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

On Dec 7, 7:59 pm, Michael wrote: > Hi - I want to put a high side shunt resistor on a grounded load > (~10ma peak) that has about 250VDC across it. I then want to have a > ground referenced voltage that is the amplified difference across the > shunt with a range of about 0-5V. I would like the voltage across the > shunt to be in the low mV range. Power supplies available to me will > be the 250V, which will be noisy, 5V, and probably some sort of

+-12V > analog (clean) supply. I would like to get 8b or more of resolution > from the ground referenced amplified shunt voltage signal. I would > also like this circuit to be fast. The faster the better, MHz > bandwidth at least. If this were a lower voltage I'd just do an op- amp > difference amplifier. Unfortunately, last I checked, Apex had the > market for op-amps that can handle such high voltages fairly well > cornered, and their op-amps are expensive, large, and not particularly > quick. High side current amplifiers seem to top out at about 100VDC. >

The resistor matching becomes nearly impossible an dthe opamp offsets and DC performance is tough, too. Back in the '70s we were retrofitting remote control and metering onto a 1956 RCA TV transmitter. To monitor high side ( and in a transmitter, high side is HIGH - several kV ) the manufacturer made a gadget that converted the current to an AC voltage, transformer coupled it to the low side and converted back to a level. Doesn't sound so good, huh ? It DID work well enough to satisfy the FCC. I'm sure there are better ways now.

Something like this help? Unfortunately it is not recommended for new designs.

GG

I suggest that you consider the benefits of a bipolar- transistor current-replicator approach, as in my ASCII circuit below. (view in a fixed font, like Courier)

high-voltage high-bandwidth high-side current monitor

10.0 +250V Iout ----+---/\\/\\----+------o ---> 0 to 10mA load | | / | \\ 49.9 | all resistors 1% / | | | Q1-Q2 PNP low-voltage e e high beta, matched b --+-- b Q1 c | c Q2 | e | +-- b | | c Q3 2n3906 | |_____| e | b --------+ Ia = 1mA c | | Q4 | 270k 1.00k | mpsa92 '--/\\/\\---+--/\\/\\--- gnd | 0.5W | | 5% \\ 4.02k | Ib / | 2.49k \\ | ,--/\\/\\---, | 2.49k | | __ | +--/\\/\\---, 0 to +5V '--+--|- \\ | 2.49k | __ | out, for | >--+--/\\/\\--+--|- \\ | 0 to 10mA gnd ---|+_/ | >--+---- gnd ---|+_/

The circuit drops 100mV across the sense resistor at full current. It has a 5V fs output, as you requested.

The transistor arrangement is designed to operate Q1 and Q2 at similar currents for matching, and at low voltages to keep junction heating low. They should be high beta types to insure accuracy at low currents.

Q4 provides the power-dissipating 250-volt voltage drop for the output current. High-voltage transistors have reduced beta, so up to 2% of the signal current will be steered away from the output, but you can add a second mpsa92, as a Darlington, to capture all the current.

Normally circuits like this are run with bias currents, Ia, that are very low compared to the measured current. But your maximum current is only 10mA, and you want a 1MHz or more bandwidth, so I chose a Q2 bias of 1mA, which means it adds 10% of fs to the measurement and has to be subtracted from the output. What's more, I set the reflected output current Ib = 1/5 Iout, to keep the Q4 transistor power dissipation down (550mW max), so we have to take away 4/5 of the Ia correction current before the subtraction.

If you want to add an offset adjustment, you can use your +/-12V supplies and a trimpot to get a current into one of the opamp summing junctions.

There are ICs to accomplish much of what I've shown, but that's not nearly as interesting this morning.

On a sunny day (Sat, 8 Dec 2007 05:49:52 -0800 (PST)) it happened Winfield wrote in :

Jan Panteltje says: I see a direct crystal path from +250 to ground via Q2, Q1, Q3,Q4, and op-amp. This sort of guarantees big bang in case Q4 has a problem, possibly taking any other logic on the same supply as 'opamp' with it. I would like to see some series resistor in collector Q4 perhaps.

Yeah, it's always a good policy to have no power-ground paths through silicon, especially as the voltage goes up. Unless the transistors are so big they can take down the power supply.

John

Could you get away with a Hall sensor? The MHz requirement makes things tricky.

If you use a cheap dc-dc converter to float a, say, +-15 volt supply on the bus, you could use a good opamp (OP-27, LT1028) to scale up the shunt voltage to 10 volts maybe, then you may be able to drive a grounded diffamp at decent accuracy. Or do Win's current pumper, but with power you can do a very precise closed-loop current source at MHz bandwidth.

PWM with fiber optics would be fun, too.

Classic DCCTs (DC current transformers) like the Danfysik things, are phenomenally accurate and would be ideal here, but the cheapest ones are something like $400.

There are some cheaper open-loop or closed-loop Hall sensors, LEM or somebody.

John

You could hang a zener on the bus with a big resistor to ground, and derive a -5 volt supply up there. Then use a r-r opamp. 9, 10 parts maybe to deliver a grounded output voltage.

John

across the

Stop right there: a "noisy" 250V supply, all of which is across the load, and you want a sense drop in the "low" mv range?

And all these "I want" , "I would like" 's,

Go away...

across the

Scrooge is back ;-)

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...Jim Thompson

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|  James E.Thompson, P.E.                           |    mens     |
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across the

Hell, this is just along the borderline between "hard" and "fun."

The MHz+ bandwidth requirement makes it interesting; otherwise, it would be easy.

John

across the

But not creative. Goes along.

John

Yes, and thanks to a lower offset voltage, more accurate as well. Here's the circuit I used in my RIS-496 +/-250-volt amplifier, adjusted for the O.P.'s specs, with Jan's extra resistor:

10.0 +250V Iout ------+---+---/\\/\\----+----o ---> 0 to 10mA load | | | / '----, | \\ 200 | | / 1% +------|--+---, | | | | | +--------|----, | | | 7.5V zener | _|_ | | _|_/ | Q1 e / -|--' | /_\\ _|_ mpsa92 b --< | | | --- 0.1uF PNP c \\__+|----' | | | | LT1783 | | | +---------+---' \\ | / \\ 330k \\ 100k / 0.5W | \\ | | | gnd | __ +-----|+ \\ | | >--+--- 0 to +5.0V for \\ ,-|-_/ | 0 to 10mA load / |________| \\ / 10.0k 1% | gnd

It does have fewer parts, 10 vs 15. The LT1783 RRIO rail-rail 300uA opamp has f_T = 1.5MHz, so the O.P. might get his desired bandwidth.

I don't know why it's less appealing to me. I designed both circuits, so that can't be it...

...

Have you thought of just an optoisolator? I'd heard that they're terribly nonlinear, but I actually sat at a bench once and measured one - a 4N something-or-other, and when I saw the graph, I almost fell out of my chair - the CTR was linear within eyeball percent.

And there are other opto tricks, like use a dual, with one for feedback; use a V/F converter, optoisolate it, then use an F/V converter; the latter two would need some kind of power supply, of course, which could be snitched from the 250V with a zener and resistor.

And just sticking the opto's LED in series with the load would also give you free overload protection! ;-)

Good Luck! Rich

Because you need a small bypass cap across the 100K series resistor otherwise you can say goodbye to your MHz BW :-)

--
Thanks,
Fred.

Yep, right, 11 parts.

9 parts exactly! 10 if you include the bottom amp.

Not through that 100K resistor. Dang, 11 parts!

You like to design circuits, whereas the momentum these days is to connect boxes.

John

--- This oughta work: (View in Courier)

. FWB REG . +----+ +-----+ .20AC>-+ +--|~ +|---+---| |--+------------+------>Vcc . P||S | | |+ +--+--+ | | . R||E | | [BFC] | | | . I||C | | | | +--|----+ | .20AC>-+ +--|~ -|-+-+------+ | | | | ADC . +----+ | | Vcc [RFB] +---+---+ .250DC>-----------+-+-[Rs]--+---+-|-\\ | | Vcc | . | | | >--+---|IN OUT1|-->ADOUT1 . +---------|---+-|+/ . . . | | GND | OUT8|-->ADOUT8 . | | | | GND | . | +--+ +---+---+ . [LOAD] | | . | +------------+ . | .250GND>--------------------+

Typical ADC output circuitry:

. Vcc . | . [R] . | . | OPTO . +-+----+ . | A C|--> To logic . | | . | K E|--> To logic ground . +-+----+ . | . D .ADOUTn>--G . S . | .250DC>-----+

Note that everything except the load is running with 250VCD as its common, so if you need to talk to the ADC you'll need to do it through optocouplers unless you can use the same common for your micro. Pretty dangerous if you don't know what you're doing!

No offense intended. :-)

-- JF

I built myself something like this for scoping mains currents (order of

10-100mA) into power supplies, and it seems to work OK. In my case I have a 4R7 series resistor and a pair of 18k:330k potential dividers, into an AD620 instrumentation amplifier with a +/- 9v supply. I'm not really sure what the frequency response is, because I don't have a better measure to compare it with. (High frequency components are probably also messed up by the fact that my -9v comes from a switched-capacitor voltage doubler!) The box also has a scaled-down voltage output and I feed both into my scope on "multiply" mode with a whole 50Hz cycle visible. It has way more frequency response than is needed for this, but whether it gets anywhere near a MHz I don't know.

I've put a picture of my contraption here:

Try it out and see if the reason jumps out at you.

I think that the best solution would involve a battery, an ADC, some fibre-optic, and either a DAC into a scope or directly capturing the digital signal on your PC. I have a very vague feeling that I may have seen a circuit like this in a applications note or something like that.

Cheers,

Phil.

Actually, my circuit above is a short-changing cheat! It has no standing bias for the critical transistor, Q1, and hence cannot perform to the desired bandwidth when Iout is near 0mA. Adding a bias like the original non-opamp BJT current-mirror circuit, boosts the parts count to 15, exactly the same as the all-BJT circuit.

So it's a tie. Go to the trouble of finding a suitable low-current high-side opamp and get better zero-current accuracy without trimming. Or something like that.

high-voltage fast high-side current monitor

10.0 +250V Iout ------+------+---/\\/\\----+----o ---> 0 to 10mA load | | | / '--+--------|---, \\ 200 | | | / 1% | 200 / | | | 1% \\ | +---------|----, / +---, | _|_ | | | | 7.5V zener Q1 e / -|--' | _|_/ | mpsa92 b ---< | | /_\\ _|_ PNP c \\__+|------+ | --- 0.1uF | LT1783 | | | | | '--------|---+---+---, / | | \\ 1k 1.0M | / 330k / / \\ 0.5W | Ib = 0.25 \\ Ia / | to 0.75mA / | | | gnd | 10.0k | +--/\\/\\---, | 10.0k | __ | +--/\\/\\---, 0 to +5V '--|- \\ | 10.0k | __ | out, for | >--+--/\\/\\--+--|- \\ | 0 to 10mA gnd ---|+_/ | >--+------ gnd ---|+_/

One more resistor if you're willing to run a bit of current through the shunt. Two more if not.

Actually, at MHz speed, to keep the bus noise from coupling through capacitances, one might decide to throw a differential current down at ground, which also fixes the zero problem. More parts on the high side, ballpark 12, and a few more near ground.

Chopamp.

Won't the 1M squirt in an error proportional to bus voltage? Are you trying to compensate the collector resistance of Q1?

You could take the standing-current offset bias thingie from the zener, or better yet use a bandgap.

John