch a pair of small mosfets, and a second transformer to bring the modulated signal back down.

Somebody (one of my competitors, actually) does thermocouple acquisition that way. We float the entire preamp/adc per channel, with a dc/dc converter powering the floating stuff and an Analog Devices logic isolator bringing the data back down,

John

Reply to
John Larkin

Yes, forward-biasing the b-c junction of Q1 won't happen, bacause the power supply would cave in first.

Still, startup is interesting, but probably OK.

John

Reply to
John Larkin

You don't what you're talking about. Deconvolution no more manufactures lost bits than pre-emphasis 'cheats' line loss. Deconvolving a linear circuit is the simplest possible use.

Reply to
Fred Bloggs

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