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I am reading an application note and find a current source circuit as the following. Why the lower AD8610 is needed since the output voltage of the upper AD8610 is determined by VIN and the two 2kΩ resistors? enter image description here

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  • \$\begingroup\$ It is a funny circuitry. It's more like; bottom one is the sensing and gain, while the upper one is error & control function. It is straightforward calculation, if you need that part. \$\endgroup\$
    – jay
    Nov 8 at 2:50
  • \$\begingroup\$ So the bottom one is to get a gain of 1/5 of the output current to attribute to the the upper one's feedback. And the purpose of the feedback is to minimize the effect of VIN's fluctuation? \$\endgroup\$ Nov 8 at 3:15
  • \$\begingroup\$ Voltage gain is 1 + 10k/1k \$\endgroup\$
    – jay
    Nov 8 at 3:24
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    \$\begingroup\$ I don't see how that works to create a constant current through (varying resistance) [LOAD] at all. \$\endgroup\$ Nov 8 at 3:32
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    \$\begingroup\$ It seems a pity to use an extra opamp, and then still have a load that's not referenced to ground! \$\endgroup\$
    – Neil_UK
    Nov 8 at 6:24
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Looks like an error.

If you eliminate the top 2k\$\Omega\$ resistor it does work as a constant current source/sink. -110mV in should give you 10mA out. Stability may not be ideal- maybe there's a series capacitor missing.

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  • \$\begingroup\$ The circuit actually comes from this AN-968 Current Sources: Options and Circuits analog.com/media/en/technical-documentation/application-notes/…. And there three more circuits after this one with 2kΩ resistor. If without the top 2k resistor, doesn't the Vout at pin 6 be 0? \$\endgroup\$ Nov 8 at 6:20
  • \$\begingroup\$ No, as I said, it is functional without that resistor. Vout is driven to whatever voltage is required to have the input of the op amp reach 0V, and thus the voltage across the sense resistor equal -Vin/11. \$\endgroup\$ Nov 8 at 11:09
  • \$\begingroup\$ I took a serious calculation today. The Iout would be -Vin/(34+RL). It can't explain itself as a current source. For an intuitive perspective I agree with you that the feedback would try its best to be a constant if you want the output to be a constant. \$\endgroup\$ Nov 9 at 7:59
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I don't think the circuit is "funny" at all. Over the years it and its variations have appeared in app notes from many manufacturers.

Because the non-inverting input of the upper opamp is grounded, the entire function of the circuit is to keep the inverting input at 0 V. The circuit is similar to the "classic" constant current circuit that has only one opamp. The lower opamp circuit has some gain, but other than that the constant-current math is the same.

The circuit works to maintain a constant voltage across a constant resistance, thus creating a constant current through anything that is in series with that resistance. While there is a circuit element named "LOAD", see things from the opamp's point of view. Voltage-mode feedback is picked off at the top of the 1 ohm resistor, making that resistor the actual load. As with any normal, linear, inverting opamp circuit, the opamp works to produce a voltage across the load that is the opposite of its input voltage, modified by the feedback network. In this circuit, the feedback network has non-inverting voltage gain, so the output current is 1/11th of the input current.

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  • \$\begingroup\$ Why all the complexity ? What is the advantage over a single opamp V-to-I converter? \$\endgroup\$
    – tobalt
    Nov 8 at 8:24
  • \$\begingroup\$ The DC negative feedback around the top op amp prevents it from working as you describe. \$\endgroup\$ Nov 8 at 11:16
  • \$\begingroup\$ An excellent analysis! I feel embarrassing to ask, but.. Could you/anyone explain what that 1Meg ohm is for? \$\endgroup\$
    – jay
    Nov 8 at 12:39
  • \$\begingroup\$ Spehro - why? The upper opamp (can't believe ADI left out reference designators - !) has two DC negative feedback paths in parallel, one with gain and one without. Complicating, yes; but not wrong. \$\endgroup\$
    – AnalogKid
    Nov 8 at 13:04
  • \$\begingroup\$ @AnalogKid The load current is a function of the load resistance rather than being (ideally) an infinite impedance source. Break the connection to the output of the top op amp and imagine it connected to a voltage source- it has equal influence to the input voltage… \$\endgroup\$ Nov 8 at 13:53

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