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Hi!
That 1A is drawn from the DUT, not from the battery/circuit Vcc. Battery only powers the control circuitry.
A small question about the efficiency of the circuit:
This circuit supposes to work on battery power, so does the 1.3W-5W constant power consumption is the best solution?
Is there a way to have a more power efficient circuit?
Thanks
Yoav
(This is my first post, so if I'm breaking any rules - SORRY !!!) -
This looks like one for a kick starter project.
I'd buy it
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I'd like to make a simple proposal to correct for the errors.
The error budget of Dave's circuit is calculated as:
Iout = Uref/Rs * (1 +/- 250ppm +/- 200ppm) + 5.75mA*Rreturn/Rs
where 250ppm and 200ppm are the specified errors of the volt reference and the shunt resistor, respectively.
The last term describes the additional error by the negative sense return current over the resistance of the cable.
This latter error annoyed Dave, by exceeding the max. expected error of +450ppm.
Rewriting for a sum of all 3 errors yields:
Iout = Uref/Rs * (1 +/- 250ppm +/- 200ppm + 5.75mA*Rreturn/Uref)
From the observed deviations of +650 ppm and +350ppm with different components ,
the return resistance on the breadboard can be estimated to be between: 0.05 Ohm < Rret. < 0.20 Ohm
As the sum of those 3 errors yields a slightly too high current, I propose to add a small amplification = (1 + eps) on the OPA376.
Then the above formula is expanded to:
Iout = Uref/Rs * (1 +/- 250ppm +/- 200ppm + 5.75mA*Rreturn/Uref - eps)
In the case of the first setup:
Iout = Uref/Rs * (1 + 650ppm - eps),
it's possible by selecting eps = 0.00065, e.g. 68 Ohm over 100kOhm, to adjust the output current to exactly 1.00000A.
With different components and another return line resistance (on a PCB), the correction resistor (eps) has to be chosen accordingly.
Rem.: Buffering the negative sense line with a 2nd OpAmp would only compensate for the return current.
Frank
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May I suggest the series gate resistor of the power fet( unspecified ) is the majic adjustment for the dominate ( 1A) current loop was left undiscussed. Please start with a value of 100k and go up or down from there. You may find the high frequency low current path can not effect the output.
Ray -
If you look at the diagram, 3,5,8 all go only to the bandgap reference and is marked "internal function". I don't see what else this could be than a Kelvin type connection to the low side of the bandgap reference. It's clear to me that the 2 mA can not go through 3,5,8. These 2 mA flow because of the resistors in the feedback loop, and come out through ground pin 4.
Let's analyze the circuit. We have a 1.25 V bandgap reference and a Kelvin current shunt of 1.25 ohm. When there's a 1.25 V voltage present over the shunt, we know 1 A flows through the resistor. This is the obvious basic operation of the circuit. But the problem then becomes, where are these things actually referenced from? I would like to refine the function of the circuit to "the goal is to keep the voltage over the shunt sense pin identical to the voltage over the bandgap reference sense pins".
This is where derive that 3,5,8 instead of all the chip's ground should be connected to the shunt sense pin. There will be a small voltage drop over the shunt's current and sense pins, and the bandgap reference's lower connection will now (presumably) be lifted from the circuit ground by that many mV. If we didn't do this, we would have an error current. This error current is not caused by the shunt's voltage drop, but by the quiescent current and the drive current for the force pin, through the metal between the chip and the lower shunt sense pin. I'm even willing to bet that this may have something to do with the oscillation.
And likewise, keeping the other two legs of the circuit identical is done by the opamp in the voltage reference chip. So what does it use as it reference? Depends on your definitions. It's ultimately using the device ground as its reference. However, if the shunt's sense pin raises the reference ground to say 0.001 V over the circuit ground, then the bandgap reference will now output 1.251 V over circuit ground which is now the amplifier's target for the shunt's higher pin (which it tries to control by driving the MOSFET.) But the goal to keep the voltage over the sense pins of the hunt at 1.25 V is still met.
So yes, I think this is the way to go. Can we get an empirical confirmation from some crazy bloke down under?
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If you look at the diagram, 3,5,8 all go only to the bandgap reference and is marked "internal function". I don't see what else this could be than a Kelvin type connection to the low side of the bandgap reference. It's clear to me that the 2 mA can not go through 3,5,8. These 2 mA flow because of the resistors in the feedback loop, and come out through ground pin 4.
Let's analyze the circuit. We have a 1.25 V bandgap reference and a Kelvin current shunt of 1.25 ohm. When there's a 1.25 V voltage present over the shunt, we know 1 A flows through the resistor. This is the obvious basic operation of the circuit. But the problem then becomes, where are these things actually referenced from? I would like to refine the function of the circuit to "the goal is to keep the voltage over the shunt sense pin identical to the voltage over the bandgap reference sense pins".
This is where derive that 3,5,8 instead of all the chip's ground should be connected to the shunt sense pin. There will be a small voltage drop over the shunt's current and sense pins, and the bandgap reference's lower connection will now (presumably) be lifted from the circuit ground by that many mV. If we didn't do this, we would have an error current. This error current is not caused by the shunt's voltage drop, but by the quiescent current and the drive current for the force pin, through the metal between the chip and the lower shunt sense pin. I'm even willing to bet that this may have something to do with the oscillation.
And likewise, keeping the other two legs of the circuit identical is done by the opamp in the voltage reference chip. So what does it use as it reference? Depends on your definitions. It's ultimately using the device ground as its reference. However, if the shunt's sense pin raises the reference ground to say 0.001 V over the circuit ground, then the bandgap reference will now output 1.251 V over circuit ground which is now the amplifier's target for the shunt's higher pin (which it tries to control by driving the MOSFET.) But the goal to keep the voltage over the sense pins of the hunt at 1.25 V is still met.
So yes, I think this is the way to go. Can we get an empirical confirmation from some crazy bloke down under?
in page 16 there is indication pin 4 is for 4 kelvin connection..
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Hi Dave
Just a simple question from a non designer.
Why would'nt you take into account the 2mA by using a resistor to pass 0.998 A so that 1A passes through the load.
Rich -
Hi Folks: Is there reasonably good alternative to using the Vishay whoopty doo VPR221Z Precision 4-terminal Z-Foil resistor? For example: using eight 10 ohm 0.1% soldered in parallel (1.25 ohm) to copper bus bars with the sense and current wire connected to the buses. Cheers, Mark
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Hi Folks: Is there reasonably good alternative to using the Vishay whoopty doo VPR221Z Precision 4-terminal Z-Foil resistor? For example: using eight 10 ohm 0.1% soldered in parallel (1.25 ohm) to copper bus bars with the sense and current wire connected to the buses.
Sure, but that's not 0.02%, and the tempco likely isn't as low. -
in page 16 there is indication pin 4 is for 4 kelvin connection..
And that's what you want. everyone connected back to the star point, which will be the current shunt sense terminal.
IIRC that's what the app notes say as well. -
When I jam TO220s and other square pins into my breadboard, it distends the contacts, and then thin leads like resistors and jumpers don't fit snugly anymore. Does anybody else find that, or do I just have junky breadboard?
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in page 16 there is indication pin 4 is for 4 kelvin connection..
And that's what you want. everyone connected back to the star point, which will be the current shunt sense terminal.
IIRC that's what the app notes say as well.
This IC meant for Voltage reference, not design perfectly for current reference application, since the expected load = high impedance
And with star conection at pin4 in this application, there is a larger current from R1,25 ohm.
starpoint at pin4 = current flow from low sens vishay resistor to pin 4
star point at resistor = current flow from pin4 to sens terminal at vishay
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in page 16 there is indication pin 4 is for 4 kelvin connection..
Yes, and that schematic also doesn't mention the other pins. I suspect that those example were made "by the book", ie not breaking the rule of connecting the various ground pins in PCB metal. But why would three pins be broken out as "internal function" instead of all being connected internally if you couldn't use them for example in the way I described?
Another relevant quote from the same page which I think supports my hypothesis: "Although there are several pins that are required to be connected to ground, Pin 4 is the actual ground for return current."
But, empirical testing beats babbling. If only I had an LTC6655, a VPR221Z and a precision multimeter. But last time Dave tested something for me, it turned out I was wrong, so maybe I should shut up... -
Hey guys, and hey just_fib_it,
btw thanks again for the hint.
Today Vishay contacted me back, they have a little subsidiary here in Germany who service all the customer requests.
So I asked for 4 pcs VPR221Z 4 pole TO-220 resistors in the 100 to 200 Ohm range.
Each one with a different value to make an inital calibration unit for PT100 temperautre meters.
I asked for 0.01% precision.
They offered each one for just under 30 EUR (40 USD / 45 AUD).
The lead time for them will be 8 weeks.
I think this is in the affordable range for what they are.
I am still not sure if I go over the top with choosing 8W TO220 ones instead of these 2512 SMD 4 pole one Dave uses on the µCurrrent. However, they just fit into the design.
For Germany you can contact the Vishay owned company called "Powertron".
Web address is www.powertron.de or http://www.vishaypg.com/powertron/
I hope this information is helpful for those who might want to build similar stuff.
Cheers,
ChipguyWhere can you actually buy the precision resistors with customer specific value ?
You can get them from Vishay directly: http://www.vishaypg.com/foil-resistors/how-to-order/
Caveat: I haven't actually ordered from them, so I don't know if they'll talk to hobbyists. -
Murphy got him ?
mabye -
I can confirm that pins 3,5,8 are not connected together internally. I measure 12ohms between pins 4 and 3, and 40K or so between pins 5 and 8 and pin 4. double that between pins 5 and 8.
I can also confirm the reference still works with pins 3,5,8 disconnected. -
I got a reply back from Linear saying that these pins are used in production, and suggesting my theory wouldn't work. But, that just makes more curious what (exactly) these pins are actually doing. If you disconnect 3, 5 and 8, what is the potential on each of the pins, with respect to pin 4?
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Couldn't leakage current in the electrolytic be a problem if it sits across the shunt resistor?
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Hmm..., 1amp current source hey?
a transistor is a current controlled current source, similarly mosfets are voltage controlled current sources by themselves and between opamps output and inverting input exists a current source, that is to say a common emitter configured transistor there exists a current source between collector and the supply voltage rail.
Now if we jam a reference voltage of 5.6 into a BJT' base, configured as a common emitter, with emitter degeneration of say 5 Ohms there will exist a approx 1 amp current source between collector and Vcc, as is ,you will find it to be a little unstable and drifty.
Now the thing is opamps are ubiquitus and are crying out to be used as PID controllers, each litttle opamp is a little controller in it's own right. So why not use it here, we can control the current through the transistor.
Connect the reference to the opamps non inverting input and the opamp's output to the base of the ce transistor and lastly connect the opamps inverting terminal to the bjt's emitter still using emitter degen. resistor of around 5Ohm or so and that's it for now. The current output is between top supply rail and the transistors collector terninal.
Using brand new expensive parts specialist resistors costing tens of dollars is a no no
:
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Now if we jam a reference voltage of 5.6 into a BJT' base, configured as a common emitter, with emitter degeneration of say 5 Ohms there will exist a approx 1 amp current source between collector and Vcc, as is ,you will find it to be a little unstable and drifty.
Now the thing is opamps are ubiquitus and are crying out to be used as PID controllers, each litttle opamp is a little controller in it's own right. So why not use it here, we can control the current through the transistor.
Connect the reference to the opamps non inverting input and the opamp's output to the base of the ce transistor and lastly connect the opamps inverting terminal to the bjt's emitter still using emitter degen. resistor of around 5Ohm or so and that's it for now. The current output is between top supply rail and the transistors collector terninal.
A reasonable suggestion; take the voltage reference out of the control loop -- I like it.Using brand new expensive parts specialist resistors costing tens of dollars is a no no
Can you suggest a reliable source for 0.02%, 0.05ppm/K resistors running less than ten dollars? :
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Using brand new expensive parts specialist resistors costing tens of dollars is a no no
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Not if:
a) You get them for nothing
or
b) Your time is infinitely more valuable than any potential potential issues with another solution.