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I want to design a 1A current source on a PCB. I have bunch of MAX9611 chips laying around. In their datasheet it says it can be configured as inrush current limiter and current source (in a closed-loop system). The problem is, there is no application example of current source in the datasheet. The only thing there is:

MAX9611 INRUSH CURRENT LIMITER

The relation for I_limit is given (correction as Neil_UK pointed out): $$I_{LIMIT} = \frac{V_{CC} \times R_{3}}{(R_{3}+R_{4})(2.5\times R_{SENSE})}$$

I know a simple current source can be built by a p-channel MOSFET:

p-channel MOSFET current source

I am failing at relating these two concepts together. Is what I want to achieve even possible? If so, can I get some clues? If not, please let me know an alternative.

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Your equation for current limit is incorrect. According the Maxim datasheet, the expression in the brackets is (R4+R3), not (R2+R3). That may or may not have been confusing you, as R2 is closely associated with the PFET.

With the correct expression, we can see that the circuit is servoing the drop across Rsense to be proportional (equal to 0.4x) to a reference voltage on R3, the Vset voltage.

The reason that this is an 'inrush current limiter' rather than a 'current source', is because the reference voltage and therefore the output current depends on VBAT, which in most applications can be expected to vary, perhaps over a 2:1 range, not accurate enough for most people's idea of a 'current source', but good enough to limit current inrush.

You can implement a current source in closed loop mode, as there is an ADC in there (there's an ADC in there!!!) to measure what you have, and adjust it.

How accurate do you need your current source? The voltage on the SET pin needs to be known (either by measurement, or by simply setting it) to better than this accuracy, to allow for errors in Rsense, the 2.5x gain factor, amplifier offsets to have their chew at the accuracy budget. I haven't read the data sheet in enough detail yet to see what sort of accuracy could be expected even if vSET was known exactly. Maybe they add up to a large figure, perhaps that's why it's an inrush limiter rather than a current source? I'm a little disturbed that when reading the SET voltage, they quote the typical error as 0.2% and the maximum error as 5% !!!

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  • \$\begingroup\$ Or, instead of closing the loop through the ADC, just use a precision voltage reference to provide a fixed voltage to the SET pin, instead of a divider depending on the inaccurate supply voltage. \$\endgroup\$ – dim lost faith in SE May 30 '16 at 13:30
  • \$\begingroup\$ Thanks for explanation. I still do not know what should I change in the schematic above. Can you please explain? \$\endgroup\$ – Sean87 May 30 '16 at 14:07
  • \$\begingroup\$ The data sheet omits all mention of the ADC's reference, which kinda suggests it has its own internal reference. One would have thought that this reference could be either brought out, or better yet, internally muxed to the SET pin? Lack of imagination on Maxim's part there. \$\endgroup\$ – Neil_UK May 30 '16 at 14:12
  • \$\begingroup\$ It depends how accurate you want the current source to be. You don't necessarily need to change anything in the schematic, you just have to understand how to use it. Either use the ADC to read the voltage that happens to be on the SET pin. Or as dim suggests, drive the SET pin with a voltage sufficiently precise for your application. Either way, your output current will be Vset/(2.5*Rsense) \$\endgroup\$ – Neil_UK May 30 '16 at 14:16
  • \$\begingroup\$ Thanks, I just found out that the new version of datasheet has corrected the mistake in the equation...the problem was in old datasheet! I will correct it in my question. \$\endgroup\$ – Sean87 May 30 '16 at 17:50
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I believe that you misunderstand the circuit. If you set the components to the values implied by the equation, the load current will be limited to the value indicated. As long as the load resistance is low enough, and the input voltage high enough, to draw the current you'll be fine. Ignore the data interface unless you want to use the 9611 to produce a digital measure of current for a display.

R3 and R4 can, in principle, be replaced by a pot for easy setting of current level, but the voltage to the top of R3 must be supplied by a regulated voltage for stable operation. Although Vcc is labelled as "2.7 to 5.5 volts", this only specifies that the chip may not work if you use more or less. If you let it wander, the current will wander, too.

The same page of the data sheet you got your equation from explains how the 9611 and the FET combined form a current source. The output of the 9611 increases with increasing current. Since increasing this voltage will reduce the gate/source voltage and reduce the transconductance of the FET, the combination will form a negative-feedback loop which will (loop dynamics and stability permitting) stabilize the output current. The result is a current source. Your simple current source example does not apply, since there is active control of the gate voltage through R2 and R1.

So you don't need to "change" anything. Just use the datasheet and the datasheet of your desired FET to get the gate drive right. Find the values of R1 to R4 which give you 1 amp.

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