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Recently, I read one paper on Libbrecht-Hall-Style circuit. As shown in the following picture. Libbrecht-Hall-Style test circuit (arxiv, AIP journal) I found that the circuit uses one LM399 as a voltage reference fed to the non-inverting input pin of Op amp(AD8671) to set the current of the circuit.

I refer to the datasheet of LM399, I found that the voltage across the Rset resistor would be 6.95V, and the voltage fed to the non-inverting pin of AD8671 could be adjusted by the value of Rset. But I do not know how to calculate the maximum current value that the circuit could deal with. It seems that the AD8671 works as a comparator, so how to maintain a stable current?

Could any one give some suggestion? Any comments are highly appreciated.

Edit

According to Neil_UK's answer, I draw a simple circuit to see if I grasp the working principle of the circuit. In the picture, I use a zener diode to represent the LM399.

current source in LT-Spice Circuit

The different value of R7 and R8 would set the non-inverting voltage, so the inverting voltage would change correspondingly. Thus the voltage across the Rsense voltage change to maintain a balance.

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2 Answers 2

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The FET IRF9Z14 is a Pchannel, used as a follower (the source at the top follows the gate) thus is noninverting. The OpAmp thus is wired, properly, to provide the inverting loop polarity.

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  • \$\begingroup\$ Sorry, I am not familiar with the P-channel MOS. And I do not get what you mean. Maybe I should search the website and find some materials related with the classic p-channel MOS application circuit. I thought the gate voltage set the max. current that would pass the P-channel MOS. \$\endgroup\$
    – Jane
    Commented Aug 29, 2017 at 4:45
  • \$\begingroup\$ Not clear why the down-vote. The opamp is not operating as a comparator; the opamp is operating in a servo-loop mode, with the Pmos providing high-current linear non-inverting buffering. Does this make sense to you? \$\endgroup\$ Commented Aug 29, 2017 at 15:34
  • \$\begingroup\$ Thanks for your comment. I do understand now. By the way, it is not me that down-vote your answer. \$\endgroup\$
    – Jane
    Commented Aug 29, 2017 at 15:48
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The current servo consists of mainly \$R_{sense}\$ and the ADC8671. However, the IRF9Z14 is added to boost the amplifier's output current.

In operation, the servo maintains the same voltage across \$R_{sense}\$ as its reference voltage. The actual voltage that the amplifier drives the FET gate to is irrelevant, it adjusts it to keep its input voltages equal.

When the amplifier runs out of output voltage range, then its output voltage becomes relevant, as that stops the servo working linearly, and we have reached one of the current limits of the circuit. Other limits include a power dissipation limit in \$R_{sense}\$ or the FET, a power supply current limit, or reaching the output voltage limit (the compliance voltage) of the load.

What limits the maximum output current of the circuit is whichever of these other limits bites first.

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  • \$\begingroup\$ Thanks for your answer. I have edited the question to see if I grasp what you mean. \$\endgroup\$
    – Jane
    Commented Aug 29, 2017 at 7:20
  • \$\begingroup\$ @Jane yes, that looks good \$\endgroup\$
    – Neil_UK
    Commented Aug 29, 2017 at 10:29

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