I'm using AD8418 to detect a LD current, but it normally keep a minimum output about 4mV when on shunt current, the Rshunt is 0.1 ohm.
The output won't go all the way to zero volts, if you are using the configuration of fig27 in the datasheet.
The specs indicate the minimum voltage around 15mV with a 25k load, so 4mV would be a reasonable value .
Fig27 grounded operation will not accurately read currents near zero in the shunt.
You could possibly bias the output up a little bit if needed, by using a 100R resistor between ref1&ref2, tying ref 1 to GND, and ref2 to a 4.9k resistor to +5 , this will give a controlled 100.00mV at the output for zero current. You lose a bit of CMMR this way, but adding a 50ohm resistor in series with input+ will restore most of the CMRR.
BTW this is a nice chip, especially with the split references, it's much better than the classical approach using IA's and driven references, and always trying to stay in the CM envelope.
How are you configuring the AD8418? To what voltages are your references tied? What supply are you using?Can you provide your circuit schematic? If you are able to remove the shunt resistor, could you tie your inputs together and connect them to ground and check what you get at your output?
I was short Rshunt, but result same. My Vcc is 5v, the circuit is like fig 35 in the datasheet, the load change to LED and no clamp diode, Vref1 and 2 tie to GND.
OK it seems you don't understand how instrumentation amplifiers or difference amplifiers work.
With these types of amplifiers the output voltage is presented as the voltage between the "output pin" and the "reference pin"
The circuit of fig 35 when wired correctly, with vref1=5v, and vref2=0v will
The AD8418 is a bidirectional current sensor.
If you have a unidirection current, and want 0volt out for 0v , then there are other chips you should look at, all of these work by driving a variable current into (sometimes external load resistor)
If you google ZXCT1010 you will see a primitive version of this concept, as a starting point.
The AD8211 improves on it with much less drift and higher accuracy and lower offset.
An excellent description is (figure1) here: http://www.analog.com/library/analogdialogue/archives/44-12/high_side.pdf
you could also use a difference amp like the AD8267 , with an external transistor , and an external load resistor
so pins 1,2 go across your shunt, and pins 5 and 1 go across a tap on your output resistor, and pin6 drives the base/gate of the transistor, the emitter source pulls up the total output resistor.
So output resistor = 10k, with 100 ohm on top of this, then with say 20mv on the shunt, 20mV gets copied over to the 100ohm, which being in series with the 10k, produces 2V across this 10k resistor. (look up Howland current source to get some idea of the topology) , this arrangement will give you 0.00v out for 0v in, and even better, you can mount the 10k resistor a long way away (even 100mtrs) and not affect the accuracy.
Thank you for your advice,
My application is high size current sense, high size switch and unidirectional operation, so I referred fig 35 and set Vref1, Vref2 to ground together.By the way, I can't find IC AD8267, may be it is wrong part number.
Oops, dyslexic part number, its an AD8276 (just a difference amplifier).
If you google images of the Howland source, you get the general idea of the topology.
if want to get within millivolts of zero out for zero in, you need an emitter follower transistor at the output of the diff amp, so when the output (= transistor base) is < 0.5v , then the voltage on your load resistor will be the emitter voltage , which will be zero (as the transistor is cutoff). You can alternateively use a MOSFET for more accuracy.
Here is a very interesting and helpful handbook by AD ,
If you browse through (at least chapter 6) it will improve your understanding of In amps and diff amps.
Its pretty much everything you need to know all in the one place.
Ahh found a schematic with an output transistor buffer;
See figure 8 for the topology.
You just need Q1 and it's base and emitter resistors , (leave off all the other diodes and Q2).
Add a 100nF cap from pin9 to output sense if the circuit oscillates.
You can move the sense signals from RL to across R1 to make a Howland pump, probably make R1 bigger also, in this case RL can then be mounted remotely.
This topology has slightly compromised CMRR, but should be acceptable with a reasonably low RL (say 500R or 1k)
Do you have other solution for AD8418 current detect, except first suggestion. I don't want use more component to build this circuit.
For single supply, how to get output zero when input zero, at least the output can keep in uV level.
Achieving zero volts output on single supply is a Holy Grail of IC design.
Also achieving uV level of output error is almost impossible , with commercial grade designs, even with dual supplies. Even getting a ground potential that is within uV of nominal ground is difficult, most PCB's and wiring in any system that draws current and/or has temperature differentials of 10C will see variations of ground potential in the 1 to 10mV range.
If you look at the AD8418 datasheet, the input offset is +/-200uV , so with a gain of 20, this is an output offset of +/-4000uV.
The CMRR is 100dB at DC , this is 100,000:1 , so a 10v common mode input results in 100uV of apparent input signal or 2000uV at the output.
Given the set of impossible design constraints you have imposed on yourself, the best option is to load the output with a 470ohm resistor, this will drag the ouput closer to gnd (but the chip will run warmer as the output increases)
Or use a Jfet (in place of the resistor) in constant current mode (e.g. J202 or J175,J176, BFR31 , look for Idss~5mA) this behaves like ~100ohm pulldown near zero volts but like constant current load with several volts applied.
Thank bobsbots, I will try it.
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