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I'm planning to use the ADA4610 in unity gain configuration as a current buffer for a high precision (16-bit) DAC chip. The chip is labeled as being stable in the unity gain configuration. I currently have the chip configured like the image shown below. Practically speaking, should I be adding a capacitance in the loop or is this configuration truly stable? In addition, are there any best practices for this approach to current buffering?

enter image description here

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    \$\begingroup\$ why do you think a capacitance will add a stability? \$\endgroup\$
    – Eugene Sh.
    Jul 7, 2016 at 15:36
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    \$\begingroup\$ If the datasheet says: "Unity gain stable" then it will be unity gain stable provided you do not violate the conditions for which this is true, for example a maximum capacitive load at the output might be specified. \$\endgroup\$
    – Bimpelrekkie
    Jul 7, 2016 at 15:43
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    \$\begingroup\$ The data sheet says it's stable in that configuration. Why don't you believe them? You probably believe the voltage and pinout they tell you to use. Caution, that's a totally useless circuit, as the opamp is driving no load, no DMM, no scope probe, no LED, no resistor to ground, no nothing. Presumably you want it to drive something? Analog devices revG data sheet page 19 The ADA4610-1/ADA4610-2/ADA4610-4 are unconditionally stable for all gain configurations, even with capacitive loads well in excess of 1 nF. What about much much larger capacitance? Probably not, few opamps are. \$\endgroup\$
    – Neil_UK
    Jul 7, 2016 at 15:45
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    \$\begingroup\$ Figures 40 and 42 of the revG datasheet helpfully give overshoot versus load capacitance for the unity gain configuration. It's a flat 10% from 0 to 100pF, then rises to 40% at 1nF, and keeps on rising to 45% at 2nF, where the graph stops. What happens beyond there? Who knows, but presumably it's not good, otherwise they'd tell you! \$\endgroup\$
    – Neil_UK
    Jul 7, 2016 at 15:51
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    \$\begingroup\$ The bigger issue is that you have this configured as a voltage follower. You said you want a current buffer, but have you got a circuit in mind? \$\endgroup\$
    – pgvoorhees
    Jul 7, 2016 at 15:57

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If it's indeed unity-gain stable, you don't need to do anything beyond the voltage follower configuration you've shown above.

That said, you should note that the short circuit output current is 63 mA, so if you expect this thing to source/sink 200mA to drive a motor, it won't.

Also, with regard to the nature of the input, when powered with +/- 5 Volt rails, the chip will only function with inputs between -2.5V and +2.5V. So, if you're powering with a single sided supply, you can go nowhere near ground and expect the circuit to work! For that, you would need a chip with rail to rail inputs. Sometimes it even makes sense to use two -1 gain inverting op amps instead of a non-inverting follower, which let's you pin the inputs to a legal value, and you don't have to worry about your common mode input range.

At some level, there's no substitute for trying something to see if it works. Build the development time into your project. Modeling in LTSpice can also be a great timesaver.

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  • \$\begingroup\$ This chip is indeed rail-to-rail, so no concerns there. Very helpful response! \$\endgroup\$
    – nick_name
    Jul 7, 2016 at 16:39
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    \$\begingroup\$ @nick_name -- the chip is rail to rail on the OUTPUT, not the input. There is a difference, and that difference has bitten me on the ass ONCE. I'm trying to save you that once. \$\endgroup\$
    – Scott Seidman
    Jul 7, 2016 at 16:40
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    \$\begingroup\$ @nick_name -- simply put, that circuit WILL NOT WORK if you power it between ground and 5V. If you power it with +/-5V, it will work for inputs between -2.5V and +2.5V, with unpredictable behavior outside of that range. \$\endgroup\$
    – Scott Seidman
    Jul 7, 2016 at 17:14
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    \$\begingroup\$ @nick_name, that will work fine. If you select a RRIO from analog.com/en/parametricsearch/11095#/p5055=Both, you might be able do away with the negative supply, which can be a PITA is this is the only reason it's there. As to where this comes from-- look on the data sheet under Input Voltage Range in the Input Characteristics section. They give them for +/- 5V supply, but you generally assume the required headroom stays the same for all supplies (an OK first approximation, based on input configuration of these things) \$\endgroup\$
    – Scott Seidman
    Jul 7, 2016 at 17:45
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    \$\begingroup\$ Other brands data sheets might specify this as Vcm, the allowable common mode voltage range on the input. \$\endgroup\$
    – Scott Seidman
    Jul 7, 2016 at 17:47

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